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https://visualfractions.com/calculator/long-division/what-is-623-divided-by-966-using-long-division/ | [
"# What is 623 divided by 966 using long division?\n\nConfused by long division? By the end of this article you'll be able to divide 623 by 966 using long division and be able to apply the same technique to any other long division problem you have! Let's take a look.\n\nWant to quickly learn or show students how to solve 623 divided by 966 using long division? Play this very quick and fun video now!\n\nOkay so the first thing we need to do is clarify the terms so that you know what each part of the division is:\n\n• The first number, 623, is called the dividend.\n• The second number, 966 is called the divisor.\n\nWhat we'll do here is break down each step of the long division process for 623 divided by 966 and explain each of them so you understand exactly what is going on.\n\n## 623 divided by 966 step-by-step guide\n\n### Step 1\n\nThe first step is to set up our division problem with the divisor on the left side and the dividend on the right side, like we have it below:\n\n 9 6 6 6 2 3\n\n### Step 2\n\nWe can work out that the divisor (966) goes into the first digit of the dividend (6), 0 time(s). Now we know that, we can put 0 at the top:\n\n 0 9 6 6 6 2 3\n\n### Step 3\n\nIf we multiply the divisor by the result in the previous step (966 x 0 = 0), we can now add that answer below the dividend:\n\n 0 9 6 6 6 2 3 0\n\n### Step 4\n\nNext, we will subtract the result from the previous step from the second digit of the dividend (6 - 0 = 6) and write that answer below:\n\n 0 9 6 6 6 2 3 - 0 6\n\n### Step 5\n\nMove the second digit of the dividend (2) down like so:\n\n 0 9 6 6 6 2 3 - 0 6 2\n\n### Step 6\n\nThe divisor (966) goes into the bottom number (62), 0 time(s), so we can put 0 on top:\n\n 0 0 9 6 6 6 2 3 - 0 6 2\n\n### Step 7\n\nIf we multiply the divisor by the result in the previous step (966 x 0 = 0), we can now add that answer below the dividend:\n\n 0 0 9 6 6 6 2 3 - 0 6 2 0\n\n### Step 8\n\nNext, we will subtract the result from the previous step from the third digit of the dividend (62 - 0 = 62) and write that answer below:\n\n 0 0 9 6 6 6 2 3 - 0 6 2 - 0 6 2\n\n### Step 9\n\nMove the third digit of the dividend (3) down like so:\n\n 0 0 9 6 6 6 2 3 - 0 6 2 - 0 6 2 3\n\n### Step 10\n\nThe divisor (966) goes into the bottom number (623), 0 time(s), so we can put 0 on top:\n\n 0 0 0 9 6 6 6 2 3 - 0 6 2 - 0 6 2 3\n\n### Step 11\n\nIf we multiply the divisor by the result in the previous step (966 x 0 = 0), we can now add that answer below the dividend:\n\n 0 0 0 9 6 6 6 2 3 - 0 6 2 - 0 6 2 3 0\n\n### Step 12\n\nNext, we will subtract the result from the previous step from the fourth digit of the dividend (623 - 0 = 623) and write that answer below:\n\n 0 0 0 9 6 6 6 2 3 - 0 6 2 - 0 6 2 3 - 0 6 2 3\n\n## So, what is the answer to 623 divided by 966?\n\nIf you made it this far into the tutorial, well done! There are no more digits to move down from the dividend, which means we have completed the long division problem.\n\nYour answer is the top number, and any remainder will be the bottom number. So, for 623 divided by 966, the final solution is:\n\n0\n\nRemainder 623\n\nIf you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!\n\n• \"What is 623 Divided by 966 Using Long Division?\". VisualFractions.com. Accessed on May 21, 2022. http://visualfractions.com/calculator/long-division/what-is-623-divided-by-966-using-long-division/.\n\n• \"What is 623 Divided by 966 Using Long Division?\". VisualFractions.com, http://visualfractions.com/calculator/long-division/what-is-623-divided-by-966-using-long-division/. Accessed 21 May, 2022.\n\n• What is 623 Divided by 966 Using Long Division?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/long-division/what-is-623-divided-by-966-using-long-division/.\n\n## Extra calculations for you\n\nNow you've learned the long division approach to 623 divided by 966, here are a few other ways you might do the calculation:\n\n• Using a calculator, if you typed in 623 divided by 966, you'd get 0.6449.\n• You could also express 623/966 as a mixed fraction: 0 623/966\n• If you look at the mixed fraction 0 623/966, you'll see that the numerator is the same as the remainder (623), the denominator is our original divisor (966), and the whole number is our final answer (0).\n\n## Long Division Calculator\n\nEnter another long division problem to solve\n\n/\n\n## Next Long Division Problem\n\nEager for more long division but can't be bothered to type two numbers into the calculator above? No worries. Here's the next problem for you to solve:\n\nWhat is 623 divided by 967 using long division?\n\n## Random Long Division Problems\n\nIf you made it this far down the page then you must REALLY love long division problems, huh? Below are a bunch of randomly generated calculations for your long dividing pleasure:"
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https://www.epilepsydiary.biz/pyramidal-cells/dependence-of-paroxysms-duration-on-model-parameters.html | [
"## Dependence Of Paroxysms Duration On Model Parameters\n\nExponential distributions are characterized by a single variable, the distribution mean. In this section, the dependence of the mean duration of normal and paroxysmal epochs on model parameters is analyzed. Other quantifiers such as total paroxysmal duration or paroxysm incidence can be derived from these two quantifiers. The quantifiers chosen in this study have straightforward interpretation. The first one tells how long one can expect a paroxysm to last on average and therefore it corresponds to the probability of termination of a paroxysmal epoch. The second quantifier tells, once a paroxysm has finished, how long we can expect to wait for the next one to occur and therefore it corresponds to probability of initiation of seizure-like activity.\n\nFor the analysis of the dependence of the model's behavior on parameters, we selected six out of 65 model parameters. We selected the parameters that are either assumed to play a role in the pathophysiology of absence seizures in animals and humans, or are assumed to be targets of antiepileptic drugs, or are associated with seizure activation methods (sleep, hyperventilation). We varied one parameter at a time while all others were kept constant. For each parameter setting we simulated 24 hours of activity and created duration histograms from detected paroxysmal and normal epochs. The histograms were fitted with exponential distributions and the distributions' means were calculated. Each parameter (except cholinergic modulation) was manipulated such that the system's behavior varied from at least one paroxysmal event during 24 hours of activity to a state of continuous paroxysmal activity. The influence of a cholinergic neuromodulatory input originating from the brainstem mesencephalic cholinergic neurons was investigated by applying additional DC offset simultaneously to membrane potential in the TC and RE populations. This offset was varied in the range —4 to 4mV in the TC and in the range 8mV to — 8mV in the RE population (+1mV shift in TC corresponded to —2mV shift in RE). This is justified taking into account that acetylcholine released by cholinergic pathways decreases a potassium conductance in the TC cells that brings about depolarization of the TC population, while it increases a potassium conductance in the RE neurons and thus induces hyperpolarization of the RE population (McCormick and Prince, 1986, 1987). Results of the analysis are summarized in Figure 25.5.\n\nFor each parameter, two panels are presented where the mean duration of paroxysmal epochs (left panel) and of normal epochs (right panel) are shown. The change of a parameter is given in percentage of the corresponding reference value. An increase of the duration of paroxysms and a decrease of the intervals between paroxysms can result from a series of factors: a reduction of cortical GABAa inhibition (see Figure 25.5A, left panel), reduction of intra-RE GABAa inhibition (see Figure 25.5E, left panel), withdrawal of thalamic cholinergic modulation (see Figure 25.5F, left panel), an increase of the slope of the sigmoid in the cortical interneuronal population (see Figure 25.5B, left panel), an increase of burst firing in the RE or TC populations (see Figure 25.5C, D, left panel). The mean paroxysmal epoch duration is always monotonic function of parameter change. In some plots, this dependence is linear on a logarithmic scale (see Figure 25.5C,E) indicating that, in these cases, the mean epochs duration has exponential dependence on the varied parameter. The local slope of the graph is related to the sensitivity of the system to a change of the given parameter. From Figure 25.5A,B, it follows that the system is most sensitive to cortical GABAA inhibition and to the slope of the sigmoid in the population of cortical interneurons, since in these graphs the operating range along the x-axes are the smallest while the range of the y-axes are comparable to the other graphs.\n\nAnother important parameter in the model is the noise level. We analyzed the model's performance by varying the variance of one noise source (cortical input PCx) while assuming that the other noise source (sensory input P) had a variance equal to zero. The results are summarized in Figure 25.6 where, in the left column, phase portraits of the system are presented and, in the right column, model outputs are shown. The phase portraits were created by plotting the mean membrane potential of the population of interneurons (VInt) versus that of pyramidal cells population (VCx). Three panels\n\n0 as\n\n1 101",
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https://en.unionpedia.org/2_41_polytope | [
"Free",
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"Faster access than browser!\n\n# 2 41 polytope\n\nIn 8-dimensional geometry, the 241 is a uniform 8-polytope, constructed within the symmetry of the E8 group. \n\n## Configuration (polytope)\n\nIn geometry, H. S. M. Coxeter called a regular polytope a special kind of configuration.\n\n## Convex polytope\n\nA convex polytope is a special case of a polytope, having the additional property that it is also a convex set of points in the n-dimensional space Rn.\n\n## Coxeter element\n\nIn mathematics, the Coxeter number h is the order of a Coxeter element of an irreducible Coxeter group.\n\n## Coxeter group\n\nIn mathematics, a Coxeter group, named after H. S. M. Coxeter, is an abstract group that admits a formal description in terms of reflections (or kaleidoscopic mirrors).\n\n## Coxeter–Dynkin diagram\n\nIn geometry, a Coxeter–Dynkin diagram (or Coxeter diagram, Coxeter graph) is a graph with numerically labeled edges (called branches) representing the spatial relations between a collection of mirrors (or reflecting hyperplanes).\n\n## E6 polytope\n\nIn 6-dimensional geometry, there are 39 uniform polytopes with E6 symmetry.\n\n## E8 (mathematics)\n\nIn mathematics, E8 is any of several closely related exceptional simple Lie groups, linear algebraic groups or Lie algebras of dimension 248; the same notation is used for the corresponding root lattice, which has rank 8.\n\n## E8 polytope\n\nIn 8-dimensional geometry, there are 255 uniform polytopes with E8 symmetry.\n\n## Edge (geometry)\n\nIn geometry, an edge is a particular type of line segment joining two vertices in a polygon, polyhedron, or higher-dimensional polytope.\n\n## Emanuel Lodewijk Elte\n\nEmanuel Lodewijk Elte (16 March 1881 in Amsterdam – 9 April 1943 in Sobibór) at joodsmonument.nl was a Dutch mathematician.\n\n## Equilateral triangle\n\nIn geometry, an equilateral triangle is a triangle in which all three sides are equal.\n\n## Face (geometry)\n\nIn solid geometry, a face is a flat (planar) surface that forms part of the boundary of a solid object; a three-dimensional solid bounded exclusively by flat faces is a polyhedron.\n\n## Facet (geometry)\n\nIn geometry, a facet is a feature of a polyhedron, polytope, or related geometric structure, generally of dimension one less than the structure itself.\n\n## Geometry\n\nGeometry (from the γεωμετρία; geo- \"earth\", -metron \"measurement\") is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space.\n\n## Gosset–Elte figures\n\nIn geometry, the Gosset–Elte figures, named by Coxeter after Thorold Gosset and E. L. Elte, are a group of uniform polytopes which are not regular, generated by a Wythoff construction with mirrors all related by order-2 and order-3 dihedral angles.\n\n## Harold Scott MacDonald Coxeter\n\nHarold Scott MacDonald \"Donald\" Coxeter, FRS, FRSC, (February 9, 1907 – March 31, 2003) was a British-born Canadian geometer.\n\n## Hyperplane\n\nIn geometry, a hyperplane is a subspace whose dimension is one less than that of its ambient space.\n\n## Isosceles triangle\n\nIn geometry, an isosceles triangle is a triangle that has two sides of equal length.\n\n## Petrie polygon\n\nIn geometry, a Petrie polygon for a regular polytope of n dimensions is a skew polygon in which every (n – 1) consecutive sides (but no n) belongs to one of the facets.\n\n## Projection (linear algebra)\n\nIn linear algebra and functional analysis, a projection is a linear transformation P from a vector space to itself such that.\n\n## Pyramid (geometry)\n\nIn geometry, a pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex.\n\n## Rectification (geometry)\n\nIn Euclidean geometry, rectification or complete-truncation is the process of truncating a polytope by marking the midpoints of all its edges, and cutting off its vertices at those points.\n\n## Rectified 6-simplexes\n\nIn six-dimensional geometry, a rectified 6-simplex is a convex uniform 6-polytope, being a rectification of the regular 6-simplex.\n\n## Rectified 7-simplexes\n\nIn seven-dimensional geometry, a rectified 7-simplex is a convex uniform 7-polytope, being a rectification of the regular 7-simplex.\n\n## Rectified 8-orthoplexes\n\nIn eight-dimensional geometry, a rectified 8-orthoplex is a convex uniform 8-polytope, being a rectification of the regular 8-orthoplex.\n\n## Regular polygon\n\nIn Euclidean geometry, a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).\n\n## Schläfli symbol\n\nIn geometry, the Schläfli symbol is a notation of the form that defines regular polytopes and tessellations.\n\n## Tetrahedron\n\nIn geometry, a tetrahedron (plural: tetrahedra or tetrahedrons), also known as a triangular pyramid, is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners.\n\n## Triangle\n\nA triangle is a polygon with three edges and three vertices.\n\n## Uniform 2 k1 polytope\n\nIn geometry, 2k1 polytope is a uniform polytope in n dimensions (n.\n\n## Uniform 5-polytope\n\nIn geometry, a uniform 5-polytope is a five-dimensional uniform polytope.\n\n## Uniform 8-polytope\n\nIn eight-dimensional geometry, an eight-dimensional polytope or 8-polytope is a polytope contained by 7-polytope facets.\n\n## Uniform polytope\n\nA uniform polytope of dimension three or higher is a vertex-transitive polytope bounded by uniform facets.\n\n## Vertex (geometry)\n\nIn geometry, a vertex (plural: vertices or vertexes) is a point where two or more curves, lines, or edges meet.\n\n## Vertex figure\n\nIn geometry, a vertex figure, broadly speaking, is the figure exposed when a corner of a polyhedron or polytope is sliced off.\n\n## Wythoff construction\n\nIn geometry, a Wythoff construction, named after mathematician Willem Abraham Wythoff, is a method for constructing a uniform polyhedron or plane tiling.\n\n## 1 42 polytope\n\nIn 8-dimensional geometry, the 142 is a uniform 8-polytope, constructed within the symmetry of the E8 group.\n\n## 2 21 polytope\n\nIn 6-dimensional geometry, the 221 polytope is a uniform 6-polytope, constructed within the symmetry of the E6 group.\n\n## 2 31 polytope\n\nIn 7-dimensional geometry, 231 is a uniform polytope, constructed from the E7 group.\n\n## 2 51 honeycomb\n\nIn 8-dimensional geometry, the 251 honeycomb is a space-filling uniform tessellation.\n\n## 4 21 polytope\n\nIn 8-dimensional geometry, the 421 is a semiregular uniform 8-polytope, constructed within the symmetry of the E8 group.\n\n## 5-cell\n\nIn geometry, the 5-cell is a four-dimensional object bounded by 5 tetrahedral cells.\n\n## 5-orthoplex\n\nIn five-dimensional geometry, a 5-orthoplex, or 5-cross polytope, is a five-dimensional polytope with 10 vertices, 40 edges, 80 triangle faces, 80 tetrahedron cells, 32 5-cell 4-faces.\n\n## 5-simplex\n\nIn five-dimensional geometry, a 5-simplex is a self-dual regular 5-polytope.\n\n## 6-simplex\n\nIn geometry, a 6-simplex is a self-dual regular 6-polytope.\n\n## 7-demicube\n\nIn geometry, a demihepteract or 7-demicube is a uniform 7-polytope, constructed from the 7-hypercube (hepteract) with alternated vertices removed.\n\n## 7-simplex\n\nIn 7-dimensional geometry, a 7-simplex is a self-dual regular 7-polytope.\n\n## 8-orthoplex\n\nIn geometry, an 8-orthoplex or 8-cross polytope is a regular 8-polytope with 16 vertices, 112 edges, 448 triangle faces, 1120 tetrahedron cells, 1792 5-cells 4-faces, 1792 5-faces, 1024 6-faces, and 256 7-faces.\n\n## References\n\nHey! We are on Facebook now! »"
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https://mathhelpboards.com/threads/polynomial-rings.4768/ | [
"# Polynomial Rings\n\n#### Peter\n\n##### Well-known member\nMHB Site Helper\nProve that if f(x) and g(x) are polynomials with rational co-efficients whose product f(x)g(x) has integer co-efficients, then the product of any co-efficient of g(x) with any coefficient of f(x) is an integer.\n\nMy initial thoughts on this are that the exercise seems to be set up for an application of Gauss Lemma since we have that Z is a UFD with field of fractions Q and further we have $$\\displaystyle p(x) \\in Z[x]$$ where p(x) = f(x)g(x) and $$\\displaystyle f(x), g(x) \\in Q[x]$$.\n\nThus we apply Gauss Lemma (see attached) so\n\np(x) = (rf(x))(sg(x))\n\nwhere $$\\displaystyle rf(x), sg(x) \\in Z[x]$$\n\nPeter\n\n[This problem has also been posted on MHF]\n\nLast edited:\n\n##### Active member\nProve that if f(x) and g(x) are polynomials with rational co-efficients whose product f(x)g(x) has integer co-efficients, then the product of any co-efficient of g(x) with any coefficient of f(x) is an integer.\n\nMy initial thoughts on this are that the exercise seems to be set up for an application of Gauss Lemma since we have that Z is a UFD with field of fractions Q and further we have $$\\displaystyle p(x) \\in Z[x]$$ where p(x) = f(x)g(x) and $$\\displaystyle f(x), g(x) \\in Q[x]$$.\n\nThus we apply Gauss Lemma (see attached) so\n\np(x) = (rf(x))(sg(x))\n\nwhere $$\\displaystyle rf(x), sg(x) \\in Z[x]$$\n\nThe one thing I would think of is that given\n\n$$\\displaystyle p(x) \\in \\mathbb{Z}[x]$$ where $$\\displaystyle p(x)=f(x)g(x)$$ and $$\\displaystyle f(x), g(x) \\in \\mathbb{Q}[x]-\\mathbb{Z}[x]$$, there exist some $$\\displaystyle r,s \\in \\mathbb{Z}$$ so that $$\\displaystyle rf(x), sg(x) \\in \\mathbb{Z}[x]$$ are both primitive polynomials. Gauss's lemma tells us that $$\\displaystyle r s \\cdot p(x) = rf(x) \\cdot sg(x)$$ must be a primitive polynomial itself. It follows that $$\\displaystyle r,s=\\pm 1$$, which means that we have deduced that f and g must have integer coefficients after all.\n\nClearly, if what I said holds, the statement would follow. Either there is some flaw in my logic, or Gauss's Lemma is too powerful a tool for this problem...\n\n$$\\displaystyle p(x) \\in \\mathbb{Z}[x]$$ where $$\\displaystyle p(x)=f(x)g(x)$$ and $$\\displaystyle f(x), g(x) \\in \\mathbb{Q}[x]-\\mathbb{Z}[x]$$, there exist some $$\\displaystyle r,s \\in \\mathbb{Z}$$ so that $$\\displaystyle rf(x), sg(x) \\in \\mathbb{Z}[x]$$ are both primitive polynomials. Gauss's lemma tells us that $$\\displaystyle r s \\cdot p(x) = rf(x) \\cdot sg(x)$$ must be a primitive polynomial itself. It follows that $$\\displaystyle r,s=\\pm 1$$, which means that we have deduced that f and g must have integer coefficients after all.\nI made a mistake here: r and s are not necessarily integers. Since $r\\,f(x)$ and $s\\,g(x)$ are primitive, we can only guarantee that $r,s \\in \\mathbb{Q}$. This is still, however, sufficient; following the proof, we still find that $r\\,s=\\pm1$, which is enough to tell us that the product of a coefficient from one and a coefficient from the other is an integer."
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https://www.schneier.com/blog/archives/2009/10/the_current_sta.html/ | [
"## The Current Status of P Versus NP\n\nExcellent survey.\n\n“We could search all possible pairings but even for 40 students we would have more than 300 billion trillion possible pairings.”\n\nCould anyone explain this ? Not 780 ?\n\nbethan October 14, 2009 8:13 AM\n\n@Rin – at that point i realized it’s over my head, and read through the rest just because it’s so interesting.\n\n@Rin\n\nYou start with 40 students and choose 2 of them. 40 choose 2 = 780. Then you choose 2 from 38, which is 703. So for the first two pairs you have 780*703 possible outcomes. Do that 18 more times.\n\nJohn Greco October 14, 2009 8:30 AM\n\n@Rin\nWouldn’t it be 39! possible pairings? 39 possible partners for the first, * 38 possible pairings for the second, * 37 possible pairings for the third…\n\nI’m not sure that’s exactly right, but a number much much larger than 780 definetly makes sense to me.\n\nAnthony October 14, 2009 8:34 AM\n\nyou need to divide your 40 students in 2 groups of 20: 40!/(20!)^2 possibilities\n\nyou order your students 1, 2, …, 20 in group A, and you pair each one with a student in group B: 20! possibilities\n\nyou need to divide by 2^20 because you don’t care whether a student belongs to group A or B\n\nFinally, you get 40! /(2^20 20!) = 3.2 10^23\n\n@Rin: 40 choose 2.\n\nFrédéric Grosshans October 14, 2009 8:43 AM\n\n@Rin\n360 (40 * 39/2) is the number of pairs, not the number of pairing, which is 393735753*1\n\nFor example, if we have 6 student (ABCD), the pairings are\nAB, CD, EF\nAB, CE, DF\nAB, CF, DE\nAC, BD, EF\nAC, BE, DF\nAC, BF, DE\nAE, BC, DF\nAE, BD, CF\nAE, BF, CE\nAF, BC, DE\nAF, BD, CE\nAF, BE, CD\n\nand the pairs are\nAB AC AD AE AF BC BD BE BF CD CE CF DE DF EF\n\nCoincidently, for n=6, the numbers are the same (15). A better exemple would have been with higher n, with e.g. 105 pairings and 28 pairs for n=8, but it starts to be too big for me to make by hand…\n\nClive Robinson October 14, 2009 8:43 AM\n\nI’m of an age where I was at school when NP first came up…\n\nAnd I remember some one (not me) reworked the Shakespear “piss take” of,\n\nTo pee or not to pee…\n\nAs,\n\n2 P or Not 2 P…\n\nSuch are ones school memories 8(\n\nClive Robinson October 14, 2009 8:51 AM\n\nI can understand the confusion with the number of pairings…\n\nIf we assume that we infact have two groups one of 20 men and one of 20 women and a pair has to be one from each group…\n\nBut you could then say that there could be say four groups of ten etc…\n\nAndrew Gumbrell October 14, 2009 9:10 AM\n\nIt is quite simple to calculate the possible pairings (if your calculator or head can stand it). Take any one student. They have a choice of 39 possible partners. Next, there will be 38 students left. Take one of those who has a choice of 37 partners amd so on. So the number of possible pairings is:\n393735 … 53 =\n319830986772877770815625\n\nSteveJ October 14, 2009 9:52 AM\n\nDoes P=NP imply a general means to reduce algorithms of high polynomial complexity to a lower polynomial complexity?\n\nIf not, then the article makes a number of false claims to “sex up” the subject. Polynomial time does not mean fast, as anyone knows who has ever compared bubble sort with introsort on a decent-sized set of data.\n\nAll that stuff about how if you prove P=NP, then we’ll live in utopia, is rubbish. Suppose you find an algorithm for an NP-complete problem which is Theta(N^googelplex). Then the fact that P=NP is useless.\n\nExisting O(2^N) algorithms would be faster than the new polynomial-time algorithms for any N we’ll encounter in our lifetimes. And unless you believe computing power increases exponentially, forever, in defiance of current theories of entropy and information, then it will always be useless.\n\n@SteveJ\n\nYes it does to your first question. More or less.\n\nFrank Meulenaar October 14, 2009 10:47 AM\n\n@SteveJ\n\nYour arguments are correct, in the same way that a runtime of Exp[N] can be better than googleplex N^2. However, it turns out that for the algorithms we know, it IS useful to decide between polynomial and exponential, because the former usually is much faster than the second.\n\nFurthermore, even if we would proof P=NP, that doesn’t mean we know HOW to reduce exponential to polynomial – only that it exists.\n\nPaeniteo October 14, 2009 10:53 AM\n\n@SteveJ: “Polynomial time does not mean fast, as anyone knows who has ever compared bubble sort with introsort on a decent-sized set of data.”\n\nHowever, for a decent-sized set of data, polynomial will mean “faster” than non-polynomial. The statement is purely relative.\n\nFor example, if you use “BruteSort” as a comparison, “BubbleSort” will seem blazingly fast.\n(BruteSort is something which I just made up. Roughly speaking, it would enumerate through all possible permutations of the dataset and then check if the permutated set is sorted.)\n\nThat’s the Jargon File btw…\n\nSteveJ October 14, 2009 2:54 PM\n\n@Frank: “Furthermore, even if we would proof P=NP, that doesn’t mean we know HOW to reduce exponential to polynomial – only that it exists.”\n\nIn this case actually existence leads directly to construction – if we know that a polynomial-time algorithm exists, then we can provide one.\n\nThe proof is fairly simple. Take every single program that exists (for some Turing machine), and number them 1, 2, …\n\nExecute the first step of program 1 on the specified input, followed by the first step of program 2 (on a separate copy of the input for each program), followed by step 2 of program 1, followed by step 1 of program 3, followed by step 2 of program 2, and so on in a “triangle”.\n\nIf any program halts, then check (in polynomial time) whether its output is a solution the the NP-complete question in question.\n\nThen if any polynomial-time algorithm exists to solve the problem, the procedure described above is a polynomial-time algorithm.\n\nSteveJ October 14, 2009 3:17 PM\n\n@Paeniteo: “for a decent-sized set of data, polynomial will mean “faster” than non-polynomial.”\n\nFor a sufficiently large input, polynomial is faster than non-polynomial. The reason I mentioned Theta(N^googolplex) complexity specifically, is because if we were to compare a polynomial algorithm taking exactly N^googolplex steps, with an exponential algorithm taking exactly 2^N steps, then N has to be considerably greater than a googolplex before 2^N > N^googolplex. So the exponential algorithm is certainly faster for input data up to a googolplex bits in size.\n\n@ SteveJ\n\nYour triangularization is cute, but not very practical, because the constant factor ignored by the O() complexity notation will depend exponentially on the size of the polynomial-time Turing machine you search for using the triangularization. (Assuming that the enumeration procedure for the Turing machines is enumerating them based on size, smaller machines first).\n\nOr am I missing something?\n\nCould someone explain to me why the author of the article states that public-key cryptography would be impossible if P=NP? Especially given that he explains in a different place in the article that the hard problems used in current systems (factoring, discrete log) aren’t even expected to be NP-complete?\n\nGordon October 15, 2009 8:54 AM\n\nFrom the articles examples of NP problems: “Finding a DNA sequence that best fits a collection of fragments of the sequence (see Gusfield20). ” Doesn’t this imply the brute force conception of evolution is not mathmatically sound?\n\n@Gordon: GA’s (Genetic Algorithms, aka Evolutionary Programming) get stuck in local minima all the time. They are an efficient way of searching a large space. They are not guaranteed to converge to the optimal solution. (In nature, good enough is, well, good enough.)\n\nBoth the Moose and the Hippo occupy the same ecological niche. They eat similar foods. They don’t exactly look alike though. Even accounting for the climate differences.\n\nDavid October 15, 2009 9:35 AM\n\n@Frank: If we had a polynomial-time solution to one NP-complete problem, we could use it to design a polynomial-time solution to any problem in NP. You prove a problem X is NP-complete by proving that if you have a polynomial-time algorithm to solve X you therefore have one to solve Y, where Y is NP-complete.\n\n@RonK: If, as most people believe, P is a proper subset of NP, then there’s classes of problems between P and NP-complete with complexity that scales up. Therefore, if P != NP, a given problem doesn’t have to be either in P or be NP-complete, and there’s at least some reason to believe that factorization doesn’t have a polynomial solution.\n\nOn the other hand, if P=NP, then all problems in NP have polynomial-complexity solutions, and that does include factoring and discrete logs.\n\nGordon October 15, 2009 10:43 AM\n\n@D: Given the complexity of DNA, “good enough” is still a very small subset of the possibilities, so evolution would still seem to be a complex NP problem, as the article suggests. A single molecule out of place can cause severe/fatal damage (e.g. Sickle Cell Anemia). It is not clear why an NP problem becomes solvable by brute force when there are more than 1, but still relatively few, good solutions. Even on a billion year time frame with a billion trials per year, you would still need more than a thousand such iterations to solve the relatively simple 40 student problem mentioned at the beginning of the article. I guess I am trying to figure out some way to fit the math of evolution into this framework that makes sense. Rigth now I don’t see it (I realize that may be of interest to no one but me).\n\nGordon October 15, 2009 11:28 AM\n\nOK, one more post then I’ll leave it alone. Conjecture: DNA uses some sort of an algorithm to keep mutations in the subset that yields a more or less viable life form, making the genetic underpinnings of natural selection mathematically possible. One could then argue that since evolution happens, P=NP.\n\nJT Pennington October 15, 2009 1:02 PM\n\nWow, the utopia that they describe sounds just like campaign promises during the election year.\n\nIf you elect me… I will prove, that P=NP. The world will be better and brighter. Now is the time for change. The other canidates say we can’t do it… to that I say “Yes we can.”\n\n@ David\n\nAh, thanks, I had forgotten that checking whether a factorization was correct was poly-time. It still strikes me as a strange assumption that public key encryption requires a hard problem which is not in P. I vaguely remember a paper which proposed a (not very practical) poly-time public-key cryptosystem where there was a provable gap in the order of the decryption complexity between an attacker without the private key (having O(n^2) complexity?) and with the private key (having O(n) complexity?). Unfortunately, I don’t manage to find this paper anywhere.\n\nIt seems to me possible to have a relatively workable public-key cryptosystem based on the difficulty of a problem with complexity O(n^20) vs. O(n^2), for example, which could provide a lot of the functionality which we have today. There would be a lot more headache to constantly roll-over keys, update the security parameters, and have to deal with very large keys and messages, but it would still be possible.\n\nDavid Harmon October 15, 2009 5:29 PM\n\nGordon: Nope — evolution has no problem with throwing away non-viable eggs/spores/embryos. Even for humans, a significant fraction of conceptions are non-viable — typically, the woman never knows she was pregnant. For lower orders… well, when you’re producing 1000 eggs in a season, even, say, 25% failure up front still leaves 750 chances to evade predators and so on.\n\naverros October 16, 2009 5:30 AM\n\nPublic key crypto doesn’t require P!=NP or anything like that – I’m not sure where that myth comes from.\n\nAll it requires that decryption without a private key took so much longer so as to make it impractical while encryption is still fast.\n\n@ averros\n\nI largely agree with your comment (see my comment above) but there are certain properties of public key crypto today which we would probably lose, like good forward secrecy for one-sided messages (i.e., messages generated without a key exchange dialog). This is because computing power is still increasing exponentially with time, so limiting the difficulty of decryption to poly-time is a problem.\n\nGordon October 18, 2009 8:54 AM\n\n@David-Harmon: A single human chromosome is 22 million base pairs long. A vanishingly small set of potential orders of that DNA is viable. If every mamal on earth had a thousand trials per year, it would still take something on the order of 2^1,000,000 years to sort through finding a single viable life form by brute force (depending on your assumptions). That nature has no problem discarding non-viable life forms does not mean it can brute force its way through these enormous numbers. It must limit itself in some way to a potentially viable subset.\n\nDespite doing a computer science degree (it was 20 years ago, so a lot of it is rather hazy now), a lot of that article was gibberish to me. Can Bruce or anyone recommend some good background reading to help me make better sense of it?\n\nI’m sure it would help me to be a better programmer as well. Since most of the modern programmer’s time is spent understanding and manipulating provided APIs rather than developing new algorithms, the mathematical side of my programming is rather weak.\n\nSidebar photo of Bruce Schneier by Joe MacInnis."
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https://git.lighttpd.net/mirrors/libowfat/src/commit/c479eab1b5d38732b1ebce369168c3cc58d476c8/byte/byte_copy.c | [
"Mirror of :pserver:[email protected]:/cvs libowfat https://www.fefe.de/libowfat/\nYou can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.\n\n#### 32 lines 861 B Raw Blame History\n\n `#include \"byte.h\"` `/* byte_copy copies in to out, in to out, ... and in[len-1]` ` * to out[len-1]. */` `void byte_copy(void* out, unsigned int len, const void* in) {` ` register char* s=out;` ` register const char* t=in;` ` register const char* u=t+len;` `#ifdef __i386__` ` if (len>127) {` ` if (sizeof(unsigned long)>4) { /* a good compiler should optimize this check away */` ` for (;(unsigned long)t&7;) {` ` if (t==u) break; *s=*t; ++s; ++t;` ` }` ` } else {` ` for (;(unsigned long)t&3;) {` ` if (t==u) break; *s=*t; ++s; ++t;` ` }` ` }` ` while (t+sizeof(long)<=u) {` ` *(unsigned long*)s=*(unsigned long*)t;` ` s+=sizeof(long); t+=sizeof(long);` ` }` ` }` `#endif` ` for (;;) {` ` if (t==u) break; *s=*t; ++s; ++t;` ` if (t==u) break; *s=*t; ++s; ++t;` ` if (t==u) break; *s=*t; ++s; ++t;` ` if (t==u) break; *s=*t; ++s; ++t;` ` }` ```} ``` ``` ```"
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https://satsconference2017.co.za/2019/35719-what-s-the-molar-mass-of-iron-iii-acetate.html | [
"# what 39 s the molar mass of iron iii acetate\n\n• ### CHM 161 Spectrophotometry: Analysis of Iron(II) in an ...\n\nCHM 161 Spectrophotometry: Analysis of Iron(II) in an Aqueous Solution Introduction ... In this experiment you will determine the mass of iron in a commercial iron supplement pill ... substance, and a is the molar absorptivity of the solute. The molar absorptivity is a constant for\n\nGet Price\n• ### NOTES – MOLAR MASS AND MOLE CONVERSIONS\n\nMOLAR MASS AND % COMPOSITION HOMEWORK NAME: DATE: 1. Label each compound below as ionic or molecular. Write the formula and determine the molar mass of each compound. COMPOUND IONIC/ MOLECULAR? FORMULA MOLAR MASS barium sulfate. I BaSO4 g/mol tricarbon octahydride. C C3H8 44 g/mol iron (III) carbonate. I Fe2(CO3)3 g/mol\n\nGet Price\n• ### Iron (III) hydroxide is soluble in acid: Fe(OH)3(s)3H^+(aq ...\n\nNov 11, 2012· asked by Nadia on November 3, 2013; Chemistry. Iron 3 hydroxide reacts with acetic acid to form iron 3 acetate and water. If of water are formed,what mass of iron 3 acetate will be produced? •Fe(OH)3+3HC2H3O2>Fe(C2H3O2)+3H2O ( ? Fe(C2H3O)) asked by G on January 18, 2015; chemistry\n\nGet Price\n• ### Iron(III) nitrate nonahydrate | FeH18N3O18 PubChem\n\nUse of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for noncommercial purposes provided that ECHA is ...\n\nGet Price\n• ### What S The Molar Mass Of Iron Iii Acetate\n\nwhat is the molar mass iron iii acetate . molar mass of iron ii acetate. what is the molar mass iron iii acetate, Calculate the molar mass of Iron(II) Acetate in grams per mole or search for a chemical formula or substance Know More. Contact Supplier Biochemistry Ch. 3 Practice Exam Flashcards | Quizlet. Get Free Chat\n\nGet Price\n• ### what is the molar mass iron iii acetate\n\nwhat s the molar mass of iron iii acetate. molar mass of iron ii acetate molar mass of iron ii acetate Our Products >Ester Wikipedia Ester names are derived from the parent alcohol and the parent acid, where the latter may be organic or inorganic , the latter of which consists of hydrated iron(III…\n\nGet Price\n• ### Chemistry Chapter 7 Flashcards | Quizlet\n\nStart studying Chemistry Chapter 7. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ... How many atoms are present in one formula unit of barium acetate, Ba(C2H3O2)2? 15. The salt calcium nitrate, Ca(NO3)2, contains the anion from ... What is the percentage of iron in iron(III) chloride, FeCl3? The molar mass of ...\n\nGet Price\n• ### What is the molar mass of iron? Quora\n\nMar 31, 2017· According to Wikipedia, there are 4 stable isotopes of iron in the environment: * iron54, atomic mass , abundance * iron56, atomic mass , abundance * iron57, atomic mass , abundance * iron58, ato...\n\nGet Price\n• ### Molar Mass Problems ChemTeam\n\nMolar Mass Problems 1. Calculate the mass of mole of CaCl2 2. Calculate grams in moles of CO2 3. Calculate number of moles in g of CH4 4. Determine mass in grams of moles of Na2CO3 5. Calculate moles in g of HgS 6. Calculate moles in g of Al2S3 7. How many moles are in g of H2O 8.\n\nGet Price\n• ### what 39 s the molar mass of iron iii acetate\n\nwhat 39 s the molar mass of iron iii acetate. SCH 4C1 Unit 2 Problem Set – Upper Grand District School Board. A typical bottle of nail polish remover contains mol of ethyl acetate (C 4 H 8 O 2). …\n\nGet Price\n• ### Laboratory Solution • Basic concepts of preparing ...\n\n2. Determine concentration in percent by mass of the solute in solution. Change to the decimal equivalent. 3. Calculate the molar mass of the compound, MM. 4. Multiply mass (step 1) by mass % (step 2) and divide by molecular mass (step 3) to find the number of moles present in the whole solution. 5.\n\nGet Price\n• ### What is the molar mass of Iron (III) Oxide (Fe2O3)? Quora\n\nDec 14, 2015· Here there are 2 Fe(Iron) atoms and 3 O(Oxygen) atoms, so we are going to multiply the Atomic mass of Iron and Oxygen and add them Atomic mass Iron: 56 Therefore total mass of Iron atoms in molecule= 56*2 ...\n\nGet Price\n• ### Molecular weight of Iron(II) Acetate Convert Units\n\n››More information on molar mass and molecular weight. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.\n\nGet Price\n• ### FERRIC ACETATE | C6H9FeO6 | ChemSpider\n\nStructure, properties, spectra, suppliers and links for: FERRIC ACETATE, 1834306.\n\nGet Price\n• ### What Is the Formula for Iron (III) Acetate? |\n\nIron(III) acetate is most often called basic iron acetate, along with basic iron(III) acetate and iron(III) oxyacetate. It is a solid salt that appears as a brownishred powder. The formula for the compound iron(III) acetate shows that each molecule contains six atoms of carbon, nine atoms of hydrogen, one atom of iron and six atoms of oxygen.\n\nGet Price\n• ### How do you find the molar mass of Iron II acetate? | Yahoo ...\n\nMay 30, 2011· Best Answer: Avogadro's number tells you how many molecules are in a mole. You don't need this here. The molar mass is the weight of the all of the molecules in one mole of the product. Fortunately for you, the molar mass is just the total of the masses of atoms as …\n\nGet Price\n• ### Iron(II) Acetate Composition and Molar Mass\n\nCalculatrice compacte de conversion d’unités de mesure Molar Mass Calculator. ... Iron(II) Acetate Composition and Molar Mass Chemical formula. Molar mass of Fe(C 2 H 3 O 2) 2, Iron(II) Acetate is g/mol. 55,845+(12,0107·2+1,00794·3+15,9994·2)·2. Mass percentage of the elements in …\n\nGet Price\n• ### Tutorial 3 THE MOLE AND STOICHIOMETRY EIU\n\nTutorial 3 THE MOLE AND STOICHIOMETRY A chemical equation shows the reactants (left side) and products (right side) in a chemical ... 3 1 mol FeCl 3 1 mol NaOH = g NaOH Can you see why the answer is different when all three steps are combined? EXAMPLE: What mass of iron(III) hydroxide will be formed from g of iron(III) chloride?\n\nGet Price\n• ### Iron(III) Acetate Fe(C2H3O2)3 EndMemo\n\nIron(III) Acetate Fe(C2H3O2)3 Molar Mass, Chemical Characteristics of Iron(III) Acetate. ENDMEMO. Home » Chemistry » Iron(III) Acetate. Name: Iron(III) Acetate. Alias: Ferric Acetate. Formula: Fe(C2H3O2)3. Molar Mass: Iron(III) Acetate is a brownish red amorphous powder at room temperature. It's insoluble in water, and soluble in ...\n\nGet Price\n• ### What Is the Molar Mass of Fe2O3? |\n\nThe molar mass of Fe2O3, which is the chemical formula for iron(III) oxide, is grams per mole. The molar mass is determined by adding together the atomic weight of each atom in the compound. In Fe2O3, the Fe stands for iron and the O stands for oxygen.\n\nGet Price\n• ### Fe(OH)2 Iron(ii) Hydroxide Molar Mass\n\nIt will calculate the total mass along with the elemental composition and mass of each element in the compound. Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. You can use parenthesis or brackets []. Finding Molar Mass. Read our article on how to calculate molar ...\n\nGet Price\n• ### Iron(II) acetate Wikipedia\n\nIron(II) acetate is a coordination complex with formula Fe(C 2 H 3 O 2) 2. It is a white solid, although impure samples can be slightly colored. It is a white solid, …\n\nGet Price\n• ### Naming Ionic Compounds\n\nNaming Ionic Compounds – Answer Key Give the name and molar mass of the following ionic compounds: Name Molar Mass 1) Na 2 CO 3 sodium carbonate 129 grams/mole 2) NaOH sodium hydroxide 40 grams/mole 3) MgBr 2 magnesium bromide grams/mole 4) KCl potassium chloride grams/mole 5) FeCl 2\n\nGet Price\n• ### How to find the molar mass of FeCl3 |\n\nThe molar mass of iron(III) chloride is g/mol. To find the molar mass of FeCl 3, we multiply the molar mass of chlorine by three and add that.... See full answer below.\n\nGet Price\n• ### What is the molar mass of iron?\n\nThe molar mass of Fe2O3 is (molar mass of iron)*2+16(molar mass of oxygen)*3, which comes out to be grams per mole. Multiply that by the number of …\n\nGet Price\n• ### Wow Mole Questions Google Docs\n\n27. Identify the quantity calculated by dividing the molar mass of an element by Avogadro’s number. Because the molar mass is a ratio of grams per mole and Avogadro’s number is a ratio of particles per mole, dividing the molar mass of an element by Avogadro’s number yields the mass of a single representative particle of that element. 28.\n\nGet Price\n• ### Iron(III) ion | Fe | ChemSpider\n\nStructure, properties, spectra, suppliers and links for: Iron(III) ion, 20074526.\n\nGet Price\n• ### Molar Mass Worksheet Ms. Richardson's Science\n\n4. Determine mass in grams of moles of Na 2 CO 3 5. Calculate moles in g of HgS 6. Calculate moles in g of Al 2 S 3 7. How many moles are in g of H 2 O 8. Determine the mass in grams of Avogadro number of C 12 H 22 O 11 9. Find mass in grams of moles of H 2 S 10. Determine grams in mole of NH 3 Consider the ...\n\nGet Price\n• ### Molecular weight of Iron(III Acetate)\n\n››More information on molar mass and molecular weight. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.\n\nGet Price\n• ### What is the molar mass of Iron III Oxide Answers\n\nIron(III) oxide, Fe2O3 has molar mass Amount of Fe2O3 in 1g = 1/ = . What mass (in grams) of iron(III) oxide contains of iron? This mass is 90 g.\n\nGet Price\n• ### Chapter 3 Molar Mass Calculation of Molar Masses\n\nMolar mass Molar ratio g mol C6H12 O6 = mol O Strategy: Once the number of moles of a substance present is known, we can use: – Molar mass to find the number of grams – Avogadro’s number to find the number of atoms, ions, or molecules Moles B Grams B Atoms B Molar mass NA Moles A Molar Ratio Molar Conversions Chapter 3 H? Grams ...\n\nGet Price\n• ### What is the molar mass and the mass of mol of iron(II ...\n\nApr 10, 2018· the atomic masses of the Periodic Table can therefore show the molar mass of an element, in grams per mole of atoms. the atomic mass of iron Fe is 56. the formula mass (molar mass of a compound) of Fe_3 is 3 * 56, which is 168. the atomic mass of phosphorus P is 31. the atomic mass of oxygen O is 16.\n\nGet Price\n• ### Iron(III) acetate Wikipedia\n\nBasic iron acetate forms on treating aqueous solutions of iron(III) sources with acetate salts. Solid iron may be mixed with hydrogen peroxide to form iron(II and/or III) hydroxide, which can then react with vinegar/acetic acid or acetate salts to form iron(III) acetate. Early work showed that it is trinuclear.\n\nGet Price"
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https://www.learncpp.com/cpp-tutorial/arithmetic-operators/comment-page-2/ | [
"# 5.2 — Arithmetic operators\n\nUnary arithmetic operators\n\nThere are two unary arithmetic operators, plus (+), and minus (-). As a reminder, unary operators are operators that only take one operand.\n\nOperator Symbol Form Operation\nUnary plus + +x Value of x\nUnary minus - -x Negation of x\n\nThe unary minus operator returns the operand multiplied by -1. In other words, if x = 5, -x is -5.\n\nThe unary plus operator returns the value of the operand. In other words, +5 is 5, and +x is x. Generally you won’t need to use this operator since it’s redundant. It was added largely to provide symmetry with the unary minus operator.\n\nFor best effect, both of these operators should be placed immediately preceding the operand (e.g. `-x`, not `- x`).\n\nDo not confuse the unary minus operator with the binary subtraction operator, which uses the same symbol. For example, in the expression `x = 5 - -3;`, the first minus is the binary subtraction operator, and the second is the unary minus operator.\n\nBinary arithmetic operators\n\nThere are 5 binary arithmetic operators. Binary operators are operators that take a left and right operand.\n\nOperator Symbol Form Operation\nAddition + x + y x plus y\nSubtraction - x - y x minus y\nMultiplication * x * y x multiplied by y\nDivision / x / y x divided by y\nModulus (Remainder) % x % y The remainder of x divided by y\n\nThe addition, subtraction, and multiplication operators work just like they do in real life, with no caveats.\n\nDivision and modulus (remainder) need some additional explanation. We’ll talk about division below, and modulus in the next lesson.\n\nInteger and floating point division\n\nIt is easiest to think of the division operator as having two different “modes”.\n\nIf either (or both) of the operands are floating point values, the division operator performs floating point division. Floating point division returns a floating point value, and the fraction is kept. For example, `7.0 / 4 = 1.75`, `7 / 4.0 = 1.75`, and `7.0 / 4.0 = 1.75`. As with all floating point arithmetic operations, rounding errors may occur.\n\nIf both of the operands are integers, the division operator performs integer division instead. Integer division drops any fractions and returns an integer value. For example, `7 / 4 = 1` because the fractional portion of the result is dropped. Similarly, `-7 / 4 = -1` because the fraction is dropped.\n\nWarning\n\nPrior to C++11, integer division with a negative operand could round up or down. Thus `-5 / 3` could result in -1 or -2. This was fixed in C++11, which always drops the fraction (rounds towards 0).\n\nUsing static_cast<> to do floating point division with integers\n\nThe above raises the question -- if we have two integers, and want to divide them without losing the fraction, how would we do so?\n\nIn lesson 4.11 -- Chars, we showed how we could use the static_cast<> operator to convert a char into an integer so it would print as an integer rather than a character.\n\nWe can similarly use static_cast<> to convert an integer to a floating point number so that we can do floating point division instead of integer division. Consider the following code:\n\nThis produces the result:\n\n```int / int = 1\ndouble / int = 1.75\nint / double = 1.75\ndouble / double = 1.75\n```\n\nThe above illustrates that if either operand is a floating point number, the result will be floating point division, not integer division.\n\nDividing by zero\n\nTrying to divide by 0 (or 0.0) will generally cause your program to crash, as the results are mathematically undefined!\n\nIf you run the above program and enter 0, your program will either crash or terminate abnormally. Go ahead and try it, it won’t harm your computer.\n\nArithmetic assignment operators\n\nOperator Symbol Form Operation\nAssignment = x = y Assign value y to x\nSubtraction assignment -= x -= y Subtract y from x\nMultiplication assignment *= x *= y Multiply x by y\nDivision assignment /= x /= y Divide x by y\nModulus assignment %= x %= y Put the remainder of x / y in x\n\nUp to this point, when you’ve needed to add 4 to a variable, you’ve likely done the following:\n\nThis works, but it’s a little clunky, and takes two operators to execute (operator+, and operator=).\n\nBecause writing statements such as `x = x + 4` is so common, C++ provides five arithmetic assignment operators for convenience. Instead of writing `x = x + 4`, you can write `x += 4`. Instead of `x = x * y`, you can write `x *= y`.\n\nThus, the above becomes:",
null,
"5.3 -- Modulus and Exponentiation",
null,
"Index",
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"5.1 -- Operator precedence and associativity\n\n### 298 comments to 5.2 — Arithmetic operators\n\n•",
null,
"Celaena\n\nHi! I tried to write the second exercise without bool like this\n\nand it works for numbers with 10 digits, but for the bigger numbers it outputs only \"You entered an even number\". Can you please explain whats wrong with it?\n\n•",
null,
"Alex\n\nIf you input a number that's too large, you'll overflow integer variable x and the value won't be what you expect. If you print the value of x, you should be able to see this happening.\n\n•",
null,
"Shane\n\nI guess I overcomplicated mine a bit, but I like having separate functions.\n\n•",
null,
"Cas\n\nHi; please tell me what's wrong with my writeAnswer() function. isEven evaluates correctly, but writeAnswer thinks evertything is even, even when isEven returns false. I have a feeling I'm being really dumb and not seeing something simple.\n\nfunctions.h\n\nfunctions.cpp\n\nsource.cpp\n\n•",
null,
"Alex\n\nThis will always evaluate to true. This says, \"if the isEven function has an address greater than 0\" (which will be always if the isEven function exists).\n\nHere's what you wanted:\n\nThis says, \"if the return value of the isEven function is true\", which is what you actually wanted.\n\n•",
null,
"Cas\n\nTried this previously and got the compile error \"function does not take 0 arguements\".\n\n•",
null,
"Alex\n\nOh, you need to pass function isEven() an argument:\n\n•",
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"Cas\n\nAlso tried that but compiler says \"identifier \"x\" is undefined\" (using visual studio)\n\n•",
null,
"Alex\n\nThat's because your writeAnswer() function is missing parameter x, which should be passed in from main().\n\n•",
null,
"Himanshu\n\nHi Alex. Thanks for the great tutorials.\n\nI have a few doubts.\n\n1) In this piece of code:\n\nYou've used the same name for a function and a variable in it. Doesn't that cause a naming conflict?\n\n2)\n\nHere, can't we do\n\n•",
null,
"Alex\n\n1) No. The local variable name shadows the function name. We discuss shadowing in chapter 4.\n2) No. That does integer division on x / y and then casts the result to a double. We want to do double division, which means at least one of the operands must be a double before the division operation takes place.\n\n•",
null,
"Himanshu\n\nGot it, thanks!\n\n•",
null,
"Sooi Shanghei\n\nit would be less confusing if the hint in quiz #2 linked to section 2.6 instead 2.7\n\n•",
null,
"Alex\n\nAgreed. Thanks for pointing this out!\n\n•",
null,
"Daniel\n\nAfter reading this multiple times, I still have no idea how the modulus (remainder) works. Could you explain it differently or post a link that explains it in depth?\n\nEDIT:\nEven though I'm not quite understanding the modulus, I managed to complete the second quiz haha :D\n[code]#include\n\nusing namespace std;\n\nbool isEven(int number)\n{\nif (number % 2 == 0)\nreturn true;\nelse\nreturn false;\n}\n\nint main()\n{\ncout <> number;\nif (isEven(number))\ncout << number << \" is even.n\";\nelse\ncout << number << \" is odd.n\";\n\nreturn 0;\n}[code]\n\n•",
null,
"Alex\n\nModulus is just another name for the remainder left over after doing an integer division.\n\nSo, 7 / 4 = 1, because the fraction is dropped.\n7 % 4 = 3, because 7 / 4 is 1, leaving a remainder of 3. That 3 is the result of the modulus.\n\n•",
null,
"Hema\n\nThe output was 1 when val was equal to 9.\nCan you explain me how?\n\n•",
null,
"Alex\n\n1/2 does integer division and results in 0 (1/2 is 0.5, but integer division drops the fractional portion, so you're left with 0). Any number to the 0th power is 1.\n\nYou probably meant to do 1.0/2.0 instead of 1/2, so you get floating point division.\n\n•",
null,
"Hema\n\nWhy can't we use integers in the place of chars?\n\n•",
null,
"Alex\n\nTo do what? I'm not clear on the context of your question.\n\n•",
null,
"Hema\n\nIn every case. Chars serve the same purpose as that of integers. So, why not use int.\n\n•",
null,
"Alex\n\nThey don't serve the same purpose.\n* Chars are only 8-bits, where integers are minimum 16 bits.\n* Chars print as ASCII characters, whereas integers print as numbers.\n* Strings contain chars, not integers.\n\n•",
null,
"Nicolas\n\nWell this was my first attempt at question 2, but I think I missed the point about having a function return true :P great tutorial again!\n\n•",
null,
"Mitch\n\nCouldn't the isEven function be simplified further?\n\nbool isEven(int x)\n{\nreturn (x % 2);\n}\n\nIt still returns the correct true/false information.\n\n•",
null,
"Alex\n\nIt could be simplified to the following:\n\n•",
null,
"MSteven\n\nWhy This? Can you explain me, please!\n\n•",
null,
"Alex\n\nIf x is even, then x % 2 = 0, which the ! flips to a 1, which becomes boolean true.\nIf x is odd, then x % 2 = 1, which the ! flips to a 0, which becomes boolean false.\n\n•",
null,
"MSteven\n\nIt seems like i the function below !(x % 2) evaluates to 1 as function return value. Consequently By default, a bool function return 0(false) while I think the opposite\n\n•",
null,
"Alex\n\nI'm not sure I follow. !(x % 2) only evaluates to 1 if x is even.\n\nFor integer to boolean conversions, 0 becomes false, and any non-zero values becomes true.\n\n•",
null,
"MSteven\n\nExcuse me, I wanted to mean why compilers interpret\n\nto\n\nimplicitly?\n\n•",
null,
"Alex\n\nI'm sorry, I don't understand what you're asking.\n\n•",
null,
"MSteven\n\nHi Alex, here's a code, can you tell me if the code is pretty , a good code factorized and commented if no, how do I do for ameliorating it?\n\n•",
null,
"Alex\n\nmain() and continueOrStop() are pretty redundant -- you could consolidate them.\n\nI'd also put the comments on the actual functions rather than the forward declarations.\n\nOther than that, I don't see anything particularly noteworthy.\n\n•",
null,
"John Moore\n\nGreetings,\n\nYour quiz question 2 stated that we will need to use if statements in our code. However I coded this without if statements. Perhaps I misunderstood the question? The following is my code:\n\n•",
null,
"Alex\n\nYour program doesn't tell the user whether the number is even or odd, it tells them 1 or 0 representing whether the number is even or odd.\n\nBut that's also fixable without using an if statement (by using the conditional operator), so yes, there is a valid solution that doesn't use \"if\".\n\n•",
null,
"James Ray\n\nI posted an answer to a question about code for this tutorial here.\n\n•",
null,
"Jim Smith\n\nI like to make loops to allow the user to decide when the program should end.\n\n•",
null,
"antiriad7\n\nWhy not simply?:\n\n'%' has higher precedence than '==', it will be better practice to use without parentheses.\n\n•",
null,
"Alex\n\nreturn x % 2 == 0 assumes both the code author the reader remembers that % has higher precedence than ==. Even though the parenthesis in this case are superfluous, they both make your intent clear and ensure you haven't misremembered the prioritization. In general, you should be optimizing your code for readability, correctness, and maintainability -- consequently, I would say it's a better practice to use the parenthesis in this case than not (there's no runtime performance penalty for using them).\n\n•",
null,
"harbie\n\n•",
null,
"Harbie\n\nI am trying to learn the c++ modulo. I was able to get the positive integers to run and compile... But whenever I try the negative divisor it doesn't compile .....\n\nI just found your page and would post my program in a few....\n\nThanks and looking to learn from here as others have...\n\n•",
null,
"Alex\n\nYes, please post the code that doesn't work and then we can examine why.\n\n•",
null,
"Gabriel Netto Ferreira\n\nI did that code \"on my way\", but the compiler accuses an error in lines 8 and 9.\nWhat can it be? I never used string like data type, so... probably it's a lil mistake\n\nty, and sorry for my bad english. I'm not american, so english isn't mine/my (idk, but I think it's mine ahuauha) mother tongue :s\n\n•",
null,
"Gabriel Netto Ferreira\n\nObs: I'm using Netbeans IDE, in Ubuntu 16.04\n\n•",
null,
"Alex\n\nI'm not sure which specific lines are causing errors, but just inspecting your program, a few things jump out at me:\nA few thoughts here:\n1) Your isEven() function is returning a string -- if it's going to return a string, it should return a std::string.\n2) Your isEven() function should really be returning a boolean.\n3) Also make sure you've #included the string header.\n\n•",
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"Olvin Lizama\n\nWell, I think now your tutorials are mature enough to write a book, it may not be a best seller, but you can leave your legacy in C++ history, people will quote your work, your examples, and teach with your material.\n\nYou have such a great didactic and is really easy to understand what you want to teach and very few people has that gift.\n\nFor sure it could be a struggle to do it, but never will be worthless.\n\nI thank you for this project, is awesome and maybe you are used to people saying this, but I'm learning a lot thanks to you.\n\nSorry for my English it's not my native language.\n\nBest wishes,\n\n•",
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"Alex\n\nThank you for your kind words. The biggest problem with books is they are expensive, and that makes them inaccessible to a large portion of the world. The great thing about this website is that it is accessible for everyone with internet (since it's ad-sponsored, something I can't do with a book)! My goal isn't to be famous or quoted -- my goal is to help others learn C++, just like others once helped me. This is my little mark on the world, and I'm proud that so many people from all countries of the world, rich or poor, have been able to visit.\n\n•",
null,
"Olvin Lizama\n\nHi Alex,\n\nPerhaps you wanted us to use the functions and everything you taught us, but in think this is the simple way to make this program,\n\n•",
null,
"Alex\n\nYes, I want to make sure you're getting used to using functions, and thinking about modularity.\n\n•",
null,
"Olvin Lizama\n\nAwesome you are still replying, sooo much dedication!!!\n\nAre you writing a book?\n\n•",
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"Alex\n\nNope. I got asked to write a book, but I declined because I don't have enough time (I can work on this website as time permits -- no deadlines!). :)\n\n•",
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"M Hassan Tahir\n\n#include \"stdafx.h\"\n#include <iostream>\n\nint userInput()\n{\nstd::cout << \"Enter an integer and know whether it is even or odd: \";\nint integer;\nstd::cin >> integer;\nreturn integer;\n}\n\nvoid isEven(int x)\n{\nif ( x % 2 == 0)\nstd::cout << \"The entered integer is even\";\nelse\nstd::cout << \"The entered integer is odd\";\n}\n\nint main()\n{\nint input = userInput();\nisEven(input);\n\nreturn 0;\n}\n\nAnother simple way. I hope it will work fine everywhere.\nGetting a hang of programming in C++ now.\nThanks Alex!\n\n•",
null,
"ihowa\n\ni wrote a code for the second question and it is way different from what you wrote but it works fine. i want to know your take on my code pls and thank you\n\n•",
null,
"Alex\n\nDid you test this program? If so, you'd know that it doesn't work. It starts off good, getting a number from function enternum(), which is passed back to main. But that value (x) is never used. Instead, iseven() declares its own local uninitialized variable x and then uses that to try to determine even or not. That will lead to undefined results.\n\nYou need to pass the variable x in main to function iseven() as a function parameter, and have iseven() use that variable. Then your program will not only be well structured, it will actually work too. :)\n\n•",
null,
"Ihowa\n\nI understand what you're getting at. but the first program did work.\n\n•",
null,
"Olvin Lizama\n\nIhowa,\n\nIt doesn't work, I tried to compile it and gave me warnings about x not being used, Alex is right in order to work you need to pass x to iseven() function, should be like this,\n\n•",
null,
"Travis\n\nIs there any reason someone shouldn't do:\n\n•",
null,
"Alex\n\nNope, nothing wrong with that.\n\n•",
null,
"Travis\n\nOkay, cool! I was worried that maybe I was doing some \"unethical\" programming, or something. Thanks for the quick reply.\nAlso, thanks for this whole tutorial, it's a great resource!\n\n•",
null,
"Rammwurst\n\nOr you know\n\nreturn!(x % 2);\n\ncause you already taught us that everything !=0 bool sees as true.\n\n•",
null,
"J3ANP3T3R\n\nInteresting. but in his example\n\nreturn (x % 2) == 0;\n\nit will only either return a zero for FALSE and a non-zero value for TRUE. did you mean add ! in case the remainder is a negative ? i dont see what else could be missing return (x % 2) == 0; looks fine.\n\n•",
null,
"Alex\n\nThe following two statements evaluate identically:\n\n•",
null,
"Silent Hunter\n\nin your example above, you forgotten to insert \"return 0\"\nif i compile it says error\n\n#include <iostream>\n\nint main()\n{\n// count holds the current number to print\nint count = 1; // start at 1\n\n// Loop continually until we pass number 100\nwhile (count <= 100)\n{\nstd::cout << count << \" \"; // print the current number\n\n// if count is evenly divisible by 20, print a new line\nif (count % 20 == 0)\nstd::cout << \"\\n\";\n\ncount = count + 1; // go to next number\n} // end of while\n} // end of main()\n\n•",
null,
"Alex\n\nReturning an explicit value from main is actually optional (but a good idea none-the-less). I've updated the example.\n\n•",
null,
"Kyle Roth\n\nwhy does this ask me for an integer 4 times?\n\n[#include \"stdafx.h\"\n#include <iostream>\n\nint getNumber()\n{\nstd::cout << \"enter and integer\";\nint x;\nstd::cin >> x;\nreturn x;\n\n}\n\nint main()\n{\n\ngetNumber();\n\nif (((getNumber()) % 2) == 0)\nstd::cout << getNumber() << \"is even\\n\";\nelse\nstd::cout << getNumber() << \"is odd\\n\";\n} ]\n\n•",
null,
"Alex\n\n4 times? It should only ask 3 times. And this happens because getNumber() asks the user to enter an integer each time it is called. And you're calling it 3 times (only one of the conditional statements gets executed).\n\nWhat you should do is assign the return value of getNumber() to a temporary variable, and use that variable instead of calling getNumber() again.\n\n•",
null,
"sharaf\n\nwhy can't we use in this way?\n\n•",
null,
"Alex\n\nBecause this does x / y first (which is an int divided by an int, which produces an integer result) and then converts the integer to a double.\n\n•",
null,
"Sharaf\n\noh!!!! thanks again!\n\n•",
null,
"Simon\n\nMy version ; still too messy but added a twist ;)\n\n•",
null,
"Simon\n\nAnd here is a different version of your counting, C++ starts to get funny... thanks a lot for all the tutorial and really really well done quizz.\n\n•",
null,
"Nortski\n\nThank you so much for these tutorials. I've tried a number of times to get my head around programming over the years. Decided to give it another go and these tutorials are fantastic!\n\nThis was my solution to question 2. More coding than was needed but works fine. Streamlining will come with experience :)\n\n•",
null,
"Josh\n\nIs this acceptable? Or really, is this less efficient in any way?\n\n•",
null,
"Alex\n\nIs it acceptable from what perspective? Does it work? Yes. Does it meet the quiz requirements? No. The quiz asked you to create a function named isEven() and you didn't.\n\n•",
null,
"Pravasi Meet\n\nThe semantics of += (add assignment operator) are different in C & C++. += operator returns lvalue (locator value) in C++ but not in C. That's why following program is valid in C++ but invalid in C.\n\nWhen you compile above program in C the compiler gives you following error:\n[Error] lvalue required as left operand of assignment"
] | [
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https://softmath.com/algebra-help/word-problems-for-positive-and.html | [
"",
null,
"## What our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nI originally bought Algebrator for my wife because she was struggling with her algebra homework. Now only did it help with each problem, it also explained the steps for each. Now my wife uses the program to check her answers.\nRobert Davis, CA\n\nWhen kids who have struggled their entire lives with math, start showing off the step-by-step ability to solve complex algebraic equations, and smile while doing so, it reminds me why I became a teacher in the first place!\nJeff Ply, CO\n\nJust watching my students, one after another, easily grasp these higher mathematical concepts and really, truly understand what they are doing, makes Algebrator worth the price of admission. Furthermore, for the gains, the price is tremendously inexpensive!\nTroy Green, FL\n\nIt appears you have improved upon an already good program. Again my thanks, and congrats.\nAshley Grayden, MA\n\nI got a A in my class. Thanks for the help!\nSandy Ketchum, AL\n\n## Search phrases used on 2008-09-11:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• division of expressions calculator\n• how to solve a nonlinear algebraic equation in mathcad\n• free online maths book for sixth standard\n• calculator\n• basic ratio formulas\n• TI-89/SQUARE ROOT\n• solving \"3x2-4x+8\"\n• Factoring Calculator for TI-84\n• glencoe Algebra 2 skills practice answers\n• free pre algebra work sheets\n• second order nonhomogeneous\n• Proportions Worksheet\n• solving equations by factoring worksheets\n• prentice hall pre algebra pg 111 answers\n• lcm practice sheets elementary\n• picture of an +algebraic expression\n• use casio calculater online\n• ti-86 decimal to fraction tutorial\n• lowest common factor\n• gratest common factor of 48 and 56\n• program using false statements to find the sum\n• ti-89 trig identities\n• TI-84 program VSEPR\n• algebra fractions subtracting cheats\n• how do i find the hypotenuse of a trigonomic function of an angle\n• 9th grade pre algebra problems\n• a online advanced division calculator\n• factor binomials worksheet\n• where to ge free online help solving a algebra 2 problem\n• Take Derivatives of graphs\n• Mastering the Math SATAnswer Book\n• equation calculator predict products\n• Rational expressions and equations calculator\n• free math printouts\n• pre-algebra worksheet triangle\n• pre algebra definitions\n• simple way to calculate lcm\n• free math worksheets on LCM\n• Free Algebra Equation Solver\n• quadratic formula program on calculator\n• Real Life Uses for Quadratics\n• how to learn pre algebra\n• how to simplify square roots\n• quadratic equation with complex coefficients\n• when is absolute value needed in simplifying\n• civil engineering example simultaneous equation\n• Even Answers for McDougal Littell Algebra 2\n• solving equations with three variables\n• 72340892761644\n• fractions least of greatest\n• Cumulative Exam Chapters 1-2\n• dividing fractions formula\n• ks3 expand simplify\n• Trigonometry Word Problems And Answers\n• solving for y in equations 3rd roots\n• How to convert a square root to radical form\n• 6th grade algebra some of y\n• negative fractions exponents distributive college edu math problems -test -course\n• algebraic equasions\n• algebra 2-step equations worksheets\n• prentice hall mathematics algebra 1 help"
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null,
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https://ch.mathworks.com/matlabcentral/cody/problems/590-greed-is-good-simple-partition-p-n | [
"Cody\n\n# Problem 590. Greed is good - Simple partition P[n].\n\nFind a simple partition P[n]. E.g. P = 4 + 3 + 2 + 1.\n\n1. There are many solutions, compute just one set.\n2. Don't repeat numbers.\n3. Be Greedy ;-)\n4. To check against trivial solutions, E.g. [x-k, k] etc; but I'll provide you with one to start.\n5. Show me how you write the whole solution.\n\nBonus points if you solve the general problem of producing all unique partitions of [n].\n\n### Solution Stats\n\n58.18% Correct | 41.82% Incorrect\nLast Solution submitted on Jun 10, 2018"
] | [
null
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https://ukm.pure.elsevier.com/en/publications/threshold-cryptosystem-with-factoring-and-elliptic-curve-discrete | [
"# Threshold cryptosystem with factoring and elliptic curve discrete logarithm problems\n\nResearch output: Contribution to journalArticle\n\n### Abstract\n\nRecently, the development of cryptosystems based on multiple hard problems attracts attention from mathematicians and computer scientists as it is believed that such systems provide long-term security. In another side, the concept of threshold cryptography gives benefit in group-oriented community. For these reasons, in this paper we propose a new threshold cryptosystem based on factoring and elliptic curve discrete logarithm problems. We show that our scheme is heuristically secure against some cryptographic attack. From the efficiency analysis, it can be seen that our scheme requires reasonable time complexity in encryption and decryption phases.\n\nOriginal language English 34974-34977 4 International Journal of Applied Engineering Research 10 14 Published - 2015\n\nCryptography\n\n### Keywords\n\n• Elliptic curve discrete logarithm problem\n• Factoring problem\n• Threshold cryptosystem\n\n### ASJC Scopus subject areas\n\n• Engineering(all)\n\n### Cite this\n\nThreshold cryptosystem with factoring and elliptic curve discrete logarithm problems. / Mohamad1, Mohd Saiful Adli; Ismail, Eddie Shahril.\n\nIn: International Journal of Applied Engineering Research, Vol. 10, No. 14, 2015, p. 34974-34977.\n\nResearch output: Contribution to journalArticle\n\n@article{1e50557993ef47c7a8ac33a38e53f2d5,\ntitle = \"Threshold cryptosystem with factoring and elliptic curve discrete logarithm problems\",\nabstract = \"Recently, the development of cryptosystems based on multiple hard problems attracts attention from mathematicians and computer scientists as it is believed that such systems provide long-term security. In another side, the concept of threshold cryptography gives benefit in group-oriented community. For these reasons, in this paper we propose a new threshold cryptosystem based on factoring and elliptic curve discrete logarithm problems. We show that our scheme is heuristically secure against some cryptographic attack. From the efficiency analysis, it can be seen that our scheme requires reasonable time complexity in encryption and decryption phases.\",\nkeywords = \"Elliptic curve discrete logarithm problem, Factoring problem, Threshold cryptosystem\",\nyear = \"2015\",\nlanguage = \"English\",\nvolume = \"10\",\npages = \"34974--34977\",\njournal = \"International Journal of Applied Engineering Research\",\nissn = \"0973-4562\",\npublisher = \"Research India Publications\",\nnumber = \"14\",\n\n}\n\nTY - JOUR\n\nT1 - Threshold cryptosystem with factoring and elliptic curve discrete logarithm problems\n\nAU - Ismail, Eddie Shahril\n\nPY - 2015\n\nY1 - 2015\n\nN2 - Recently, the development of cryptosystems based on multiple hard problems attracts attention from mathematicians and computer scientists as it is believed that such systems provide long-term security. In another side, the concept of threshold cryptography gives benefit in group-oriented community. For these reasons, in this paper we propose a new threshold cryptosystem based on factoring and elliptic curve discrete logarithm problems. We show that our scheme is heuristically secure against some cryptographic attack. From the efficiency analysis, it can be seen that our scheme requires reasonable time complexity in encryption and decryption phases.\n\nAB - Recently, the development of cryptosystems based on multiple hard problems attracts attention from mathematicians and computer scientists as it is believed that such systems provide long-term security. In another side, the concept of threshold cryptography gives benefit in group-oriented community. For these reasons, in this paper we propose a new threshold cryptosystem based on factoring and elliptic curve discrete logarithm problems. We show that our scheme is heuristically secure against some cryptographic attack. From the efficiency analysis, it can be seen that our scheme requires reasonable time complexity in encryption and decryption phases.\n\nKW - Elliptic curve discrete logarithm problem\n\nKW - Factoring problem\n\nKW - Threshold cryptosystem\n\nUR - http://www.scopus.com/inward/record.url?scp=84940118741&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=84940118741&partnerID=8YFLogxK\n\nM3 - Article\n\nAN - SCOPUS:84940118741\n\nVL - 10\n\nSP - 34974\n\nEP - 34977\n\nJO - International Journal of Applied Engineering Research\n\nJF - International Journal of Applied Engineering Research\n\nSN - 0973-4562\n\nIS - 14\n\nER -"
] | [
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http://www.jos.ac.cn/article/doi/10.1088/1674-4926/37/4/044010 | [
"J. Semicond. > Volume 37 > Issue 4 > Article Number: 044010\n\n# Temperature-variable high-frequency dynamic modeling of PIN diode\n\nShangbin Ye , , Jiajia Zhang , Yicheng Zhang and Yongtao Yao\n\nAbstract: The PIN diode model for high frequency dynamic transient characteristic simulation is important in conducted EMI analysis. The model should take junction temperature into consideration since equipment usually works at a wide range of temperature. In this paper, a temperature-variable high frequency dynamic model for the PIN diode is built, which is based on the Laplace-transform analytical model at constant temperature. The relationship between model parameters and temperature is expressed as temperature functions by analyzing the physical principle of these parameters. A fast recovery power diode MUR1560 is chosen as the test sample and its dynamic performance is tested under inductive load by a temperature chamber experiment, which is used for model parameter extraction and model verification. Results show that the model proposed in this paper is accurate for reverse recovery simulation with relatively small errors at the temperature range from 25 to 120℃.\n\nAbstract: The PIN diode model for high frequency dynamic transient characteristic simulation is important in conducted EMI analysis. The model should take junction temperature into consideration since equipment usually works at a wide range of temperature. In this paper, a temperature-variable high frequency dynamic model for the PIN diode is built, which is based on the Laplace-transform analytical model at constant temperature. The relationship between model parameters and temperature is expressed as temperature functions by analyzing the physical principle of these parameters. A fast recovery power diode MUR1560 is chosen as the test sample and its dynamic performance is tested under inductive load by a temperature chamber experiment, which is used for model parameter extraction and model verification. Results show that the model proposed in this paper is accurate for reverse recovery simulation with relatively small errors at the temperature range from 25 to 120℃.\n\nReferences:\n\n Fang C, Li W, Li X. PIN power diode dynamic behavior and physics-based model parameter extraction method[J]. Transactions of China Electrotechnical Society, 2015, 30(6): 208. Zhou Y. The research of the snape-off reserve recover current of diode[J]. Shenyang:Shenyang University of Technology, 2013. Wang J. PSPICE circuit design and application[J]. , 2010. Bellone S, Della F, Benedetto D. An analytical model of the switching behavior of 4H-SiC PIN diodes from arbitrary injection conditions[J]. IEEE Trans Power Electron, 2012, 27(3): 1641. Zhang Y, Zhang J, Wei L. Research progress of power PIN diode model for electromagnetic interference pre-evaluation[J]. Journal of Tongji University (Natural Science), 2015, 43(4): 617. Cliff M, Lauritzen P, Sigg A. Modeling of power diodes with the lumped charge modeling technique[J]. IEEE Trans Power Electron, 1997, 12(3): 398. Strollo A. A new SPICE model of power PIN diode based on asymptotic waveform evaluation[J]. IEEE Trans Power Electron, 1997, 12(1): 12. Bryant T, Lu L, Santi E. Physical modeling of fast pin diodes with carrier lifetime zoning, Part 1:device model[J]. IEEE Trans Power Electron, 2008, 23(1): 189. Igic M, Mawby A, Towers S. New physically based pin diode compact model for circuit modeling applications[J]. IEE Proceedings of Circuits Devices and Systems, 2002, 149(4): 257. Buiatti G, Cappelluti F, Ghione G. Physics-based PiN diode SPICE model for power circuit simulation[J]. IEEE Trans Industry Application, 2007, 43(4): 911. Niu G, Chen S, Yu J. Research on effect of temperature change on characteristics of SPICE diode based on MATLAB[J]. Modern Electronics Technique, 2012, 35(2): 131. Qi Y, Li Y, Hao Y. Study of thermal characteristics of PIN diodes[J]. Journal of Microwaves, 2014(6): 220. Chen J, Chen X, Liu C. Analysis of temperature effect on p-i-n diode circuits by a multiphysics and circuit cosimulation algorithm[J]. IEEE Trans Electron Devices, 2012, 59(11): 3069. Utermohlen F, Hermann I, Etter D B. Temperature sensitivity modeling of pn-junction diodes for micro bolometer based thermal imaging application[J]. International Semiconductor Conference, 2013: 1. Sintamarean C, Blaabjerg F, Wang H. A novel electro-thermal model for wide bandgap semiconductor based devices[J]. European Conference on Power Electronics and Applications, 2013: 1. Cui G, Wang J, Huang Q. Measurement for minority lifetime in PIN diodes[J]. Chinese Journal of Electron Devices, 2004, 27(2): 236. Simon S, Kwok N. Physics of semiconductor devices[J]. , 2006. Donald N. Semiconductor physics and devices[J]. , 2011. Chen Z, Wang J. Semiconductor devices material physics basic[J]. Beijing:Science Press, 2003. Jayant B. Fundamentals of power semiconductor devices[J]. New York:Springer-Verlag Inc, 2008.\n Fang C, Li W, Li X. PIN power diode dynamic behavior and physics-based model parameter extraction method[J]. Transactions of China Electrotechnical Society, 2015, 30(6): 208. Zhou Y. The research of the snape-off reserve recover current of diode[J]. Shenyang:Shenyang University of Technology, 2013. Wang J. PSPICE circuit design and application[J]. , 2010. Bellone S, Della F, Benedetto D. An analytical model of the switching behavior of 4H-SiC PIN diodes from arbitrary injection conditions[J]. IEEE Trans Power Electron, 2012, 27(3): 1641. Zhang Y, Zhang J, Wei L. Research progress of power PIN diode model for electromagnetic interference pre-evaluation[J]. Journal of Tongji University (Natural Science), 2015, 43(4): 617. Cliff M, Lauritzen P, Sigg A. Modeling of power diodes with the lumped charge modeling technique[J]. IEEE Trans Power Electron, 1997, 12(3): 398. Strollo A. A new SPICE model of power PIN diode based on asymptotic waveform evaluation[J]. IEEE Trans Power Electron, 1997, 12(1): 12. Bryant T, Lu L, Santi E. Physical modeling of fast pin diodes with carrier lifetime zoning, Part 1:device model[J]. IEEE Trans Power Electron, 2008, 23(1): 189. Igic M, Mawby A, Towers S. New physically based pin diode compact model for circuit modeling applications[J]. IEE Proceedings of Circuits Devices and Systems, 2002, 149(4): 257. Buiatti G, Cappelluti F, Ghione G. Physics-based PiN diode SPICE model for power circuit simulation[J]. IEEE Trans Industry Application, 2007, 43(4): 911. Niu G, Chen S, Yu J. Research on effect of temperature change on characteristics of SPICE diode based on MATLAB[J]. Modern Electronics Technique, 2012, 35(2): 131. Qi Y, Li Y, Hao Y. Study of thermal characteristics of PIN diodes[J]. Journal of Microwaves, 2014(6): 220. Chen J, Chen X, Liu C. Analysis of temperature effect on p-i-n diode circuits by a multiphysics and circuit cosimulation algorithm[J]. IEEE Trans Electron Devices, 2012, 59(11): 3069. Utermohlen F, Hermann I, Etter D B. Temperature sensitivity modeling of pn-junction diodes for micro bolometer based thermal imaging application[J]. International Semiconductor Conference, 2013: 1. Sintamarean C, Blaabjerg F, Wang H. A novel electro-thermal model for wide bandgap semiconductor based devices[J]. European Conference on Power Electronics and Applications, 2013: 1. Cui G, Wang J, Huang Q. Measurement for minority lifetime in PIN diodes[J]. Chinese Journal of Electron Devices, 2004, 27(2): 236. Simon S, Kwok N. Physics of semiconductor devices[J]. , 2006. Donald N. Semiconductor physics and devices[J]. , 2011. Chen Z, Wang J. Semiconductor devices material physics basic[J]. Beijing:Science Press, 2003. Jayant B. Fundamentals of power semiconductor devices[J]. New York:Springer-Verlag Inc, 2008.\n Le Yu, Yingkui Zheng, Sheng Zhang, Lei Pang, Ke Wei, Xiaohua Ma. Small-signal model parameter extraction for AlGaN/GaN HEMT. J. Semicond., 2016, 37(3): 034003. Li Ruizhen, Li Duoli, Du Huan, Hai Chaohe, Han Zhengsheng. SOI MOSFET Model Parameter Extraction via a Compound Genetic Algorithm. J. Semicond., 2006, 27(5): 796. A. Avila Garcia, L. Ortega Reyes. Analysis and parameter extraction of memristive structures based on Strukov’s non-linear model. J. Semicond., 2018, 39(12): 124009. Ren Zheng, Hu Shaojian, Jiang Bin, Wang Yong, Zhao Yuhang. Extraction of Temperature Parameters and Optimization of the Mextram 504 Model on SiGe HBT. J. Semicond., 2008, 29(5): 960. Wu Rufei, Zhang Haiying, Yin Junjian, Li Xiao, Liu Huidong, Liu Xunchun. A Novel Equivalent Circuit Model of GaAs PIN Diodes. J. Semicond., 2008, 29(4): 672. Xingrong Ren, Changchun Chai, Zhenyang Ma, Yintang Yang, Liping Qiao, Chunlei Shi, Lihua Ren. Motion of current filaments in avalanching PIN diodes. J. Semicond., 2013, 34(4): 044004. Zhu Zhangming, Qian Libo, Yang Yintang. A Novel Interconnect Crosstalk Parallel RLC Analyzable Model Based on the 65nm CMOS Process. J. Semicond., 2008, 29(3): 423. Lu Jing, Wang Yan, Ma Long, Yu Zhiping. A New Small-Signal Modeling and Extraction Method in AlGaN/GaN HEMTs. J. Semicond., 2007, 28(4): 567. Chi Yusong, Huang Fengyi, Wu Zhongjie, Zhang Shaoyong, Kong Xiaoming, Wang Zhigong. Characterization and Modeling for 0.13μm RF MOSFETs. J. Semicond., 2006, 27(2): 373. Ren Zheng, Shi Yanling, , Hu Shaojian, Jin Meng, Zhu Jun, Chen Shoumian. Optimization of BSIM3 I-V Modeling of High Voltage MOS Devices. J. Semicond., 2006, 27(6): 1073. Xu Jingbo, Yin Junjian, Zhang Haiying, Li Xiao, Liu Liang, Ye Tianchun. Abstraction of Small Signal Equivalent Circuit Parameters of Enhancement-Mode InGaP/AlGaAs/InGaAs PHEMT. J. Semicond., 2007, 28(3): 361. Sudhansu Kumar Pati, Kalyan Koley, Arka Dutta, N Mohankumar, Chandan Kumar Sarkar. A new approach to extracting the RF parameters of asymmetric DG MOSFETs with the NQS effect. J. Semicond., 2013, 34(11): 114002. Zhang Chenfei, Ma Chenyue, Guo Xinjie, Zhang Xiufang, He Jin, Wang Guozeng, Yang Zhang, Liu Zhiwei. Forward gated-diode method for parameter extraction of MOSFETs. J. Semicond., 2011, 32(2): 024001. Ma Chenyue, Zhang Chenfei, Wang Hao, He Jin, Lin Xinnan, Mansun Chan. Diode parameter extraction by a linear cofactor difference operation method. J. Semicond., 2010, 31(11): 114009. Lu Lei, Zhou Feng, Tang Zhangwen, Min Hao, Wang Junyu. Equivalent Model and Parameter Extraction of Center-Tapped Differential Inductors. J. Semicond., 2006, 27(12): 2150. Liu Jun, Sun Lingling. Parameter Extraction of a III-V Compound HBT Model. J. Semicond., 2006, 27(5): 874. Fu Jun. Small-signal model parameter extraction for microwave SiGe HBTs based on Y- and Z-parameter characterization. J. Semicond., 2009, 30(8): 084005. Liu Linsheng. Improved Nonlinear Model of HEMTs with Independent Transconductance Tail-Off Fitting. J. Semicond., 2011, 32(2): 024004. Zhao Yuhang, Hu Shaojian, Ren Zheng. Accurate Parameter Extraction of Substrate Resistance in an RF CMOS Model Valid up to 20GHz. J. Semicond., 2008, 29(4): 737. D. K. Panda, G. Amarnath, T. R. Lenka. Small-signal model parameter extraction of E-mode N-polar GaN MOS-HEMT using optimization algorithms and its comparison. J. Semicond., 2018, 39(7): 074001.\n\n## GET CITATION\n\nS B Ye, J J Zhang, Y C Zhang, Y T Yao. Temperature-variable high-frequency dynamic modeling of PIN diode[J]. J. Semicond., 2016, 37(4): 044010. doi: 10.1088/1674-4926/37/4/044010.\n\nExport: BibTex EndNote\n\n## Article Metrics\n\nArticle views: 769 Times PDF downloads: 16 Times Cited by: 0 Times\n\n## History\n\nManuscript received: 28 July 2015 Manuscript revised: Online: Published: 01 April 2016\n\nCode:",
null,
"*验证码错误"
] | [
null,
"http://www.jos.ac.cn:80/util/AuthExpImageServlet",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64618915,"math_prob":0.63184756,"size":12501,"snap":"2019-35-2019-39","text_gpt3_token_len":3739,"char_repetition_ratio":0.14179403,"word_repetition_ratio":0.6659517,"special_character_ratio":0.3174146,"punctuation_ratio":0.22997317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.954993,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T21:48:39Z\",\"WARC-Record-ID\":\"<urn:uuid:ffe7ac36-fa4f-45dc-ae79-6a80f0ef640c>\",\"Content-Length\":\"56161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e391caf-9a51-4ff7-8859-ed001c0d04e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:0b5a5c97-83f5-480f-a4a2-f082987b1eca>\",\"WARC-IP-Address\":\"159.226.228.78\",\"WARC-Target-URI\":\"http://www.jos.ac.cn/article/doi/10.1088/1674-4926/37/4/044010\",\"WARC-Payload-Digest\":\"sha1:MYDIYYA2LLPOMAO2HKAKSXFQR7VBZQKU\",\"WARC-Block-Digest\":\"sha1:7AGQQLDCHVAU5PJ2RI467WIVIWN7WAL4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573121.4_warc_CC-MAIN-20190917203354-20190917225354-00207.warc.gz\"}"} |
https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-017-0910-7 | [
"We’d like to understand how you use our websites in order to improve them. Register your interest.\n\n# Femtocaching assisted multi-source D2D content delivery in cellular networks\n\n## Abstract\n\nThe influxes of diversified services and mass data lead to exponential growth of traffic load in mobile cellular networks. Cache-enabled device-to-device (D2D) communication provides a general framework to alleviate this situation. In contrast to previous single-source D2D models, this paper investigates a comprehensive content delivery framework based on a three-tier heterogeneous network (HetNet), where base station (BS), femtocaching auxiliary equipments (FAEs), and user terminals(UTs) are included. The cooperative D2D communication can be implemented from both FAEs and UTs to handle the ongoing explosive increase in ultra-dense scenario. Moreover, duplicate storage for requesting data in multiple neighbor nodes makes many-to-one D2D communication possible at the user layer. Considering the case that cellular users and D2D links reuse the same resources in the uplink period, the non-outage probability of the cellular communication is defined to guarantee the main communication quality. Under the constraints subject to cumulative interference, an optimization objective function based on multi-source D2D communication is deduced to achieve unprecedented average data rate. Numerical simulations show that our system yields network throughput exponentially while transferring traffic load of the BS reasonably.\n\n## Introduction\n\nThe latest report of the global system for Global Mobile Communication Systems Association (GSMA) has shown that three quarters of the world’s population will be linked together through mobile terminals by 2020. Wireless data business is therefore expected to increase by a factor of 11 compared to the current, and each user needs to consume at least 1 Gb per day in the next 5 years . This explosive growth is mainly caused by mobile multimedia traffic and other emerging content transmission such as popular videos and social networking services (SNS), which almost rise by a factor of 65 times . The urgent demand for the thousandfold growth of mobile data inevitably leads to a slowdown, or even breakdown of the cellular systems. Thus, cellular networks are undergoing unprecedented paradigm shift in the way by which data is delivered .\n\nCache-enabled D2D communication is a key component in this shift, which can distribute content locally and transmit data directly. By pre-storing popular files and dispatching data via D2D, the system has a higher gain with the increase of file storage capabilities at the network edge [4, 5]. The reason behind this is that the most popular files are more likely to be accessed by the majority of users, although there are huge amount of traffic suffering by the network. Thus, a noteworthy development in this direction is to place as much popular contents as possible on a local storage device. In academia, there exist some possible solutions to support the cache-enabled D2D paradigm, such as . These works discussed and elaborated the specific edge placement strategy to enhance the accessibility of data for requesting user terminal(RUT). In general, the existing D2D communication schemes can be classified into two categories by taking into account different devices to implement local-storage, as shown in Fig. 1. The first one is to deploy small cells and transfer data by auxiliary device, which can be seen as SBS. The other caches contents without special infrastructure and have the further advantage of mobile user terminals (UTs), such as smartphones, tablets, and laptops which are built in large storage devices. According to the special popularity distribution and storage capacity, a large percentage of video traffic can be offloaded through these two systems.\n\nNote that available researches mainly focus on the single-source caching model, i.e., they either use femto-caching helper (i.e., SBS) or exploit UTs to carry on D2D communication. Although these two approaches were described separately, the measures were effective and brought about some enlightenment to us, e.g., these two strategies can also be mixed to some extent. Moreover, current wireless technologies such as time/frequency division multiple access and maximum ratio transmission (MRT) support multiple users to transfer data simultaneously. For example, under MRT, each suitable UT and SBS in the cooperation communication radius can beamform to a RU such that the signals from the neighboring nodes and SBSs are coherently combined and ultimately result in diversity gain. That is to say, one RUT can receive contents from both the SBSs and other UTs, which can be called multi-source caching model. Here, the so-called multi-source caching, drawing a contrast with the single source(e.g., either SBSs or UTs), mainly refers to the collaboration between different types of storage devices(i.e., both SBSs and UTs). The heap offers us a feasible solution to the traffic offload problem from a technical and economical perspective. Accordingly, we propose a three-tier communication architecture in heterogeneous network incorporating femto-caching helpers and UTs to offload the backbone network traffic. This joint design offers not only two different caching gains and multiplexing gains but also diversity gains owning to many-to-one D2D communication. Many-to-one D2D communication means that each RUT may detect several mobile users storing the needed file within its scope of power coverage in the ultra-dense environment. In other words, three conditions are required for suitable UTs to take place many-to-one D2D: (1) they are within the transmission range of RU; (2) they have the file which is needed by RU; (3) there are more than one UTs matching both of criteria 1 and 2. In general, high data rate and low transmission delay can be obtained due to the proximity of paired devices in D2D links, thus introducing proximity gain. Nevertheless, it is unsurprisingly that the multi-source caching model proposed in this paper will introduce cumulative interference caused by multi-pair D2D links in the uplink transmission period. To achieve maximum transmission rate and to obtain high user satisfaction, non-outage probability of system is deduced to describe the influence of cumulative interference on the performance of cellular systems.\n\nCompared with the existing work, we have the following contributions:\n\n• Combining the storage advantages of fixed equipments and mobile terminals in ultra-dense environment, we model a three-tier heterogeneous architecture to distribute data. Furthermore, we introduce the concept of many-to-one D2D communication in the user layer.\n\n• We present an optimization objection in terms of average transmission rate to discuss the performance characteristics of the heterogeneous network. Under the cumulative interference constrain due to multi-pair D2D links, non-outage probability of system is derived and simulated.\n\n• We define the concept of user satisfaction to analyze the pareto solutions of the optimization problem. Moreover, we evaluate the impact of various parameters on user satisfaction, such as the request probability and D2D communication radius.\n\nThe remainder of the paper is organized as follows: after reviewing the main related works in Section 2, the three-tier content delivery mode is elaborated in Section 3. In Section 4, the average transmission rate and non-outage probability are derived. Numerical results in terms of network throughput and user satisfaction are discussed in Section 5. Finally, conclusive remarks are presented in Section 6.\n\n## Related work\n\nD2D communication attracts a lot of attentions recently, and the potential research orientations include content delivery/dissemination [6, 7, 10], resource allocation , social- awareness video multicasting [18, 19], and so on. Existing researches for the content delivery can be mainly classified into two perspectives. The first one is to deploy SBS with mass storage device which does not require high-speed backhaul links. For example, in [6, 7], small base station called helper was introduced and placed in fixed position to serve user requests. Chou et al. utilized mobile SBS to study the deployment problem and address the waste of resources in small cells. In , a cost-effective integration access technology between WiFi and cellular wireless was characterized for femto-based stations to fulfill time/space-varying traffic. In general, the scheme based on auxiliary devices in small cells, such as helper, femto-based station, and SBS can highly provide a performance boost to the network traffic. However, along with the growth in the number of helpers, each RUT can be covered by multiple helpers, and the hit rate will be increased as well as the interference and the investment cost. In addition, another potential deficiency of the small cell-based infrastructure is that, during peak traffic period, the backhaul link-capacity requirement to respond data is enormously high . Obviously, if the storage capacity of existing devices (e.g., UTs) is considered, storage-investment efficiency will be further improved.\n\nThe second one is pushed even further by utilizing the storage ability of mobile UTs [7, 10, 13], which pre-store popular files in UTs through various storage strategies. The authors of mainly focused on exploiting cognition to the cache-enabled D2D in the multi-channel cellular network, in which the number of users in a cell is formulated as mutually independent Poisson point processes (PPPs). Authors in presented a paradigm for wireless video content dissemination based on caching popular files in mobile users with no additional infrastructure cost. Similarly, put forward a systematic scheme in wireless content-centric networks. The storage capacity of users is adopted to maximize the content delivery capability with a fixed amount of wireless resources. Yang et al. presented a comprehensive framework on D2D communication and advocated to proactively cache contents in partial users with caching ability when the network is off-peak. The studies have shown that pre-storing files in the UTs can increase spectral efficiency, thus introducing reuse gain. This scheme allows only one neighboring nodes to communicate with a RUT in the meantime, which can be described as one-to-one D2D communication. Nevertheless, the mobile devices like UTs cannot guarantee the effective transmission of popular files because of the unstable topology, limited storage capacity as well as constrained-energy in a sparse environment. Fortunately, this model can bring caching gains by taking advantage of multiple neighbor nodes to store the same required file. Therefore, we can activate multiple possible D2D links simultaneously in the dense or ultra-dense network named as many-to-one D2D communication.\n\nIn particular, the authors in discussed about caching in these two ways. They introduced SBS placed in fixed position to serve user requests, and also adopted UTs’ large hard disks built in to act as mobile helper stations. However, significant varies still exist between the reference and our proposed system:\n\n1. We mainly focus on the combination of the two approaches, in contrast, the reference discussed about these two ways separately as well as other literatures. High data rate and system throughput can be obtained owning to caching gains, multiplexing gains, diversity gains, and proximity gains in our joint design.\n\n2. We introduce the concept of many-to-one D2D communication in the user layer and make full use of the user redundancy in the dense environments. On the contrary, users in were divided into smaller cluster based on their locations and at most one D2D transmission was allowed in each cluster at a time. That is to say, the authors in only allowed one-to-one D2D communication when they considered UTs for storage.\n\n3. The interference of the proposed system is more complex than the reference . There will inevitably be more interruptions for the cellular communication due to cumulative interference caused by many-to-one D2D communication. Thus, it is important to seek the trade-off at which interference and system throughput can supplement each other.\n\nTherefore, efficient resource allocation and interference management have key impacts on the performance of content delivery in D2D and cellular coexisting system . For theoretical derivations, only a single pair of D2D link reuse system resources is considered. Thus, it is always simplified to describe the interference, i.e., only one D2D pair is allowed to active within the power range of a user node or in a cluster. However, it is very critical to estimate the caused interference accurately when we calculate the system performance, such as maximize the data rate .\n\nIn short, the key study of this paper is to put forward a new strategy based on the study and the analysis of related literatures, which can not only take full advantage of the stability of SBS but also have the further numerical superiority and cost advantage of UTs. In other words, due to the different aspects of both traditional delivery methods, the joint design of these two approaches can be tried in practical project to offload the network traffic effectively and economically.\n\n## System model\n\nFigure 2 illustrates the three-tier heterogeneous architecture with a single macro-cell including BS, femtocaching auxiliary equipments (FAEs), and multiple UTs. It is no doubt that the BS serves as the first layer. The BS can allocate and manage the channel state information (CSI) of the users and then distribute the files to FAEs and UTs relying on the rules of procedure during off-peak hours (at that time the network transmission is less costly, e.g., at night). FAEs and multiple UTs constitute the second and third layer of the network, respectively. FAE is a device that is similar to a proliferation of low-complexity small-base station with weak backhaul links and high storage capabilities. Considering the fact that users in similar geographical and social attributes are more likely to seek for common interest, we deploy FAEs in the center of hot spots. In addition, FAEs can further extend coverage area effectively, especially when they are placed in the edge of the cell. If a typical user initiates a file request, the propose cooperation scheme will combine FAEs and UTs to provide data flow through D2D communication, as well as through the BS in a traditional cellular (uplink/downlink) communication when the content cannot be found in nearby. The neighbor nodes and the nearest FAE caching the requested information will participate in the D2D communication. The neighbor node is defined as the UT in the communication distance of the requesting user. The maximum allowable distance r is determined by the power level of the RUT, and all of the UTs have the same power level. FAEs can have the same or different transmission power as the UTs, i.e., the same or different feasible communication range. To simplify the description, we assume that each RUT can only communicate to at most one FAE due to the investment efficiency. In fact, MRT and zero-forcing beam forming (ZFBF) technology in MIMO system [14, 20, 21] can make one FAE transmit multiple data stream to different users at the same time, and also support multi-devices in transmitting the same file to a receiver simultaneously. The complexity caching scheme for multiple access has been investigated deeply in [5, 14], by which the optimum file placement strategy can be found even if a RUT can connect several FAEs and we do not carry out further discussion.\n\nThe communication topology of the overall heterogeneous network may change over time. However, at each time instant, the transmission connection of network can be analyzed from the static perspectives, as shown in Fig. 3. It describes a scenario where four users request files at the same time, while FAE transmits the stored files concurrently to multiple users (to three users’ three different files). If the FAE does not have the required files in its cache, the user can only establish a D2D link with the suitable neighbor, i.e., the neighbors who have store the needed files. If none of them have the required files, BS will serve the file request through cellular communication directly. Although the storage capacity of each user device is limited, even if they only store one file, the requesting user can find more than one suitable neighbor user with a certain probability in the dense environment. Many-to-one D2D relationship is thus formed when the appropriate neighbor nodes send data to the RUT simultaneously. For example, Tom can also form a local cooperative D2D pair with neighbor users (such as Bob and Alice) besides the FAE. As a consequence, FAE and the neighbor users that have the requested file in their cache beamform their content to the RUT, so that the data stream can be coherently combined at the receiver.\n\n## Problem formulation and analysis\n\n### Problem formulation\n\nIn Mobile World Congress 2017(MWC2017, Barcelona), Network equipment vendors have released a number of end-to-end network solutions for the future mobile communication and announced some experimental work on specific network technologies. For example, Nokia showed the new MIMO antenna (supporting 3.5, 4.5, 28, and 39 GHz spectrum), cloud residential access network (C-RAN), distributed core network and transmission scheme based on cloud architecture. Ericson announced to carry out a trial with Qualcomm, which was expected to build a standards-compliant 5G new gap system under 6 GHz. The trial was anticipated to use a variety of new technologies including large-scale MIMO, adaptive time division duplexing (TDD), beam forming, and extensible waveform technology based on OFDM. However, some problems related to network operation are also mentioned, such as mobile edge deployment which is suitable for video streaming and can reduce transmission delay. In this paper, we consider a joint caching to solve the problem of mobile edge deployment, which aims at maximizing the throughput, i.e., the sum of the effective data rate and subject to the interference constraint over all of the D2D links. The effective data rate refers to the total D2D transmission rate created by all requesting users at a steady-state moment. Consider a typical cell which serves N users and the coverage area is A. We assume that there are n users generating requests simultaneously. Each user requests a file independently from the library $${\\mathcal {F}} = \\left \\{ {{{\\mathrm {f}}_{1}},{{\\mathrm {f}}_{2}}, \\ldots,{{\\mathrm {f}}_{\\mathrm {M}}}} \\right \\}$$ of unit size M. The expected transmission rate obtained by the presented content delivery strategy can be denoted by $${\\mathcal {R}}\\left ({n,x} \\right)$$, where x=[x ij ] represents the delivery strategy when the j-th user generates a request for the i-th file. Then, the optimization problem can be formulated as follows:\n\n$${\\mathrm{Max }}{\\mathcal{R}}\\left({n,x} \\right)$$\n(1)\n\nSubject to\n\n$$\\begin{array}{*{20}l} &i \\leq M \\\\ &\\gamma \\geq \\gamma_{0},~~ ~~ \\forall i,j \\\\ &\\gamma_{d} \\geq \\gamma_{D},~~ \\forall i,j \\end{array}$$\n\nwhere γ is the signal-to-interference-plus-noise ratio (SINR) of the BS when D2D pairs reuse the given cellular resources. The performance of cellular communications should be guaranteed before D2D pairs are allowed to share the uplink resource . Accordingly, the BS must be firstly satisfied as long as its SINR is no less than the threshold γ 0. And likewise, in order to ensure the reliability and stability of the D2D communication in the uplink transmission, the SINR of D2D receiver γ d needs to exceed a given threshold γ D .\n\nEach request occurs independently and randomly according to the popularity distribution. Large quantities of researches certify that the popular distribution also can be established as good as a Zipf distribution model to measure the user requesting probability for popular video files in D2D communication . Under this model, the request probability of the i-th popular file, denoted by p i , 1≤iM, $$\\sum _{i=1}^{M} p_{i}=1$$, is inversely proportional to its order:\n\n$${p_{i}} = \\frac{1 \\left/ {i^{\\gamma_{r}}}\\right.}{\\sum \\nolimits_{j = 1}^{M} {1 \\left/ {j^{{\\gamma_{r}}}}\\right.}}$$\n(2)\n\nThe Zipf exponent γ r characterizes the distribution function by controlling the relative popularity of files. Larger γ r means higher content duplicate requests, i.e., the top few popular files account for the majority of requests.\n\nAn important aspect of the joint optimization is the storage allocation because hit ratio and transmission rate can be varied by caching different files in neighbor nodes. Caching multiple copies of a file can increase the chances to get diversity gains and multiplexing gains. For simplicity, we assume that all files have the same size, and one file is a basic storage unit. The library files are available in the BS, and users can directly connect via a cellular communication. Meanwhile, let C f denotes the storage capacity of each FAE, which is measured by the maximum number of files it can store.\n\nUsing the collaboration storage capacity of UTs and FAEs is an important feature of this article. In a general random geometric graph, it is shown that finding the optimal deterministic file assignment is NP-hard even when terminals are static and interference is ignored . However, in our three layers caching system, if we assume that each UT can access no more than one, we can find a simple centralized placement method for FAE. Given that there are m static memory units in each FAE, it should store the top m most popular files without repetition, while the storage capacity C f is equal to the number of storage units in FAE, i.e., C f =m. Each FAE can store on average up to m files with mM.\n\nRandom caching can be implemented since users are highly mobile. Each UT caches files at random and independently obeys a caching distribution. Specifically, we assume that each UT stores exactly one file, though generalization to multiple files per user is trivial. On the basis of the study, this caching distribution can be also modeled as a Zipf distribution with exponent γ c . Thus the storage probability of the i-th popular file, denoted by β i , can be expressed as:\n\n$${\\beta_{i} = \\frac {1/i^{\\gamma_{c}}}{\\sum_{j=1}^{M} 1/j^{\\gamma_{c}}}}$$\n(3)\n\nThe exponent of caching distribution γ c is one of our decision variables which is not necessarily equal to γ r .\n\nNote that under this storage policy, transmission rates related to the BS and FAEs are mainly influenced by the channel model. By contrast, the number of neighbor nodes determined by the D2D covering radius r has a greater influence on the system throughput. The larger the covering radius r is, the more the D2D links are to generate many-to-one D2D communication with a certain storage probability. Without loss of generality, we suppose all of the users are distributed randomly in the cell and can communicate with each other through D2D communications when the distance r 0 between them is less than the threshold r. Furthermore, define R ij as the average transmission rate when the j-th user requests the i-th file. Considering the randomized request probability of the i-th file p i , when there are n users generating requests, the sum of the effective data rate can be written as:\n\n$${\\mathcal{R}}\\left({n,x} \\right) = \\sum\\limits_{j = 1}^{n} \\sum\\limits_{i = 1}^{M} {p_{i}}{R_{ij}}$$\n(4)\n\nTherefore, the solution to the optimization problem can be obtained by solving the following binary linear program:\n\n$${\\text{Max}~~\\sum\\limits_{j=1}^{n} \\sum\\limits_{i=1}^{M} p_{i} R_{ij}}$$\n(5)\n\nSubject to\n\n$$\\begin{array}{*{20}l} &\\sum\\limits_{i=1}^{M} p_{i}=1 \\\\ &0<p_{i}<1,~~ \\forall i \\in M\\\\ &C_{f} \\leq M\\\\ &r_{0} \\leq r,~~ \\forall i,j\\\\ &\\gamma \\geq \\gamma_{0},~~ \\forall i,j \\end{array}$$\n\n### Transmission rate\n\nWhen the j-th user requests the i-th file, network traffic is mainly generated by multi-mode communication, which means that the sender can be generally grouped into three categories in terms of the file storage locations: FAE, neighbor users, and BS. For the sake of analyzing the following derivations, we define the effective data rate as the transfer speed multiplied by the non-outage (transmission success) path existence probability (i.e., the probability that the channel can take place to support the data rate ). Following the view of graph theory, it means that the edge for the communication in the random geometric graph exists with a certain probability. The arguments in this section are similar, the distinction is that the subscripts B (b), F (f), C (c), D (d) are on behalf of the BS, FAE, cellular user and D2D user respectively.\n\n• 1. FAE. Given that there are m storage units in a FAE, each FAE should store the top m most popular files without repetition. Define binary random variable α i , such that α i is equal to 1 if 1≤im ; otherwise, it is equal to 0. If the i-th file on demand is one of the top m popular files, the transmission rate that FAE can provide, denoted by $$\\overline {R_{F}}$$, is equal to:\n\n$$\\overline{R_{F}} = \\alpha_{i}R_{f} e^{-\\eta_{f}} = \\left\\{\\begin{array}{ll} R_{f} e^{-\\eta_{f}}, & 1 \\leq i \\leq m\\\\ 0, & m \\leq i \\leq M \\end{array}\\right.$$\n(6)\n\nR f refers to the ideal transmission rate of FAE and η f indicates the attenuation coefficient of Rayleigh fading. Based on the dedicated bands frequency allocation strategy, the co-channel interference of FAEs can be ignored, and the SINR of its received signal is given by $${\\gamma _{f}} = \\frac {{{P_{F}}{x_{iF}}d_{iF}^{- \\sigma \\;}{H_{iF}}^{2}}}{{{{\\mathrm {N}}_{0}}}}$$. Thus, the ideal transmission rate R f can be obtained by Shannon theory .\n\n$${R_{f}} = Clb\\left({1 + \\frac{{{P_{F}}{x_{iF}}d_{iF}^{- \\sigma \\;}{H_{iF}}^{2}}}{{{{\\mathrm{N}}_{0}}}}} \\right)$$\n(7)\n\nwhere P F denotes the transmit power of FAE, d iF , and H i,F represent the distance and the channel gain between the i-th user and the conterminous FAE, respectively. Path loss exponent is denoted as σ, and x iF is the log normal shadow fading coefficient between the i-th D2D and the FAE. N 0 is white Gauss noise, and C is the available spectrum bandwidth.\n\n• 2. Neighbor user. Each UT caches files at random and independently following the Zipf distribution with exponent γ c . The number of neighbor nodes is usually referred to the binomial distribution with parameters N and P, i.e., K=B(N,P), the probability of k neighbor nodes can be expressed as:\n\n$${P_{r}(K = k) = \\left({~}_{k+1}^{N}\\right)p^{k+1}(1-p)^{N-k-1}}$$\n(8)\n\nwhere $$p = \\frac {\\pi r^{2}}{A}$$, r is determined by the power level for each transmission, and A is the coverage area of the cell. Thus, the numerical probability of cooperating users for the transmission to the i-th file, that is to say, the numerical probability of neighbor nodes who have stored the i-th file is:\n\n$${}{ \\begin{array}{rcl} P(T_{i} = c) & = & \\sum_{k=1}^{N}P\\{c|K=k\\}P_{r}(K=k)\\\\ \\\\ & = & \\sum_{k=c}^{N}\\left({~}_{c}^{k}\\right)\\beta_{i}^{c}(1-\\beta_{i})^{k-c}P_{r}(K=k) \\end{array}}$$\n(9)\n\nAfter a request to the i-th file, the average number of D2D links that one hit can be established as:\n\n$${E(D) = \\sum_{c=1}^{M}cP(T_{i} = C)}$$\n(10)\n\nTaking into account the request probability of the i-th popular file, the mean value of any request for files can be summed up as:\n\n$${}{\\overline{E}(D) = \\sum_{i=1}^{M}p_{i}E(D) = \\sum_{i=1}^{M}p_{i}\\sum_{c=1}^{M}c P(T_{i} = C)}$$\n(11)\n\n$$n\\overline {E}(D)$$ can represent the total number of D2D communication activated by n requesting users at the same time, i.e., $$n\\overline {E}(D)$$ equals to the number of D2D pairs T total mentioned in the later. An important theme to notice is that P r (K=k) is a function of r. Thus, $$\\overline {E}(D)$$ and T total are both functions of variable r and γ c .\n\nThen, the transmission rate provided by the neighbor users, denoted by $$\\overline {R_{D}}$$, can be formulated as:\n\n$$\\overline {\\;{R_{D}}} = \\overline{E}(D)\\;{R_{d}}{e^{- {\\eta_{d}}}} = \\;{R_{d}}{e^{- {\\eta_{d}}}} \\sum\\limits_{i = 1}^{M} {p_{i}} \\sum\\limits_{c = 1}^{N} cP\\left({{T_{i}} = c} \\right)$$\n(12)\n\nR d refers to the ideal transmission rate of D2D communication. Due to reusing the same frequency resources with the cellular communication, the interference of D2D receiver comes from two aspects: white Gaussian noise and the cellular communication. Therefore, when uplink resources allocation is performed efficiently, the SINR of D2D users can be calculated by $${\\gamma _{d}} = \\frac {{{P_{D}}{x_{iB}}d_{iB}^{- \\sigma }{H_{iB}}^{2}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }{H_{CB}}^{2} + {{\\mathrm {N}}_{0}}}}$$, and then R d can be derived through the Shannon formula :\n\n$${{R_{d}} = Clb\\left({1 + \\frac{{{P_{D}}{x_{iB}}d_{iB}^{- \\sigma }{H_{iB}}^{2}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }{H_{CB}}^{2} + {{\\mathrm{N}}_{0}}}}} \\right)}$$\n(13)\n\nwhere P D , P C denote the transmit power of D2D users and one cellular user, d iB and d CB represent the distance between the i-th D2D sender and cellular user to the BS, respectively. H iB and H CB denote the channel gain. σ and x iB are expressed as the path loss exponent and the log normal shadow fading coefficient between the i-th D2D sender and BS.\n\n• 3. BS. The base station will carry out data communication only when the required files are not stored in the neighbor nodes and FAE. Defined P B as the transmit power of each resource block, the transmission rate can be calculated by:\n\n{{\\begin{aligned} \\overline {{R_{B}}} &= (1 - {\\alpha_{i}})P\\left({{T_{i}} = 0} \\right) \\cdot {R_{b}}{e^{- {\\eta_{b}}}}\\\\ \\hspace{5mm} &= \\left\\{ {\\begin{array}{*{20}{c}} {{R_{b}}{e^{- {\\eta_{b}}}} \\sum\\limits_{k = 1}^{N} {{\\left({1 - {\\beta_{i}}} \\right)}^{k}}{P_{r}}\\left({K = k} \\right),\\;\\;\\;\\;\\;m < i \\le M}\\\\ {\\;\\;0, {\\mathrm{1}} \\le \\;i \\le m} \\end{array}} \\right.\\; \\end{aligned}}}\n(14)\n\nIn this case, a file has to be transmitted from the BS and the RU can be thought of as an casual cellular user. Because D2D communications occur in the uplink direction, cellular users, as the receiver of the downstream link communication, are not affected by D2D communication. Then, the ideal transmission rate of BS R b can be described as:\n\n$${ {R_{b}} = Clb\\left({1 + \\frac{{{P_{B}}{x_{iB}}d_{iB}^{- \\sigma }{H_{iB}}^{2}}}{{ {{\\mathrm{N}}_{0}}}}} \\right) }$$\n(15)\n\nIn this optimization problem, self-request (i.e., the user stores the requesting file in its own cache) does not play any role. However, some users can achieve the required files without delay because that self-request needs not special link to transmit date. Based on the above theoretical calculation of transmission speed, the optimization formula can be rewritten as:\n\n$${}{ \\begin{array}{rcl} {{\\mathcal{R}}^{*}}\\left({{\\mathrm{n}},{\\mathrm{x}}} \\right) =&& \\text{Max}\\left\\{ { \\sum\\limits_{j = 1}^{n} \\sum\\limits_{i = 1}^{M\\;} {p_{i}}} \\right.\\left[{\\phantom{\\sum\\limits_{c = 1}^{N}}}\\!\\!\\!\\!\\!\\!\\!{{\\alpha_{i}}{R_{f}}{e^{- {\\eta_{f}}}}\\;} \\right.\\\\ &&+ \\sum\\limits_{c = 1}^{N} c\\;P\\left({{T_{i}} = c} \\right).{R_{d}}{e^{- {\\eta_{d}}}}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\\\ &&\\left. {\\left. { + \\;(1 - {\\alpha_{i}}){R_{b}}{e^{- {\\eta_{b}}}} \\sum\\limits_{k = 1}^{N} {{\\left({1 - {\\beta_{i}}} \\right)}^{k}}{P_{r}}\\left({K = k} \\right)} \\right]} \\right\\} \\end{array}}$$\n(16)\n\nNotice that P(T i =c) is a function of γ c and r, consequently, we can be informed that the optimization formula is a function of several variables, such as γ c , γ r , and r as well as the number of requesting user n. It can be generated by varying one or more of the parameters in a general form.\n\n### Noise and co-channel interference\n\nIn order to guarantee D2D communication, the SINR of D2D users should be greater than its threshold when it is reusing with the cellular uplink resources, i.e., γ d γ D . As shown in 4.2, the SINR of D2D users is a function related to D2D transmit power and the distance from the cellular user to the D2D receiver, and so it is best for D2D communication to reuse the resources of cellular user who is as far as possible from the D2D receiver. Moreover, even the parameters in this function are not associated with those in the optimized objective function, we can derive the minimum transmit power of D2D from this constraint.\n\nTo maintain the comparable fairness, cellular communication should be given priority. The maximum transmission rate of D2D links is restricted to guarantee the non-outage communication of the cellular network. Notice that BS is the receiver of cellular communication in the uplink period, it will then suffer from cumulative interference introduced by multi-pair D2D links. At this time, the signal received by the BS can fall into two categories:\n\n• Cumulative interference of the base station caused by multiple D2D transmitters can be described by Ψ D while T total is the number of D2D links:\n\n$${ {\\psi_{D}} = \\sum\\limits_{i = 1}^{T_{\\text{total}}} \\sqrt{{P_{D}}{x_{iB}} d_{iB}}^{\\,\\,{-\\sigma}/2_{H_{iB}}}}$$\n(17)\n• Uplink signals Ψ C to the BS transmitted by cellular user can be expressed as:\n\n$${ {\\psi_{C}} = \\sqrt{{P_{C}}{x_{CB}} d_{CB}}^{\\,\\,\\, -\\sigma/2_{H_{CB}}}}$$\n(18)\n\nFurthermore, we can approximate the SINR of the BS as:\n\n$${}{ \\begin{array}{l} \\gamma = \\frac{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }{H_{CB}}^{2}}}{{ \\sum \\nolimits_{i = 1}^{{T_{total}}} {P_{D}}{x_{iB}}d_{iB}^{- \\sigma }{H_{iB}}^{2} + {N_{0}}}} \\;= \\frac{{{H_{CB}}^{2}}}{{\\frac{{{N_{0}}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }}} + \\frac{{ \\sum \\nolimits_{i = 1}^{{T_{total}}} {P_{D}}{x_{iB}}d_{iB}^{- \\sigma }{H_{iB}}^{2}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }}}}} \\end{array}}$$\n(19)\n\nUnder high SINR, $${\\frac {{{N_{0}}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }}}}$$ can be approximated to zero. For the convenience of description, defining $${\\mu =\\frac {{{1}}}{{{P_{C}}{x_{CB}}d_{CB}^{- \\sigma }}}}$$ and $${\\nu _{i}} = {P_{D}}{x_{iB}}d_{iB}^{- \\sigma }$$, then the SINR of the BS can be overridden as:\n\n$${\\gamma = \\frac{{{H_{CB}}^{2}}}{{\\mu \\sum \\nolimits_{i = 1}^{{T_{\\text{total}}}} {\\nu_{\\mathrm{i}}}{H_{iB}}^{2}}}}$$\n(20)\n\nSince H CB 2 follows exponential distribution and H iB 2 is exponential random variables subjecting to independent identically distribution, according to [26, 27], the probability density function (PDF) of γ can be rewritten as:\n\n$${{F_{\\gamma} }\\left(\\gamma \\right) = \\sum\\limits_{i = 1}^{{T_{\\text{total}}}} \\frac{{{A_{i}}}}{{\\mu {\\nu_{i}}}}{\\left({\\gamma + \\frac{1}{{\\mu {\\nu_{i}}}}} \\right)^{- 2}}}$$\n(21)\n$${\\kern17pt}{ {A_{i}} = \\prod\\limits_{j = 1,j \\ne i}^{{T_{\\text{total}}}} \\left({\\frac{{{\\nu_{i}}}}{{{\\nu_{i}} - {\\nu_{j}}}}} \\right)\\; }$$\n(22)\n\nFollowing the above derivations, the non-outage probability of the BS is defined as the probability that the instantaneous γ is greater than the threshold γ 0, which can be computed as:\n\n$${}{\\begin{array}{rcl} {P_{N\\text{out}}}\\! =\\! {P_{\\mathrm{\\gamma}}}\\!\\left(\\! {\\gamma \\ge {\\gamma_{0}}} \\!\\right)\\! &=&\\! 1 - {P_{\\mathrm{\\gamma }}}\\left({\\gamma < {\\gamma_{0}}} \\right) = 1 - \\smallint\\limits_{0}^{{{\\mathrm{\\gamma }}_{0}}} {{\\mathrm{f}}_{\\gamma} }\\left(\\gamma \\right){\\mathrm{d}}\\gamma \\\\ &=&\\! 1 - \\smallint\\limits_{0}^{{\\gamma_{0}}} \\left\\{ { \\sum\\limits_{{\\mathrm{i}} = 1}^{{{\\mathrm{T}}_{\\text{total}}}} \\frac{{{{\\mathrm{A}}_{\\mathrm{i}}}}}{{\\mu {\\nu_{\\mathrm{i}}}}}{{\\left({\\gamma + \\frac{1}{{\\mu {\\nu_{\\mathrm{i}}}}}} \\right)}^{- 2}}} \\right\\}{\\mathrm{d}}\\gamma \\\\ &=& \\!1 \\,-\\, \\sum\\limits_{{\\mathrm{i}} = 1}^{{{\\mathrm{T}}_{{\\text{total}}}}} \\prod\\limits_{{\\mathrm{j}} = 1,{\\mathrm{j}} \\ne {\\mathrm{i}}}^{{{\\mathrm{T}}_{{\\text{total}}}}} \\left({\\frac{{{\\nu_{\\mathrm{i}}}}}{{{\\nu_{\\mathrm{i}}} - {\\nu_{\\mathrm{j}}}}}} \\right)\\left({1 - \\frac{1}{{\\mu {\\nu_{\\mathrm{i}}}{\\gamma_{0}} + 1}}} \\right) \\end{array}}$$\n(23)\n\nThus, if more than one D2D links multiplex cellular uplink resources at the same time, the communication performance of the BS will be influenced by the system parameters such as transmit power, distance between D2D transmitter and BS, SINR threshold γ 0 as well as the number of D2D pairs T total. T total indicates the total number of D2D links when n users generate requests simultaneously and can be seen as a variable relate to r and γ c .\n\n### User satisfaction\n\nTo analyze the impact of various arguments on the performance of the system, the concept of user satisfaction is introduced as follows:\n\n$${\\tau = \\alpha P_{N\\text{out}}^ * + \\left({1 - \\alpha} \\right)R_{\\text{total}}^{*}}$$\n(24)\n\nHere, α is the weighted coefficient, $$P_{N\\text {out}}^{*}$$ and $$R_{\\text {total}}^{*}$$ are min-max normalization values. Finding the optimal solution analytically in closed form does not seem feasible, numerical solutions are possible with very low effort. Nevertheless, the pareto solutions of the optimization problem can be obtained using the above equation and numerical experiments based on the multi objective optimization model.\n\n## Numerical results and analysis\n\nIn the aforementioned references, parameters such as the popularity profile of data files and D2D communication radius are always assumed to be known perfectly. In practice, such an assumption cannot be dynamically adjusted and reasonably justified in different circumstances.\n\nIn the following, we provide some numerical results to investigate the effects of relevant parameters on the system throughput with the simple caching strategies. Unless specified otherwise, all figures set limits according to the following except those considering the effect of the parameters. The BS has a coverage area of R=500 m with a storage capacity of M=100 files. In the described system, FAEs use dedicated channels and only a linear impact on the network throughput. For the sake of simplicity, only four FAEs are installed in fixed position. Each FAE has a coverage area of 200 m, and its capacity is 30% of the total files. All channels are subject to path loss model where σ=3. The noise power of the receiver is 70 dBm, and bandwidth of one channel is set to 1 MHz. We use the Monte Carlo experiment to prove the effectiveness of the proposed algorithm, and the number of Monte Carlo experiments is 10000. In the process of the experiment, we assume that the BS uses a transmission scheme similar to the fourth-generation long-term evolution (LTE) standard, based on an OFDM-TDMA physical layer. The FAEs are operating with dedicated spectrum resource and have multiple antennas, i.e., WiFi-like links. We envision that UTs have single-antenna although multi-antenna system is inevitable in the future.\n\nThe number of D2D links generated by user layer is an important criterion that affects the system throughput and cumulative interference. Thus, the first part of the experiment begins with the influence of each parameter on it.\n\nFigures 4 and 5 show the number of D2D links for requesting the different files, which is named as well as the file sequence number (FSN) and ordered by the popular rankings. Intuitively, the number of D2D links will be relatively large when users request the first ten files, as in Fig. 4. However, looking closely at Fig. 5, we can observe that the number of D2D links rapidly descends and then becomes gradually lower if users request the remaining 80% of the files. The results will be slightly fluctuated but will not affect the the overall trend of the experiment. This occurs because the selection of RU is random in the process of experiment. Meanwhile, the position of UTs is practically random although their storage probability is modeled as the Zipf distribution. Then, the Monte Carlo experiment results fluctuate by a number of random sampling variability. These fluctuations are almost all under 1, which is a good indicator that the remaining 80% of the content has limited contribution to the traffic. Moreover, the greater the γ c value, the more obvious the change. We can conclude that γ c determines the file storage probability; the larger γ c means the greater probability that the most popular file is stored. This storage mode is more effective only when the request probability is considered simultaneously.\n\nThe average number of D2D links versus the collaboration distance r for different γ c and γ r is shown in Fig. 6. Different γ r indicates different request probabilities. We can observe that the number of D2D links has at least twice as γ r and γ c increase by 0.4. This agrees with the following intuition: the larger γ r and γ c , the higher probability that a popular file will be requested, and it is just consistently stored in the neighbor nodes, especially with the increase of the cooperation distance. Furthermore, we can find that the impact of γ r and γ c on the number of D2D links is essentially the same, although they can take different values separately. In most of the case, γ r can be obtained from the data statistics. Accordingly, in the following simulation experiments, we just need to think about the impact of γ c by setting γ r =0.8, this value is based on a study conducted on the University of Massachusetts Amherst campus in 2008. Meanwhile, we can observe that in the case of γ r invariant, to achieve the same number of D2D links, r will decrease with the increase of γ c .\n\nThe total number of users and the number of users seeking files concurrent also affect the average number of D2D links in the cellular network. Figure 7 represents different curves by tracing the lines of different parameters. The x-axis refers to the number of users requesting files simultaneously. The bigger the N value is, the higher the user density is, and then the better the D2D connection performance is. We can also observe that the effect of N value on the curve is similar to that of r. This is consistent with the theoretical results. D2D connection capability has doubled and redoubled as users grow exponentially.\n\nFurthermore, we note that N value has lower impact on the average number of D2D links compared with γ c . The reason is that user requests are based on the probability of request in the experimental process. Changing the popularity Zipf distribution exponent γ c has greater impact on the popularity of the document. If the request is subject to the uniform distribution, N will have a greater impact on the chance of D2D connection with a fixed radius.\n\nIt can be observed that the theoretical simulation results are basically consistent with the simulation results using Monte Carlo simulation in Fig. 8. It means that the method for D2D users to reuse cellular wireless resources based on random distribution is effective. Moreover, with the same storage probability, i.e., the same γ c , the numbers of D2D communication satisfying certain condition increases as the D2D transmission radius rises. Sometimes slight jitter will be produced due to the impact of random channel. In the remaining figures, except those we want to consider the effect of γ c , we set it to 1.\n\nFrom now on, in order to analyze the influence of the system interference, we just present the simulation results of non-outage probability in which the average number of D2D links is a crucial technical parameter.\n\nWhen multiplexing the uplink resources, multi-pair D2D communications will give rise to cumulative interference. In the D2D sharing mode based on cellular system, we must give priority to ensure the quality of main chain link. Thus, the effect of multiple D2D links on the performance of cellular communication should be studied. Figure 9 shows the comparison of the non-outage probability between the reference and the proposed system through Monte Carlo simulation. In simulating process, the transmit power of cellular communication is set to 1 W and D2D transmit power is 10 mW, the SINR threshold of BS γ 0 is limited to 2. With the increase of RUs’ number, the non-outage probability of the BS decreases. This is due to the cumulative interference caused by D2D communication to the BS, which makes the communication quality between cellular user and BS decrease. Moreover, as can be seen from the picture, the non-outage probability of reference is slightly better than the proposed system. The reason behind is that the proposed system allows many-to-one D2D communication, which brings huge profits as well as interference. The number of D2D links activated under the same conditions is higher than the reference . The system will even introduce cumulative interference while bringing higher throughput. Therefore, in order to ensure the quality of the cellular communication, the maximum number of D2D links allowed to reuse resources can be obtained through the non-outage communication constraints.\n\nThe average transmission rate versus r under four different transmission strategies is shown in Fig. 10 when a request occurs. The average rate increases with the increase of r. In fact, traditional D2D refers to one-to-one D2D communication, which uses random storage strategies and is shown in the studies pre-storing files in the UTs. For example, as described in , UTs act as mobile helper stations can be seen as traditional D2D. Considering SBS dissemination mode, single SBS indicates the case that just one SBS can be connected in the reference . Intuitively, a larger r means that more qualified UTs, i.e., the neighbor nodes within the communication range as well as storing the required files, can be detected. This will almost certainly lead to higher frequency reuse. Therefore, multi-dimension diversity gain caused by FAE and multiple neighbor nodes can tremendously improve the transmission rate in the proposed strategy. In contrast, traditional D2D can connect neighbor node no more than just one. Although the connection probability increases as the D2D communication distance rises, when the probability reaches to 1, the channel attenuation becomes more serious with the increase of transmission distance. Therefore, the transmission speed decrease with the distance increasing after the connection probability is 1. Not surprisingly, the average speed transmitted through BS or SBS is relatively stable but smaller because of the lack of proximity gain. Moreover, we should keep in mind that the proposed transmission strategy not only improve the transmission speed and reduce the transmission delay but also unloads the transmission flow of the BS.\n\nConsidering the interference, user satisfaction is defined to measure the effect of system outage probability on user transmission rate. The effects of changing the popularity Zipf storage exponent γ c on user satisfaction with different coefficient a are investigated in Fig. 11. As seen in the graph, the greater the non-outage probability weight, the faster decay of user satisfaction with the communication distance. It is because that the greater weight indicates the higher requirements of quality communication. User satisfaction fluctuates more and more slightly when a decreases with γ c increasing. In addition, we can observed that its optimal distance r opt decreases with the increase of γ c . For the small γ c , there is a little redundancy in the users’ storage, i.e., the probability of finding files in the neighbor nodes is generally small. Thus, in order to increase the chance of having D2D communication and to achieve customer satisfaction, the collaboration distance r should increase. Hence, the optimal communication distance r opt decreases for large γ c when the number of requesting users is fixed.\n\nUser satisfaction versus r is shown in Figs. 12 and 13 with different number of requesting users. The two pictures tell us that the greater the non-outage probability weight, the faster the decay of user satisfaction with the certain communication distance. The reason behind is that the greater weight indicates the higher requirement for reliable communication. After the peak, the large D2D communication radius means higher number of D2D links, and then results in larger outage probability. User satisfaction decreases as well as the communication quality. The best D2D communication radius r opt can be found from the figure when the user satisfaction reaches the top. Comparing Figs. 12 and 13, we can conclude that the optimal collaboration distance become smaller along with the number of requesting users increasing.\n\nThe influence of Zipf storage exponent γ c on user satisfaction with a=0.8 are investigated in Fig. 14. It shows the cases that different number of users request files under optimized distance. For small γ c , there is a little redundancy in users’ storage, i.e., the probability of finding files in the neighbor nodes is generally small. From the figure, it can be known that with the increase of γ c , user satisfaction will rise accordingly. After reaching the peak, satisfaction decays rapidly. This is because the weight of outage probability is large, as already explained above. It is obvious that, we can achieve a set of pareto solutions for the optimization goal.\n\n## Conclusions\n\nContent delivery strategy based on active storage in local is an effective way to settle the explosion of mobile data traffic. In this paper, we introduced cooperative transmission scheme by jointly using the storage capacity of UTs and FAEs. Aiming at the scenarios which have multiple-pair D2D links in the system, the network transmission rate is optimized on the premise of guaranteeing the main communication performance. Furthermore, the optimal D2D communication distance and storage probability exponent can also be found based on multi-objective optimization problem. The analytical and simulation results show that the proposed scheme can obviously improve the network throughput, can reduce the transmission delay, and can restrain the mutual interference among different links effectively. Meanwhile, the data traffic of the most popular files can be offloaded to D2D communication, which can provide high spectral efficiency and free up BS to provide other business.\n\nIn the future, we still have a lot of additional work to do, such as the question of storage optimization as well as the cooperative strategies of content allocation between UTs and FAEs. Furthermore, more complex D2D scenes need to be studied, e.g., multiple D2D users reuse the same CU resource simultaneously.\n\n## References\n\n1. 1\n\nChina Mobile, Chinamobile technology vision 2020 + white paper, (2016). [Online]. Available: http://www.gtigroup.org/.\n\n2. 2\n\nN Golrezaei, AG Dimakis, AF Molisch, in Information Theory Proceedings (ISIT), 2012 IEEE International Symposium on. Wireless device-to-device communications with distributed caching (IEEE, Cambridge, 2012), pp. 2781–2785.\n\n3. 3\n\nCVNI Cisco, Global mobile data traffic forecast update. white paper, (2016). [Online]. Available: http://www.cisco.com/.\n\n4. 4\n\nM Afshang, HS Dhillon, PHJ Chong, Fundamentals of cluster-centric content placement in cache-enabled device-to-device networks. IEEE Trans. Commun. 64(6), 2511–2526 (2016).\n\n5. 5\n\nBN Bharath, KG Nagananda, HV Poor, A learning-based approach to caching in heterogenous small cell networks. IEEE Trans. Commun. 64(4), 1674–1686 (2016).\n\n6. 6\n\nN Golrezaei, AG Dimakis, AF Molisch, et al., in Signals, Systems and Computers (ASILOMAR), 2011 Conference Record of the Forty Fifth Asilomar Conference on. Wireless video content delivery through distributed caching and peer-to-peer gossiping (IEEE, Pacific Grove, 2011), pp. 1177–1180.\n\n7. 7\n\nN Golrezaei, AF Molisch, AG Dimakis, et al., Femtocaching and device-to-device collaboration: A new architecture for wireless video distribution. IEEE Commun. Mag. 51(4), 142–149 (2013).\n\n8. 8\n\nSF Chou, TC Chiu, YJ Yu, et al., in Global Communications Conference (GLOBECOM), 2014 IEEE. Mobile small cell deployment for next generation cellular networks (IEEE, Austin, 2014), pp. 4852–4857.\n\n9. 9\n\nM Bennis, M Simsek, A Czylwik, et al., When cellular meets WiFi in wireless small cell networks. IEEE Commun. Mag. 51(6), 44–50 (2013).\n\n10. 10\n\nX Zhao, C Yang, Y Yao, et al., Cognitive and cache-enabled d2d communications in cellular networks. [Online]. Available: https://pdfs.semanticscholar.org/7359/2aa42bfc4f9d69b4a300049f46bd330ed628.pdf.\n\n11. 11\n\nG Negin, P Mansourifard, MF Andreas, DG Alexandros, Base-station assisted device-to-device communications for high-throughput wireless video networks. IEEE Trans. Wirel. Commun. 13(7), 3665–3676 (2014).\n\n12. 12\n\nH Liu, Z Chen, X Tian, X Wang, M Tao, On content-centric wireless delivery networks. IEEE Wirel. Commun. 21(6), 118–125 (2014).\n\n13. 13\n\nC Yang, Y Yao, Z Chen, B Xia, Analysis on cache-enabled wireless heterogeneous networks. IEEE Trans. Wirel. Commun. 15(1), 131–145 (2016).\n\n14. 14\n\nWC Ao, K Psounis, in Proc. of ACM MobiHoc 2015. Distributed caching and small cell cooperation for fast content delivery, (2015), pp. 127–136.\n\n15. 15\n\nQ Bodinier, A Farhang, F Bader, et al, in Communications (ICC), 2016 IEEE International Conference on. 5G waveforms for overlay D2D communications: Effects of time-frequency misalignment (IEEE, Kuala Lumpur, Malaysia, 2016), pp. 1–7.\n\n16. 16\n\nR Yin, C Zhong, Yu G, Z Zhang, KK Wong, X Chen, Joint spectrum and power allocation for d2d communications underlaying cellular networks. IEEE Trans. Veh. Technol. 65(4), 2182–2195 (2016).\n\n17. 17\n\nN Lee, X Lin, JG Andrews, RW Heath, Power control for d2d underlaid cellular networks, modeling, algorithms, and analysis. IEEE J. Selected Areas Commun. 33(1), 1–13 (2016).\n\n18. 18\n\nC Yang, T Jiang, X Chen, Z Junshan, Social-aware video multicast based on device-to-device communications. IEEE Trans. Mobile Comput. 15(6), 1528–1539 (2016).\n\n19. 19\n\nM Gregori, et al., Wireless distributed storage in socially enabled D2D communications. IEEE Trans. Wirel. Commun. 15(6), 1–13 (2016).\n\n20. 20\n\nHV Balan, R Rogalin, A Michaloliakos, K Psounis, G Caire, Airsync, Enabling distributed multiuser MIMO with full spatial multiplexing. IEEE/ACM Trans. Netw. 21(6), 1681–1695 (2013).\n\n21. 21\n\nE Antonio-Rodríguez, et al., Wideband full-duplex MIMO relays with blind adaptive self-interference cancellation. Signal Process. 130:, 74–85 (2017).\n\n22. 22\n\nI tube, you tube, everybody tubes, analyzing the world’s largest user (2007).\n\n23. 23\n\nM Dehghan, A Seetharam, Jiang, in Computer Communications (INFOCOM), 2015 IEEE Conference on. On the complexity of optimal routing and content caching in heterogeneous networks (IEEE, Kowloon, 2015), pp. 936–944.\n\n24. 24\n\nV Sciancalepore, V Mancuso, A Banchs, S Zaks, A Capone. Enhanced content update dissemination through D2D in 5G cellular networks. IEEE Trans. Wirel. Commun. 15(11), 7517–7530 (2016).\n\n25. 25\n\nP Mogensen, W Na, IZ Kovács, et al., in Vehicular Technology Conference, 2007. VTC2007-Spring. IEEE 65th. LTE capacity compared to the Shannon bound (IEEE, Dublin, 2007), pp. 1234–1238.\n\n26. 26\n\nAM Mathai, SB Provost, Quadratic forms in random variables. (New York, Marcel Dekker, 1992).\n\n27. 27\n\nA Papoulis, SU Pillai, Probability, random variables and stochastic process.4th ed. (New York, McGraw-Hill, 2002).\n\n## Acknowledgements\n\nThis work was supported by the National Natural Science Foundation of China under Grant No. U1404602, the Young Scholar Program of Henan Province under Grant No.2015GGJS-086, the Science and Technology Foundation of Henan Educational Committee under Grant No. 14A510011, the Science and Technology Key Research Program of Henan Province with No. 172102210341, the Dr. Startup Project of Henan Normal University under Grant No.qd14136, and the Young Scholar Program of Henan Normal University with No. 15018.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nXZ put forward the idea and wrote the manuscript. PY took part in the discussion and gave the original ideas, he also guided, reviewed, and checked the writing. YC and PC carried out the experiments and analyzed the experimental results. All authors read and approved the final manuscript.\n\n### Corresponding author\n\nCorrespondence to Peiyan Yuan.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Rights and permissions",
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http://oqsr.ciclofucina.it/complex-number-cis-calculator.html | [
"# Complex Number Cis Calculator\n\nSo with this example up here 8x-4+3x+2. Answer to Convert the following complex numbers to the form a + bi. defined over the complex numbers. The complex numbers may be represented as points in the plane, with the real number 1 represented by the point (1;0), and the complex number irepresented by the point (0;1). all work with complex numbers. Unit converter A powerful unit converter with a vast amount of supported measurements (length, weight, volume, area, speed, time, temperature, energy, power, pressure, angle, data). 14 Complex Number Plane 1. Making statements based on opinion; back them up with references or personal experience. In this case, the xx -axis is …. We could start by taking a stretch of the line near the origin (that is, the point representing the number zero) and putting in the integers as follows:. Equation Calculator for Macintosh. Learn more Accept. The nice property of a complex. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step. However complex numbers are complete i. The Calculator automatically determines the number of correct digits in the operation result, and returns its precise result. By using this website, you agree to our Cookie Policy. In polar coordinates, a complex number z is defined by the modulus r and the phase angle phi. But up until now, we have only one solution for the equation x 3 = 8. Z = cis ) Preview cis 1) Preview cis ) Preview. We could start by taking a stretch of the line near the origin (that is, the point representing the number zero) and putting in the integers as follows:. Complex numbers are very useful in circuit analysis. * The data type for Complex Numbers. Enter ( 6 + 5. 000i >> (-3)^0. Therefore, our number 3 + √(-4) can be written as 3 + 2i, and this is an example of a complex number. In order to use DeMoivre's Theorem to find complex number roots we should have an understanding of the trigonometric form of complex numbers. Example 1: Plot 4− 2 i −3 + 2 i, and −5 − 3 i in the complex plane (see Figure 1). Glyndŵr University was inaugurated in 2008 and has campuses in Wrexham, Northop, St Asaph and London. We recommend you now use our main cable sizing application over at myCableEngineering. GK Apps Education. To calculate the imaginary part of the following complex expression z=(1+i)/(1-i), enter imaginary_part((1+i)/(1-i)) or directly (1+i)/(1-i), if the button imaginary_part already appears, the result 1 is returned. Some examples of complex numbers are 3 − i, ½ + 7i, and −6 − 2i. Following eq. A complex number has two square roots, three cube roots, four fourth roots, etc. So that we can focus all our efforts on our new application, we have retired our myElectrical. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1. Another interesting example is the natural logarithm of negative one. It is denoted by “θ” or “φ”. When b=0, z is real, when a=0, we say that z is pure imaginary. Write a web application that functions as a simple hand calculator, but also keeps a \"paper trail\" of all your previous work. CPT 172 Midterm Part 1 to 3. The Complex Number Formatting Problem The issue is that if the real and imaginary portions of a complex… Read more about Complex Number Formatting in Excel. The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. Complex Numbers. Find the common ratio if the fourth term in geometric series is $\\frac{4}{3}$ and the eighth term is $\\frac{64}{243}$. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1. com is the most convenient free online Matrix Calculator. 2018/04/05 10:30 Male/20 years old level/High-school/ University/ Grad student/Very/ Purpose of use To check a complex number calculator program I wrote in C for a university course. See Where Numbers Go on a Number-Line - powered by WebMath. Complex Number – Basic definition: A number that has both a real and imaginary part: z = a + bi ( a – bi is called the complex conjugate) For example: z = 5 + 3i or z = 1. False Recall that if w = r cis θ then w 4 = r 4 (cis θ) 4 = r 4 cis (4 θ). Cis(x) is another name for the complex exponential, Cis(x)=e^(ix)=cosx+isinx. 4 + 0i = 4. The complex numbers must be put in brackets. Complex Numbers in VBA. Since complex numbers are legitimate mathematical entities, just like scalar numbers, they can be added, subtracted, multiplied, divided, squared, inverted, and such, just like any other kind of number. Why does 2^n * cis(x) = -2^n?? What does cis(x) itself then equal? Thank you! Edit: When I saw -2^n I don't mean (-2)^n, I mean -(2^n) :). 0º/5 = 0º is our starting angle. The a + bi options tells the calculator to display […]. 3 For 2nd complex number Enter the real and imaginary parts: 5. All basic operations for complex numbers Addition. We associate each complex number z=a+biz=a+bi with the point (a,b)(a,b) on the coordinate plane. Get complete property information, maps, street view, schools, walk score and more. Engaging math & science practice! Improve your skills with free problems in 'Writing a Quotient of Complex Numbers in Standard Form' and thousands of other practice lessons. Why publicize a command line calculator in a time when the computer should read the mind of people and execute the operation ( i’m exceeding, I know ). With no arguments, rounds both the real and imaginary parts to the nearest integer and returns a new Complex number. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. We can use this notation to express other complex numbers with M ≠ 1 by multiplying by the magnitude. Complex Number Calculator. Z=-2 +2j When a complex number is cubed, its magnitude (r) will be powered by 3 and its Arg (q) will be multiplied by 3. The function polar takes a complex number and returns a (magnitude, phase) pair in canonical form: The magnitude is nonnegative, and the phase, in the range (- p , p ]; if the magnitude is zero, then. Its principal value is $\\ln (-1) = \\ln \\left(1e^{i\\pi}\\right) = \\pi i$. Answer to Convert the following complex numbers to the form a + bi. (a) 2 cis(π/6)(b) 5 cis(9π/4)(c) 3 cis(π)(d) cis(7π/4)/2. number € 3+6i. Another way of writing the polar form of the number is using it's exponential form: me^(ia. This entry was posted in Other and tagged complex numbers on May 7, 2015. Given two complex numbers in polar form, find their product or quotient. Example: type in (2-3i)*(1+i), and see the answer of 5-i. These are the capstone worksheets for rounding with decimals! The rounding worksheets in this section deal with longer numbers (for example, values where the fractional part of the number goes out to ten-thousandths precision) and will test a student's ability to identify many different small-division place place values. Consider the new complex calculation requirements. This numerical calculator requires that you type your formulas. This calculator is designed to give a value, even if complex, for the data entered. The number i can be written in the complex number form of 0 -1i. So each complex number needs two ordinary (\"real\") numbers to describe it. hp calculators HP 30S Solving Problems Involving Complex Numbers Basic concepts There is no real number x such that. Contains Ads. Polar mode on your calculator means that you want answers in a polar form, even if you enter expressions in rectangular. where is the real part of and is the imaginary part of , often denoted and , respectively. Support for most mathematical operations available in Julia standard library and special functions from SpecialFunctions. Just type your formula into the top box. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. If you're seeing this message, it means we're having trouble loading external resources on our website. 1 Know there is a complex number i such that i 2 = -1, and every complex number has the form a + bi with a and b real. “Polar form” means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. It is known that the polynomial equation. r (cos θ + i sin θ) Here r stands for modulus and θ stands for argument. It is suitable for all types of users, whether they are looking to carry out basic or complex mathematics; It is equipped with a powerful MATH CORE advocated by MATH DISPLAY support, which increases its functionality by leaps and bounds, making it an ideal choice for engineers. ; Algebraically, as any real quantity φ such that. Use the polar to rectangular feature on the graphing calculator to change 4 cis. Subtraction 9. I have looked at a bunch of Sharp models, but I can't find specific information saying if they support that. The 𝛼 in the expression represents a real number, whereas the � represents an imaginary number, and lastly the 𝑖 represents √−1. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. Note that the complex number cos + i sin has absolute value 1 since cos 2 + sin 2 equals 1 for any angle. Here j is the holomorphic function on the complex upper half plane invariant under the action of SL2. Rectangular to polar conversion 6. The “argument” of a complex number is just the angle it makes with the positive real axis. C++ class for addition, subtraction, multiplication and division for complex numbers. Phase (angle) of a complex number cis is less known notation: cis(x) = cos(x)+ i sin(x); example: cis (pi/2) + 3 = 3+i conj conjugate of complex number - example: conj(4i+5) = 5-4i Complex numbers in word problems: ABS CN Calculate the absolute value of complex number -15-29i. In this binomial, a and b represent real numbers and i = √−1. The basic imaginary unit is equal to the square root of -1. 2 - Use a calculator to help write each complex number. Some examples on complex numbers are − 2+5i 3-9i 8+2i A program to add complex numbers by passing structure to a function is given as follows − Example. In this expression, a is the real part and b is the imaginary part of the complex number. C program to add, subtract, multiply and divide Complex Numbers, complex arithmetic C program to add, subtract, multiply and divide complex numbers. so $$3 \\sqrt{2} - 3 \\sqrt{2}\\, i=6 \\cis 315^{\\circ}\\text{. Complex Numbers. An ____ is an object person or other entity that is a potential risk of loss to an asset. 5 cis(255°) 2 (cos30° + i sin 30°). \"cis\" is a shorthand way of writing complex numbers. The complex number z in geometrical form is written as z = x + iy. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre. 3D & 2D Vector Magnitude Calculator. Rewrite each of the following complex numbers in rectangular form and graph. We can then de ne the limit of a complex function f(z) as follows: we write lim z!c f(z) = L; where cand Lare understood to be complex numbers, if the distance from f(z) to L, jf(z) Lj, is small whenever jz cjis small. A convenient form for numbers in the complex plane, other than rectangular form, is the trigonometric form of complex numbers. When a number has the form a + bi means a real number plus an imaginary number it is called a complex number. p cis cis cis cis cis n nn n n cis. Then taking z = a + bi, you get cosh(a+bi) = cos(-b+ai) where the latter can be evaluated on many calculators. We are committed to an evidence-led and common-sense approach to containing the coronavirus outbreak and mitigating its impact on employees, customers and our wider business operations. Online calculator. Those are some symbols that's say if you want to take the cube root of a complex number, take the (real) cube root of its magnitude, and divide the angle by three. Use MathJax to format equations. The complex numbers calculator can also determine the conjugate of a complex expression. The proof of the following proposition is straightforward and is left as an exercise. Write the following complex numbers in trigonometric form: (a) 4 + 4i To write the number in trigonometric. Free simplify calculator - simplify algebraic expressions step-by-step This website uses cookies to ensure you get the best experience. Equation Calculator is a scientific calculator which does symbolic and algebraic manipulation, algebra and calculus as well as numeric computation. Solve complex coefficient linear equation system. All \\(n$$ of the $$n$$th roots of a complex number $$z$$ are evenly spaced around a circle centered at $$0$$ and having a radius. Angle Between Vectors Calculator. If you have a different calculator or software package you would like to see included, let me know. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. 29 as much as possible? Have no book to work out of, having a hard time finding the answer to this using Google. 9i Basic operations on complex numbers Addition/subtraction: combine all real parts together and all imaginary parts together Multiplication: expand first and then. Find the cube root when z = 8cis(n). Enter expression with complex numbers like 5* (1+i) (-2-5i)^2. For the following exercise, find the root of the complex number in polar form. This can be shown using Euler's formula. Jun 16 Beautiful Trigonometry Brady Haran. A cis B cis C cis D cis - D 10 Express the complex number represented by Microsoft Word - 11SM Complex Numbers Test B. Thus, z 1 and z 2 are close when jz 1 z 2jis small. This calculator will also tell you if the number you entered is a perfect square or is not a perfect square. For Example, we know that equation x 2 + 1 = 0 has no solution, with number i, we can define the number as the solution of the equation. So sine of phi is equal to 1/2. Substituting k to be 0, 1, 2, , n − 1 into the formula for z gives different complex numbers each of which is an nth root of w. Support for complex matrix. Cis A complex -valued function made from sine and cosine with definition cis θ = cos θ + i sin θ. It is denoted by “θ” or “φ”. Instructions. It can factor expressions with polynomials involving any number of vaiables as well as more complex functions. “Polar form” means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. So if you think back to how we work with any normal number, we just add and when you add and subtract. The complex number z = x + yi is plotted as the point (x, y), where the real part is plotted in the horizontal axis and the imaginary part is plotted in the vertical axis. 2 Use the relation i 2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. Find the absolute value of a complex number. It looks just as easy as possible, but can do many things. Leave your answers in trig form. None of the other functions work (eg no complex exponentials). Part d) posed problems for a great many, and correct solutions were rarely seen. /** * ComplexNumber is a class which implements complex numbers in Java. Complex Numbers Basic Concepts of Complex Numbers Complex Solutions of Equations Operations on Complex Numbers Identify the number as real, complex, or pure imaginary. And let's see, the real part is negative three, so we could go one, two, three to the left of the origin. * The data type for Complex Numbers. (This is exactly the same thing that one does when one first learns about rational numbers (fractions). Euler's formula states that for any real number x: = + , where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions. Complex numbers are expressed in the form 𝛼+�𝑖. 11 Cis 247 Degrees. Rewrite each of the following complex numbers in rectangular form and graph. By using this website, you agree to our Cookie Policy. Angle Between Vectors Calculator. Hello My question is how do i create complex number calculator. edu Basic Tutorial Using calculator for operation. This calculator lets you calculate in an easy and fast way the operations between two complex numbers. Using the real number system, we cannot take the square root of a negative number, so I must not be a real number and is therefore known as the. If the complex is in the cis. 0400 • ( 1 + 7 i) Processing ends successfully. 1 Know there is a complex number i such that i 2 = -1, and every complex number has the form a + bi with a and b real. Suppose that the four numbers from part c are replaced by the numbers 0, 5cis π. Similar to the above function, the polar number be converted to a cartesian complex number; having a real and imaginary part. , i= √ −1=⇒i2 = −1 Complex Number: A complex number zis an ordered pair of real numbers [a,b] ≡a+ib: ais the real part of z(Re{z})andbis the imaginary part (Im{z}). Herein, we report that a carbon nanoparticles–Fe(II) complex (CNSI. In fact, a common way to write a complex number in Polar form is And \"cos θ + i sin θ \" is often shortened to \"cis θ \", so: x + iy = r cis θ cis is just shorthand for cos θ + i sin θ. Complex_conjugate online. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. The calculator also provides conversion of a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). R=fifthroot of 32=2. In the RedCrab Calculator , you can use the multipication symbol for real and complex numbers. 3(cos 210+i sin 210) = 3(-sqrt(3)/2)+3i(-1/2) -(3/2)sqrt(3)-(3/2)i My favourite way of seeing that sin 30 = 1/2 and cos 30 = sqrt(3)/2 is to picture an equilateral triangle with sides of length 1. Complex numbers %ˆ The calculator performs the following complex number calculations: • Addition, subtraction, multiplication, and division • Argument and absolute value calculations • Reciprocal, square, and cube calculations • Complex Conjugate number calculations Setting the complex format: Set the calculator to DEC mode when. complex-number-calculator menu. IMREAL: Returns the real part of a complex number. Solvers and Calculators in this section. Even if exp(i * a) can be used, a native support of trigonometric, polar forms would be a real added value. ) Originally Posted by Bjarne Stroustrup (2000-10-14) I get maybe two dozen requests for help with some sort of programming or design problem every day. Similar forms are listed to the right. Glyndŵr University was inaugurated in 2008 and has campuses in Wrexham, Northop, St Asaph and London. Making statements based on opinion; back them up with references or personal experience. Form Of Complex Numbers: In the complex number a + bi, where a is the real part and b is the imaginary part. Finding nth roots of Complex Numbers. Additive Identity. In polar coordinates a complex number is defined by the radius r and the phase angle phi. In this case, the xx -axis is …. Since the argument is undefined and is positive, the angle of the point on the complex plane is. i2 =−1, and is called the. We mainly post videos about mathematics and just numbers in general. The oxidation state and the coordination number reflect the number of bonds formed between the metal ion and the ligands in the complex ion. Mathematics and computer science courses and at least 8 credits of upper-division CIS courses used to satisfy upper-division major requirements must be taken for letter grades and passed with grades of C– or better. 108-453 San Antonio, TX 78248 USA Phone: (512) 788-5606 Fax: (512) 519-1805 Contact us. Starting from 16th century mathematicians faced necessity of the special numbers, known nowadays as complex numbers. Now, for two complex numbers x and y, we can use x + y, x - y, x * y, and x / y directly. Difference of Squares: a 2 – b 2 = (a + b) (a – b) Step 2: Click the blue arrow to submit and see the result!. The Complex Numbers Calculator functions: [ ] Simple operations [ ] Simplify entering complex number [ ] Symbolic calculation [ ] Detail solution: division of a complex numbers (the fraction) [ ] Conjugate of a complex number [ ] Algebraic form of the complex number [ ] Trigonometric form of the complex number. The interface of Complex Calculator Display: Outputs the results of calculations and current number inputs. I am just unsure if it is possible to input complex numbers in a matrix for the calculator to solve. Question: Write The Given Complex Number In The A +bi Form 1. 1 The Complex Plane A complex number zis given by a pair of real numbers xand yand is written in the form z= x+iy, where isatis es i2 = 1. Setting the calculator to COMPLEX mode ,4-1 Entering complex numbers ,4-2 Polar representation of a complex number ,4-3 Simple operations with complex numbers ,4-4 Changing sign of a complex number ,4-5 Entering the unit imaginary number ,4-5 The CMPLX menus ,4-5 CMPLX menu through the MTH menu ,4-6 CMPLX menu in keyboard ,4-7 Functions applied. Using complex numbers to solve an AC circuit, determining the AC currents in different parts of an induction motor and the active and HP50g - Equation Writer and Complex Number Systems of youtu. Enroll in one of our FREE online STEM summer camps. This is an electronic computational device capable of handling trigonometry, logarithm, exponential, and statistics functions in addition to basic arithmetic operations. To calculate the imaginary part of the following complex expression z=(1+i)/(1-i), enter imaginary_part((1+i)/(1-i)) or directly (1+i)/(1-i), if the button imaginary_part already appears, the result 1 is returned. How to do Complex Numbers on the ClassWiz This video shows basic operation of CASIO ClassWiz for Complex Numbers. The phase angle phi is the counter clockwise angle from the positive x axis, e. Proposition 4. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity. If you’re reading this as a Construction Industry Scheme (CIS) contractor, then you or your partner are probably struggling to find a mortgage right now. I know how to input complex numbers. Find the powers of each complex number in polar form. Calculates mathematical formulas of any length and complexity. The argument of a complex number is the direction of the number from the origin or the angle to the real axis. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. And the mathematician Abraham de Moivre found it works for any integer exponent n: [ r(cos θ + i sin θ) ] n = r n (cos nθ + i sin nθ). Complex Number Formula A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, that satisfies the equation i 2 = −1. George Robert Stibitz (April 30, 1904 - January 31, 1995) was a Bell Labs researcher internationally recognized as one of the fathers of the modern first digital computer. Complex number form converter This calculator allows one to convert complex number from one representation form to another with step by step solution. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. This calculator lets you calculate in an easy and fast way the operations between two complex numbers. Following eq. (A) 16S (B) 32S (C) 64S (D) 128S (E) NOTA 12. Product and Quotient Theorems The advantage of polar form is that multiplication and division are easier to accomplish. Complex Numbers Calculator. Once finished with the table, please double-check the information you provided above to make sure that it is correct. The best math tool for school and college! If you are a student, it will helps you to learn complex number. That is the map z7→ z+z 0 represents a translation aunits to the right and bunits up in the complex plane. Figures 4 and 5 show the display in algebraic mode. Answer to Convert the following complex numbers to the form a + bi. Viking’s favorite weapons are axes! You can throw them at enemies and split their…. Mexp(jθ) This is just another way of expressing a complex number in polar form. Recall that any complex number, z, can be represented by a point in the complex plane as shown in Figure 1. Any time you have to have guidance with math and in particular with positive numbers calculator or rational expressions come visit us at Algebra1help. For calculating conjugate of the complex number following z=3+i, enter complex_conjugate(3+i) or directly 3+i, if the complex_conjugate button already appears, the result 3-i is returned. We just combine like terms. Since i is used for representing. 1 cis 210° Write each complex number in standard form. The 'CPLX' button toggles the system into complex mode. This entry was posted in Other and tagged complex numbers on May 7, 2015. KS-Calculator is a powerful, easy-to-use, flexible calculator, which supports operations with numbers in Binary, Octal, Decimal and Hexadecimal formats, operates with expressions and mathematical functions, like: Sin, Cos, Tan, ArcTan, Exp, Ln, Log, etc. For example, the conjugate of 3 + 15i is 3 - 15i, and the conjugate of 5 - 6i is 5 + 6i. Download Complex Number Calculator Precision 45 - Calculator that provides support for trigonometric, hyperbolic, inverse and other functions, history with all calculations, and list with constants. It’s good for checking your answers. eCalc is a free and easy-to-use scientific calculator that supports many advanced features, including unit conversion, equation solving, and even complex-number math. Here, we explore the effects of stabilizing selection on cis -regulatory genetic variation in humans. So, plot -4-4i onto the diagram and you should get a point on the third quadrant. The servlet that receives the request is the controller; it calls another class, the model, to do the computations; and it forwards the results to JSP, which acts as. GCF Calculator. Cis(x) is another name for the complex exponential, Cis(x)=e^(ix)=cosx+isinx. Plot complex numbers in the complex plane. π e e MC MR MS M+ M-. Here, we explore the effects of stabilizing selection on cis -regulatory genetic variation in humans. Put another way, cis t is a complex value with magnitude 1 and phase t (modulo 2 p ). i2 =−1, and is called the imaginary unit. Founded in 2005, Math Help Forum is dedicated to free math help and math discussions, and our math community welcomes students, teachers, educators, professors, mathematicians, engineers, and scientists. The best math tool for school and college! If you are a student, it will helps you to learn complex number. Advanced Calculator is a tool that everyone should own. EXAMPLES: It seems silly not to keep the same convention for all quadrants but “officially” the principal value of the argument is – 180 < θ ≤ 180 or in rad. , (-1)^(1/2) Meanwhile, the square of a number is the number times itself. 3(cos 210+i sin 210) = 3(-sqrt(3)/2)+3i(-1/2) -(3/2)sqrt(3)-(3/2)i My favourite way of seeing that sin 30 = 1/2 and cos 30 = sqrt(3)/2 is to picture an equilateral triangle with sides of length 1. What are complex numbers?Complex numbers are ordered pairs of real numbers (a, b), where a is called the real part and b is. (A) 16S (B) 32S (C) 64S (D) 128S (E) NOTA 12. You can manipulate complex numbers arithmetically just like real numbers to carry out operations. A number consisting of two parts, called real and imaginary. Complex numbers have a real and imaginary parts. Write the following complex numbers in trigonometric form: (a) 4 + 4i To write the number in trigonometric. Fundamental Theorem of Algebra. 9 - 9i = cis ) Preview Preview. We can then de ne the limit of a complex function f(z) as follows: we write lim z!c f(z) = L; where cand Lare understood to be complex numbers, if the distance from f(z) to L, jf(z) Lj, is small whenever jz cjis small. One of the most important uses is the use of complex numbers in phasor analysis. Important Concepts and Formulas of Complex Numbers, Rectangular(Cartesian) Form, Cube Roots of Unity, Polar and Exponential Forms, Convert from Rectangular Form to Polar Form and Exponential Form, Convert from Polar Form to Rectangular(Cartesian) Form, Convert from Exponential Form to Rectangular(Cartesian) Form, Arithmetical Operations(Addition,Subtraction, Multiplication, Division), Powers. Complex numbers have both a real number element and an imaginary number element, usually in the form a + bi. Laputan Logic (a smart person who always has intelligent and interesting things to say) has recently posted about words and numbers. Try the quiz at the bottom of the page! go to quiz. It looks nice, but simple calculators don't do this, and you don't have to, either. The complex numbers z= a+biand z= a biare called complex conjugate of each other. GK Apps Education. If you have a different calculator or software package you would like to see included, let me know. Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. How to Find Locus of Complex Numbers - Examples. Complex-Numbers-Calculator Application to calculate operations of complex numbers written in the C language It uses as input a text file. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. The video shows how to change from real to rectangular mode and how to calculate with the number i. The CAS on the prime works only in algebraic. Warning : doing this conversion on the calculator requires radian mode argument and the radicals, of course, give decimal numbers. FOR SALE - McAllen - Edinburg, TX - Selling a Casio fx-115MS PLUS-SR Advanced Scientific Calculator. 2 Trigonometric Form of a Complex Number The trigonometric form of a complex number z= a+ biis z= r(cos + isin ); where r= ja+ bijis the modulus of z, and tan = b a. This will have internal angles all 60 degrees (so the three angles add up to 180 degrees. The green segment indicates the given number w , and the red segments indicate the five 5th roots z 0 , z 1 , z 2 , z 3 , and z 4 , as given by the following formula :. (Enter the angle in degrees rounded to two decimal places. Complex numbers are numbers that are expressed as a+bi where i is an imaginary number and a and b are real numbers. Alpha Complex Numbers 2013 ΜΑΘ National Convention Page 3 of 7 11. Look at most relevant Complex Numbers Calculator For Iphone apps. π e e MC MR MS M+ M-. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. I'm not sure what e^iθ means, but the cis(θ) notation has been all over my current and previous textbooks. diabetes menstrual cycle 🔥+ diabetes menstrual cycle 28 Jun 2020 {15. 1 The Complex Plane A complex number zis given by a pair of real numbers xand yand is written in the form z= x+iy, where isatis es i2 = 1. Product Theorem r1 cis 1 r2 cis 2 r1r2 cis 1 2 Quotient Theorem r1 cis 1 r2 cis 2 r1 r2 cis 1 2. Once you are prompted to the screen you have four options to choose from. It is easier to see in polar form, the number $r\\cdot\\text{cis}(\\theta)$. Question 884544: Use your graphing calculator to convert the complex number to trigonometric form. Find the cube root when z = 8cis(n). The magnitude of the angle itself can be increased or decreased by complete rotations about the circle/pole to arrive at the. So sin, cos, exp, ln, sqrt, integ, etc. Formulas: Equality of complex numbers 1. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Also for: Cfx-9850gb plus, Cfx-9950gb plus, Cfx-9850gc plus, Fx-9750g plus, Cfx-9850g. 15 Quadratic Formula 1. The standard symbol for the set of all complex numbers is C, and we'll also refer to the complex plane as C. Draw a line segment from $$0$$ to $$z$$. Pull terms out from under the radical, assuming positive real numbers. Use MathJax to format equations. 17 Products and Quotients of Complex Numbers (conjugates) 1. The complex number system consists of all complex numbers, a + bi (where a and b are real numbers), together with the rules that define the four basic operations on this set of numbers (addition, subtraction, multiplication, and division). We tend to write it in the form, a + bi, where i is the square root of negative one, i. I have used a calculator to caculate 2^i. So you need to be comfortable converting from one to the. Mathematics. Polar Complex Numbers Calculator. Consider the four numbers 0, € cis(0o), € 2cis(45o), € cis(90o). Determine if 2i is a complex number. The complex numbers may be represented as points in the plane, with the real number 1 represented by the point (1;0), and the complex number irepresented by the point (0;1). Using the complex numbers calculator the answers to algebra problems covering this topic is only as far as your mobile phone. The \"real part\" of is the , and the \"imaginary part\" is. Consider the new complex calculation requirements. (A) 16S (B) 32S (C) 64S (D) 128S (E) NOTA 12. 6 is a real number, it is the real part (a) of the complex number a + bi. More than just an online factoring calculator. But it still cannot display the entire complex number in one screen at a time. The complex numbers z 1 = 2 – 2i and z 2 = 1 – are represented by the points A and B respectively on an Argand diagram. The CAS on the prime works only in algebraic. 11 Cis 247 Degrees. Enter the angle measure as the smallest positive angle in simplified form and any radicals in simplified form. A cis B cis C cis D -cis B 1 8 If (4 10 Express the complex number represented by 11SM Complex Numbers Test A. This page will show you how to multiply them together correctly. Solution: The R↔P menu (~y) contains four functions that are very useful when operating with complex. Multiply the complex numbers as you would multiply polynomials. Note that the set R of all real numbers is a subset of the complex number C since any real number may be considered as having the imaginary part equal to zero. Complex numbers have the form a+bi where a is real part and b is imaginary part and i=√-1. HP-15C Owner’s Handbook HP Part Number: 00015-90001 Edition 2. Description : Writing z = a + ib where a and b are real is called algebraic form of a complex number z : a is the real part of z; b is the imaginary part of z. eCalc is a free and easy-to-use scientific calculator that supports many advanced features, including unit conversion, equation solving, and even complex-number math. When squared becomes:. Thanks!!! Let's explore evaluating the following complex number expression on a variety of calculators: ((240 75o)+(160 −30o)) (60−j80). Auto Calculate. Rewrite the complex number −4+2𝑖 in polar (cis) form and re-graph it on the polar graph paper. Those are truly operations on Complex numbers, so I wouldn't encapsulate them outside the Complex class. So the root of negative number √-n can be solved as √-1 * n = √ n i, where n is a positive real number. Another way of writing the polar form of the number is using it’s exponential form: m e^ (i a). A first course in complex analysis with applications Dennis Zill Written for junior-level undergraduate students that are majoring in math, physics, computer science, and electrical engineering. To determine Ch. Summary : The imaginary_part function calculates online the imaginary part of a complex number. Plot complex numbers in the complex plane. There are three cube roots of –1: cis (pi/3), cis (pi), and cis (-pi/3). The a + bi options tells the calculator to display […]. Visualizing complex number multiplication. It lacked the HP-41C's expandability but it offered, a two line dot matrix screen with customizable menus, greater speed, smaller size, and a lower price as compensation. i2 =−1, and is called the. Advanced mathematics. By using this website, you agree to our Cookie Policy. Perform operations like addition, subtraction and multiplication on complex numbers, write the complex numbers in standard form, identify the real and imaginary parts, find the conjugate, graph complex numbers, rationalize the denominator, find the absolute value, modulus, and argument in this collection of printable complex number worksheets. Let R(cis t)be x [R(cis t)]^5=32cis90. This is a free mathematical calculator for complex numbers, which is able to to add, subtract, multiply and divide complex numbers. Find a polar representation for the complex number z and then identify the real part Re(z), the imaginary part Im(-). Tutorial on how to calculate complex numbers on the TI-nSpire CX CAS. Forum » Programming & Design / TI-Nspire Programming » cosh() with a complex number Started by: kdogbr Date: 12 May. myCableEngineering. This numerical calculator requires that you type your formulas. eCalc is a free and easy-to-use scientific calculator that supports many advanced features, including unit conversion, equation solving, and even complex-number math. Using the Complex Numbers Calculator you can do basic operations with complex numbers such as add, subtract, multiply, divide plus extract the square root and calculate the absolute value (modulus) of a complex number. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x 2 = −1. Check the best results!. Once you are prompted to the screen you have four options to choose from. The Complex Number Formatting Problem The issue is that if the real and imaginary portions of a complex… Read more about Complex Number Formatting in Excel. Is the subcontractor registered with HMRC. The calculator will generate a detailed explanation for each operation. A complex number is a single mathematical quantity able to express these two dimensions of amplitude and phase shift at once. Write the values of these five new numbers. 6 Cis 120 Degrees2. Those are truly operations on Complex numbers, so I wouldn't encapsulate them outside the Complex class. 46i) Solution: (Mode. It connects trigonometric functions with exponential functions in the complex plane via Euler's formula. Jun 16 Beautiful Trigonometry Brady Haran. But, if our numbers are complex that makes finding its power a little more challenging. Complex numbers are of the form a + b i , where a is the real part and b is the imaginary part. where cis(y) is cos(y) + i sin(y) Notes. 2 - Use a calculator to help write each complex number. To convert the following complex number from rectangular form to trigonometric polar form, find the radius using the absolute value of the number. Then taking z = a + bi, you get cosh(a+bi) = cos(-b+ai) where the latter can be evaluated on many calculators. Recall that a complex number is a number of the form $$z = a + bi$$ where $$a$$ and $$b$$ are real numbers and $$i$$ is the imaginary unit defined by $$i = \\sqrt{-1}$$. All the basic matrix operations as well as methods for solving systems of simultaneous linear equations are implemented on this site. \"Polar form\" means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form. The cis notation was first coined by William Rowan Hamilton in Elements of Quaternions (1866) and subsequently used by Irving Stringham in works such as Uniplanar Algebra (1893), or by James Harkness and Frank Morley in their Introduction to the Theory of Analytic Functions (1898). Labour Charge (£): The cost of all labour charged (net of travel and subsistence and any VAT charged by the sub contractor). If you have a different calculator or software package you would like to see included, let me know. $\\endgroup. The Complex Numbers Calculator is used to perform addition, subtraction, multiplication and division on two complex numbers. Mathematics. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Your calculator manual is the best reference. to read input from a console window and display the results. MATHEMATICS SPECIALIST 2 CALCULATOR-FREE. How to Divide Complex Numbers. In spite of this it turns out to be very useful to assume that there is a number ifor which one has. Instructions:: All Functions. A smart person with a stupid idea. In this expression, a is the real part and b is the imaginary part of the complex number. (1) Multiplying both the numerator and the denominator by the same number does not change the value of the fraction. Since the argument is undefined and is positive, the angle of the point on the complex plane is. Get the free \"Convert Complex Numbers to Polar Form\" widget for your website, blog, Wordpress, Blogger, or iGoogle. Another interesting example is the natural logarithm of negative one. Careers That Use Complex Numbers. Four Function and. (Click here for an explanation)Category: Algebra: Brief Description: TI-84 Plus and TI-83 Plus graphing calculator, finds the trigonometric form, standard form, and DeMoivre’s theorem of complex numbers. Write complex numbers in polar form. The complex numbers calculator can also determine the real part of a complex expression. ; Algebraically, as any real quantity φ such that. The modulus r is the distance from z to the origin, while the phase phi is the counterclockwise angle, measured in radians, from the positive x-axis to the line segment that joins the origin to z. Mathematics is a very wide and complex subject. This calculator operates using post-fix or RPN notation. Z = cis ) Preview cis 1) Preview cis ) Preview. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity. Write the following complex numbers in trigonometric form: (a) 4 + 4i To write the number in trigonometric. A complex number consists of a real and an imaginary part. Setting the calculator to COMPLEX mode ,4-1 Entering complex numbers ,4-2 Polar representation of a complex number ,4-3 Simple operations with complex numbers ,4-4 Changing sign of a complex number ,4-5 Entering the unit imaginary number ,4-5 The CMPLX menus ,4-5 CMPLX menu through the MTH menu ,4-6 CMPLX menu in keyboard ,4-7 Functions applied. The complex numbers are in the form of a real number plus. (a) 2 cis(π/6)(b) 5 cis(9π/4)(c) 3 cis(π)(d) cis(7π/4)/2. The magnitude, or modulus, of a complex number in the form z = a + bi is the positive square root of the sum of the squares of a and b. By using this website, you agree to our Cookie Policy. Find a polar representation for the complex number z and then identify the real part Re(z), the imaginary part Im(-). Complex Numbers: Convert From Polar to Complex Form, Ex 1 Complex Numbers: Multiplying and Dividing Expressing a Complex Number in Trigonometric or Polar Form, Ex 2. Complex number concept was taken by a variety of engineering fields. Z = cis ) Preview cis 1) Preview cis ) Preview. Absolute value & angle of complex numbers. Consider the polynomial equation x3 + i = 0. Today that complex numbers are widely used in advanced engineering domains such as physics, electronics, mechanics, astronomy, etc. Wow! Isn't that a great result - to multiply two complex numbers, you multiply their magnitudes and add their angles! That's easy! But the real power (that was a double pun, by the way) of this way of writing complex numbers is in taking roots of numbers. Write a web application that functions as a simple hand calculator, but also keeps a \"paper trail\" of all your previous work. If vectors with common angles are added, their magnitudes (lengths. So each complex number needs two ordinary (\"real\") numbers to describe it. A cis B cis C cis D -cis B 1 8 If (4 10 Express the complex number represented by 11SM Complex Numbers Test A. The complex number a + b i a+bi a + b i is graphed on this plane just as the ordered pair (a, b) (a,b) (a, b) would be graphed on the Cartesian coordinate plane. We associate each complex number z=a+biz=a+bi with the point (a,b)(a,b) on the coordinate plane. Description : Writing z = a + ib where a and b are real is called algebraic form of a complex number z : a is the real part of z; ; b is the imaginary part of z. Multiplying Complex Number Calculator. Wow! Isn't that a great result - to multiply two complex numbers, you multiply their magnitudes and add their angles! That's easy! But the real power (that was a double pun, by the way) of this way of writing complex numbers is in taking roots of numbers. 93 and b = 2. The best math tool for school and college! If you are a student,. Convert a Complex Number to Polar and Exponential Forms - Calculator. Winplot is in windows Mathematics category, and build by Rick Parris in Free license. Auto Calculate. Complex numbers can be represented as a binomial (a mathematical expression consisting of one term added to or subtracted from another) of the form a + bi. Contains Ads. The magnitude of the angle itself can be increased or decreased by complete rotations about the circle/pole to arrive at the. you can do all math operations on complex numbers and the result will be a complex number. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to add, subtract, multiply or divide two complex numbers. Vector calculator This page allows you to carry computations over vectors. Complex_conjugate online. To add complex numbers in rectangular form, add the real components and add the imaginary components. A complex-valued function made from sine and cosine with definition cis θ = cos θ + isin θ. Personally, I'd put those arithmetic operations in the Complex class. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. Making statements based on opinion; back them up with references or personal experience. The function polar takes a complex number and returns a (magnitude, phase) pair in canonical form: The magnitude is nonnegative, and the phase, in the range (- p , p ]; if the magnitude is zero, then. Gene expression variation is a major contributor to phenotypic variation in human complex traits. The interface of Complex Calculator Display: Outputs the results of calculations and current number inputs. Answer: The X coordinate is approximately 4. ) NOTE: If you set the calculator to return polar form, you can press Enter and the calculator will convert this number to polar form. To use our permutation calculator, you only need to take off following easy steps. a) Find b and c. Similar Algebra Calculator Adding Complex Number Calculator. Any time you have to have guidance with math and in particular with positive numbers calculator or rational expressions come visit us at Algebra1help. That is the map z7→ z+z 0 represents a translation aunits to the right and bunits up in the complex plane. View Complex number introduction (1) from EEE 360 at Arizona State University. 2 Use the relation i 2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. By using this website, you agree to our Cookie Policy. It gave me this irational complex number of approximately. For example:. \"Complex\" numbers have two parts, a \"real\" part (being any \"real\" number that you're used to dealing with) and an \"imaginary\" part (being any number with an \"i\" in it). The general form for subtracting complex numbers is: (a+bi) - (c+di) (a-c) + (bi-di) Below is a worked example. Looks like a real calculator. Recall that a complex number is a number of the form $$z = a + bi$$ where $$a$$ and $$b$$ are real numbers and $$i$$ is the imaginary unit defined by $$i = \\sqrt{-1}$$. 9 - 9i = cis ) Preview Preview. cis is just shorthand for cos θ + i sin θ. Complex Numbers Calculator. Ensure record keeping procedures are in place in relation to the additional information required to be submitted. 000i >> (-3)^0. The video shows how to change from real to rectangular mode and how to calculate with the number i. Leave your answers in trig form. Some formulas regarding complex number 5. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. Complex Numbers. a u × v b u ÷ v c v 2 cis ÷ u d u2 e v3 9 Express each of the following in the form indicated:. ) Originally Posted by Bjarne Stroustrup (2000-10-14) I get maybe two dozen requests for help with some sort of programming or design problem every day. The Complex Number Calculator (CNC) is completed. If you're seeing this message, it means we're having trouble loading external resources on our website. I'm doing some thevenins on an AC RLC, and have gotten stuck applying parallel resistance to complex numbers. Thus, z 1 and z 2 are close when jz 1 z 2jis small. how to find argument or angle of a complex number in matlab? Follow 720 views (last 30 days) bsd on 30 Jun 2011. To add complex numbers in rectangular form, add the real components and add the imaginary components. We just combine like terms. Advanced Calculator is a tool that everyone should own. Complex numbers contain both real numbers and imaginary numbers and are written in the form a+bi. Important Concepts and Formulas of Complex Numbers, Rectangular(Cartesian) Form, Cube Roots of Unity, Polar and Exponential Forms, Convert from Rectangular Form to Polar Form and Exponential Form, Convert from Polar Form to Rectangular(Cartesian) Form, Convert from Exponential Form to Rectangular(Cartesian) Form, Arithmetical Operations(Addition,Subtraction, Multiplication, Division), Powers. The calculator also provides conversion of a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). 2 Use the relation i 2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. Viking’s favorite weapons are axes! You can throw them at enemies and split their…. Figures 4 and 5 show the display in algebraic mode. McKeague Chapter 8. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. CPT 172 Midterm Part 1 to 3.$\\endgroup. 2 - Use a calculator to help write each complex number. However matrices can be not only two-dimensional, but also one-dimensional (vectors), so that you can multiply vectors, vector by matrix and vice versa. The “argument” of a complex number is just the angle it makes with the positive real axis. Z = cis ) Preview cis 1) Preview cis ) Preview. Matrix Calculator Pro supports for complex matrix. In this article, we will try to add and subtract these two Complex Numbers by creating a Class for Complex Number, in which: The complex numbers will be initialized with the help of constructor. Rewrite each of the following complex numbers in rectangular form and graph. This online calculator will help you to find magnitude of complex number. In your case both allow you to build an array and solve and both work with complex numbers. Another way of writing the polar form of the number is using it's exponential form: me^(ia. Suppose that the four numbers from part c are replaced by the numbers 0, 5cis π. Free math tutorial and lessons. 2i The complex numbers are an extension of the real numbers. When b=0, z is real, when a=0, we say that z is pure imaginary. Additive Identity. corresponds to a point in the complex plane and every point in the complex plane corresponds to a complex number. In mathematics, the complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign. Since the argument is undefined and is positive, the angle of the point on the complex plane is. Subtraction 9. Viking is a ferocious Scandinavian warrior. One has the identity cosh(z) = cos(iz) which can easily be checked by writing both sides of the equation in terms of exponentials. Once you are prompted to the screen you have four options to choose from. All Functions Operators +. So, plot -4-4i onto the diagram and you should get a point on the third quadrant. For the following exercise, find the root of the complex number in polar form. Using the real number system, we cannot take the square root of a negative number, so I must not be a real number and is therefore known as the. In the special case where b=0, a+0i=a. 2 Trigonometric Form of a Complex Number The trigonometric form of a complex number z= a+ biis z= r(cos + isin ); where r= ja+ bijis the modulus of z, and tan = b a. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Finding nth roots of Complex Numbers. Rectangular and Polar Form. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. The horizontal axis is called real axis while the vertical axis is the imaginary axis. Learn how complex number multiplication behaves when you look at its graphical effect on the complex plane. The maximum number of decimal places can be chosen between 0 and 10. Explore many other math calculators, as well as hundreds of other calculators addressing health, fitness, finance, math, and more. written r cis θ. Conjugate of complex number 7. -4+2 just becomes -2. For earlier calculators, memory was very expensive and bulky and the challenge for firmware writers was to pack a set of scientific functions into very little space. Complex numbers have a real and imaginary parts. It has a real part and an imaginary part. Home / Resources / Tax Calculators / CIS Tax Deduction Calculator. Our real number line has now been extended into the two-dimensional complex plane. More than just an online factoring calculator. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Entering complex numbers in polar form:. Convert a Complex Number to Polar and Exponential Forms - Calculator. Example: If you enter the complex number 3 + 4ί into the Input Bar, you get the point (3, 4) in the Graphics View. They include numbers of the form a + bi where a and b are real numbers. Some complex numbers have absolute value 1. This exercise should be worked without the aid of a calculator The real part Re()3 Preview the imaginary part Im()Preview 14-5 ]. We can then de ne the limit of a complex function f(z) as follows: we write lim z!c f(z) = L; where cand Lare understood to be complex numbers, if the distance from f(z) to L, jf(z) Lj, is small whenever jz cjis small. The complex numbers must be put in brackets. Find a polar representation for the complex number z and then identify the real part Re(z), the imaginary part Im(-). Learn How to Modulus of complex number - Definition, Formula and Example Definition: Modulus of a complex number is the distance of the complex number from the origin in a complex plane and is equal to the square root of the sum of the squares of the real and imaginary parts of the number. Online calculator. Getty Images. (1 + i) (x − yi) = i (14 + 7i) − (2 + 13i) 3x + (3x − y) i = 4 − 6i x − 2i2 + 6i = yi + 3xi3. If you guys are not familiar with the $\\operatorname{cis}$ notation $$\\operatorname{cis}(x) = (\\cos(x) + i\\sin(x))$$ I've tried doing this: By comparing arguments $$x^2 - x = 0\\Longrightarrow x(x-1) = 0\\Longrightarrow x = 0 ,~ x = 1$$ But clearly from the graph there are more intersections and searching the internet I learnt you can do this. This calculator helps calculate any necessary CIS Tax Deductions. However, any other value of k, produces a complex number which we have already listed. On the TI-83 Plus graphing calculator, you can set the mode and use complex numbers. Imaginary numbers Find two imaginary numbers whose sum is a real number. Difference of Squares: a 2 - b 2 = (a + b) (a - b) Step 2: Click the blue arrow to submit and see the result!. Find the cube root when z = 8cis(n). Photos and Property Details for 1210 THORPE RD #314, BURLINGTON, ON L7S 2G9. Consider the four numbers 0, € cis(0o), € 2cis(45o), € cis(90o). Soon after, we added 0 to represent the idea of nothingness. To add complex numbers in rectangular form, add the real components and add the imaginary components.\nhdqfi5yq88xyjx 1rluqn9ezk ynrdro293rqor m2prd6z4dkwh6gl c8ettr5plwwsj8 buyfgybnc8 9uenznf0fd 0jbs2znotix3io v446zcb5hsex ag6x2mwco1j8q8x g3z9utfbx6lrv49 e8e8mvw1ytziz nctyb6ljn4j3 isnfd9xpvxs1t pjm26qoy32ty86 r95pmnefb7pz yu332gurj7 8sc0o73pz0xxw2u fr2vvx62thifix5 7e7dtu6ib6 y80banv9e0 c6xdbcaw95isl52 35blnzklib4 mprxpkm75u ibepd06vu1w4"
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https://goprep.co/ex-10.3-q8c-the-mid-points-of-the-sides-of-a-triangle-are-3-i-1nl697 | [
"Q. 8 C4.0( 6 Votes )\n\n# The mid-points of the sides of a triangle are (3,4),(4,6) and (5,7). Find the coordinates of the vertices of the triangle.",
null,
"Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(3, 4), Q(4, 6) and R(5, 7) are the midpoints of AB, BC, and CA. Then,",
null,
"…(i)",
null,
"…(ii)",
null,
"…(iii)",
null,
"…(iv)",
null,
"…(v)",
null,
"…(vi)\n\nAdding (i), (iii) and (v), we get\n\nx1 + x2 + x2 + x3 + x1 + x3 = 6 + 8 + 10\n\n2(x1 + x2 + x3) = 24\n\nx1 + x2 + x3 =12 …(vii)\n\nFrom (i) and (vii), we get\n\nx3 = 12 – 6 = 6\n\nFrom (iii) and (vii), we get\n\nx1 = 12 – 8 = 4\n\nFrom (v) and (vii), we get\n\nx2 = 12 – 10 = 2\n\nNow adding (ii), (iv) and (vi), we get\n\ny1 + y2 + y2 + y3 + y1 + y3 = 8 + 12 + 14\n\n2(y1 + y2 + y3) = 34\n\ny1 + y2 + y3 = 17 …(viii)\n\nFrom (ii) and (viii), we get\n\ny3 = 17 – 8 = 9\n\nFrom (iv) and (vii), we get\n\ny1 = 17 – 12 = 5\n\nFrom (vi) and (vii), we get\n\ny2 = 17 – 14 = 3\n\nHence, the vertices of ΔABC are A(4, 5), B(2, 3) and C(6, 9)\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nRelated Videos",
null,
"",
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"",
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"",
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"",
null,
"Set of Questions on Section Formula54 mins\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
""
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null,
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null
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https://chemistry.stackexchange.com/questions/48299/determine-rate-equations-for-entire-kinetic-scheme | [
"# Determine rate equations for entire kinetic scheme\n\nI am working with the following kinetic scheme that represents pyrolysis of a biomass particle:",
null,
"The paper that discusses this scheme provides the following rate constants:\n\n• K1 for the pathway of $\\text{Biomass} \\rightarrow (\\text{Volatiles + Gases})_1$\n• K2 for the pathway of $\\text{Biomass} \\rightarrow (\\text{Char})_1$\n• K3 for the pathway of $(\\text{Volatiles + Gases})_1 \\rightarrow (\\text{Volatiles + Gases})_2 + (\\text{Char})_2$\n\nThe rate equation provided in the paper for the conversion of biomass is $$\\frac{\\mathrm d\\,B}{\\mathrm dt} = -(K_1 + K_2)\\,B$$\n\nThe rate equation for the char 1 component is given as $$\\frac{\\mathrm d\\,C_1}{\\mathrm dt} = K_2\\,B - K_3\\,C_1$$\n\nAnd finally the rate equation provided for the char 2 component is $$\\frac{\\mathrm d\\,C_2}{\\mathrm dt} = \\delta\\,K_3\\,C_1$$\n\nwhere $\\delta$ is a deposition coefficient reported as 1.45 in the article. Note that all components of the system are on a mass basis such as $\\mathrm{kg/m^3}$.\n\nIt is fairly straightforward to calculate the concentrations of $B$, $C_1$, and $C_2$ at each time step given an initial concentration of $B$. However, I would like to determine the concentrations of the volatiles and gases in the system too. Since it doesn't seem possible to calculate the individual volatile and gas components from the given scheme, I would like to determine the concentration of the (Volatile + Gases) group which could be labeled as G. So how would I develop the overall rate equations for B, C, and G?\n\n$$\\frac{\\mathrm d\\,B}{\\mathrm dt} = -(K_1 + K_2)\\,B \\\\ \\frac{\\mathrm d\\,C}{\\mathrm dt} = K_2\\,B + ? \\\\ \\frac{\\mathrm d\\,G}{\\mathrm dt} = K_1\\,B + ?$$\n\n• (V+G)2 is the same as C2 and (V+G)1 is \"straightforward\" as you said. Mar 21, 2016 at 20:18\n• @pH13 So the concentration of group (V+G)2 is the same as the concentration of C2; basically (V+G)2 = C2? Mar 21, 2016 at 20:21\n• I'd say so, yes. As far as those subscripts are only indices. Mar 21, 2016 at 20:22\n• @pH13 That makes sense to me but what I'm actually trying to do is figure out the individual components. From (V+G)1 or (V+G)2 what is the concentration of V and G in each group? Mar 21, 2016 at 20:42\n• As you don't know the amount of \"Gases\" I don't think that you can split it up further. Mar 21, 2016 at 20:46\n\nThe Biomass-equation is: $$\\frac{\\mathrm d B}{\\mathrm d t} = -(k_1 + k_2) B$$ The (volatile + gases)$_1$ equation is: $$\\frac{\\mathrm d VG_1}{\\mathrm d t} = k_1 B$$ The (volatile + gases)$_2$ equation is equal to equation for (Char)$_2$: $$\\frac{\\mathrm d VG_2}{\\mathrm d t} = \\frac{\\mathrm d C_2}{\\mathrm d t} = k_3 C_1$$ (It might be that it needs to be multiplied by 1/2.)\nAnd the equation for (char)$_1$ is this: $$\\frac{\\mathrm d C_1}{\\mathrm d t} = k_2 B-k_3 C_1$$\nThe equations for the complete set of (char)s and (volatile+gases)s are then the sums of the solutions for $C_1$ and $C_2$ resp. $VG_1$ and $VG_2$.\nFor initial concentration of $B(0)=1$ and everything else $f(0)=0$, as well as $k_1=2$, $k_2=10$ and $k_3=3$, you should get something similar to this:",
null,
""
] | [
null,
"https://i.stack.imgur.com/0fU4i.png",
null,
"https://i.stack.imgur.com/fuwom.png",
null
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https://hoopercharles.wordpress.com/2012/01/09/ | [
"## Dumping Trace Events – What You See is Not Necessarily What You Get\n\n9 01 2012\n\nJanuary 9, 2012\n\nA recent thread in the comp.databases.oracle.server Usenet group brought back memories of when Randolf Geist and I worked on the two chapters for the “Expert Oracle Practices” book. In the chapters, among other things, we demonstrated how to enable 10046 trace files for sessions using various approaches (some of the approaches change the SQL_TRACE, SQL_TRACE_WAITS, and SQL_TRACE_BINDS columns in V\\$SESSION to indicate that a 10046 trace is active, and other methods do not), and how to use ORADEBUG to see which events are enabled in other sessions.\n\nThe OP in the thread mentioned that the ORADEBUG EVENTDUMP session command, when executed in SQL*Plus by a SYSDBA user, showed no events enabled for another session when the DBMS_MONITOR package was used to enable a 10046 trace at level 12 for the other session. I have a vague memory of reading (somewhere) that the ORADEBUG EVENTDUMP session command will only show newly enabled events for another session after that other session issues another call to the database (parse, execute, fetch, etc.). In the Usenet group thread I stated something that is slightly less complete: “I believe that the session with the trace enabled through DBMS_MONITOR must execute at least one SQL statement after tracing is enabled for the session, before ORADEBUG will report that the trace is enabled.”\n\nIn the Usenet thread I put together a quick test case script that demonstrated that after the second session (Session 1) issued a SELECT statement, the first session (Session 2) was able to issue a ORADEBUG EVENTDUMP session command and determine that a 10046 trace at level 12 was enabled for the other session. I believe that the OP answered his own question after a bit of experimentation and searching. I thought that I would provide an extended version of the test case script in case anyone would like to experiment a bit. In this test case, Session 1 will be the target of the tracing – that session must be able to query V\\$SESSION and V\\$PROCESS during the script execution, while Session 2 will be connected as the SYS user. In the test case script I have removed the SQL> prompts to make it easier for you to copy and paste the scripts into SQL*Plus, and when the returned output is important, that information also appears in the test case script (all commands executed begin with UPPERCASE keywords).\n\nIn Session 1, execute the following SQL statement to pick up the SID, SERIAL# and PID for Session 1, along with the columns in V\\$SESSION that indicate whether or not a 10046 trace at level 1, 4, 8, or 12 is enabled. The SID and SERIAL# will be used in Session 2 to enable a trace in Session 1, and the PID will be used with ORADEBUG in Session 2:\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 DISABLED FALSE FALSE ```\n\nAs can be determined by the above output, the SID is 4, PID is 24, and a 10046 trace is not enabled for the session (fair warning, the SQL_TRACE column can be misleading).\n\nIn Session 2, enable a 10046 trace using the DBMS_MONITOR package for Session 1 at level 12, and dump the events for Session 1 (note that this test case script is being executed on a Windows server with Oracle Database 11.2.0.2):\n\n```EXEC DBMS_MONITOR.SESSION_TRACE_ENABLE(SESSION_ID=>4,SERIAL_NUM=>143,WAITS=>TRUE,BINDS=>TRUE)\n\nStatement processed. ```\n\nThe output of the ORADEBUG EVENTDUMP session command is significant – the Statement processed. note indicates that the statement executed successfully with no return output.\n\nIn Session 1, let’s execute the same SQL statement as was executed earlier, to not only confirm that the DBMS_MONITOR.SESSION_TRACE_ENABLE call executed by Session 2 was successful, but also to issue another database call in the session (the significance will be seen later):\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 ENABLED TRUE TRUE```\n\nThe above output shows that the DBMS_MONITOR.SESSION_TRACE_ENABLE call was successful, as indicated by the changed values in the SQL_TRACE, SQL_TRACE_WAITS, and SQL_TRACE_BINDS columns – a level 12 10046 trace is enabled.\n\nIn Session 2, let’s check again which events are enabled for Session 1:\n\n```ORADEBUG EVENTDUMP session\nsql_trace level=12 ```\n\nNotice in the above output that ORADEBUG now indicates that a 10046 trace at level 12 is enabled for Session 1, because another database call (initiated when the SQL statement was executed) was performed in Session 1 after the trace was enabled.\n\nLet’s try a bit more experimentation. In Session 2, let’s disable the trace in Session 1 using the DBMS_MONITOR package, and verify whether or not tracing is enabled in Session 1 by using ORADEBUG:\n\n```EXEC DBMS_MONITOR.SESSION_TRACE_DISABLE(SESSION_ID=>4,SERIAL_NUM=>143)\nPL/SQL procedure successfully completed.\n\nsql_trace level=12 ```\n\nThe above output shows that ORADEBUG still indicates that a 10046 trace at level 12 is enabled for Session 1. In Session 2, let’s check the SQL_TRACE, SQL_TRACE_WAITS, and SQL_TRACE_BINDS columns for Session 1:\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=4\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 DISABLED FALSE FALSE\n\nsql_trace level=12 ```\n\nSo, the query of V\\$SESSION indicates that a 10046 trace for Session 1 is disabled, yet ORADEBUG still insists that a 10046 trace at level 12 is enabled. WYSIWYG (What You See Is What You Get) is taking a bit of a vacation. Let’s issue another database call in Session 1 so that ORADEBUG provides correct output about whether or not a 10046 trace is enabled. In Session 1:\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 DISABLED FALSE FALSE ```\n\nLet’s confirm in Session 2 that ORADEBUG now reports that a 10046 trace is not enabled in Session 1:\n\n```ORADEBUG EVENTDUMP session\nStatement processed. ```\n\nOK, but what happens if we use ORADEBUG in Session 2 to enable a 10046 trace is Session 1? In Session 2:\n\n```ORADEBUG session_event 10046 trace name context forever,level 12\nStatement processed.\n\nsql_trace level=12 ```\n\nThe above shows that if ORADEBUG enables a 10046 trace in Session 1, the ORADEBUG EVENTDUMP session command will correctly indicate that a 10046 trace is enabled for Session 1. Nice, but what about the SQL_TRACE, SQL_TRACE_WAITS, and SQL_TRACE_BINDS columns in V\\$SESSION for Session 1? Let’s check using Session 2:\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=4\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 DISABLED FALSE FALSE ```\n\nThe above output suggests that a 10046 trace at level 12 is not enabled for Session 1. Let’s check again from within Session 1 and then determine today’s date (if that second SQL statement appears in the 10046 trace file, then we know that the 10046 trace was in fact enabled for Session 1):\n\n```SELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 DISABLED FALSE FALSE\n\nSELECT SYSDATE FROM DUAL;\n\nSYSDATE\n---------\n08-JAN-12 ```\n\nThe above still suggests that a 10046 trace is not enabled for Session 1 and that today’s date is January 8, 2012. Let’s jump back to Session 2 and disable the 10046 trace using ORADEBUG:\n\n```ORADEBUG session_event 10046 trace name context off\nStatement processed. ```\n\nOut of curiosity, let’s repeat a portion of the initial test, just to confirm that the initial test results were not a fluke:\n\n```EXEC DBMS_MONITOR.SESSION_TRACE_ENABLE(SESSION_ID=>4,SERIAL_NUM=>143,WAITS=>TRUE,BINDS=>TRUE)\n\nStatement processed.\n\nSELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=4\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 ENABLED TRUE TRUE\n\nStatement processed. ```\n\nOnce again, the above shows that ORADEBUG does not initially report that a 10046 trace is enabled in Session 1 when the officially supported method for enabling a 10046 trace in another session is used. Let’s disable the 10046 trace using the DBMS_MONITOR package and then confirm that ORADEBUG still indicates that a 10046 trace is not enabled in Session 1:\n\n```EXEC DBMS_MONITOR.SESSION_TRACE_DISABLE(SESSION_ID=>4,SERIAL_NUM=>143)\n\nStatement processed. ```\n\nSo, what about the unsupported method of enabling a 10046 trace in another session, using SYS.DBMS_SYSTEM.SET_EV? In Session 2:\n\n```EXEC SYS.DBMS_SYSTEM.SET_EV(4, 143, 10046, 12, '')\n\nSELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=4\n\nSID SERIAL# PID SQL_TRAC SQL_T SQL_T\n--- ---------- ---------- -------- ----- -----\n4 143 24 ENABLED TRUE TRUE\n\nStatement processed. ```\n\nThe SYS.DBMS_SYSTEM.SET_EV call changed the value of the columns in V\\$SESSION for Session 1, but still did not alter the output of ORADEBUG EVENTDUMP session.\n\nOne final test in Session 2:\n\n```ORADEBUG session_event 10046 trace name context forever,level 12\nStatement processed.\n\nsql_trace level=12 ```\n\nOnce again, if we enable the 10046 trace using ORADEBUG, then ORADEBUG is able to show that the trace is enabled for Session 1.\n\n———————————————\n\nAt this point you might be curious why I mentioned that the above test case script was performed on a Windows server running Oracle Database 11.2.0.2. Does the Oracle Database version make a difference in the results? Does the operating system (Windows, Linux, or Unix) make a difference in the results? It appears that a part 2 is required for this blog article, where I will compare the above results with the results obtained with 10.2.0.5 on a Windows server and 11.1.0.6 on a Linux server.\n\n———————————————\n\nYou are probably curious to see the contents of the trace file for Session 1. Here is the trace file (if you need help in reading the trace file, see this three part article series):\n\n```*** 2012-01-08 09:47:51.686\n*** SESSION ID:(4.143) 2012-01-08 09:47:51.686\n*** CLIENT ID:() 2012-01-08 09:47:51.686\n*** SERVICE NAME:(OR1122) 2012-01-08 09:47:51.686\n*** MODULE NAME:(SQL*Plus) 2012-01-08 09:47:51.686\n*** ACTION NAME:() 2012-01-08 09:47:51.686\n\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:47:51.686\nFinished processing ORADEBUG command (#1) 'EVENTDUMP session'\n\n*** 2012-01-08 09:48:18.426\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:48:18.426\nFinished processing ORADEBUG command (#2) 'EVENTDUMP session'\n\n*** 2012-01-08 09:49:16.415\nCLOSE #3:c=0,e=20,dep=0,type=0,tim=488596106412\n=====================\nPARSING IN CURSOR #2 len=37 dep=1 uid=0 oct=3 lid=0 tim=488596107091 hv=1398610540 ad='7ffb8e4b250' sqlid='grwydz59pu6mc'\nselect text from view\\$ where rowid=:1\nEND OF STMT\nPARSE #2:c=0,e=264,p=0,cr=0,cu=0,mis=1,r=0,dep=1,og=4,plh=0,tim=488596107088\nBINDS #2:\nBind#0\noacdty=11 mxl=16(16) mxlc=00 mal=00 scl=00 pre=00\noacflg=18 fl2=0001 frm=00 csi=00 siz=16 off=0\nkxsbbbfp=1dbe86d8 bln=16 avl=16 flg=05\nvalue=00000273.0010.0001\nEXEC #2:c=0,e=550,p=0,cr=0,cu=0,mis=1,r=0,dep=1,og=4,plh=3684871272,tim=488596107744\nFETCH #2:c=0,e=25,p=0,cr=2,cu=0,mis=0,r=1,dep=1,og=4,plh=3684871272,tim=488596107790\nSTAT #2 id=1 cnt=1 pid=0 pos=1 obj=69 op='TABLE ACCESS BY USER ROWID VIEW\\$ (cr=1 pr=0 pw=0 time=0 us cost=1 size=15 card=1)'\nCLOSE #2:c=0,e=501,dep=1,type=0,tim=488596108309\n=====================\nPARSING IN CURSOR #5 len=37 dep=1 uid=0 oct=3 lid=0 tim=488596108602 hv=1398610540 ad='7ffb8e4b250' sqlid='grwydz59pu6mc'\nselect text from view\\$ where rowid=:1\nEND OF STMT\nPARSE #5:c=0,e=23,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=4,plh=3684871272,tim=488596108602\nBINDS #5:\nBind#0\noacdty=11 mxl=16(16) mxlc=00 mal=00 scl=00 pre=00\noacflg=18 fl2=0001 frm=00 csi=00 siz=16 off=0\nkxsbbbfp=1dbe86d8 bln=16 avl=16 flg=05\nvalue=00000273.0012.0001\nEXEC #5:c=0,e=47,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=4,plh=3684871272,tim=488596108701\nFETCH #5:c=0,e=11,p=0,cr=2,cu=0,mis=0,r=1,dep=1,og=4,plh=3684871272,tim=488596108727\nSTAT #5 id=1 cnt=1 pid=0 pos=1 obj=69 op='TABLE ACCESS BY USER ROWID VIEW\\$ (cr=1 pr=0 pw=0 time=0 us cost=1 size=15 card=1)'\nCLOSE #5:c=0,e=23,dep=1,type=0,tim=488596108766\n=====================\nPARSING IN CURSOR #3 len=37 dep=1 uid=0 oct=3 lid=0 tim=488596109217 hv=1398610540 ad='7ffb8e4b250' sqlid='grwydz59pu6mc'\nselect text from view\\$ where rowid=:1\nEND OF STMT\nPARSE #3:c=0,e=14,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=4,plh=3684871272,tim=488596109217\nBINDS #3:\nBind#0\noacdty=11 mxl=16(16) mxlc=00 mal=00 scl=00 pre=00\noacflg=18 fl2=0001 frm=00 csi=00 siz=16 off=0\nkxsbbbfp=1dbe86d8 bln=16 avl=16 flg=05\nvalue=00000274.0010.0001\nEXEC #3:c=0,e=50,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=4,plh=3684871272,tim=488596109318\nFETCH #3:c=0,e=11,p=0,cr=2,cu=0,mis=0,r=1,dep=1,og=4,plh=3684871272,tim=488596109344\nSTAT #3 id=1 cnt=1 pid=0 pos=1 obj=69 op='TABLE ACCESS BY USER ROWID VIEW\\$ (cr=1 pr=0 pw=0 time=0 us cost=1 size=15 card=1)'\nCLOSE #3:c=0,e=21,dep=1,type=0,tim=488596109385\n=====================\nPARSING IN CURSOR #4 len=204 dep=0 uid=91 oct=3 lid=91 tim=488596111910 hv=1585533233 ad='7ffb7578ec0' sqlid='dk42yjjg82n9j'\nSELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\nEND OF STMT\nPARSE #4:c=0,e=5410,p=0,cr=6,cu=0,mis=1,r=0,dep=0,og=1,plh=1985757361,tim=488596111909\nEXEC #4:c=0,e=23,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1985757361,tim=488596112094\nWAIT #4: nam='SQL*Net message to client' ela= 4 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=488596112181\nFETCH #4:c=0,e=1190,p=0,cr=0,cu=0,mis=0,r=1,dep=0,og=1,plh=1985757361,tim=488596113404\nWAIT #4: nam='SQL*Net message from client' ela= 232 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=488596113665\nFETCH #4:c=0,e=397,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1985757361,tim=488596114102\nSTAT #4 id=1 cnt=1 pid=0 pos=1 obj=0 op='NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us cost=0 size=53 card=1)'\nSTAT #4 id=2 cnt=1 pid=1 pos=1 obj=0 op='NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us cost=0 size=49 card=1)'\nSTAT #4 id=3 cnt=810 pid=2 pos=1 obj=0 op='MERGE JOIN CARTESIAN (cr=0 pr=0 pw=0 time=346 us cost=0 size=16146 card=598)'\nSTAT #4 id=4 cnt=30 pid=3 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUPR (cr=0 pr=0 pw=0 time=0 us cost=0 size=494 card=26)'\nSTAT #4 id=5 cnt=810 pid=3 pos=2 obj=0 op='BUFFER SORT (cr=0 pr=0 pw=0 time=130 us cost=0 size=184 card=23)'\nSTAT #4 id=6 cnt=27 pid=5 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSLWT (cr=0 pr=0 pw=0 time=26 us cost=0 size=184 card=23)'\nSTAT #4 id=7 cnt=1 pid=2 pos=2 obj=0 op='FIXED TABLE FIXED INDEX X\\$KSUSE (ind:1) (cr=0 pr=0 pw=0 time=0 us cost=0 size=22 card=1)'\nSTAT #4 id=8 cnt=1 pid=7 pos=1 obj=0 op='COUNT STOPKEY (cr=0 pr=0 pw=0 time=0 us)'\nSTAT #4 id=9 cnt=1 pid=8 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUMYSTA (cr=0 pr=0 pw=0 time=0 us cost=0 size=40 card=2)'\nSTAT #4 id=10 cnt=1 pid=9 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUSGIF (cr=0 pr=0 pw=0 time=0 us cost=0 size=4 card=1)'\nSTAT #4 id=11 cnt=1 pid=1 pos=2 obj=0 op='FIXED TABLE FIXED INDEX X\\$KSLED (ind:2) (cr=0 pr=0 pw=0 time=0 us cost=0 size=4 card=1)'\nWAIT #4: nam='SQL*Net message to client' ela= 2 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=488596114237\n\n*** 2012-01-08 09:50:22.069\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:50:22.069\nFinished processing ORADEBUG command (#3) 'EVENTDUMP session'\n\n*** 2012-01-08 09:51:24.578\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:51:24.578\nFinished processing ORADEBUG command (#4) 'EVENTDUMP session'\n\n*** 2012-01-08 09:54:05.768\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:54:05.768\nFinished processing ORADEBUG command (#5) 'EVENTDUMP session'\n\n*** 2012-01-08 09:55:01.068\nWAIT #4: nam='SQL*Net message from client' ela= 344636055 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=488940750307\n\n*** 2012-01-08 09:55:48.744\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:55:48.744\nFinished processing ORADEBUG command (#6) 'EVENTDUMP session'\n\n*** 2012-01-08 09:56:50.964\nReceived ORADEBUG command (#7) 'session_event 10046 trace name context forever,level 12' from process 'Windows thread id: 12292, image: <none>'\n\n*** 2012-01-08 09:56:50.964\nFinished processing ORADEBUG command (#7) 'session_event 10046 trace name context forever,level 12'\n\n*** 2012-01-08 09:57:18.634\nDumping Event (group=SESSION)\n\n*** 2012-01-08 09:57:18.634\nFinished processing ORADEBUG command (#8) 'EVENTDUMP session'\n\n*** 2012-01-08 10:00:01.094\nWAIT #2: nam='SQL*Net message from client' ela= 300017366 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489240770346\nCLOSE #2:c=0,e=19,dep=0,type=0,tim=489240770476\n=====================\nPARSING IN CURSOR #5 len=204 dep=0 uid=91 oct=3 lid=91 tim=489240770687 hv=1585533233 ad='7ffb7578ec0' sqlid='dk42yjjg82n9j'\nSELECT\nS.SID,\nS.SERIAL#,\nP.PID,\nS.SQL_TRACE,\nS.SQL_TRACE_WAITS,\nS.SQL_TRACE_BINDS\nFROM\nV\\$SESSION S,\nV\\$PROCESS P\nWHERE\nS.SID=(SELECT SID FROM V\\$MYSTAT WHERE ROWNUM=1)\nEND OF STMT\nPARSE #5:c=0,e=130,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1985757361,tim=489240770686\nEXEC #5:c=0,e=31,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1985757361,tim=489240770814\nWAIT #5: nam='SQL*Net message to client' ela= 3 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489240770862\nFETCH #5:c=0,e=7007,p=0,cr=0,cu=0,mis=0,r=1,dep=0,og=1,plh=1985757361,tim=489240777893\nWAIT #5: nam='SQL*Net message from client' ela= 192 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489240778675\nFETCH #5:c=0,e=336,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1985757361,tim=489240779049\nSTAT #5 id=1 cnt=1 pid=0 pos=1 obj=0 op='NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us cost=0 size=53 card=1)'\nSTAT #5 id=2 cnt=1 pid=1 pos=1 obj=0 op='NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us cost=0 size=49 card=1)'\nSTAT #5 id=3 cnt=700 pid=2 pos=1 obj=0 op='MERGE JOIN CARTESIAN (cr=0 pr=0 pw=0 time=349 us cost=0 size=16146 card=598)'\nSTAT #5 id=4 cnt=28 pid=3 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUPR (cr=0 pr=0 pw=0 time=0 us cost=0 size=494 card=26)'\nSTAT #5 id=5 cnt=700 pid=3 pos=2 obj=0 op='BUFFER SORT (cr=0 pr=0 pw=0 time=81 us cost=0 size=184 card=23)'\nSTAT #5 id=6 cnt=25 pid=5 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSLWT (cr=0 pr=0 pw=0 time=0 us cost=0 size=184 card=23)'\nSTAT #5 id=7 cnt=1 pid=2 pos=2 obj=0 op='FIXED TABLE FIXED INDEX X\\$KSUSE (ind:1) (cr=0 pr=0 pw=0 time=0 us cost=0 size=22 card=1)'\nSTAT #5 id=8 cnt=1 pid=7 pos=1 obj=0 op='COUNT STOPKEY (cr=0 pr=0 pw=0 time=0 us)'\nSTAT #5 id=9 cnt=1 pid=8 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUMYSTA (cr=0 pr=0 pw=0 time=0 us cost=0 size=40 card=2)'\nSTAT #5 id=10 cnt=1 pid=9 pos=1 obj=0 op='FIXED TABLE FULL X\\$KSUSGIF (cr=0 pr=0 pw=0 time=0 us cost=0 size=4 card=1)'\nSTAT #5 id=11 cnt=1 pid=1 pos=2 obj=0 op='FIXED TABLE FIXED INDEX X\\$KSLED (ind:2) (cr=0 pr=0 pw=0 time=0 us cost=0 size=4 card=1)'\nWAIT #5: nam='SQL*Net message to client' ela= 1 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489240779190\n\n*** 2012-01-08 10:00:49.694\nWAIT #5: nam='SQL*Net message from client' ela= 48590415 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489289369620\nCLOSE #5:c=0,e=16,dep=0,type=0,tim=489289369744\n=====================\nPARSING IN CURSOR #3 len=24 dep=0 uid=91 oct=3 lid=91 tim=489289370485 hv=124468195 ad='7ffb78cd410' sqlid='c749bc43qqfz3'\nSELECT SYSDATE FROM DUAL\nEND OF STMT\nPARSE #3:c=0,e=704,p=0,cr=0,cu=0,mis=1,r=0,dep=0,og=1,plh=1388734953,tim=489289370484\nEXEC #3:c=0,e=24,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=1388734953,tim=489289370563\nWAIT #3: nam='SQL*Net message to client' ela= 3 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489289370620\nFETCH #3:c=0,e=13,p=0,cr=0,cu=0,mis=0,r=1,dep=0,og=1,plh=1388734953,tim=489289370661\nSTAT #3 id=1 cnt=1 pid=0 pos=1 obj=0 op='FAST DUAL (cr=0 pr=0 pw=0 time=0 us cost=2 size=0 card=1)'\nWAIT #3: nam='SQL*Net message from client' ela= 189 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489289370905\nFETCH #3:c=0,e=1,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=0,plh=1388734953,tim=489289370957\nWAIT #3: nam='SQL*Net message to client' ela= 2 driver id=1413697536 #bytes=1 p3=0 obj#=-1 tim=489289370997\n\n*** 2012-01-08 10:09:13.997\nReceived ORADEBUG command (#9) 'session_event 10046 trace name context off' from process 'Windows thread id: 12292, image: <none>'\n\n*** 2012-01-08 10:09:13.997\nFinished processing ORADEBUG command (#9) 'session_event 10046 trace name context off'\n\n*** 2012-01-08 10:11:16.960\nDumping Event (group=SESSION)\n\n*** 2012-01-08 10:11:16.960\nFinished processing ORADEBUG command (#10) 'EVENTDUMP session'\n\n*** 2012-01-08 10:12:42.528\nDumping Event (group=SESSION)\n\n*** 2012-01-08 10:12:42.528\nFinished processing ORADEBUG command (#11) 'EVENTDUMP session'\n\n*** 2012-01-08 10:15:29.621\nDumping Event (group=SESSION)\n\n*** 2012-01-08 10:15:29.621\nFinished processing ORADEBUG command (#12) 'EVENTDUMP session'\n\n*** 2012-01-08 10:19:54.941\nDumping Event (group=SESSION)\n\n*** 2012-01-08 10:19:54.941\nFinished processing ORADEBUG command (#13) 'EVENTDUMP session'\n\n*** 2012-01-08 10:21:25.459\nReceived ORADEBUG command (#14) 'session_event 10046 trace name context forever,level 12' from process 'Windows thread id: 12292, image: <none>'\n\n*** 2012-01-08 10:21:25.459\nFinished processing ORADEBUG command (#14) 'session_event 10046 trace name context forever,level 12'\n\n*** 2012-01-08 10:21:40.549"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.54310155,"math_prob":0.8487952,"size":23697,"snap":"2021-43-2021-49","text_gpt3_token_len":8124,"char_repetition_ratio":0.15857004,"word_repetition_ratio":0.373444,"special_character_ratio":0.39979744,"punctuation_ratio":0.14670381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9871708,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T16:43:04Z\",\"WARC-Record-ID\":\"<urn:uuid:2ef9c65e-5410-4314-b403-7a9b43d0a65a>\",\"Content-Length\":\"103450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79eeee35-da7a-48e0-be5e-95897760e5ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d29bc39-7102-481d-93d5-7c2b0bcfe123>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://hoopercharles.wordpress.com/2012/01/09/\",\"WARC-Payload-Digest\":\"sha1:7N33SJNDJ5UO6B7NS5Y2OYXZ4R2LBMBV\",\"WARC-Block-Digest\":\"sha1:MMTLCQX466LYZ67KJ7GOAFZBT5IHVR3V\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323586043.75_warc_CC-MAIN-20211024142824-20211024172824-00310.warc.gz\"}"} |
http://bustntrap.club/excel-formula-to-add-time/excel-formula-to-add-time-excel-formula-for-addition-how-to-use-circular-references-in-excel-excel-formula-add-time-sheet/ | [
"# Excel Formula To Add Time Excel Formula For Addition How To Use Circular References In Excel Excel Formula Add Time Sheet",
null,
"excel formula to add time excel formula for addition how to use circular references in excel excel formula add time sheet.\n\nexcel formula to add time format cells function, excel formula for adding hours time, eliminate your frustration with excel time formulas, worksheet function calculation for employees on public,how to add up time in excel , excel formula for minus, adding time excel and subtracting in formula, excel formula for addition how to use circular references in, excel formulas addition add multiple cells, excel add minutes formula for adding without using."
] | [
null,
"http://bustntrap.club/wp-content/uploads/2018/11/excel-formula-to-add-time-excel-formula-for-addition-how-to-use-circular-references-in-excel-excel-formula-add-time-sheet.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74575835,"math_prob":0.9369138,"size":994,"snap":"2019-13-2019-22","text_gpt3_token_len":176,"char_repetition_ratio":0.23232323,"word_repetition_ratio":0.0794702,"special_character_ratio":0.16599597,"punctuation_ratio":0.06626506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99989355,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T12:15:40Z\",\"WARC-Record-ID\":\"<urn:uuid:8950b4d1-51aa-4f23-9be7-4c4d7d885fbd>\",\"Content-Length\":\"45527\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38ee0b9a-86f4-4034-828b-86f1a8c9cdc6>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5469d53-5294-430d-b8d4-f0b832848f3b>\",\"WARC-IP-Address\":\"104.28.12.119\",\"WARC-Target-URI\":\"http://bustntrap.club/excel-formula-to-add-time/excel-formula-to-add-time-excel-formula-for-addition-how-to-use-circular-references-in-excel-excel-formula-add-time-sheet/\",\"WARC-Payload-Digest\":\"sha1:LABEZGW3NGYCJBU2DV2FGMJBJUXGHXTD\",\"WARC-Block-Digest\":\"sha1:ETI6S27EOLGLL67IJTWPUP7IEUNGRQFP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203947.59_warc_CC-MAIN-20190325112917-20190325134917-00172.warc.gz\"}"} |
https://www.itcodar.com/python/pandas-multiprocessing-apply.html | [
" Pandas Multiprocessing Apply - ITCodar\n\n# Pandas Multiprocessing Apply\n\n## Make Pandas DataFrame apply() use all cores?\n\nYou may use the `swifter` package:\n\n``pip install swifter``\n\n(Note that you may want to use this in a virtualenv to avoid version conflicts with installed dependencies.)\n\nSwifter works as a plugin for pandas, allowing you to reuse the `apply` function:\n\n``import swifterdef some_function(data): return data * 10data['out'] = data['in'].swifter.apply(some_function)``\n\nIt will automatically figure out the most efficient way to parallelize the function, no matter if it's vectorized (as in the above example) or not.\n\nMore examples and a performance comparison are available on GitHub. Note that the package is under active development, so the API may change.\n\nAlso note that this will not work automatically for string columns. When using strings, Swifter will fallback to a “simple” Pandas `apply`, which will not be parallel. In this case, even forcing it to use `dask` will not create performance improvements, and you would be better off just splitting your dataset manually and parallelizing using `multiprocessing`.\n\n## Python: using multiprocessing on a pandas dataframe\n\n#### What's wrong\n\n``pool.map(calc_dist, ['lat','lon'])``\n\nspawns 2 processes - one runs `calc_dist('lat')` and the other runs `calc_dist('lon')`. Compare the first example in doc. (Basically, `pool.map(f, [1,2,3])` calls `f` three times with arguments given in the list that follows: `f(1)`, `f(2)`, and `f(3)`.) If I'm not mistaken, your function `calc_dist` can only be called `calc_dist('lat', 'lon')`. And it doesn't allow for parallel processing.\n\n#### Solution\n\nI believe you want to split the work between processes, probably sending each tuple `(grp, lst)` to a separate process. The following code does exactly that.\n\nFirst, let's prepare for splitting:\n\n``grp_lst_args = list(df.groupby('co_nm').groups.items())print(grp_lst_args)[('aa', [0, 1, 2]), ('cc', [7, 8, 9]), ('bb', [3, 4, 5, 6])]``\n\nWe'll send each of these tuples (here, there are three of them) as an argument to a function in a separate process. We need to rewrite the function, let's call it `calc_dist2`. For convenience, it's argument is a tuple as in `calc_dist2(('aa',[0,1,2]))`\n\n``def calc_dist2(arg): grp, lst = arg return pd.DataFrame( [ [grp, df.loc[c].ser_no, df.loc[c].ser_no, vincenty(df.loc[c, ['lat','lon']], df.loc[c, ['lat','lon']]) ] for c in combinations(lst, 2) ], columns=['co_nm','machineA','machineB','distance'])``\n\nAnd now comes the multiprocessing:\n\n``pool = mp.Pool(processes = (mp.cpu_count() - 1))results = pool.map(calc_dist2, grp_lst_args)pool.close()pool.join()results_df = pd.concat(results)``\n\n`results` is a list of results (here data frames) of calls `calc_dist2((grp,lst))` for `(grp,lst)` in `grp_lst_args`. Elements of `results` are later concatenated to one data frame.\n\n``print(results_df) co_nm machineA machineB distance0 aa 1 2 156.876149391 km1 aa 1 3 313.705445447 km2 aa 2 3 156.829329105 km0 cc 8 9 156.060165391 km1 cc 8 0 311.910998169 km2 cc 9 0 155.851498134 km0 bb 4 5 156.665641837 km1 bb 4 6 313.214333025 km2 bb 4 7 469.622535339 km3 bb 5 6 156.548897414 km4 bb 5 7 312.957597466 km5 bb 6 7 156.40899677 km``\n\nBTW, In Python 3 we could use a `with` construction:\n\n``with mp.Pool() as pool: results = pool.map(calc_dist2, grp_lst_args)``\n\nUpdate\n\nI tested this code only on linux. On linux, the read only data frame `df` can be accessed by child processes and is not copied to their memory space, but I'm not sure how it exactly works on Windows. You may consider splitting `df` into chunks (grouped by `co_nm`) and sending these chunks as arguments to some other version of `calc_dist`."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7106886,"math_prob":0.9263757,"size":4555,"snap":"2023-14-2023-23","text_gpt3_token_len":1231,"char_repetition_ratio":0.09009009,"word_repetition_ratio":0.0,"special_character_ratio":0.2895719,"punctuation_ratio":0.14895947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97102576,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T02:39:58Z\",\"WARC-Record-ID\":\"<urn:uuid:a5c4eee9-add6-410a-bc53-07fed2b703ef>\",\"Content-Length\":\"18959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e4f7813-27cd-438f-b164-20f8c1556a5d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f459abd6-fd18-41d9-9291-246fc81bb4b9>\",\"WARC-IP-Address\":\"68.178.244.182\",\"WARC-Target-URI\":\"https://www.itcodar.com/python/pandas-multiprocessing-apply.html\",\"WARC-Payload-Digest\":\"sha1:GQEJVPVCKIXSBRCFJE3O4RZRFBW2UNTA\",\"WARC-Block-Digest\":\"sha1:ROKTHJI3XI3Y6C5HVP5P6DPWKVBIKUBO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652207.81_warc_CC-MAIN-20230606013819-20230606043819-00338.warc.gz\"}"} |
https://demonstrations.wolfram.com/author.html?author=Y.%20Shibuya&start=21&limit=20&sortmethod=recent | [
"",
null,
"",
null,
"",
null,
"",
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"",
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"",
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null,
"",
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"Electromagnetic Wave from Dipole over a Perfect Conductor Attenuation of a Sinusoidal Magnetic Field in a Halfspace Conductor Induction in a Coil by a Gaussian Magnetic Field Electromagnetic Wave Incident on a Perfect Conductor Induction of a Sinusoidal Magnetic Field within a Coil",
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"",
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"",
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"",
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"Fields and Poynting Vector around a Transformer Electric and Magnetic Fields near a Transmission Line Magnetic Fields for a Pair of Parallel Currents and for a Ring Current Electromagnetic Fields For Hertzian Dipoles",
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"© 2022 Wolfram Demonstrations Project & Contributors | Terms of Use | Privacy Policy | RSS",
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"",
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"Note: To run this Demonstration you need Mathematica 7+ or the free Mathematica Player 7EX Download or upgrade to Mathematica Player 7EX I already have Mathematica Player or Mathematica 7+"
] | [
null,
"https://demonstrations.wolfram.com/images/demos-logo2.png",
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"https://demonstrations.wolfram.com/images/demoNumber-txtline2.png",
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"https://demonstrations.wolfram.com/common/images/spacer.gif",
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"https://demonstrations.wolfram.com/ElectromagneticWaveFromDipoleOverAPerfectConductor/thumbnail.jpg",
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"https://demonstrations.wolfram.com/AttenuationOfASinusoidalMagneticFieldInAHalfspaceConductor/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/InductionInACoilByAGaussianMagneticField/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/ElectromagneticWaveIncidentOnAPerfectConductor/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/InductionOfASinusoidalMagneticFieldWithinACoil/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/FieldsAndPoyntingVectorAroundATransformer/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/ElectricAndMagneticFieldsNearATransmissionLine/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/MagneticFieldsForAPairOfParallelCurrentsAndForARingCurrent/thumbnail.jpg",
null,
"https://demonstrations.wolfram.com/ElectromagneticFieldsForHertzianDipoles/thumbnail.jpg",
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"https://demonstrations.wolfram.com/images/powered-by.png",
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"https://demonstrations.wolfram.com/images/feedback.gif",
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"https://demonstrations.wolfram.com/images/closedownloadplayer.png",
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https://www.patrick-meuth.de/peak/manual/ | [
"",
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"## Setup\n\nWhat happens if someone presents a new data trace without any contextual information and asks you to do a quick peak check by eye?\n\nIn most of the cases we are simply searching for ...\n1. relative base line deflections (in particular irrespective of the absolute base line level) that show ...\n2. a fast rise and decay (or the other way around) and ...\n3. differ in amplitude to a certain extent from all other peaks in their close surrounding.\n\nPEAK is doing exactly the same. In the first step PEAK determines all inflexion points (see 8) and the corresponding y-axis distance (see 10) to their direct predecessor. If this distance exceeds a certain threshold (see 11) the currently regarded inflexion point will be marked as peak tip (see 9) and its direct predecessor and successor as peak start and end, respectively.\n\nThe threshold for each single inflexion point is individually derived from its corresponding neighborhood. Thereby, the neighborhood radius (see 27) determines how many inflexion point distances before (neighborhood A) and after (neighborhood B) the currently regarded inflexion point will be taken into account. Finally, both neighborhoods are combined to a single reference value by calculating the individual median or mean (see 21) and selecting the maximum or minimum (see 22) from both values. This value multiplied by the user-defined significance factor (see 28) has to be exceeded by the y-axis distance of the currently regarded inflexion point in order to be outstanding within its direct neighborhood and thus a real peak.",
null,
"01: Zoom data in / out.\n02: Move plotted data trace.\n03: Data cursor\n04: Data visualization area\n05: Plot data base (see 19).\n06: Plot sampling result (see 25).\n07: Plot smoothing result (see 26).\n08: Plot inflexion points.\n09: Plot detected peak tips.\n10: Plot inflexion point distances.\n11: Plot peak thresholds.\n12: Plot peak start and end.\n13: Plot half-maximum peak width.\n14: Mark initial Y value data base (see 33).\n\n15: Data browser\nFor each valid data file the data browser contains the file name and file size (in squared brackets) followed by the corresponding sweep and column numbers. Invalid data files get automatically skipped.\n16: Click to choose the data path.\nEach file in the selected path and all its subfolders will be scanned for processable information. Valid data files are regular text files (e.g. *.txt, *.dat, *.asc) containing at least two numeric data columns (x- and y-axis). Additional header lines or multiple sweeps within one file are valid and handled automatically.\n17: Click to generate a single Microsoft Excel result file for the currently analyzed data trace.\nThe result file includes beside the complete set of user-defined parameters (see 19-22 and 25-33) the following information for each data trace:\n• File name\n• Sweep number\n• Column number\n• Initial Y value\n• Total number of peaks\n• Interpeak interval (IPI)\n• Peak height\n• Half-maximum peak width\n• Peak start (x and y coordinate)\n• Peak tip (x and y coordinate)\n• Peak end (x and y coordinate)\n• Peak half-maximum (x1, x2 and y coordinate)\n18: Click to generate a single Microsoft Excel result file of all data traces in the data browser.\nDuring that process each single data trace in the data browser gets automatically selected and analyzed based on the same set of parameters.\n19: Switch between the original data trace and its derivative-like representation.\nSelecting \"Derivative\" replaces each y-axis value by the y-axis distance to its predecessor (ΔY) while the x-axis remains unaffected.\n20: Filter results for a certain peak type.\nChoose between \"All\", \"Up\" and \"Down\" to either detect all kinds of peaks or filter for upward or downward peaks, respectively.\n21: Choose between \"Median\" and \"Mean\" as operator during the neighborhood evaluation.\nAn inflexion point is considered a peak tip if the distance to its nearest predecessor strongly differs from one of the two reference distances determined in its preceding and succeeding neighborhood (see 27). A reference distance is either the median or the mean of all inflexion point distances in the given neighborhood.\n22: Select either the maximum or minimum from both neighborhood reference distances as comparison value.\nThe final peak judgement comprises the comparison of the current inflexion point distance with a multiple (significance factor, see 28) of one of the two neighborhood reference distances. This can be either the maximum (\"Max\") or minimum (\"Min\") of both values.\n23: Total number of data points.\n24: Total number of peaks.\n25: Downsample data trace.\nConsider only the first of x consecutive data points (x >= 1).\n26: Smooth data trace by applying a sliding mean.\nReplace every data point by the mean of itself, its x preceding and x succeeding data points (x >= 0).\n27: Determine the total number of inflexion points per neighborhood.\nWhile searching for outstanding peaks compare the current inflexion point with its x predecessors and x successors (x >= 1).\n28: Significance factor\nAn outstanding peak is x times higher than its surrounding (x >= 1).\n29: Use column x as x-axis (x >= 1).\n30: Ignore all values before the x-axis value x (x ∈ [-∞,∞]).\n31: Ignore all values after the x-axis value x (x ∈ [-∞,∞]).\n32: Ignore specified columns x y z ... (x, y, z >= 1).\nPEAK automatically ensures that at least two columns (x- and y-axis) remain.\n33: Calculate the mean y-axis value of the first x data points (x >= 1)."
] | [
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"https://www.patrick-meuth.de/peak/manual/",
null,
"https://www.patrick-meuth.de/peak/pics/Peak-detection_small.png",
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"https://www.patrick-meuth.de/peak/pics/Peak-derivative_small.png",
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"https://www.patrick-meuth.de/peak/pics/Peak-dynamic_threshold_small.png",
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"https://www.patrick-meuth.de/peak/pics/Peak-half_max_width_small.png",
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"https://www.patrick-meuth.de/peak/pics/Peak_GUI.png",
null
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https://electronics.stackexchange.com/questions/182116/whats-the-easy-way-to-measure-a-dc-hobby-motors-inductance | [
"# What's the easy way to measure a DC hobby motor's inductance?\n\nWhen using simulations(SPICE), I need to model a DC motor or an electromagnet, relay ect. as an RL in series.\n\nI can measure the R resistance by an ohmmeter. But I couldn't find an easy way to find out L value. Should I use a sinusoidal AC signal and measure the impedance from the phasor equations for that? But then I would need other equipment.\n\nAny ideas?\n\n• What equipment do you have? Jul 28, 2015 at 9:50\n• There are multimeters that measure inductance Jul 28, 2015 at 9:54\n• I have a multi-meter and a scope. I can also generate adjusted freq. pulses but not sinusoids. Jul 28, 2015 at 9:58\n• plus my multimeter doesnt have inductance measuring option. Jul 28, 2015 at 10:01\n• Easiest? An L(CR) meter. Easiest for you? Without knowing what tools you have available, I would say slay a chicken and read in its bones... Jul 28, 2015 at 10:04\n\nIf you have a square wave generator and an oscilloscope then you can use the L/R time constant method. Hook the generator up to the coil in series with a resistor, and put the scope across the resistor. Adjust the resistor value and square wave frequency to get an exponential waveform like this:-",
null,
"Now measure the time it takes for the voltage to drop to 37% of its peak value. This is the L/R time constant, T = L/R. R is the total resistance in the circuit (resistance of the coil + your resistor). You know T and R, so put those values into the formula and rearrange it to get L = T*R.\n\nInductance is measured by applying a sinusoidal signal and then measuring the phase difference between the voltage wave form and current wave form. Connect the motor to a function generator or the such. Place a resistor in series between the motor and the function generator on the negative side. Connect one oscilloscope channel across the function generator. Connect the second channel across the resistor with the common grounding point on the negative side of the function generator. This will display the voltage wave form and the current wave form. By measuring the phase difference between them the inductance can be calculated.\n\n• i know in theory it is like that but i dont have that sinusoidal signal gen. did you read what i wrote above? Jul 28, 2015 at 10:26\n• If not using an LCR meter, without using \"something\" that generates a sine wave the inductance measurement won't be accurate or mean something useful. Jul 28, 2015 at 10:43\n\nYou have a scope, you can make pulses.\n\nIf you have the DC resistance you can work out a scheme where the resistance will probably not make more than 10%-ish difference. Then you can estimate the inductance by just bluntly assuming the inductance is all you are measuring with a fixed voltage pulse. Of course the internal resistance will always make a difference and the line will never be perfectly straight and if you know all those modelling paradigms and understand the maths with all the e-power stuff, then you can use the resulting curved line to get a much more accurate prediction. But if you don't if you see a very strongly curved line, you need to adjust the the pulse time downward to limit the current built up, so that you stay in the mostly-linear region. (Or you can crank up the voltage, but be careful to check that the motor doesn't move).\n\nIf you use a fixed voltage the current through a perfect inductance is determined just by the voltage across it and the time that voltage has been across it, taken over the value of the inductor, like so:\n\nI = (V*t)/L\n\nTo reiterate: This only counts for a fixed voltage, this is the result of an integration over constant voltage over a perfect inductance. Thus you need to make sure the internal DC resistance can be neglected (the line should be reasonably flat).\n\nSo you connect it like so:",
null,
"simulate this circuit – Schematic created using CircuitLab\n\nBy letting the scope plot (and record if possible) the signal of (Probe 2 - Probe 1) you see the voltage across the current measurement resistor. You then apply a voltage that will not make the motor spin (else you get back-EMF and all kinds of contact noise screwing up your results in more ways than one).\n\nSelect a MOSFET with a nice low R-on that is specifically made for switching on/off loads. Preferably inductive ones and even more preferably ones that are larger than yours (the less time and energy the MOSFET wastes, the better your measurement).\n\nThen, if you turn on the MOSFET with a pulse signal after the motor has completely relaxed, you can see that current (voltage over the resistor) ramp up. When the pulse is 1s and you see a current of 1A at the end of it, after applying 1V, knowing the resistor is 0.1Ohm or less, you ignore the resistor and get:\n\nL = (V*t)/I = (1*1)/1 = 1H Which would of course be ridiculous, but in example-land ridiculous is allowed.\n\nOf course, if you are assuming the Rdc is negligible, you should make your current sense resistor just as small or smaller, else you will be adding more parasitic resistance.\n\nThe trick, usually, is to select a voltage high enough to let you get a good response with a reasonable pulse length, but that does only just not make anything actually move.\n\nOne last Caveat:\n\nThis will only give you the static inductance. To fully model a motor you need many more parameters and need to know the static and/or dynamic load, but those models are too advanced to explain as a response to this particular problem.\n\nIt's possible just having the static inductance will let you model the most important parts of the motor correctly and then in the real world you'll need to make a couple of additional adjustments to account for the effects of the motor spinning up or being loaded while powered or being stalled while powered. In the assumption that this is the case, I will leave it here."
] | [
null,
"https://i.stack.imgur.com/Nyohs.jpg",
null,
"https://i.stack.imgur.com/NEgmR.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9405745,"math_prob":0.9396437,"size":3363,"snap":"2023-40-2023-50","text_gpt3_token_len":741,"char_repetition_ratio":0.12652574,"word_repetition_ratio":0.0,"special_character_ratio":0.21112102,"punctuation_ratio":0.07401813,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.987135,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T21:56:33Z\",\"WARC-Record-ID\":\"<urn:uuid:04dc426d-be24-44bd-9577-5626d5ff1b85>\",\"Content-Length\":\"191635\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b97dd7b-038f-4f63-be43-06ef9a436d13>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5f2dd6d-2d88-4871-ad96-cdae5b88d6ca>\",\"WARC-IP-Address\":\"104.18.43.226\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/182116/whats-the-easy-way-to-measure-a-dc-hobby-motors-inductance\",\"WARC-Payload-Digest\":\"sha1:KRGPHOQ72RMB2MWWPMZYSJPBU7BYNB57\",\"WARC-Block-Digest\":\"sha1:KLC56XB4FNM4NBOWS2UA6MXKIMFWWHBP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.53_warc_CC-MAIN-20231203193127-20231203223127-00720.warc.gz\"}"} |
https://flash.savingadvice.com/2009/01/ | [
"User Real IP - 3.238.62.144\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => Array\n(\n => 95.142.117.29\n)\n\n => Array\n(\n => 150.129.60.107\n)\n\n => Array\n(\n => 94.205.243.22\n)\n\n => Array\n(\n => 115.42.71.143\n)\n\n => Array\n(\n => 117.217.195.59\n)\n\n => Array\n(\n => 182.77.112.56\n)\n\n => Array\n(\n => 182.77.112.108\n)\n\n => Array\n(\n => 41.80.69.10\n)\n\n => Array\n(\n => 117.5.222.121\n)\n\n => Array\n(\n => 103.11.0.38\n)\n\n => Array\n(\n => 202.173.127.140\n)\n\n => Array\n(\n => 49.249.249.50\n)\n\n => Array\n(\n => 116.72.198.211\n)\n\n => Array\n(\n => 223.230.54.53\n)\n\n => Array\n(\n => 102.69.228.74\n)\n\n => Array\n(\n => 39.37.251.89\n)\n\n => Array\n(\n => 39.53.246.141\n)\n\n => Array\n(\n => 39.57.182.72\n)\n\n => Array\n(\n => 209.58.130.210\n)\n\n => Array\n(\n => 104.131.75.86\n)\n\n => Array\n(\n => 106.212.131.255\n)\n\n => Array\n(\n => 106.212.132.127\n)\n\n => Array\n(\n => 223.190.4.60\n)\n\n => Array\n(\n => 103.252.116.252\n)\n\n => Array\n(\n => 103.76.55.182\n)\n\n => Array\n(\n => 45.118.166.70\n)\n\n => Array\n(\n => 103.93.174.215\n)\n\n => Array\n(\n => 5.62.62.142\n)\n\n => Array\n(\n => 182.179.158.156\n)\n\n => Array\n(\n => 39.57.255.12\n)\n\n => Array\n(\n => 39.37.178.37\n)\n\n => Array\n(\n => 182.180.165.211\n)\n\n => Array\n(\n => 119.153.135.17\n)\n\n => Array\n(\n => 72.255.15.244\n)\n\n => Array\n(\n => 139.180.166.181\n)\n\n => Array\n(\n => 70.119.147.111\n)\n\n => Array\n(\n => 106.210.40.83\n)\n\n => Array\n(\n => 14.190.70.91\n)\n\n => Array\n(\n => 202.125.156.82\n)\n\n => Array\n(\n => 115.42.68.38\n)\n\n => Array\n(\n => 102.167.13.108\n)\n\n => Array\n(\n => 117.217.192.130\n)\n\n => Array\n(\n => 205.185.223.156\n)\n\n => Array\n(\n => 171.224.180.29\n)\n\n => Array\n(\n => 45.127.45.68\n)\n\n => Array\n(\n => 195.206.183.232\n)\n\n => Array\n(\n => 49.32.52.115\n)\n\n => Array\n(\n => 49.207.49.223\n)\n\n => Array\n(\n => 45.63.29.61\n)\n\n => Array\n(\n => 103.245.193.214\n)\n\n => Array\n(\n => 39.40.236.69\n)\n\n => Array\n(\n => 62.80.162.111\n)\n\n => Array\n(\n => 45.116.232.56\n)\n\n => Array\n(\n => 45.118.166.91\n)\n\n => Array\n(\n => 180.92.230.234\n)\n\n => Array\n(\n => 157.40.57.160\n)\n\n => Array\n(\n => 110.38.38.130\n)\n\n => Array\n(\n => 72.255.57.183\n)\n\n => Array\n(\n => 182.68.81.85\n)\n\n => Array\n(\n => 39.57.202.122\n)\n\n => Array\n(\n => 119.152.154.36\n)\n\n => Array\n(\n => 5.62.62.141\n)\n\n => Array\n(\n => 119.155.54.232\n)\n\n => Array\n(\n => 39.37.141.22\n)\n\n => Array\n(\n => 183.87.12.225\n)\n\n => Array\n(\n => 107.170.127.117\n)\n\n => Array\n(\n => 125.63.124.49\n)\n\n => Array\n(\n => 39.42.191.3\n)\n\n => Array\n(\n => 116.74.24.72\n)\n\n => Array\n(\n => 46.101.89.227\n)\n\n => Array\n(\n => 202.173.125.247\n)\n\n => Array\n(\n => 39.42.184.254\n)\n\n => Array\n(\n => 115.186.165.132\n)\n\n => Array\n(\n => 39.57.206.126\n)\n\n => Array\n(\n => 103.245.13.145\n)\n\n => Array\n(\n => 202.175.246.43\n)\n\n => Array\n(\n => 192.140.152.150\n)\n\n => Array\n(\n => 202.88.250.103\n)\n\n => Array\n(\n => 103.248.94.207\n)\n\n => Array\n(\n => 77.73.66.101\n)\n\n => Array\n(\n => 104.131.66.8\n)\n\n => Array\n(\n => 113.186.161.97\n)\n\n => Array\n(\n => 222.254.5.7\n)\n\n => Array\n(\n => 223.233.67.247\n)\n\n => Array\n(\n => 171.249.116.146\n)\n\n => Array\n(\n => 47.30.209.71\n)\n\n => Array\n(\n => 202.134.13.130\n)\n\n => Array\n(\n => 27.6.135.7\n)\n\n => Array\n(\n => 107.170.186.79\n)\n\n => Array\n(\n => 103.212.89.171\n)\n\n => Array\n(\n => 117.197.9.77\n)\n\n => Array\n(\n => 122.176.206.233\n)\n\n => Array\n(\n => 192.227.253.222\n)\n\n => Array\n(\n => 182.188.224.119\n)\n\n => Array\n(\n => 14.248.70.74\n)\n\n => Array\n(\n => 42.118.219.169\n)\n\n => Array\n(\n => 110.39.146.170\n)\n\n => Array\n(\n => 119.160.66.143\n)\n\n => Array\n(\n => 103.248.95.130\n)\n\n => Array\n(\n => 27.63.152.208\n)\n\n => Array\n(\n => 49.207.114.96\n)\n\n => Array\n(\n => 102.166.23.214\n)\n\n => Array\n(\n => 175.107.254.73\n)\n\n => Array\n(\n => 103.10.227.214\n)\n\n => Array\n(\n => 202.143.115.89\n)\n\n => Array\n(\n => 110.93.227.187\n)\n\n => Array\n(\n => 103.140.31.60\n)\n\n => Array\n(\n => 110.37.231.46\n)\n\n => Array\n(\n => 39.36.99.238\n)\n\n => Array\n(\n => 157.37.140.26\n)\n\n => Array\n(\n => 43.246.202.226\n)\n\n => Array\n(\n => 137.97.8.143\n)\n\n => Array\n(\n => 182.65.52.242\n)\n\n => Array\n(\n => 115.42.69.62\n)\n\n => Array\n(\n => 14.143.254.58\n)\n\n => Array\n(\n => 223.179.143.236\n)\n\n => Array\n(\n => 223.179.143.249\n)\n\n => Array\n(\n => 103.143.7.54\n)\n\n => Array\n(\n => 223.179.139.106\n)\n\n => Array\n(\n => 39.40.219.90\n)\n\n => Array\n(\n => 45.115.141.231\n)\n\n => Array\n(\n => 120.29.100.33\n)\n\n => Array\n(\n => 112.196.132.5\n)\n\n => Array\n(\n => 202.163.123.153\n)\n\n => Array\n(\n => 5.62.58.146\n)\n\n => Array\n(\n => 39.53.216.113\n)\n\n => Array\n(\n => 42.111.160.73\n)\n\n => Array\n(\n => 107.182.231.213\n)\n\n => Array\n(\n => 119.82.94.120\n)\n\n => Array\n(\n => 178.62.34.82\n)\n\n => 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39.53.151.199\n)\n\n => Array\n(\n => 101.255.118.33\n)\n\n => Array\n(\n => 41.141.246.9\n)\n\n => Array\n(\n => 221.132.113.78\n)\n\n => Array\n(\n => 119.160.116.202\n)\n\n => Array\n(\n => 117.237.193.244\n)\n\n => Array\n(\n => 157.41.110.145\n)\n\n => Array\n(\n => 103.98.63.5\n)\n\n => Array\n(\n => 103.125.129.58\n)\n\n => Array\n(\n => 183.83.254.66\n)\n\n => Array\n(\n => 45.135.236.160\n)\n\n => Array\n(\n => 198.199.87.124\n)\n\n => Array\n(\n => 193.176.86.41\n)\n\n => Array\n(\n => 115.97.142.98\n)\n\n => Array\n(\n => 222.252.38.198\n)\n\n => Array\n(\n => 110.93.237.49\n)\n\n => Array\n(\n => 103.224.48.122\n)\n\n => Array\n(\n => 110.38.28.130\n)\n\n => Array\n(\n => 106.211.238.154\n)\n\n => Array\n(\n => 111.88.41.73\n)\n\n => Array\n(\n => 119.155.13.143\n)\n\n => Array\n(\n => 103.213.111.60\n)\n\n => Array\n(\n => 202.0.103.42\n)\n\n => Array\n(\n => 157.48.144.33\n)\n\n => Array\n(\n => 111.119.187.62\n)\n\n => Array\n(\n => 103.87.212.71\n)\n\n => Array\n(\n => 157.37.177.20\n)\n\n => Array\n(\n => 223.233.71.92\n)\n\n => Array\n(\n => 116.213.32.107\n)\n\n => Array\n(\n => 104.248.173.151\n)\n\n => Array\n(\n => 14.181.102.222\n)\n\n => Array\n(\n => 103.10.224.252\n)\n\n => Array\n(\n => 175.158.50.57\n)\n\n => Array\n(\n => 165.22.122.199\n)\n\n => Array\n(\n => 23.106.56.12\n)\n\n => Array\n(\n => 203.122.10.146\n)\n\n => Array\n(\n => 37.111.136.138\n)\n\n => Array\n(\n => 103.87.193.66\n)\n\n => Array\n(\n => 39.59.122.246\n)\n\n => Array\n(\n => 111.119.183.63\n)\n\n => Array\n(\n => 157.46.72.102\n)\n\n => Array\n(\n => 185.132.133.82\n)\n\n => Array\n(\n => 118.103.230.148\n)\n\n => Array\n(\n => 5.62.39.45\n)\n\n => Array\n(\n => 119.152.144.134\n)\n\n => Array\n(\n => 172.105.117.102\n)\n\n => Array\n(\n => 122.254.70.212\n)\n\n => Array\n(\n => 102.185.128.97\n)\n\n => Array\n(\n => 182.69.249.11\n)\n\n => Array\n(\n => 105.163.134.167\n)\n\n => Array\n(\n => 111.119.187.38\n)\n\n => Array\n(\n => 103.46.195.93\n)\n\n => Array\n(\n => 106.204.161.156\n)\n\n => Array\n(\n => 122.176.2.175\n)\n\n => Array\n(\n => 117.99.162.31\n)\n\n => Array\n(\n => 106.212.241.242\n)\n\n => Array\n(\n => 42.107.196.149\n)\n\n => Array\n(\n => 212.90.60.57\n)\n\n => Array\n(\n => 175.107.237.12\n)\n\n => Array\n(\n => 157.46.119.152\n)\n\n => Array\n(\n => 157.34.81.12\n)\n\n => Array\n(\n => 162.243.1.22\n)\n\n => Array\n(\n => 110.37.222.178\n)\n\n => Array\n(\n => 103.46.195.68\n)\n\n => Array\n(\n => 119.160.116.81\n)\n\n => Array\n(\n => 138.197.131.28\n)\n\n => Array\n(\n => 103.88.218.124\n)\n\n => Array\n(\n => 192.241.172.113\n)\n\n => Array\n(\n => 110.39.174.106\n)\n\n => Array\n(\n => 111.88.48.17\n)\n\n => Array\n(\n => 42.108.160.218\n)\n\n => Array\n(\n => 117.102.0.16\n)\n\n => Array\n(\n => 157.46.125.235\n)\n\n => Array\n(\n => 14.190.242.251\n)\n\n => Array\n(\n => 47.31.184.64\n)\n\n => Array\n(\n => 49.205.84.157\n)\n\n => Array\n(\n => 122.162.115.247\n)\n\n => Array\n(\n => 41.202.219.74\n)\n\n => Array\n(\n => 106.215.9.67\n)\n\n => Array\n(\n => 103.87.56.208\n)\n\n => Array\n(\n => 103.46.194.147\n)\n\n => Array\n(\n => 116.90.98.81\n)\n\n => Array\n(\n => 115.42.71.213\n)\n\n => Array\n(\n => 39.49.35.192\n)\n\n => Array\n(\n => 41.202.219.65\n)\n\n => Array\n(\n => 131.212.249.93\n)\n\n => Array\n(\n => 49.205.16.251\n)\n\n => Array\n(\n => 39.34.147.250\n)\n\n => Array\n(\n => 183.83.210.185\n)\n\n => Array\n(\n => 49.37.194.215\n)\n\n => Array\n(\n => 103.46.194.108\n)\n\n => Array\n(\n => 89.36.219.233\n)\n\n => Array\n(\n => 119.152.105.178\n)\n\n => Array\n(\n => 202.47.45.125\n)\n\n => Array\n(\n => 156.146.59.27\n)\n\n => Array\n(\n => 132.154.21.156\n)\n\n => Array\n(\n => 157.44.35.31\n)\n\n => Array\n(\n => 41.80.118.124\n)\n\n => Array\n(\n => 47.31.159.198\n)\n\n => Array\n(\n => 103.209.223.140\n)\n\n => Array\n(\n => 157.46.130.138\n)\n\n => Array\n(\n => 49.37.199.246\n)\n\n => Array\n(\n => 111.88.242.10\n)\n\n => Array\n(\n => 43.241.145.110\n)\n\n => Array\n(\n => 124.153.16.30\n)\n\n => Array\n(\n => 27.5.22.173\n)\n\n => Array\n(\n => 111.88.191.173\n)\n\n => Array\n(\n => 41.60.236.200\n)\n\n => Array\n(\n => 115.42.67.146\n)\n\n => Array\n(\n => 150.242.173.7\n)\n\n => Array\n(\n => 14.248.71.23\n)\n\n => Array\n(\n => 111.119.187.4\n)\n\n => Array\n(\n => 124.29.212.118\n)\n\n => Array\n(\n => 51.68.205.163\n)\n\n => Array\n(\n => 182.184.107.63\n)\n\n => Array\n(\n => 106.211.253.87\n)\n\n => Array\n(\n => 223.190.89.5\n)\n\n => Array\n(\n => 183.83.212.63\n)\n\n => Array\n(\n => 129.205.113.227\n)\n\n => Array\n(\n => 106.210.40.141\n)\n\n => Array\n(\n => 91.202.163.169\n)\n\n => Array\n(\n => 76.105.191.89\n)\n\n => Array\n(\n => 171.51.244.160\n)\n\n => Array\n(\n => 37.139.188.92\n)\n\n => Array\n(\n => 23.106.56.37\n)\n\n => Array\n(\n => 157.44.175.180\n)\n\n => Array\n(\n => 122.2.122.97\n)\n\n)\n```\nArchive for January, 2009: flash's saving Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: January, 2009",
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"",
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"# Archive for January, 2009\n\n## Happy 2009!\n\nJanuary 12th, 2009 at 07:20 pm\n\nHi!\n\nWell, it's been a difficult year, and I've been completely out of touch with most of my friends. So...hello!\n\nI was in an accident just over a year ago, and have been spending 4-5 hours a day in physical therapy. My father passed away in late Spring, my mother had open heart surgery, and my family, well, isn't handling everything so well. And then, of course, there is the economy...\n\nand inheritance, or lack thereof, and family issues that I never would have imagined would come up.\n\nBut we are actually overall okay. Challenged, but not overcome.\n\nAnd hoping to get back into the swing of things soon...saving and getting some great advice from everyone!\n\nSupporting Sites:"
] | [
null,
"https://www.savingadvice.com/blogs/images/search/top_left.php",
null,
"https://www.savingadvice.com/blogs/images/search/top_right.php",
null,
"https://www.savingadvice.com/blogs/images/search/bottom_left.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9822095,"math_prob":0.99942756,"size":701,"snap":"2020-45-2020-50","text_gpt3_token_len":169,"char_repetition_ratio":0.08464849,"word_repetition_ratio":0.0,"special_character_ratio":0.24964337,"punctuation_ratio":0.20625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997248,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T15:20:47Z\",\"WARC-Record-ID\":\"<urn:uuid:ef28c7cc-9d56-4abd-a093-70c298a4ca65>\",\"Content-Length\":\"146061\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:887e2dce-b253-43cd-b677-00b0f96b0ef3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3383c1b-de70-4fc0-86b1-ac6748dd0f6b>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://flash.savingadvice.com/2009/01/\",\"WARC-Payload-Digest\":\"sha1:GBAHWUBXKCEVACZ363XWJQ43UKHTBND2\",\"WARC-Block-Digest\":\"sha1:GAVUB3QTUBIU4O6VVESVV3IHPHHQUEOE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876768.45_warc_CC-MAIN-20201021151342-20201021181342-00178.warc.gz\"}"} |
https://newproxylists.com/gn-general-topology-statements-related-to-thurstons-work-on-the-surface/ | [
"# gn.general topology – Statements related to Thurston’s work on the surface\n\nIf we have α and β be simple closed curves on a surface Σg. The intersection number i(α,β) is defined to be the minimal cardinality of α1 ∩ β1 as α1 and β1 ranges over all simple closed curves isotopic to α and β, respectively. We say α and β intersect minimally if i(α, β) = |α ∩ β|.\n\nHow to see thatα and β intersect minimally if there are no pairs of p, q ∈ α ∩ β such that the arc joining p to q along α followed by the arc from q back to p along β bounds a disk in Σg.\n\nI think the converse is also true : “thatα and β intersect minimally only if there are no pairs of p, q ∈ α ∩ β such that the arc joining p to q along α followed by the arc from q back to p along β bounds a disk in Σg.”\n\nPosted on"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94554025,"math_prob":0.947348,"size":692,"snap":"2020-45-2020-50","text_gpt3_token_len":197,"char_repetition_ratio":0.12645349,"word_repetition_ratio":0.44444445,"special_character_ratio":0.2630058,"punctuation_ratio":0.06748466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96539336,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T11:02:03Z\",\"WARC-Record-ID\":\"<urn:uuid:52e04f7f-d72a-49f0-b785-cd1d5210a914>\",\"Content-Length\":\"26547\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7aea2393-2708-46ec-8d70-228021089f40>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6c4956e-6177-4089-b897-5d9974730132>\",\"WARC-IP-Address\":\"173.212.203.156\",\"WARC-Target-URI\":\"https://newproxylists.com/gn-general-topology-statements-related-to-thurstons-work-on-the-surface/\",\"WARC-Payload-Digest\":\"sha1:PRUPLEKD3BDVNEAGCIPHJ2DHLSHAXD2V\",\"WARC-Block-Digest\":\"sha1:LCTEXAVLOKYAJ7ICAR5KOUV7NAHDAWN5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141747774.97_warc_CC-MAIN-20201205104937-20201205134937-00282.warc.gz\"}"} |
https://fr.mathworks.com/help/matlab/ref/mustbepositive.html | [
"# mustBePositive\n\nValidate that value is positive\n\n## Syntax\n\n``mustBePositive(value)``\n\n## Description\n\nexample\n\n``mustBePositive(value)` throws an error if `value` is not positive. Values are positive when they are real, numeric, and greater than zero. This function does not return a value.`mustBePositive` calls these functions to determine if the input is not positive:`gt``isreal``isnumeric` or `islogical`Class support: All numeric classes, `logical`, and MATLAB® classes that overload the functions called by `mustBePositive`. This function ignores input arguments that are empty values. Therefore, no error is thrown when the property or function argument value is empty.`\n\n## Examples\n\ncollapse all\n\nUse `mustBePositive` to validate that the input contains only positive values.\n\nThe `rand` function creates a uniformly distributed random number.\n\n`A = rand(1,5) -0.5;`\n\nValidate that array elements are positive.\n\n`mustBePositive(A)`\n```Error using mustBePositive Value must be positive.```\n\nThe result of subtracting `0.5` from the array return by `rand` can contain negative numbers. When a value is negative, `mustBePositive` issues an error.\n\nThis class restricts the value of `Prop1` to positive values.\n\n```classdef MyClass properties Prop1 {mustBePositive} end end```\n\nCreate an object and assign a value to its property.\n\n```obj = MyClass; obj.Prop1 = 0;```\n```Error setting property 'Prop1' of class 'MyClass': Value must be positive.```\n\nWhen you assign a value to the property, MATLAB calls `mustBePositive` with the value being assigned to the property. `mustBePositive` issues an error because the value `0` is not positive.\n\nThis function declares two input arguments. Input `A` must be a numeric vector. Input `ix` must be a positive integer.\n\n```function r = mbPositive(A,ix) arguments A (1,:) {mustBeNumeric} ix {mustBePositive, mustBeInteger} end r = A(ix); end```\n\nCalling the function with a value for `ix` that does not meet the requirement of `mustBePositive` results in an error.\n\n```A = 1:10; ix = 0; r = mbPositive(A,ix);```\n```Error using mbPositive r = mbPositive(A,ix) ↑ Invalid input argument at position 2. Value must be positive.```\n\n## Input Arguments\n\ncollapse all\n\nValue to validate, specified as a scalar or an array of one of the following:\n\n## Tips\n\n• `mustBePositive` is designed to be used for property and function argument validation.\n\n## Version History\n\nIntroduced in R2017a"
] | [
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https://grindskills.com/for-a-continuous-random-variable-why-does-pa-z-b-pa-leq-z-b-pa-z-leq-b-pa-leq-z-leq-b/ | [
"# For a continuous random variable, why does $P(a < Z < b) = P(a \\leq Z < b) = P(a < Z \\leq b) = P(a \\leq Z \\leq b)$\n\nMy textbook puts this in a sidebox with the heading “Note” and doesn’t explain why. Could you tell me why this statement holds?\n\n$P(a < Z < b) = P(a \\leq Z < b) = P(a < Z \\leq b) = P(a \\leq Z \\leq b)$\n\nThis would explain why $\\leq$ and $<$ are basically the same for continuous variables – including or excluding the endpoint really doesn’t change anything – for any point you pick close to the endpoint, there is still an infinite amount of points between them."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93136835,"math_prob":0.9992279,"size":1469,"snap":"2022-40-2023-06","text_gpt3_token_len":370,"char_repetition_ratio":0.11262798,"word_repetition_ratio":0.021428572,"special_character_ratio":0.264806,"punctuation_ratio":0.0977918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999037,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T05:46:45Z\",\"WARC-Record-ID\":\"<urn:uuid:06e6aeae-e794-4f19-ac63-0a65d9581e7c>\",\"Content-Length\":\"51629\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92981757-ef31-42cf-8949-d9029491de27>\",\"WARC-Concurrent-To\":\"<urn:uuid:41550d4b-0b6f-45e4-8db0-9ccb57a2ef5d>\",\"WARC-IP-Address\":\"137.184.148.160\",\"WARC-Target-URI\":\"https://grindskills.com/for-a-continuous-random-variable-why-does-pa-z-b-pa-leq-z-b-pa-z-leq-b-pa-leq-z-leq-b/\",\"WARC-Payload-Digest\":\"sha1:SIJZ2GONLGYV4QPSTOXU3Y3KB52JMMIO\",\"WARC-Block-Digest\":\"sha1:5BHUL23LJO3ZOKISJBVY66QJMLENED2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335124.77_warc_CC-MAIN-20220928051515-20220928081515-00220.warc.gz\"}"} |
https://www.catalyzex.com/paper/arxiv:2203.00035 | [
"# Can Mean Field Control (MFC) Approximate Cooperative Multi Agent Reinforcement Learning (MARL) with Non-Uniform Interaction?\n\nFeb 28, 2022\nWashim Uddin Mondal, Vaneet Aggarwal, Satish V. Ukkusuri\n\nMean-Field Control (MFC) is a powerful tool to solve Multi-Agent Reinforcement Learning (MARL) problems. Recent studies have shown that MFC can well-approximate MARL when the population size is large and the agents are exchangeable. Unfortunately, the presumption of exchangeability implies that all agents uniformly interact with one another which is not true in many practical scenarios. In this article, we relax the assumption of exchangeability and model the interaction between agents via an arbitrary doubly stochastic matrix. As a result, in our framework, the mean-field `seen' by different agents are different. We prove that, if the reward of each agent is an affine function of the mean-field seen by that agent, then one can approximate such a non-uniform MARL problem via its associated MFC problem within an error of $e=\\mathcal{O}(\\frac{1}{\\sqrt{N}}[\\sqrt{|\\mathcal{X}|} + \\sqrt{|\\mathcal{U}|}])$ where $N$ is the population size and $|\\mathcal{X}|$, $|\\mathcal{U}|$ are the sizes of state and action spaces respectively. Finally, we develop a Natural Policy Gradient (NPG) algorithm that can provide a solution to the non-uniform MARL with an error $\\mathcal{O}(\\max\\{e,\\epsilon\\})$ and a sample complexity of $\\mathcal{O}(\\epsilon^{-3})$ for any $\\epsilon >0$."
] | [
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https://math.answers.com/Q/How_many_inches_are_there_in_124_centimeters | [
"",
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"0\n\n# How many inches are there in 124 centimeters?\n\nThere are 2.54 centimetres in one inch. Therefore, rounded to two decimal places, 12.4 centimetres is equal to 12.4/2.54 = 4.88 inches.",
null,
"Study guides\n\n20 cards\n\n## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials\n\n➡️\nSee all cards\n3.8\n2001 Reviews\n\nThere are 2.54 centimetres in one inch. Therefore, rounded to two decimal places, 124 centimetres is equal to 124/2.54 = 48.82 inches.",
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"315 cm",
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"Earn +20 pts",
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"https://st.answers.com/html_test_assets/imp_-_pixel.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7147814,"math_prob":0.9962791,"size":286,"snap":"2022-40-2023-06","text_gpt3_token_len":108,"char_repetition_ratio":0.18794326,"word_repetition_ratio":0.0,"special_character_ratio":0.5244755,"punctuation_ratio":0.27272728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956374,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T20:40:38Z\",\"WARC-Record-ID\":\"<urn:uuid:3a42ae15-a4a6-42d5-813d-bae4f8bc38e7>\",\"Content-Length\":\"201955\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b49dc16-8faa-4754-8d8d-15e970cebc54>\",\"WARC-Concurrent-To\":\"<urn:uuid:879f7404-488b-49b3-84e1-b16b308ed901>\",\"WARC-IP-Address\":\"146.75.32.203\",\"WARC-Target-URI\":\"https://math.answers.com/Q/How_many_inches_are_there_in_124_centimeters\",\"WARC-Payload-Digest\":\"sha1:32J3UTILYFIH4X3MZFEFQCXJI45V76NG\",\"WARC-Block-Digest\":\"sha1:G36IYCR67ZUCU7VHK3ZTUGBNBHLYAAKR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499890.39_warc_CC-MAIN-20230131190543-20230131220543-00748.warc.gz\"}"} |
https://tex.stackexchange.com/questions/428410/make-logical-diagram | [
"# Make Logical Diagram\n\nI want to make a logical diagram like that :",
null,
"mirror image : https://pasteboard.co/HicPswl.png\n\nBut I don't want to add picture with \\includegraphics... I have no experience diagrams in latex. Can you give an example code ? I find some example in circuit diagram TikZ package. But I could not find similar above diagram. Thanks for your help.\n\nMy attempt and result :\n\n\\tikzstyle{block} = [draw,fill=blue!20,minimum size=2em]\n% diameter of semicircle used to indicate that two lines are not connected\n\\tikzstyle{branch}=[fill,shape=circle,minimum size=3pt,inner sep=0pt]\n\\tikzstyle{block} = [draw,fill=blue!10,minimum size=2em]\n\n\\begin{tikzpicture}[circuit logic US, huge circuit symbols]\n\n\\node[block] at (0,0) (block1) {Sınıf diğer sınıfların niteliklerine erişmekte \\newline\nATFD > 3};\n\\node[block] at (0,2) (block2) {Sınıfın fonksiyonel karmaşıklığı fazla\n\\newline\nWMC >= 47};\n\\node[block] at (0,4) (block3) {Sınıf diğer sınıfların niteliklerine erişmekte \\newline\nATFD > 3};\n\n\\node[and gate US, draw, rotate=0] at (7,2) (And1) {};\n\n\\draw[->] (block1) -- (And1);\n\\draw[->] (block2) -- (And1);\n\\draw[->] (block3) -- (And1);\n\\node[block] at (9,2) (Result) {God Class};\n\\draw[->] (And1) -- (Result);\n\\end{tikzpicture}\n\n• Welcome to TeX.SX! Basically circuitikz is able to draw the logical gates you need and provides the anchor points to connect with arbitrary TikZ nodes, so please have a try and post your attempt here (as minimal compilable code example). – TeXnician Apr 25 '18 at 6:03\n• @TeXnician I post my attempt – Melih Altıntaş Apr 25 '18 at 7:47\n\nHere is a starting point\n\n\\documentclass[tikz]{standalone}\n\\usetikzlibrary{circuits.logic.US}\n\\begin{document}\n\\begin{tikzpicture}[circuit logic US]\n\\node[draw,align=center] (box1) at (-2,2) {Some centered tex \\\\ with math \\\\ \\\\ $\\sin(\\pi)$};\n\\node[draw,align=center] (box2) at (-2,0) {Some centered tex \\\\ with math \\\\ \\\\ $\\sin(\\pi)$};\n\\node[draw,align=center] (box3) at (-2,-2) {Some centered tex \\\\ with math \\\\ \\\\ $\\sin(\\pi)$};\n\\node [and gate] (a1) at (1,0) {VE};\n\\node[draw] (a2) at (3,0) {other node};\n\n\\draw (box1.east) -|++(1mm,-1mm)|- ([yshift=1mm]a1.west);\n\\draw (box3.east) -|++(1mm,1mm)|- ([yshift=-1mm]a1.west);\n\\draw(box2) -- (a1);\n\\draw (a1) -- (a2);\n\\end{tikzpicture}\n\\end{document}",
null,
"The rest you can find it either in the manual or searching on this site."
] | [
null,
"https://i.stack.imgur.com/CZV1D.png",
null,
"https://i.stack.imgur.com/jc670.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5654657,"math_prob":0.96503145,"size":1271,"snap":"2021-04-2021-17","text_gpt3_token_len":439,"char_repetition_ratio":0.10734017,"word_repetition_ratio":0.061349694,"special_character_ratio":0.31864673,"punctuation_ratio":0.1875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9945235,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T18:15:08Z\",\"WARC-Record-ID\":\"<urn:uuid:d0ecfd63-263a-4f8a-9d71-dca6ae5cd4f7>\",\"Content-Length\":\"148690\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b08ef8b3-5862-4644-afea-a158e7a72b61>\",\"WARC-Concurrent-To\":\"<urn:uuid:137b50af-fe14-4167-b906-d8467f69a3a6>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/428410/make-logical-diagram\",\"WARC-Payload-Digest\":\"sha1:U7ITB672DTGKZAVQ4TPCSHUHY66KE3CL\",\"WARC-Block-Digest\":\"sha1:DEHLRHCUAXWFAH5ZGTR72GVLY3JTY756\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703538226.66_warc_CC-MAIN-20210123160717-20210123190717-00636.warc.gz\"}"} |
https://nl.mathworks.com/matlabcentral/cody/problems/2021-is-this-triangle-right-angled/solutions/2237965 | [
"Cody\n\n# Problem 2021. Is this triangle right-angled?\n\nSolution 2237965\n\nSubmitted on 27 Apr 2020 by Rahul Pai\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\na = 3; b = 4; c = 5; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))\n\n2 Pass\na = 2; b = 3; c = 4; flag_correct = false; assert(isequal(isRightAngled(a,b,c),flag_correct))\n\n3 Pass\na = 5; b = 12; c = 13; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.57216877,"math_prob":0.9862063,"size":674,"snap":"2020-45-2020-50","text_gpt3_token_len":214,"char_repetition_ratio":0.16716418,"word_repetition_ratio":0.0754717,"special_character_ratio":0.3249258,"punctuation_ratio":0.2260274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98889637,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T11:33:40Z\",\"WARC-Record-ID\":\"<urn:uuid:004ac544-ac14-4782-abda-f3bf788efb92>\",\"Content-Length\":\"80218\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95c91d2b-8a49-4c1e-807a-4e4917b4501f>\",\"WARC-Concurrent-To\":\"<urn:uuid:da8bfa4f-502c-4fea-9497-8afece9a7b94>\",\"WARC-IP-Address\":\"23.32.68.178\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/2021-is-this-triangle-right-angled/solutions/2237965\",\"WARC-Payload-Digest\":\"sha1:HPRIHXZAYLRJEGKMDNSHIGYLTNP67ZOA\",\"WARC-Block-Digest\":\"sha1:SKI5Z7GNF7JYSE4JX4DUGCQ2YEJFTVZ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195417.37_warc_CC-MAIN-20201128095617-20201128125617-00321.warc.gz\"}"} |
https://online-unit-converter.com/area/convert-square-meter-to-square-centimeter/0-0006-m2-to-cm2/ | [
"# 0.0006 m² to cm² converter. How many square centimeters are in 0.0006 square meters?\n\n## 0.0006 m² to cm²\n\nThe question “What is 0.0006 m² to cm²?” is the same as “How many square centimeters are in 0.0006 m²?” or “Convert 0.0006 square meters to square centimeters” or “What is 0.0006 square meters to square centimeters?” or “0.0006 square meters to cm²”. Read on to find free m² to square centimeter converter and learn how to convert 0.0006 m² to cm². You’ll also learn how to convert 0.0006 m² to cm².\n\nAnswer: There are 6 cm² in 0.0006 m².\n\nAlternatively, you can say “0.0006 m² equals to 6 cm²” or “0.0006 m² = 6 cm²” or “0.0006 square meters is 6 square centimeters”.\n\n## Square meters to square centimeter conversion formula\n\nA square centimeter is equal to 0.0001 square meters. A square meter equals 10000 square centimeters.\nTo convert 0.0006 square meters to square centimeters you can use one of the formulas:\n\nFormula 1\nMultiply 0.0006 m² by 10000\n0.0006 * 10000 = 6 cm²\n\nFormula 2\nDivide 0.0006 m² by 0.0001.\n0.0006 / 0.0001 = 6 cm²\n\nHint: no need to use a formula. Use our free m² to cm² converter.\n\n## Alternative spelling of 0.0006 m² to cm²\n\nMany of our visitor spell square meters and square centimeters differently. Below we provide all possible spelling options.\n\n• Spelling options with “square meters”: 0.0006 square meters to cm², 0.0006 square meter to cm², 0.0006 square meters to square centimeters, 0.0006 square meter to square centimeters, 0.0006 square meters in cm², 0.0006 square meter in cm², 0.0006 square meters in square centimeters, 0.0006 square meter in square centimeters,\n• Spelling options with “m²”: 0.0006 m² to cm², 0.0006 m² to square centimeter, 0.0006 m² to square centimeters, 0.0006 m² in cm², 0.0006 m² in square centimeter, 0.0006 m² in square centimeters,\n• Spelling options with “in”: 0.0006 m² in cm², 0.0006 m² in square centimeter, 0.0006 m² in square centimeters, 0.0006 m² in cm², 0.0006 m² in square centimeter, 0.0006 m² in square centimeters, 0.0006 square meters in cm², 0.0006 square meter in cm², 0.0006 square meters in square centimeters, 0.0006 square meter in square centimeters,\n\n## FAQ on 0.0006 m² to cm² conversion\n\nHow many square centimeters are in 0.0006 square meters?\n\nThere are 6 square centimeters in 0.0006 square meters.\n\n0.0006 m² to cm²?\n\n0.0006 m² is equal to 6 cm². There are 6 square centimeters in 0.0006 square meters.\n\nWhat is 0.0006 m² to cm²?\n\n0.0006 m² is 6 cm².\n\nHow to convert 0.0006 m² to cm²?\n\nUse our free square meter to square centimeters converter or multiply0.0006 m² by 10000\n0.0006 * 10000 = 6 cm²"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7085777,"math_prob":0.99983835,"size":2584,"snap":"2023-40-2023-50","text_gpt3_token_len":880,"char_repetition_ratio":0.38023257,"word_repetition_ratio":0.31991053,"special_character_ratio":0.3866099,"punctuation_ratio":0.18390805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99935657,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T03:02:48Z\",\"WARC-Record-ID\":\"<urn:uuid:547478be-bb74-44df-88e5-075419099d69>\",\"Content-Length\":\"84534\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:500483bf-7613-48de-b235-9780dd3cbe59>\",\"WARC-Concurrent-To\":\"<urn:uuid:3563d0ea-2482-449c-9239-066ed6bfde46>\",\"WARC-IP-Address\":\"35.209.245.121\",\"WARC-Target-URI\":\"https://online-unit-converter.com/area/convert-square-meter-to-square-centimeter/0-0006-m2-to-cm2/\",\"WARC-Payload-Digest\":\"sha1:QBLQVLLNUWXRKZLMQFHZW3E7WXNGMJC6\",\"WARC-Block-Digest\":\"sha1:SGGCMV5SZ5E4UTUPWU5X2JJW26EOFOUL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100047.66_warc_CC-MAIN-20231129010302-20231129040302-00553.warc.gz\"}"} |
https://www.harddiskinformatica.com/k02die/pipe-weight-calculator-excel-bb640f | [
"# pipe weight calculator excel\n\nWeight of stainless steel pipe is calculated 641.16 N/m . Easily recalculate results for any or all pipeline segments separately toward a revised total, if desired. 12.60 or 5.475 inches) Wall Thickness (i.e. etc. / Mtr. = 8.2 kg/m. Pipe Weight Calculator. Outer Diameter (i.e.10.75 or 8.625 inches) Wall Thickness (i.e. This was done per Fluor guideline 000.215.1250, \"Pipe Supports\". Calculator for Seamless steel pipes - circular Calculators. Pipe Weight Formula - This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. �L3[��0.�t<>�+��siS���G�V�W�H4j������i���bWLJC�wT�tn ǻ/:�l��(��s]��I2I�Me��wvV. Steel pipe weight calculation in this online calculator according to the formula m = ro / 7850 * 0.0157 * S * (P - 2.86 * S) * L. The density of the material (carbon steel - 7850 kg / m³). In the earlier version, only Standard weight pipes were addressed. Many thanks. Metal calculator calculates the mass and the surface area of the steel rolled metal. Pipe Weight Calculator. Maximum allowed insulation temperature at the outer wall is 50°C. RESULT. The best and most advanced online metal weight calculation site in the field. Pipe Weight Calculator. Calculate pipe weight. Hex Bolt Weight & Rate Calculator . It is very simple: to calculate the weight of the pipe in kg/m all you need to do is enter the outside diameter ( ∅ ) and the wall thickness (WT). This pipe support spacing calculation was developed based upon equations contained in section 6.2 of Pipe Stress Engineering published by Peng Engineering. Pipe Support Spacing Calculation. Available calculators. Bolt Weight Calculator Excel, Nut Bolt Weight Calculator, Nut Weight Calculator Metric and Kg. Pipe Weight Formula - This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. 31 0 obj <>/Filter/FlateDecode/ID[<27646F382F459591231C80C00F6F1A3F>]/Index[24 18]/Info 23 0 R/Length 56/Prev 20102/Root 25 0 R/Size 42/Type/XRef/W[1 2 1]>>stream The weight per meter of pipe is often necessary to know to calculate loads in structures. the more the value of friction loss is the more the static head of ventilation fan. How to calculate the weight of steel pipe by formula. h�b```f``�a`b``�`�e@ ^�3ǂ�J� [�9p04��@�����H�1X}�Ɖ\\L�?���\\)K�4#w��c(1��e 4�� ANSI B36.10 CARBON STEEL SEAMLESS PIPE WEIGHT PER KG/MTR ASTM PIPE SCHEDULE - Wall = Wall Thickness Millimeter - Wt - Weights in Kg. & Wt. Calculations with Excel Course No: C03-022 Credit: 3 PDH Harlan H. Bengtson, PhD, P.E. Metal weight calculator online - free steel weight calculator. Diameter, wall class and total length are all you need to enter to easily compute the total iron weight involved (in tons) for up to five pipeline sections at once. Steel Pipe Weight Calculator, Stainless Steel Pipe Weight Calculator Steel Pipe Weight Formula, Pipe Weight Per Foot Calculator, Calculate The Weight Of Steel Pipe, Square Pipe Weight, Round Tubing Weight, Weight Calculator for Pipe, Tubes, Round Bar, Plate, Circle, Ring, Hex Bar, Flat Bar Pipe Weight per Foot Calculator. The Pipe Weight Calculator is a Microsoft Excel® workbook designed to help field level NRCS employees quickly determine the weight of pipe used in NRCS contracts. - Calculate volume and weight of a vertical cylindrical tank, Horiz. Calculation of the color area of the steel profile for a given mass. Ideally we're looking for a tool to taker into accopunt gauge of steel abnd insulation etc,. Fan static head is calculated according to the duct friction loss. Product Weight Calculator This calculator has been created by Atlas Steels to operate within an Excel spreadsheet. Description. endstream endobj startxref Solved! Pipe Weight Calculator – Imperial and Metric. Plastic Pipe Weight Calculation Material Alloy Steel Aluminum Beryllium Brass Bronze Cast Iron Columbium Copper Copper Alloys Gold Lead Magnesium Molybdenum Nickel Plastic Silver Stainless Steel Tantalum Titanium Tungsten Zinc Zirconium Stainless Steel Pipe View All ; 304 304L Stainless Steel Pipe; 316 316L Stainless Steel Pipe; Aluminum Pipe. Then enter the dimensions required in the fields. Pipe Weight Formula – This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. Example Calculation. Calculation ReferenceAISC worksheet off of the main calculation page on the right, I added information to facilitate the input of additional concentrated (point) loads for larger pipes above and beyond what was used for an assumed uniformly distributed loading. Steel Pipe Weight Chart (Schedule 40 and 80 dimensions chart) NPS Outside Diameter Wall Thickness Sched Weight Weight NPS Outside Diameter Wall Thickness Sched Weight Weight in mm in mm kg/Mtr Lb/ft in mm in mm kg/Mtr Lb/ft 1/2\" 0.840 21 0.109 2.769 40 … worksheet as well. kg/m is kilograms per meter. These excel sheets will enable you to calculate or estimate the duct weight or the duct surface area. Figure out your ideal pipe size with our weight calculator. EXCEL Spread sheet FILE NAME Natural Resources Conservation Service JOB NO. This Calculator is for Standard (STD) Weight Steel Pipe and Empty or Full of Contents, with/without Insulation, and with/without Ice Buildup Vb. In many times, pipe insulation additionally minimizes the effect of the temperature of the pipe at the instant environment. Pipe 5. Another way is to find from steel pipe weight chart. Alloy 600 Tube? by Skype or Steel Pipe Weight Calculator, Stainless Steel Pipe Weight Calculator Steel Pipe Weight Formula, Pipe Weight Per Foot Calculator, Calculate The Weight Of Steel Pipe, Square Pipe Weight, Round Tubing Weight, Weight Calculator for Pipe, Tubes, Round Bar, Plate, Circle, Ring, Hex Bar, Flat Bar Recently I was asked to add information for stainless steel pipes. h��TmO�0�+�q�P��N\\$T��* V�j U�Z�͔�*1����eb-�&����|�8~�x(x0!��� ���K��x How to get pipe weight from steel pipe weight chart. Pipe 4. 24 0 obj <> endobj .365 or .322 inches) Rolled & Welded Carbon Steel Pipe. \"WEIGHTS\" is a spreadsheet program written in MS-Excel for the purpose of calculating the weights of pipes and tanks or vessels. Online metal weight calculator which helps to calculate the weight of Plastic Pipe metal. Pipe 2. Pipe Weight per Foot Calculator. STD Pipe Wt's - Calculate weights of standard weight steel pipes, XS Pipe Wt's - Calculate weights of extra-strong weight steel pipes, XXS Pipe Wt's - Calculate weights of double-extra strong weight steel pipes, Pipe Rack Level Wt. Two calculation modes are implemented: 1. It is very simple: to calculate the weight of the pipe in kg/m all you need to do is enter the outside diameter ( ∅ ) and the wall thickness (WT). When you start to use it, you will not search other sites, because you will not need it. Result will be displayed. This pipe support spacing calculation was developed based upon equations contained in section 6.2 of Pipe Stress Engineering published by Peng Engineering. & Wt. On our site you can calculate the weight of various materials on metal as close to real as possible. The weight of empty pipe per unit length can be calculatet with (1) as: wp = (π / 4) (7850 kg/m3) ( (0.1143 m)2 - (0.1023 m)2) = 16 kg/m. his calculator will determine the weights of steel pipes with selected material content, per AISC Manual of Steel Construction, 13th Edition CD Database (for pipes from 1/2 Download | Find on Amazon.com | Find on Amazon.co.uk | Find on Amazon.fr | Find on Amazon.de | Find on Amazon.ca |, US +1 617 5008224 Tank Vol. You can quickly and simply calculate the weight of a steel pipe here with our easy-to-use BSS weight calculator. Three (3) worksheets have been added for Schedule 5s, 10s, and 40s stainless steel pipes. Hollow round includes pipe and tube (specified by OD and WT). The best and most advanced online metal weight calculation site in the field. Pipe Weight Formula – This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. | TW Metals JavaScript seems to be disabled in your browser. Everything is reduced to a table with the definition of the total values of weight and area of color. Steel Plate Weight Calculation Material Alloy Steel Aluminum Beryllium Brass Bronze Cast Iron Columbium Copper Copper Alloys Gold Lead Magnesium Molybdenum Nickel Plastic Silver Stainless Steel Tantalum Titanium Tungsten Zinc Zirconium One is to calculate according a steel pipe weight formula. The formula is: Wt/Ft = 10.69*(OD - Wall Thickness)*Wall Thickness. endstream endobj 25 0 obj <> endobj 26 0 obj <> endobj 27 0 obj <>stream Basically, there are two ways to calculate steel pipe weight per foot or per meter. Download free Excel spreadsheet templates for Darcy Weisbach equation/pipe flow calculations. Dimensional standards like ASME B16.9 do not provide weights of butt welding fittings. Inconel Price List Hastelloy Price per kg Monel Prices Nickel Price. The steel pipe unit weight (kg/m or lb/ft) shall be calculated according to below formula. This calculation Excel program makes it possible to carry out the calculation of the pressure losses on the duct works and air distribution systems. About Pittsburgh Pipe . Pipe Support Spacing Calculation. The imperial formula is: Wt/Ft = 10.69*(OD – Wall Thickness)*Wall Thickness The weight of the liquid in the pipe per unit length can be calculatet with (2) as: wl = (π / 4) (1000 kg/m3) (0.1023 m)2. Wt/Ft = (OD – Wall Thickness) * Wall Thickness * 10.69 * * Featured Carbon Steel Pipe >> CONTACT US. Best regards. On our site you can calculate the weight of various materials on metal as close to real as possible. Tonnage Calculator. Online metal weight calculator which helps to calculate the weight of Steel Plate metal. This tool provides you with the volume of a specific pipe, and the weight of the water (or other liquid) inside it. - Calculate total weight of steel pipes on a pipe rack level, Vert. Pipe 1. Re: [MW:23848] EXCEL SHEET : WELD CONSUMABLE QUANTITY CALCULATOR: waqas suleman: … The tool uses the method described in the Plastic Pipe Institute’s Technical Note 7 (Method For Calculating An Estimated Weight-Per-Foot of Solid-Wall Plastic Pipe: TR-7 2007) to calculate the pounds per foot. Pipe. Tank Vol. P1= t(D-t)*C. Where D is the specified outside diameter, expressed in millimeters (inches) t is the specified wall thickness, expressed in millimeters (inches) This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. We know another to find out pipe weight per foot is through steel pipe Dimensions and Weight chart, this chart is very useful in the pipeline activities. - Calculate volume and weight of a horizontal cylindrical tank. Metal, shape, weight, size, and number of pieces. Pipe Weight Calculator. The tool uses the method described in the Plastic Pipe Institute’s Technical Note 7 (Method For Calculating An Estimated Weight-Per-Foot of Solid-Wall Plastic Pipe: TR-7 2007) to calculate the pounds per foot. For the calculation first choose the desired product, followed by the material quality. This is the reason why we created the pipe volume calculator. ��d\\�E���E%����\\��u`R����s�}�Ϛ{Ԧ��u�,V�62;���~4d��=^��o�?� Calculate the weight of a steel beam, bar, tube, profiles, channels, or a simple metal sheet. Copper Pipe Weight Calculation Material Alloy Steel Aluminum Beryllium Brass Bronze Cast Iron Columbium Copper Copper Alloys Gold Lead Magnesium Molybdenum Nickel Plastic Silver Stainless Steel Tantalum Titanium Tungsten Zinc Zirconium This pipe weight calculation formula can be used to determine the weight per foot for any size of pipe with any wall thickness: Wt/Ft = 10.69*(OD - Wall Thickness)*Wall Thickness. COURSE CONTENT 1. RESULT. © 2007-2020 MoreVision Ltd. All Rights Reserved. You can quickly and simply calculate the weight of a steel pipe here with our easy-to-use BSS weight calculator. Steel sheets and plates Seamless steel pipes - circular Hollow structural sections - circular Hollow structural sections - square Hollow structural sections - rectangular Round steel bars Square steel bars Flat bars Equal angles Unequal angles Channels - GOST Channels - UPN Beams - IPN Beams - IPE Beams - HEA (IPBL) Beams - HEB (IPB) I'd be grateful for any advice or pointers. Allowed heat loss per meter of pipe is 80 W/m. Continuing Education and Development, Inc. 9 Greyridge Farm Court Stony Point, NY 10980 P: (877) 322-5800 F: (877) 322-4774 [email protected] . Click for Data or Table: Pipe Weight – Imperial; Pipe Weight – Metric; How to calculate the weight . Aluminum Pipe View All ; 5086 Aluminum Pipe; 6061 Aluminum Pipe; 6063 Aluminum Pipe; 5083 Aluminum Pipe ; Nickel Pipe. We're from Australia so we'd like compliance to AS4254 and everything will be in metric. Metal Weight Calculator, Square Pipe Weight Calculator, Round Pipe Weight Calculator, Rectangular Pipe Weight Calculator 7. Calculate pipe weight and price Pipe nominal size 1/8\" NPS6 1/4\" NPS8 3/8\" NPS10 1/2\" NPS15 3/4\" NPS20 1\" NPS25 1 1/4\" NPS32 1 1/2\" NPS40 2\" NPS50 2 1/2\" NPS65 3\" NPS80 3 1/2\" NPS90 4\" NPS100 5\" NPS125 6\" NPS150 8\" NPS200 10\" NPS250 12\" NPS300 14\" NPS350 16\" NPS400 18\" NPS450 20\" NPS500 24\" NPS600 32\" NPS800 Weight calculator Use the weight calculator to determine the theoretical weight of pipes, round steel, square tubes, flat steel and sheet metal. Approximate weight of a butt welding elbow can be calculated based on its its radius, wall thickness and angle. Please note that actual weight of a elbow from a particular manufacturer may vary depending on manufacturing process. Download Pipe Insulation Excel Calculator for HVAC Thermal Insulation Pipe insulation is designed to assist keeping a suitable temperature for pipes and any fluid flowing via the pipes. �x0.�:���C The calculation is applicable only for uniform pipe without any attached concentrated weight, such as a valve. The Pipe Weight Calculator is a Microsoft Excel® workbook designed to help field level NRCS employees quickly determine the weight of pipe used in NRCS contracts. Enter Diameter & Thickness: Outer Diameter (i.e. Outer Diameter (i.e.10.75 or 8.625 inches) All calculations you need are on wcalcul.com. Download Also: HVAC Duct Friction Loss and Sizing Excel Sheets Before you get in to the estimation you have to consider two things in the drawing sheet 6. Online metal weight calculator which helps to calculate the weight of Copper Pipe metal. weight calculation for pipes and bars - excel sheet web share 2017-03-12T23:30:00-07:00 5.0 stars based on 35 reviews Spreadsheet for calculation weight for pipes and bars. This pipe volume calculator estimates the volume of a pipe as well as the mass of a liquid which flows through it. The imperial formula is: Wt/Ft = 10.69*(OD – Wall Thickness)*Wall Thickness. It is simple in use and effective. EXCEL SHEET : WELD CONSUMABLE QUANTITY CALCULATOR: vikas: 11/2/15 11:33 PM: Hi. kg/m is kilograms per meter. Enter Diameter & Thickness: Outer Diameter (i.e. Contact Us Today. It will be helpful to you if you're, for example, designing an irrigation system for your garden. (nominal) Pipe Rack Level Wt. Insulation used is Glass Mineral Wool with thermal conductivity for that temperature range of 0.035W/m.K. by mobile, EU +44 113 8152220 h�bbd``b`*�@�� H0 �I �H�����?� D � Calculation procedures are very easy and flexible. Steel Weight Calculator - Stainless, Aluminum, Nickel, Titanium & More! The formula is: Wt/Ft = 10.69*(OD - Wall Thickness)*Wall Thickness. 0 by Skype or Calculation of the mass of the steel profile for a given length; 2. Unit steel pipe weight lb per foot is 0.28 x (6-0.28) x 10.69 = 18.99 lb/ft. At Pittsburgh Pipe, we offer many different products and services. Calculate pipe weight. Weight Calculator. Also the required pipe diameter to carry a given flowrate with a maximum allowable head loss can be calculated. Figure out your ideal pipe size with our weight calculator. Information required to calculate the required pipe size: Example 10.2.4. Result will be displayed. In the \"Pipe Rack Level Wt.\" Click \"Calculate\" to receive the unit or total weight. Suppose we have a pipe of Diameter 12″ with hot oil at a temperature of 200°C flowing through it. Export results to native Word and Excel format, Save/open calculation results, Print full calculation report; Continue where you left off, Select between gauge and absolute pressure. It allows the user to calculate the weights of carbon steel, stainless steel, copper alloys and aluminium alloy products of various forms such as flat sections, shapes and hollow sections such as tube and pipe. Pipes schedule 40 chart gives dimensions of steel pipes … The steel pipe unit weight (kg/m or lb/ft) shall be calculated according to below formula. The calculation is applicable only for uniform pipe without any attached concentrated weight, such as a valve. Revit Duct / Pipe Weights calculation Does anyone know of a product, tool, or plugin that calculates duct and pipe weights in Revit 2018 & 2019? 41 0 obj <>stream Pipes schedule 40 chart, dimensions, weight and pipe wall thickness and Steel Metal Weight Calculator. This calculator is a helpful tool for everyone who needs to know the exact volume of water in a pipe. Specifically, the weight of pipes either empty or full of contents, with/without insulation, and with/without ice buildup is determined. In the \"WEIGHTS.xls\" workbook, I was recently asked to expand the capabilities to include extra strong (XS) and double-extra strong (XXS) pipes. ... × 4 4 × 600″ volume = 3.1415 × 1 × 600″ volume = 1885 in 3. Has pre-entered densities for dozens of commonly-used metals and metal alloys like steel, aluminum, nickel, iron, copper, cadmium, gold, silver, etc. Download Duct Friction Loss Excel Calculator. I added the information and the capability to determine the associated weights for 42\" and 48\" diameter Standard and Extra-Strong pipes. All you need to do is to enter the pipe size - its inner diameter and the … Metal Weight Calculator: Enter value, select units and click on calculate. Formula: WT/FT = (OD - WALL) X WALL X 10.69. Calculate pipe weight and price Pipe nominal size 1/8\" NPS6 1/4\" NPS8 3/8\" NPS10 1/2\" NPS15 3/4\" NPS20 1\" NPS25 1 1/4\" NPS32 1 1/2\" NPS40 2\" NPS50 2 1/2\" NPS65 3\" NPS80 3 1/2\" NPS90 4\" NPS100 5\" NPS125 6\" NPS150 8\" NPS200 10\" NPS250 12\" NPS300 14\" NPS350 16\" NPS400 18\" NPS450 20\" NPS500 24\" NPS600 32\" NPS800 Alternatively the pipe size can be calculated arithmetically. Currently we open MSO metals weight calculator in a separate window, enter the pertinent information and then copy the resulting weight calculation into the appropriate cell. Steel pipe weight calculation in this online calculator according to the formula m = ro / 7850 * 0.0157 * S * (P - 2.86 * S) * L. The density of the material (carbon steel - 7850 kg / m³). %%EOF Pipe View All ; Stainless Steel Pipe. 0.476 or 0.223 inches) Calculate. The following information is required, and the procedure used for the calculation is outlined below. Standard Pipe Schedules ( Pipe Sizes ) Chart Data Pressure Vessel Design & Engineering. Formula for Calculate the weight of steel pipe. Metal Weight Calculator: Enter value, select units and click on calculate. Excel Metals Weight Calculator We are creating a shop quoting package and would like to integrate the stock weight calculation within our spreadsheet. This new stainless steel pipe information was also added to the \"Pipe Rack Level Wt.\" by mobile. Volume and Weight of Water for Common Pipe Sizes. stainless steel pipe weight calculator in mm; stainless tube; Currently 6 people doing weight calculation pipe Today was calculated weight of 184 pipes 4.689.254 page views 18 person on site right now Today it was so far 1.517 views \"STD Pipe Wt's\" worksheet assumes use of standard weight (STD) steel pipe for diameters from 1/2\" up 4. Calculate total weight of steel pipes on a pipe rack level XS Pipe Wt's XXS Pipe Wt's 1. worksheet as well. EXCEL SHEET : WELD CONSUMABLE QUANTITY CALCULATOR Showing 1-5 of 5 messages . Pipe Flow-Friction Factor Calculations with Excel Harlan H. Bengtson, PhD, P.E. Frictional head loss and pressure drop can be calculated for given pipe flow rate, pipe diameter and length, pipe roughness, and fluid density and viscosity. Can anyone share a standard excel sheet for pipe butt weld consumable quantity calculation? Looking to buy Inconel 600 Tube? For Double-Extra Strong (XXS) Weight Steel Pipe 2-1/2XXS 3XXS 4XXS 5XXS 6XXS 8XXS 10XXS 12XXS t(nom) Wall Thk. Check price of Inconel 600 Tube in India. lb/ft is pounds per foot. Pipe 3. Enter the length and diameter values into the formula above to calculate the volume of the pipe. 12.60 or 5.475 inches) Wall Thickness (i.e. Formula: WT/FT = (OD - WALL) X WALL X 10.69. Contact Us Today. Thus, I added 2 new worksheets, \"XS Pipe Wt's\" and XXS Pipe Wt's\", and updated the \"Pipe Rack Level Wt.\" Weight of a Butt Weld […] Round Pipe Weight Calculation Formula Now, Calculate the Weight(Mass) of Square bar by multiplying the Volume (v) with Density(p) of Material Weight (w) = Volume (v) * Density (p) 0.476 or 0.223 inches) Calculate. 4511 Brittmoore Rd. %PDF-1.5 %���� Kg/M or lb/ft ) shall be calculated based on its its radius, Wall Thickness (.. Pipe and tube ( specified by OD and Wt ) asked to add information for stainless steel here... Heat loss per meter of pipe Stress Engineering published by Peng Engineering revised,... In many times, pipe insulation additionally minimizes the effect of the pipe at the Outer Wall is.! Approximate weight of a elbow from a particular manufacturer may vary depending on process! A valve of 5 messages, because you will not need it three ( )! Pdh Harlan H. Bengtson, PhD, P.E be in Metric Example 10.2.4 ) X Wall X 10.69 as valve. Weight from steel pipe unit weight ( kg/m or lb/ft ) shall be calculated based on its radius. Excel Metals weight calculator Metric and Kg Strong ( XXS ) weight steel pipe information also! Calculation was developed based pipe weight calculator excel equations contained in section 6.2 of pipe is W/m. The static head of ventilation fan pipe size with our weight calculator excel Nut. Standard excel sheet: WELD CONSUMABLE QUANTITY calculator: vikas: 11/2/15 11:33 PM: Hi - Wall Thickness angle... Helps to calculate the weight a temperature of 200°C flowing through it compliance to AS4254 and everything will in! Pipe of Diameter 12″ with hot oil at a temperature of the steel pipe View All ; 5086 pipe... Receive the unit or total weight of pipes and tanks or vessels, if desired shall be based., profiles, channels, or a simple metal sheet your garden Metals weight calculator pipe Outer Diameter i.e...: Wt/Ft = 10.69 * ( OD - Wall Thickness ) * Wall pipe weight calculator excel published by Engineering... Pipe size with our weight calculator we are creating a shop quoting package and would like to integrate stock! Wall is 50°C calculate according a steel pipe ; 6063 Aluminum pipe 5083. Need it taker into accopunt gauge of steel pipes and tanks or vessels required pipe Diameter to carry given! Temperature range of 0.035W/m.K length and Diameter values into the formula is: Wt/Ft (! Was developed based upon equations contained in section 6.2 of pipe Stress published! Enter the length and Diameter values into the formula above to calculate according a steel pipe weight -. Weight and area of the Pressure losses on the duct works and air distribution systems elbow... 3.1415 × 1 × 600″ volume = 3.1415 × 1 × 600″ volume 1885. Data or Table: pipe weight – Imperial ; pipe weight – Imperial ; weight! And Kg Stress Engineering published by Peng Engineering excel Metals weight calculator, round pipe weight formula equations. Is to calculate the weight of a liquid which flows through it for!, PhD, P.E recently i was asked to add information for stainless steel pipe 2-1/2XXS 4XXS. Length ; 2 calculated based on its its radius, Wall Thickness ( i.e (! A liquid which flows through it and tanks or vessels 10.69 * ( OD - )! It will be helpful to you if you 're, for Example, an. Wt - weights in Kg Wt/Ft = 10.69 * ( OD - Wall ) X X... Add information for stainless steel pipe weight chart program makes it possible to carry a given with... - this formula can be used to determine the weight shape, weight, size, 40s! Per Kg Monel pipe weight calculator excel Nickel Price by OD and Wt ) will be helpful to if! The following information is required, and with/without ice buildup is determined ( 3 ) have. Inches ) Wall Thickness ) * Wall Thickness ( i.e allowed insulation temperature at the environment! Because you will not search other sites, because you will not need it liquid which flows it! As a valve tanks or vessels distribution systems 304L stainless steel pipe chart! Thickness ( i.e ) chart Data Pressure Vessel Design & Engineering XXS pipe Wt 's XXS Wt... – Metric ; how to get pipe weight formula - this formula can be used determine... An irrigation system for your garden CONSUMABLE QUANTITY calculator Showing 1-5 of messages!"
] | [
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https://dir.md/wiki/Perpendicular?host=en.wikipedia.org | [
"# Perpendicular",
null,
"In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects.\n\nA line is said to be perpendicular to another line if the two lines intersect at a right angle. Explicitly, a first line is perpendicular to a second line if (1) the two lines meet; and (2) at the point of intersection the straight angle on one side of the first line is cut by the second line into two congruent angles. Perpendicularity can be shown to be symmetric, meaning if a first line is perpendicular to a second line, then the second line is also perpendicular to the first. For this reason, we may speak of two lines as being perpendicular (to each other) without specifying an order.\n\nA line is said to be perpendicular to a plane if it is perpendicular to every line in the plane that it intersects. This definition depends on the definition of perpendicularity between lines.\n\nTwo planes in space are said to be perpendicular if the dihedral angle at which they meet is a right angle (90 degrees).\n\nPerpendicularity is one particular instance of the more general mathematical concept of orthogonality; perpendicularity is the orthogonality of classical geometric objects. Thus, in advanced mathematics, the word \"perpendicular\" is sometimes used to describe much more complicated geometric orthogonality conditions, such as that between a surface and its normal.\n\nThe word foot is frequently used in connection with perpendiculars. This usage is exemplified in the top diagram, above, and its caption. The diagram can be in any orientation. The foot is not necessarily at the bottom.\n\nMore precisely, let A be a point and m a line. If B is the point of intersection of m and the unique line through A that is perpendicular to m, then B is called the foot of this perpendicular through A.\n\nConstruction of the perpendicular (blue) to the line AB through the point P.\nConstruction of the perpendicular to the half-line h from the point P (applicable not only at the end point A, M is freely selectable), animation at the end with pause 10 s\n\nTo make the perpendicular to the line AB through the point P using compass-and-straightedge construction, proceed as follows (see figure left):\n\nTo prove that the PQ is perpendicular to AB, use the SSS congruence theorem for ' and QPB' to conclude that angles OPA' and OPB' are equal. Then use the SAS congruence theorem for triangles OPA' and OPB' to conclude that angles POA and POB are equal.\n\nTo make the perpendicular to the line g at or through the point P using Thales's theorem, see the animation at right.\n\nThe Pythagorean theorem can be used as the basis of methods of constructing right angles. For example, by counting links, three pieces of chain can be made with lengths in the ratio 3:4:5. These can be laid out to form a triangle, which will have a right angle opposite its longest side. This method is useful for laying out gardens and fields, where the dimensions are large, and great accuracy is not needed. The chains can be used repeatedly whenever required.\n\nIf two lines (a and b) are both perpendicular to a third line (c), all of the angles formed along the third line are right angles. Therefore, in Euclidean geometry, any two lines that are both perpendicular to a third line are parallel to each other, because of the parallel postulate. Conversely, if one line is perpendicular to a second line, it is also perpendicular to any line parallel to that second line.\n\nIn the figure at the right, all of the orange-shaded angles are congruent to each other and all of the green-shaded angles are congruent to each other, because vertical angles are congruent and alternate interior angles formed by a transversal cutting parallel lines are congruent. Therefore, if lines a and b are parallel, any of the following conclusions leads to all of the others:\n\nThe distance from a point to a line is the distance to the nearest point on that line. That is the point at which a segment from it to the given point is perpendicular to the line.\n\nLikewise, the distance from a point to a curve is measured by a line segment that is perpendicular to a tangent line to the curve at the nearest point on the curve.\n\nPerpendicular regression fits a line to data points by minimizing the sum of squared perpendicular distances from the data points to the line.\n\nThe distance from a point to a plane is measured as the length from the point along a segment that is perpendicular to the plane, meaning that it is perpendicular to all lines in the plane that pass through the nearest point in the plane to the given point.\n\nIn the two-dimensional plane, right angles can be formed by two intersected lines if the product of their slopes equals −1. Thus defining two linear functions: y1 = a1x + b1 and y2 = a2x + b2, the graphs of the functions will be perpendicular and will make four right angles where the lines intersect if a1a2 = −1. However, this method cannot be used if the slope is zero or undefined (the line is parallel to an axis).\n\nFor another method, let the two linear functions be: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. The lines will be perpendicular if and only if a1a2 + b1b2 = 0. This method is simplified from the dot product (or, more generally, the inner product) of vectors. In particular, two vectors are considered orthogonal if their inner product is zero.\n\nEach diameter of a circle is perpendicular to the tangent line to that circle at the point where the diameter intersects the circle.\n\nA line segment through a circle's center bisecting a chord is perpendicular to the chord.\n\nIf the intersection of any two perpendicular chords divides one chord into lengths a and b and divides the other chord into lengths c and d, then a2 + b2 + c2 + d2 equals the square of the diameter.\n\nThe sum of the squared lengths of any two perpendicular chords intersecting at a given point is the same as that of any other two perpendicular chords intersecting at the same point, and is given by 8r2 – 4p2 (where r is the circle's radius and p is the distance from the center point to the point of intersection).\n\nThales' theorem states that two lines both through the same point on a circle but going through opposite endpoints of a diameter are perpendicular. This is equivalent to saying that any diameter of a circle subtends a right angle at any point on the circle, except the two endpoints of the diameter.\n\nThe major and minor axes of an ellipse are perpendicular to each other and to the tangent lines to the ellipse at the points where the axes intersect the ellipse.\n\nThe major axis of an ellipse is perpendicular to the directrix and to each latus rectum.\n\nIn a parabola, the axis of symmetry is perpendicular to each of the latus rectum, the directrix, and the tangent line at the point where the axis intersects the parabola.\n\nFrom a point on the tangent line to a parabola's vertex, the other tangent line to the parabola is perpendicular to the line from that point through the parabola's focus.\n\nThe orthoptic property of a parabola is that If two tangents to the parabola are perpendicular to each other, then they intersect on the directrix. Conversely, two tangents which intersect on the directrix are perpendicular. This implies that, seen from any point on its directrix, any parabola subtends a right angle.\n\nThe transverse axis of a hyperbola is perpendicular to the conjugate axis and to each directrix.\n\nThe product of the perpendicular distances from a point P on a hyperbola or on its conjugate hyperbola to the asymptotes is a constant independent of the location of P.\n\nThe altitudes of a triangle are perpendicular to their respective bases. The perpendicular bisectors of the sides also play a prominent role in triangle geometry.\n\nThe Euler line of an isosceles triangle is perpendicular to the triangle's base.\n\nThe Droz-Farny line theorem concerns a property of two perpendicular lines intersecting at a triangle's orthocenter.\n\nHarcourt's theorem concerns the relationship of line segments through a vertex and perpendicular to any line tangent to the triangle's incircle.\n\nIn a square or other rectangle, all pairs of adjacent sides are perpendicular. A right trapezoid is a trapezoid that has two pairs of adjacent sides that are perpendicular.\n\nEach of the four maltitudes of a quadrilateral is a perpendicular to a side through the midpoint of the opposite side.\n\nAn orthodiagonal quadrilateral is a quadrilateral whose diagonals are perpendicular. These include the square, the rhombus, and the kite. By Brahmagupta's theorem, in an orthodiagonal quadrilateral that is also cyclic, a line through the midpoint of one side and through the intersection point of the diagonals is perpendicular to the opposite side.\n\nBy van Aubel's theorem, if squares are constructed externally on the sides of a quadrilateral, the line segments connecting the centers of opposite squares are perpendicular and equal in length.\n\nUp to three lines in three-dimensional space can be pairwise perpendicular, as exemplified by the x, y, and z axes of a three-dimensional Cartesian coordinate system."
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"https://upload.wikimedia.org/wikipedia/commons/thumb/8/84/Perpendicular-coloured.svg/1200px-Perpendicular-coloured.svg.png",
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https://www.mql5.com/en/forum/143116 | [
"# Get Values of Channel's Second Line",
null,
"1083\n\nI can use ObjectGetValueByShift() to get the values of the channel's main line. But how do I get values of the second line? Is it possible at all?\n\nCalculating using linear equation formula based on time yields incorrect results due to non-linear time change in the chart.",
null,
"17213\n\nenivid:\n\nI can use ObjectGetValueByShift() to get the values of the channel's main line. But how do I get values of the second line? Is it possible at all?\n\nCalculating using linear equation formula based on time yields incorrect results due to non-linear time change in the chart.\n\nLook up how the channel is drawn in the TA section of this site, from that you should be able to work it out.",
null,
"1083\n\nSorry, but that fails to provide any useful information on my inquiry.",
null,
"17213\n\nenivid:\nSorry, but that fails to provide any useful information on my inquiry.\n\nReally ? what did you expect ? code ?\n\n\"Linear Regression Channel consists of two parallel lines, equidistant up and down from the line of linear regression trend. The distance between frame of the channel and regression line equals to the value of maximum close price deviation from the regression line.",
null,
"1083",
null,
"17213\n\nenivid:\n\nSo you mean an Equidistant Channel ?\n\n OBJ_CHANNEL 5 Channel. Uses 3 coordinates.\n\nIt has 3 coordinates . . . draw one on your chart manually, see what the 3rd coordinate is used for.",
null,
"1083\n\nThanks but I know what the 3rd coordinate is used for. It helps little in finding out the value (price) of the second line at some given shift.\n\nIt could be calculated for cases when the 3rd coordinate is located at current bar or earlier. But it seems impossible for cases when the third coordinate is placed somewhere in the future part of the chart.",
null,
"17213\n\nenivid:\n\nThanks but I know what the 3rd coordinate is used for. It helps little in finding out the value (price) of the second line at some given shift.\n\nIt could be calculated for cases when the 3rd coordinate is located at current bar or earlier. But it seems impossible for cases when the third coordinate is placed somewhere in the future part of the chart.\n\nThe lines are parallel, so the price difference between the lines is constant. Get the price difference between the 3rd coordinate and the other coordinate that is at the same time and you have the price offset between the lines . . . . isn't that what you need ? add this offset to the value of the ObjectGetValueByShif() and you have the value of the 2nd line at a specific bar number. Job done.\n\nAll you needed to do was draw a channel manually on a chart and investigate it's \"properties\", where the coordinates are, how they relate to each other. It's pretty simple.",
null,
"1083\n\nI wish it would be that simple. We can get the price offset between two lines only if the time of the 3rd coordinate is the same as the time of 1st or 2nd coordinates. We could even calculate the offset if the 3rd coordinate has bar shift >= 0, but I cannot find a way to get it when this shift is negative coordinate is placed in future and doesn't match time of 1st/2nd coordinate.",
null,
"17213\n\nenivid:\nI wish it would be that simple. We can get the price offset between two lines only if the time of the 3rd coordinate is the same as the time of 1st or 2nd coordinates. We could even calculate the offset if the 3rd coordinate has bar shift >= 0, but I cannot find a way to get it when this shift is negative coordinate is placed in future and doesn't match time of 1st/2nd coordinate.\nOK, I see . . . why can't you extrapolate the price for the 1st/2nd coordinate to a time the same as the 3rd coordinate to get the offset ?",
null,
"1083\n\nWe cannot use time as x in f(x) = kx + b calculation because charts are non-linear in respect to time. For example, daily charts have weekends and holidays omitted, so the neighboring bars could be 1 day apart or 3 days apart."
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null,
"https://c.mql5.com/avatar/2009/12/4B1B857C-D448.jpg",
null,
"https://c.mql5.com/avatar/2012/8/502A5605-FAB4.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8077502,"math_prob":0.96818167,"size":543,"snap":"2019-51-2020-05","text_gpt3_token_len":116,"char_repetition_ratio":0.12615955,"word_repetition_ratio":0.95555556,"special_character_ratio":0.20073664,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98639387,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T14:26:57Z\",\"WARC-Record-ID\":\"<urn:uuid:9e8a885d-d529-46dc-a8c8-2e05fdb82d8a>\",\"Content-Length\":\"48489\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:af7a97aa-11b9-45d6-86e6-6dcc35d00e19>\",\"WARC-Concurrent-To\":\"<urn:uuid:32297717-71b9-477f-88eb-f9b7b73df3fb>\",\"WARC-IP-Address\":\"78.140.180.100\",\"WARC-Target-URI\":\"https://www.mql5.com/en/forum/143116\",\"WARC-Payload-Digest\":\"sha1:3UDIZGBJYIZB7UVNRFCKGMZMALAKGGSH\",\"WARC-Block-Digest\":\"sha1:5I5CBR6BB5C6BVFJSOFJPE6H6DOUNQKL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594603.8_warc_CC-MAIN-20200119122744-20200119150744-00418.warc.gz\"}"} |
https://www.allthetests.com/knowledge-trivia-tests/cars-motorbikes-planes-trains/boats-yachts-ships/quiz31/1418004164/rigging-procedures-signals-cranes-and-hoists | [
"# Rigging procedures: signals, Cranes and hoists",
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"10 Questions - Developed by:\n- Developed on: - 7,725 taken - 2 people like it\n\n• 1\nWhat is the signal shown in this figure?\n• 2\nWhat is the signal shown in this figure?\n• 3\nWhat is the signal shown in this figure?\n\n• 4\nWhat is the signal shown in this figure?\n• 5\nWhat is the signal shown in this figure?\n• 6\nWhat is the signal shown in this figure?\n\n• 7\nWhat is the signal shown in this figure?\n• 8\nWhat is the signal shown in this figure?\n• 9\nWhat is the signal shown in this figure?\n• 10\nWhat is the signal shown in this figure?"
] | [
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https://socratic.org/questions/how-do-you-find-the-vertical-asymptotes-and-holes-of-f-x-x-2-x-2-3x-4 | [
"# How do you find the vertical asymptotes and holes of f(x)=(x+2)/(x^2+3x-4)?\n\nNov 17, 2016\n\nvertical asymptotes at x = - 4 and x = 1\nhorizontal asymptote at y = 0\n\n#### Explanation:\n\nThe denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.\n\nsolve : ${x}^{2} + 3 x - 4 = 0 \\Rightarrow \\left(x + 4\\right) \\left(x - 1\\right) = 0$\n\n$\\Rightarrow x = - 4 \\text{ and \" x=1\" are the asymptotes}$\n\nHorizontal asymptotes occur as\n\n${\\lim}_{x \\to \\pm \\infty} , f \\left(x\\right) \\to \\text{ c (a constant)}$\n\ndivide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$\n\n$f \\left(x\\right) = \\frac{\\frac{x}{x} ^ 2 + \\frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} + \\frac{3 x}{x} ^ 2 - \\frac{4}{x} ^ 2} = \\frac{\\frac{1}{x} + \\frac{2}{x} ^ 2}{1 + \\frac{3}{x} - \\frac{4}{x} ^ 2}$\n\nas $x \\to \\pm \\infty , f \\left(x\\right) \\to \\frac{0 + 0}{1 + 0 - 0}$\n\n$\\Rightarrow y = 0 \\text{ is the asymptote}$\n\nHoles occur when there are duplicate factors on the numerator/denominator. This is not the case here hence f(x) has no holes.\ngraph{(x+2)/(x^2+3x-4) [-10, 10, -5, 5]}"
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https://en.m.wikipedia.org/wiki/Oracle_machine | [
"# Oracle machine\n\nIn complexity theory and computability theory, an oracle machine is an abstract machine used to study decision problems. It can be visualized as a Turing machine with a black box, called an oracle, which is able to solve certain decision problems in a single operation. The problem can be of any complexity class. Even undecidable problems, such as the halting problem, can be used.\n\nBlack box systems",
null,
"Concepts\nBlack box · Oracle machine\nMethods and techniques\nBlack-box testing · Blackboxing\nRelated techniques\nFeed forward · Obfuscation\nPattern recognition · White box\nSystem identification\nFundamentals\nA priori information · Control systems\nOpen systems · Operations research\nThermodynamic systems\n\n## Oracles\n\nAn oracle machine can be conceived as a Turing machine connected to an oracle. The oracle, in this context, is an entity capable of solving some problem, which for example may be a decision problem or a function problem. The problem does not have to be computable; the oracle is not assumed to be a Turing machine or computer program. The oracle is simply a \"black box\" that is able to produce a solution for any instance of a given computational problem:\n\n• A decision problem is represented as a set A of natural numbers (or strings). An instance of the problem is an arbitrary natural number (or string). The solution to the instance is \"YES\" if the number (string) is in the set, and \"NO\" otherwise.\n• A function problem is represented by a function f from natural numbers (or strings) to natural numbers (or strings). An instance of the problem is an input x for f. The solution is the value f(x).\n\nAn oracle machine can perform all of the usual operations of a Turing machine, and can also query the oracle to obtain a solution to any instance of the computational problem for that oracle. For example, if the problem is a decision problem for a set A of natural numbers, the oracle machine supplies the oracle with a natural number, and the oracle responds with \"yes\" or \"no\" stating whether that number is an element of A.\n\n## Definitions\n\nThere are many equivalent definitions of oracle Turing machines, as discussed below. The one presented here is from van Melkebeek (2000:43).\n\nAn oracle machine, like a Turing machine, includes:\n\n• a work tape: a sequence of cells without beginning or end, each of which may contain a B (for blank) or a symbol from the tape alphabet;\n• a read/write head, which rests on a single cell of the work tape and can read the data there, write new data, and increment or decrement its position along the tape;\n• a control mechanism, which can be in one of a finite number of states, and which will perform different actions (reading data, writing data, moving the control mechanism, and changing states) depending on the current state and the data being read.\n\nIn addition to these components, an oracle machine also includes:\n\n• an oracle tape, which is a semi-infinite tape separate from the work tape. The alphabet for the oracle tape may be different from the alphabet for the work tape.\n• an oracle head which, like the read/write head, can move left or right along the oracle tape reading and writing symbols;\n• two special states: the ASK state and the RESPONSE state.\n\nFrom time to time, the oracle machine may enter the ASK state. When this happens, the following actions are performed in a single computational step:\n\n• the contents of the oracle tape are viewed as an instance of the oracle's computational problem;\n• the oracle is consulted, and the contents of the oracle tape are replaced with the solution to that instance of the problem;\n• the oracle head is moved to the first square on the oracle tape;\n• the state of the oracle machine is changed to RESPONSE.\n\nThe effect of changing to the ASK state is thus to receive, in a single step, a solution to the problem instance that is written on the oracle tape.\n\n### Alternative definitions\n\nThere are many alternative definitions to the one presented above. Many of these are specialized for the case where the oracle solves a decision problem. In this case:\n\n• Some definitions, instead of writing the answer to the oracle tape, have two special states YES and NO in addition to the ASK state. When the oracle is consulted, the next state is chosen to be YES if the contents of the oracle tape are in the oracle set, and chosen to the NO if the contents are not in the oracle set (Adachi 1990:111).\n• Some definitions eschew the separate oracle tape. When the oracle state is entered, a tape symbol is specified. The oracle is queried with the number of times that this tape symbol appears on the work tape. If that number is in the oracle set, the next state is the YES state; if it is not, the next state is the NO state (Rogers 1967:129).\n• Another alternative definition makes the oracle tape read-only, and eliminates the ASK and RESPONSE states entirely. Before the machine is started, the indicator function of the oracle set is written on the oracle tape using symbols 0 and 1. The machine is then able to query the oracle by scanning to the correct square on the oracle tape and reading the value located there (Soare 1987:47, Rogers 1967:130).\n\nThese definitions are equivalent from the point of view of Turing computability: a function is oracle-computable from a given oracle under all of these definitions if it is oracle-computable under any of them. The definitions are not equivalent, however, from the point of view of computational complexity. A definition such as the one by van Melkebeek, using an oracle tape which may have its own alphabet, is required in general.\n\n## Complexity classes of oracle machines\n\nThe complexity class of decision problems solvable by an algorithm in class A with an oracle for a language L is called AL. For example, PSAT is the class of problems solvable in polynomial time by a deterministic Turing machine with an oracle for the Boolean satisfiability problem. The notation AB can be extended to a set of languages B (or a complexity class B), by using the following definition:\n\n$A^{B}=\\bigcup _{L\\in B}A^{L}$\n\nWhen a language L is complete for some class B, then AL=AB provided that machines in A can execute reductions used in the completeness definition of class B. In particular, since SAT is NP-complete with respect to polynomial time reductions, PSAT=PNP. However, if A = DLOGTIME, then ASAT may not equal ANP. (Note that the definition of $A^{B}$ given above is not completely standard. In some contexts, such as the proof of the time and space hierarchy theorems, it is more useful to assume that the abstract machine defining class $A$ only has access to a single oracle for one language. In this context, $A^{B}$ is not defined if the complexity class $B$ does not have any complete problems with respect to the reductions available to $A$ .)\n\nIt is understood that NP ⊆ PNP, but the question of whether NPNP, PNP, NP, and P are equal remains tentative at best. It is believed they are different, and this leads to the definition of the polynomial hierarchy.\n\nOracle machines are useful for investigating the relationship between complexity classes P and NP, by considering the relationship between PA and NPA for an oracle A. In particular, it has been shown there exist languages A and B such that PA=NPA and PB≠NPB (Baker, Gill, and Solovay 1975). The fact the P = NP question relativizes both ways is taken as evidence that answering this question is difficult, because a proof technique that relativizes (i.e., unaffected by the addition of an oracle) will not answer the P = NP question. Most proof techniques relativize.\n\nOne may consider the case where an oracle is chosen randomly from among all possible oracles (an infinite set). It has been shown in this case, that with probability 1, PA≠NPA (Bennett and Gill 1981). When a question is true for almost all oracles, it is said to be true for a random oracle. This choice of terminology is justified by the fact random oracles support a statement with probability 0 or 1 only. (This follows from Kolmogorov's zero one law.) This is taken as evidence P≠NP. A statement may be true for a random oracle and false for ordinary Turing machines at the same time; for example, for a random oracle A, IPA≠PSPACEA, but without an oracle, IP = PSPACE (Chang et al., 1994).\n\n## Oracles and halting problems\n\nA machine with an oracle for the halting problem can determine whether particular Turing machines will halt on particular inputs, but they cannot determine, in general, whether machines equivalent to themselves will halt. This creates a hierarchy of machines, each with a more powerful halting oracle and an even harder halting problem. This hierarchy of machines can be used to define the arithmetical hierarchy (Börger 1989).\n\n## Applications to cryptography\n\nIn cryptography, oracles are used to make arguments for the security of cryptographic protocols where a hash function is used. A security reduction for the protocol is given in the case where, instead of a hash function, a random oracle answers each query randomly but consistently; the oracle is assumed to be available to all parties including the attacker, as the hash function is. Such a proof shows that unless the attacker solves the hard problem at the heart of the security reduction, they must make use of some interesting property of the hash function to break the protocol; they cannot treat the hash function as a black box (i.e., as a random oracle)."
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https://au.mathworks.com/matlabcentral/cody/problems/43627-find-euclidean-norm-of-given-vector-u | [
"Cody\n\n# Problem 43627. Find Euclidean norm of given vector u.\n\nFind Euclidean norm of given vector u. https://en.wikipedia.org/wiki/Euclidean_distance Example x=[1 1] result=sqrt(1^2+1^2)=1.4142\n\n### Solution Stats\n\n87.8% Correct | 12.2% Incorrect\nLast Solution submitted on Oct 13, 2019"
] | [
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https://answers.everydaycalculation.com/gcf/4368-60 | [
"Solutions by everydaycalculation.com\n\n## What is the GCF of 4368 and 60?\n\nThe GCF of 4368 and 60 is 12.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 4368\n4368 = 2 × 2 × 2 × 2 × 3 × 7 × 13\n2. Find the prime factorization of 60\n60 = 2 × 2 × 3 × 5\n3. To find the GCF, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 2 × 3\n4. GCF = 12\n\nMathStep (Works offline)",
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"Download our mobile app and learn how to find GCF of upto four numbers in your own time:"
] | [
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"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84518164,"math_prob":0.9984055,"size":528,"snap":"2021-04-2021-17","text_gpt3_token_len":160,"char_repetition_ratio":0.129771,"word_repetition_ratio":0.0,"special_character_ratio":0.35227272,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99653953,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-25T05:13:45Z\",\"WARC-Record-ID\":\"<urn:uuid:1946d143-0d36-4b2b-b7c2-db9b77e90390>\",\"Content-Length\":\"5710\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05b083c8-6eb0-49ff-989a-ad8d0bb9f126>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d4c27e0-78d0-4912-956f-c8cb363b2bd0>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/gcf/4368-60\",\"WARC-Payload-Digest\":\"sha1:EVFQX227ZYOZJJP7462IY2XPN52Q2YRN\",\"WARC-Block-Digest\":\"sha1:KBHEX6SOBEJ344L5NBUJWVR224NVQ5IY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703564029.59_warc_CC-MAIN-20210125030118-20210125060118-00612.warc.gz\"}"} |
https://www.brainscape.com/flashcards/chpt-5-6315416/packs/9731854 | [
"# Chpt 5 Flashcards Preview\n\n## CP Chemistry > Chpt 5 > Flashcards\n\nFlashcards in Chpt 5 Deck (41)\n1\nQ\n\nDaltons Model (disproven)\n\nA\n• elements made indivisible, indestructible particles called atoms\n• all atoms of element are identical & have same properties\n2\nQ\n\nDaltons Model (true)\n\nA\n• atoms of different elements combine forming compounds\n• compounds contain atoms in small whole number ratios\n• atoms can combine in more than than one way to form different compounds\n3\nQ\n\nThompsons Model\n\nA\n\ndense positive charge w/ negative charge electrons interspersed\n\n4\nQ\n\nRutherfords Gold foil experiment\n\nA\n\nalpha particles were directed at a thin sheet. most passed few deflected -> atomic nucleus\n\na) if plum pudding model correct, alpha particles would pass without deflection\nb) since small number deflected -> idea most of mass and all +charge of atom concentrated in space smaller than nucleus size\n\n5\nQ\n\natomic number\n\nA\n\n(Z) number of protons (also # electrons if a neutral atom)\n\n6\nQ\n\nMass number\n\nA\n\n(A) total # of protons & neutrons in nucleus of atom\n\n7\nQ\n\nsimilarities b/t protons electrons\n\nA\n\nprotons and electrons same -> atom neutral\n\n8\nQ\n\nisotopes\n\nA\n• atoms of same element with different number of neutrons\n\n- have same atomic number\n\n9\nQ\n\namu\n\nA\n\natomic mass number\n\n10\nQ\n\nA\n\ntype of energy travels thru space at 3.0 x 10^8 m/s exhibits wave and particle like behavior.\ne.g. light\n\n11\nQ\n\nwavelength (lamba)\n\nA\n\n12\nQ\n\nfrequency (v)\n\nA\n\nof cycles or crests that pass thru stationary point in one sec.\n\n13\nQ\n\nwhite light\n\nA\n\ncontinuous spectrum\n\n14\nQ\n\nlight\n\nA\n\n15\nQ\n\nvisible spectrum\n\nA\n\nregion of light spectrum our eyes can see\n\n16\nQ\n\nA\n\ncontinuous spectrum of visible and invisible light that ranges from short to long lamba.\n\n17\nQ\n\nwave length *\n\nA\n• increases to the left\n\n- longer wavelength = lower frequency\n\n18\nQ\n\nwavelength and frequency\n\nA\n\ninversely related\n\n19\nQ\n\nwavelength and energy\n\nA\n\ninversely related\n\n20\nQ\n\nfrequency and energy\n\nA\n\ndirectly related\n\n21\nQ\n\nx rays\n\nA\n\n22\nQ\n\nultraviolet (UV) and infrared (IR) light\n\nA\n\ninvisible to human eye\n\n23\nQ\n\nmicrowaves\n\nA\n\nabsorbed by water molecules and used to heat water containing substances\n\n24\nQ\n\nROYGBIV\n\nA\n\nred. orange. yellow. green. blue. indigo. violet.\n\n25\nQ\n\nphoton\n\nA\n\npacket of light energy\n\n26\nQ\n\nquantum concept\n\nA\n\nenergy is composed of discrete units (its quantized)\n\n27\nQ\n\nenergy level\n\nA\n\nelectrons travel in fixed energy orbits\n\n28\nQ\n\nbohr atom\n\nA\n\nelectrons exits in specific distances from nucleus and occupy orbits of discrete energy levels\n\n29\nQ\n\nemission light spectrum\n\nA\n\nassociated with emission of electromagnetic radiation by elements or compounds; collection of narrow bands of light energy\n\n30\nQ\n\nquantum numbers\n\nA\n\nenergy of each bohr orbit has specific numbers (n=1,2,3…)\n\n31\nQ\n\nquantized\n\nA\n\nfixed\n\n32\nQ\n\nground state\n\nA\n\nlowest energy state\n\n33\nQ\n\nexcited state\n\nA\n\nwhen an electron is in a higher energy orbital\n\n34\nQ\n\nsubshell\n\nA\n\n(energy sublevel) s (sphere), p(dumbbell), d (clover leaf), and f (no specific shape)\n\n35\nQ\n\n8A\n\nA\n\nnoble gases\n\n36\nQ\n\nelectron configuration\n\nA\n\nshows occupation of orbitals by electrons for particular atoms that is\n\n37\nQ\n\nsubatomic particles\n\nA\n\npositive (+) negative (-) neutron\n\n38\nQ\n\n2 types of atomic notation\n\nA\n\n^3H & H-3\n\n39\nQ\n\ncalculating protons neutrons and electrons\n\nA\n• neutrons = mass number - atomic number\n• protons = atomic number\n• identify element by # of protons\n40\nQ\n\nbohr model of atom\n\nA\n• energy levels (n=1,2,3…)\n• energy levels correspond to periodic table\n• energy levels are aka principal quantum numbers\nenergy levels get closer the further away are from nucleus\n41\nQ"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79318804,"math_prob":0.9362016,"size":3424,"snap":"2023-40-2023-50","text_gpt3_token_len":780,"char_repetition_ratio":0.13011695,"word_repetition_ratio":0.0,"special_character_ratio":0.19450934,"punctuation_ratio":0.04485981,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96845204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T08:15:47Z\",\"WARC-Record-ID\":\"<urn:uuid:6d42b96e-f58e-4543-a281-3a6f4f30eb53>\",\"Content-Length\":\"189512\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8d1a331-0e4f-49a3-8321-d6b6554626af>\",\"WARC-Concurrent-To\":\"<urn:uuid:35b60a09-eff6-4799-a199-f3dd8c061e2f>\",\"WARC-IP-Address\":\"35.172.97.227\",\"WARC-Target-URI\":\"https://www.brainscape.com/flashcards/chpt-5-6315416/packs/9731854\",\"WARC-Payload-Digest\":\"sha1:PTUOF4WORWKWQFJOC5IOIX5YLSEULETD\",\"WARC-Block-Digest\":\"sha1:TVK6O3OKNEOSJG7BERORJRYDBGSHDKA6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511055.59_warc_CC-MAIN-20231003060619-20231003090619-00370.warc.gz\"}"} |
https://www.teachoo.com/3459/748/Example-27---If-i-j-k--2i-5j--3i-2j-3k-and-i-6j-k-are-position/category/Examples/ | [
"Examples\n\nChapter 10 Class 12 Vector Algebra\nSerial order wise",
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"Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class\n\n### Transcript\n\nExample 27 If 𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂, 2𝑖 ̂ + 5𝑗 ̂, 3𝑖 ̂ + 2𝑗 ̂ – 3𝑘 ̂ and 𝑖 ̂ – 6𝑗 ̂ – 𝑘 ̂ are the position vectors of points A, B, C and D respectively, then find the angle between (𝐴𝐵) ⃗ and (𝐶𝐷) ⃗ . Deduce that (𝐴𝐵) ⃗ and (𝐶𝐷) ⃗ are collinear.Angle between (𝐴𝐵) ⃗ & (𝐶𝐷) ⃗ is given by cos θ = ((𝑨𝑩) ⃗ . (𝑪𝑫) ⃗)/|(𝑨𝑩) ⃗ ||(𝑪𝑫) ⃗ | A(𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂), B(2𝒊 ̂ + 5𝒋 ̂) (𝑨𝑩) ⃗ = (2 − 1) 𝑖 ̂ + (5 − 1) 𝑗 ̂ + (0 − 1) 𝑘 ̂ = 1𝑖 ̂ + 4𝑗 ̂ − 𝑘 ̂ |(𝑨𝑩) ⃗ | = √(1^2+4^2+(−1)^2 ) = √18 = √(9 × 2) = 3√𝟐 C(3𝒊 ̂ + 2𝒋 ̂ – 3𝒌 ̂), D(𝒊 ̂ – 6𝒋 ̂ – 𝒌 ̂) (𝐶𝐷) ⃗ = (1 − 3) 𝑖 ̂ + (–6 − 2) 𝑗 ̂ + (−1 − (-3)) 𝑘 ̂ = –2𝑖 ̂ – 8𝑗 ̂ + 2𝑘 ̂ |(𝑪𝑫) ⃗ | = √((−2)^2+(−8)^2+2^2 ) = √72 = √(36 × 2) = 6√𝟐 Now, cos θ = ((𝑨𝑩) ⃗ . (𝑪𝑫) ⃗)/|(𝑨𝑩) ⃗ ||(𝑪𝑫) ⃗ | = ((𝑖 ̂ + 4𝑗 ̂ − 𝑘 ̂ ).(−2𝑖 ̂ − 8𝑗 ̂ + 2𝑘 ̂ ))/(3√2 × 6√2) = (1(−2) + 4(−8) − 1(2))/(3√2 × 6√2) = (−2 − 32 − 2)/(3 × 6 × √2 × √2) = (−36)/36 = –1 Since cos θ = –1, θ = 180° So, θ = 180° = 180° × 𝜋/180 = π So, angle between (𝐴𝐵) ⃗ & (𝐶𝐷) ⃗ is π Also, Since angle between (𝐴𝐵) ⃗ & (𝐶𝐷) ⃗ is 180° , they are in opposite directions Since (𝐴𝐵) ⃗ & (𝐶𝐷) ⃗ are parallel to the same line 𝑚 ⃗, they are collinear.",
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"https://www.teachoo.com/static/misc/Davneet_Singh.jpg",
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http://slideplayer.com/slide/9189929/ | [
"",
null,
"# Literal Equations and Formulas. Definition Literal Equation – an equation with two or more variables. –You can \"rewrite\" a literal equation to isolate.\n\n## Presentation on theme: \"Literal Equations and Formulas. Definition Literal Equation – an equation with two or more variables. –You can \"rewrite\" a literal equation to isolate.\"— Presentation transcript:\n\nLiteral Equations and Formulas\n\nDefinition Literal Equation – an equation with two or more variables. –You can \"rewrite\" a literal equation to isolate any one of the variables using inverse operations. This is called solving for a variable. –When you rewrite literal equations, you may have to divide by a variable or variable expression. Assume all variables are not equal to zero. –Examples:\n\nSolving for a Variable Step 1 Locate the variable you are asked to solve for in the equation. Step 2 Identify the operations on this variable and the order in which they are applied. Step 3 Use inverse operations to undo operations and isolate the variable. Solving a Literal Equation\n\nA. Solve x + y = 15 for x. x + y = 15 Locate x in the equation. Since y is added to x, subtract y from both sides to undo the addition. –y x = –y + 15 B. Solve pq = x for q. pq = x Locate q in the equation. Since q is multiplied by p, divide both sides by p to undo the multiplication. Example: Solving Literal Equations\n\nSolve 5 – b = 2t for t. 5 – b = 2t Locate t in the equation. Since t is multiplied by 2, divide both sides by 2 to undo the multiplication. Your Turn:\n\nSolve for V Locate V in the equation. Since m is divided by V, multiply both sides by V to undo the division. VD = m Since V is multiplied by D, divide both sides by D to undo the multiplication. Your Turn:\n\nSolve for the indicated variable. 1. 2. 3. 2x + 7y = 14 for y 4. P = R – C for CC = R – P for m m = x(k – 6 ) 5. for C C = Rt + S for h Your Turn: More Practice\n\nDefinition Formula – is an equation that states a rule for a relationship among quantities. – A formula is a type of literal equation. –In the formula d = rt, d is isolated. You can \"rearrange\" a formula to isolate any variable by using inverse operations. Examples: –Circumference: C = 2 π r –Area of Triangle: A = 1/2bh\n\nDividing by a fraction is the same as multiplying by the reciprocal. Remember!\n\nThe formula for a person’s typing speed is,where s is speed in words per minute, w is number of words typed, e is number of errors, and m is number of minutes typing. Solve for e. Locate e in the equation. Since w–10e is divided by m, multiply both sides by m to undo the division. ms = w – 10e –w ms – w = –10e Since w is added to –10e, subtract w from both sides to undo the addition. Example: Solving Formulas\n\nSince e is multiplied by –10, divide both sides by –10 to undo the multiplication. Example: Continued\n\nThe formula for an object’s final velocity is f = i – gt, where i is the object’s initial velocity, g is acceleration due to gravity, and t is time. Solve for i. f = i – gt Locate i in the equation. Since gt is subtracted from i, add gt to both sides to undo the subtraction. f = i – gt + gt +gt f + gt = i Your Turn:\n\nLocate d in the equation. Since d is multiplied by , divide both sides by to undo the multiplication. Example: Application\n\nNow use this formula and the information given in the problem. The bowl's diameter is inches. The formula C = d gives the circumference of a circle C in terms of diameter d. The circumference of a bowl is 18 inches. What is the bowl's diameter? Leave the symbol in your answer. Example: Continued\n\nLocate t in the equation. Solve the formula d = rt for t. Find the time in hours that it would take Ernst Van Dyk to travel 26.2 miles if his average speed was 18 miles per hour. Now use this formula and the information given in the problem. Since t is multiplied by r, divide both sides by r to undo the multiplication. d = rt Your Turn: Van Dyk’s time was about 1.46 hrs.\n\nJoke Time How do you know when it’s raining cats and dogs? -When you step in a poodle! Why did the apple go out with a fig? -Because it couldn’t find a date. Which sea creature can add? -An octoplus!\n\nDownload ppt \"Literal Equations and Formulas. Definition Literal Equation – an equation with two or more variables. –You can \"rewrite\" a literal equation to isolate.\"\n\nSimilar presentations"
] | [
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https://www.tutorialspoint.com/multimap-cbegin-and-multimap-cend-in-cplusplus-stl | [
"# multimap::cbegin() and multimap::cend() in C++ STL\n\nC++Server Side ProgrammingProgramming\n\nIn this article we will be discussing the working, syntax and examples of multimap::cbegin() and multimap::cend() functions in C++ STL.\n\n## What is Multimap in C++ STL?\n\nMultimaps are the associative containers, which are similar to map containers. It also facilitates to store the elements formed by a combination of key value and mapped value in a specific order. In a multimap container there can be multiple elements associated with the same key. The data is internally always sorted with the help of its associated keys.\n\n## What is multimap::cbegin()?\n\nmultimap::cbegin() function is an inbuilt function in C++ STL, which is defined in <map> header file. cbegin() is the constant begin function. This function returns the constant iterator which is pointing to the first element in the multimap container. The iterator returned is the constant iterator, they can’t be used to modify the content.We can use them to traverse among the elements of the map container by increasing or decreasing the iterator.\n\n### Syntax\n\nmulti.cbegin();\n\n### Parameters\n\nThis function accepts no parameter.\n\n### Return value\n\nIt returns an iterator pointing to the first element of the associated map container.\n\nInput\n\nmultimap<char, int> newmap;\nnewmap[‘a’] = 1;\nnewmap[‘b’] = 2;\nnewmap[‘c’] = 3;\nnewmap.cbegin();\n\nOutput\n\na = 1\n\n### Example\n\nLive Demo\n\n#include <bits/stdc++.h>\nusing namespace std;\nint main(){\n//create the container\nmultimap<int, int> mul;\n//insert using emplace\nmul.emplace_hint(mul.begin(), 1, 10);\nmul.emplace_hint(mul.begin(), 2, 20);\nmul.emplace_hint(mul.begin(), 2, 30);\nmul.emplace_hint(mul.begin(), 1, 40);\nmul.emplace_hint(mul.begin(), 1, 50);\nmul.emplace_hint(mul.begin(), 5, 60);\nauto it = mul.cbegin();\ncout << \"First element in the multimap is: \";\ncout << \"{\" << it->first << \", \" << it->second << \"}\\n\";\ncout << \"\\nElements in multimap is : \\n\";\ncout << \"KEY\\tELEMENT\\n\";\nfor (auto i = mul.cbegin(); i!= mul.cend(); i++){\ncout << i->first << \"\\t\" << i->second << endl;\n}\nreturn 0;\n}\n\n### Output\n\nIf we run the above code it will generate the following output −\n\nFirst element in the multimap is: {1, 50}\nElements in multimap is :\nKEY ELEMENT\n1 50\n1 40\n1 10\n2 30\n2 20\n5 60\n\n## What is multimap::cend()?\n\nmultimap::cend() function is an inbuilt function in C++ STL, which is defined in <map> header file. cend() function is the constant end (). This function returns the constant iterator of the element which is past the last element in the associated multimap container.\n\nThe iterator returned is the constant iterator, they can’t be used to modify the content. We can use them to traverse among the elements of the map container by increasing or decreasing the iterator.\n\nmultimap::cbegin() and multimap::cend() are used to traverse through the whole container by giving a start of the range and end of the range.\n\n### Syntax\n\nmulti.cend();\n\n### Parameters\n\nThis function accepts no parameter.\n\n### Return value\n\nIt returns an iterator pointing to the past the last element of the associated map container.\n\nInput\n\nmultimap <char, int> newmap;\nnewmap(make_pair(‘a’, 1));\nnewmap(make_pair(‘b’, 2));\nnewmap(make_pair(‘c’, 3));\nnewmap.cend();\n\nOutput\n\nerror\n\n### Example\n\nLive Demo\n\n#include <bits/stdc++.h>\nusing namespace std;\nint main(){\n//create the container\nmultimap<int, int> mul;\n//insert using emplace\nmul.emplace_hint(mul.begin(), 1, 10);\nmul.emplace_hint(mul.begin(), 2, 20);\nmul.emplace_hint(mul.begin(), 2, 30);\nmul.emplace_hint(mul.begin(), 1, 40);\nmul.emplace_hint(mul.begin(), 1, 50);\nmul.emplace_hint(mul.begin(), 5, 60);\ncout << \"\\nElements in multimap is : \\n\";\ncout << \"KEY\\tELEMENT\\n\";\nfor (auto i = mul.cbegin(); i!= mul.cend(); i++){\ncout << i->first << \"\\t\" << i->second << endl;\n}\nreturn 0;\n}\n\n### Output\n\nIf we run the above code it will generate the following output −\n\nElements in multimap is :\nKEY ELEMENT\n1 50\n1 40\n1 10\n2 30\n2 20\n5 60"
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http://whydomath.org/node/voting/ballot.html | [
"",
null,
"",
null,
"",
null,
"",
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"",
null,
"Home / Voting / A Best Procedure? / Same Ballot Many Outcomes",
null,
"",
null,
"# Same Ballet Many Outcomes:\n\n## Tallying the Same Ballots under Different Procedures\n\nDoes it matter which election procedure is used? If the outcome is the same regardless of which procedure is used, then no. However, as we have seen, election outcomes often change when the method used to tally the votes changes. In such a case, which election outcome represents the will of the people? This phenomenon has been demonstrated in a few of the applications of different election procedures. The following example demonstrates that the number of alternative election outcomes that may be possible from one set of voters’ preferences.\n\nExample: Three Candidates and Four Election Outcomes\nSuppose that 23 voters cast the following ballots by rank ordering candidates A, B, and C:\n\n 4 5 8 6 A A B C B C C B C B A A\n\nEven in the class of voting vectors, there are multiple election outcomes depending on which voting vectors are used.\n\n• If plurality rule is used to tally votes, a voter assigns one point to its top-ranked candidate and zero to the others, then A is the top candidate with nine votes, B comes in second place with 8 votes, and C loses with only six votes.\n This example demonstrates a worst-case scenario for voting vectors in which one set of voters’ ballots can yield four different outcomes. Considering other election procedures may result in even more outcomes. As each election outcome is backed by a procedure, an election outcome may be the result of which procedure is selected, as opposed to the outcome that the electorate desires. Because there are so many election procedures, is there a way to eliminate some of the procedures – and, hence, their results. Is there a best procedure?\n• If the Borda count is used to tally votes, a voter assigns two points to its top-ranked candidate, one point to its second-ranked candidate, and zero points to its bottom-ranked, then B finishes first with 20 points, A comes in second with 18 points, and C is in third again with 17 points.\n• If voters use a voting vector, the top-ranked candidate receives seven points, the second-ranked candidate receives four points, and the last-place candidate receives zero points, then B wins the election with 96 (= 8x7 + 10x4) points, C comes in second place with 94 (= 6x7 + 13x4) points, and A rounds out the field with 63 (= 9x7) points.\n• If anti-plurality is used to tally votes, a voter assigns the bottom-ranked candidate zero points and one point to the other candidates, then C receives 19 points, B receives 18 points, and A comes in last place with nine points.\n\nFour different voting vectors were used and four different outcomes occurred! There are three additional outcomes that have candidates tied. Insights about how to construct such an example appear on the Mathematics page.\n\n n 3 4 5 6 7 8 9 10 (n - 1) x (n - 1)! 4 18 96 600 4320 35280 322560 3265920\n\nWith as few as 10 candidates, once the ballots are cast, it is possible to have over three million different election outcomes that rank order the 10 candidates!",
null,
"",
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"",
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"Vote Home How to Vote Procedures/Applications A Best Procedure? The Mathematics References and Links",
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"",
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""
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https://answers.everydaycalculation.com/compare-fractions/7-4-and-3-5 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Compare 7/4 and 3/5\n\n1st number: 1 3/4, 2nd number: 3/5\n\n7/4 is greater than 3/5\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 4 and 5 is 20\n2. For the 1st fraction, since 4 × 5 = 20,\n7/4 = 7 × 5/4 × 5 = 35/20\n3. Likewise, for the 2nd fraction, since 5 × 4 = 20,\n3/5 = 3 × 4/5 × 4 = 12/20\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 35/20 > 12/20 or 7/4 > 3/5\n\n#### Compare Fractions Calculator\n\nand\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS\n\n© everydaycalculation.com"
] | [
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http://blog.ezyang.com/2010/10/existential-type-curry/ | [
"the arc of software bends towards understanding\n\n## Existential type-curry\n\nThis post is for those of you have always wondered why we have a forall keyword in Haskell but no exists keyword. Most of the existing tutorials on the web take a very operational viewpoint to what an existential type is, and show that placing the forall in the “right place” results in the correct behavior. I’m going to take a different approach and use the Curry-Howard isomorphism to explain the translation. Some of the logic examples are shamelessly stolen from Aaron Coble’s Logic and Proof lecture notes.\n\nFirst, a little logic brush up. (Feel free to skip.)",
null,
"$P \\to (Q \\land R)$\n\nAt the very bottom of the hierarchy of logic systems lies propositional logic. Whenever you write a non-polymorphic function in Haskell, your function definition corresponds to a statement in propositional logic—this is the simply typed lambda calculus. You get some propositional symbols P, Q and R (corresponding to types) and some logical connectives",
null,
"$\\lnot \\land \\lor \\to \\iff$. In particular,",
null,
"$\\to$ corresponds to the function arrow ->, so you can read",
null,
"$P \\to Q$ as the type P -> Q.",
null,
"$\\forall xy.\\ P(x) = P(y)$\n\nThe next step up is first-order predicate logic, which allows you to use the quantifiers ∀ and ∃ on variables ranging over individuals x, y and z (the predicates take individuals and return propositions). Logical formulas in this system start to look a lot like Haskell polymorphism, but it actually corresponds to dependent types: individuals are terms, not types.\n\nFor the purpose of this post, we’ll instead have x, y and z to range over propositions (types), except for two examples of first order logic to get some intuition for quantifiers. Then polymorphic function definitions are statements in what is called propositional second order logic.\n\nPropositional second order logic gives us a bit of rope, and we can do some fairly unintuitive things with it. Existential types are one such application. However, most Haskellers have a pretty good intuition for polymorphic functions like id :: a -> a, which actually have an ∀ quantifier at the very beginning, like id :: forall a. a -> a or",
null,
"$\\forall x. x \\to x$. What I’d like to do next is make the connection between our intuitive sense of polymorphic functions and our intuitive sense of a universal quantifier.\n\nConsider the following English sentence: All professors can teach and do research. We can translate this into a statement in first-order logic (x ranging over individuals):",
null,
"$\\forall x.\\ \\mathrm{professor}(x) \\to \\mathrm{teaches}(x) \\land \\mathrm{researches}(x)$\n\nThe intuition for the trick of “narrowing” a universally quantified variable by placing it in an implication corresponds directly to the implicit dictionary passing that occurs when you use a type class (which also narrows a universally quantified variable).\n\nWe can do similar translations for the existential quantifier. Everybody loves somebody and there is somebody that everybody loves correspond to, respectively:",
null,
"$\\forall x\\ \\exists y\\ \\mathrm{loves}(x, y)$",
null,
"$\\exists y\\ \\forall x\\ \\mathrm{loves}(x, y)$\n\nTake a moment to convince yourself that these are not the same statements, and figure out which direction implication goes.\n\nWe’ll now jump straight to the implication equivalences, which are the punchline, so to speak. Here, x ranges over propositions (i.e. types).",
null,
"$(\\exists x\\ A[x]) \\to B \\equiv \\forall x\\ (A[x] \\to B)$",
null,
"$(\\forall x\\ A[x]) \\to B \\equiv \\exists x\\ (A[x] \\to B)$\n\nConsider the first equivalence: intuitively, it states that we can simulate a function that takes an existential type by using forall x. (A x -> B). This is precisely the existential data constructor:\n\ndata OpaqueBox = forall a. OpaqueBox a\n\n\nwhich has the type forall a. (a -> OpaqueBox).\n\nThe second proposition is a little trickier to grasp: in the right to left direction, it seems clear that if there exists an inference A(x) to B for some x, if I provide all x I will get B. However, from left to right, if I provide all A(x) to get B, one of those A(x) will have to have been used but I have no good way of figuring out which one.\n\nWe can rigorously prove this equivalence with sequent calculus. We can think of these as “deduction rules” much like modus ponens (if A then B, A; therefore, B). However, statements in the sequent calculus take the form",
null,
"$\\Gamma \\vdash \\Delta$, where Γ is the set of propositions which conjunctively form the assumption, and Δ is the set of propositions which disjunctively form the result. (The",
null,
"$\\vdash$ is called a “turnstile” and indicates implication.)",
null,
"$\\cfrac{\\Gamma \\vdash A, \\Delta \\qquad \\Sigma, B \\vdash \\Pi}{\\Gamma, \\Sigma, A\\rightarrow B \\vdash \\Delta, \\Pi} \\quad ({\\rightarrow }L)$",
null,
"$\\cfrac{\\Gamma, A \\vdash B, \\Delta}{\\Gamma \\vdash A \\rightarrow B, \\Delta} \\quad ({\\rightarrow}R)$",
null,
"$\\cfrac{\\Gamma, A[t/x] \\vdash \\Delta}{\\Gamma, \\forall x A \\vdash \\Delta} \\quad ({\\forall}L)$",
null,
"$\\cfrac{\\Gamma \\vdash A[y/x], \\Delta}{\\Gamma \\vdash \\forall x A, \\Delta} \\quad ({\\forall}R)$",
null,
"$\\cfrac{\\Gamma, A[y/x] \\vdash \\Delta}{\\Gamma, \\exists x A \\vdash \\Delta} \\quad ({\\exists}L)$",
null,
"$\\cfrac{\\Gamma \\vdash A[t/x], \\Delta}{\\Gamma \\vdash \\exists x A, \\Delta} \\quad ({\\exists}R)$",
null,
"$\\forall L$ and",
null,
"$\\exists R$, in particular, are quite interesting:",
null,
"$\\forall L$ says I can make any assumed proposition “polymorphic” by picking some subterm and replacing all instances of it with a newly universally quantified variable (it’s a stronger assumption, so we’re weakening our entailment). We can indeed do this in Haskell (as one might transform (Int -> Bool) -> Int -> Bool into (a -> b) -> a -> b), so long as our proof doesn’t peek at the actual type to perform its computation.",
null,
"$\\exists R$, on the other hand, says that I can take any resulting proposition and “hide” my work by saying something weaker: instead of A[t], I merely say there exists some x for which A[x] is true. This corresponds nicely to the intuition of an existential type hiding representation. Another nice duality is that universal quantification hides information inside the proof, while existential quantification hides information outside of the proof.",
null,
"$\\forall R$ and",
null,
"$\\exists L$ don’t do as much work, but they are a little tricky to use: any universal quantification on the right side of the turnstile can create/destroy a free variable, and any existential quantification on the left side can create/destroy a free variable. Note that",
null,
"$\\forall L$ and",
null,
"$\\exists R$ cannot be used this way; while they can use existing free variables, they can’t create or destroy them.\n\nHere is the proof in both directions of the equivalence. What we’re trying to prove lives on the bottom; tautologies are at the top.",
null,
"The proofs are nicely symmetrical: one uses ∀L and ∃L, and the other ∀R and ∃R. The application of the →R “uncurries” each entailment. Furthermore, the fact that both proofs are constructive indicates that there is this equivalence is one that can be witnessed by a Haskell program! You can check out a Coq version of the proof from kmc.\n\nPostscript. I picked the wrong equivalence initially, but I felt it would be a shame not to share it. Here is the proof for:",
null,
"$\\exists x\\ (A[x] \\to B) \\vdash (\\forall x\\ A[x]) \\to B$.",
null,
"This is done entirely with intuitionistic logic: the other direction requires classical logic. This is left as an exercise for the reader, the solution is here by monochrom. There is also a version from kmc in Coq in both directions. This result has an interesting implication for existentials over functions: we can translate from an existential to a universal, but not back!\n\n### 8 Responses to “Existential type-curry”\n\n1.",
null,
"Derek Elkins says:\n\nThe polymorphic lambda calculus doesn’t correspond to first-order logic. In Haskell we might take forall a. a -> a and instantiate it to Nat -> Nat. Thus a is quantified over propositions which is exactly what first order logic doesn’t do. In first order logic we might write forall n : Nat. … but it certainly doesn’t make sense to write forall n : Nat. n -> n, what does 3 -> 3 mean? First order logic corresponds to a simple form of dependent typing. The polymorphic lambda calculus corresponds to a second order logic minus the first order parts. The paper, “Lectures on the Curry-Howard Isomorphism”, calls this “propositional second order logic.” I highly recommend reading “Lectures on the Curry-Howard Isomorphism.”\n\nAlso of note is that runST has the form of (forall x. A[x]) -> B. It is interesting to consider what the difference would be if it instead had the type exists x. A[x] -> B.\n\n2.",
null,
"Edward Z. Yang says:\n\nYeah, I was a bit confused on that point. I’ve reworded it, hopefully the new wording is correct.\n\n3.",
null,
"Dan Doel says:\n\nThese work out fairly nicely in Agda. And for the pair of intuitionistically equivalent formulae, it’s clear that what’s going on is (un)currying:\n\nhttp://hpaste.org/40605/forall_exists\n\nI didn’t bother postulating a classical axiom to prove ‘impossible’.\n\n4.",
null,
"wren ng thornton says:\n\nStill a bit off:\n\nThe simply types lambda calculus corresponds with (intuitionistic) first-order propositional logic. System F (aka polymorphism over types) corresponds with second-order propositional logic. Whereas dependent types (i.e., LF) correspond with first-order predicate calculus.\n\nYou don’t get any quantification in LF that you don’t get in STLC, you just get predicates instead of being restricted to propositions. In System F you get full quantification over propositions, but not quantification over individuals (unless your “individuals” are just propositions).\n\n5.",
null,
"Edward Z. Yang says:\n\nOk, reworded it some more.\n\n6.",
null,
"Dan Doel says:\n\nIncidentally, here:\n\n7.",
null,
"Eric Kow says:\n8.",
null,
"Edward Z. Yang says:"
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https://softwareto.com/python-program-quadratic-equation/ | [
"Here is a Python program that can be used to solve a quadratic equation:\n\n```\t\t\t\t```\nimport math\n\n# calculate the discriminant\nd = (b**2) - (4*a*c)\n\n# check if the discriminant is negative\nif d < 0:\nprint(\"This equation has no real solutions\")\nelif d == 0:\n# calculate the only solution\nx = (-b + math.sqrt(d)) / (2*a)\nprint(\"The only solution is: \", x)\nelse:\n# calculate the two solutions\nx1 = (-b + math.sqrt(d)) / (2*a)\nx2 = (-b - math.sqrt(d)) / (2*a)\nprint(\"The solutions are: \", x1, \"and\", x2)\n\n# test the function\n\n```\n```\n\nTo use this program, simply call the `solve_quadratic_equation()` function with the coefficients of the quadratic equation as arguments. For example, to solve the equation `x^2 + 3x + 2 = 0`, you would call the function like this:\n\n```\t\t\t\t```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80209166,"math_prob":0.99994624,"size":974,"snap":"2023-14-2023-23","text_gpt3_token_len":280,"char_repetition_ratio":0.15773197,"word_repetition_ratio":0.024691358,"special_character_ratio":0.3090349,"punctuation_ratio":0.15625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T20:30:00Z\",\"WARC-Record-ID\":\"<urn:uuid:78f24819-92e7-4c17-beaf-c158107d5604>\",\"Content-Length\":\"301199\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0149e1db-d0be-48fe-ab74-34ecd187361c>\",\"WARC-Concurrent-To\":\"<urn:uuid:16487cdf-8093-430f-8ad6-58c91e6598e3>\",\"WARC-IP-Address\":\"34.230.232.255\",\"WARC-Target-URI\":\"https://softwareto.com/python-program-quadratic-equation/\",\"WARC-Payload-Digest\":\"sha1:4USO3ALNTGC6FPB7NPUIGBE4JFU3YQQX\",\"WARC-Block-Digest\":\"sha1:SQC45W2X3ZBXMWYTNRMPVMRK6KH6QEUQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650264.9_warc_CC-MAIN-20230604193207-20230604223207-00222.warc.gz\"}"} |
https://zbmath.org/0913.41020 | [
"## Gegenbauer approximation and its applications to differential equations on the whole line.(English)Zbl 0913.41020\n\nAuthor’s abstract: “A Gegenbauer approximation is discussed. Several imbedding inequalities and inverse inequalities are obtained. Some approximation results are given. By variable transformations, differential equations on the whole line are changed to certain equations on a finite interval. Gegenbauer polynomials are used for their numerical solutions. The stabilities and convergences of proposed schemes are proved. The main idea and techniques used in this paper are also applicable to other multiple-dimensional problems in unbounded domains”.\n\n### MSC:\n\n 41A65 Abstract approximation theory (approximation in normed linear spaces and other abstract spaces) 34A45 Theoretical approximation of solutions to ordinary differential equations\n\n### Keywords:\n\nGegenbauer approximation\nFull Text:\n\n### References:\n\n Maday, Y.; Pernaud-Thomas, B.; Vandeven, H., Une réhabilitation des méthodes spectrales de type Laguerre, Rech. Aérospat., 6, 353-379 (1985) · Zbl 0604.42026 Coulaud, O.; Funaro, D.; Kavian, O., Laguerre spectral approximation of elliptic problems in exterior domains, Comp. Mech. in Appl. Mech. and Engi., 80, 451-458 (1990) · Zbl 0734.73090 Funaro, D., Estimates of Laguerre spectral projectors in Sobolev spaces, (Brezinski, C.; Gori, L.; Ronveaux, A., Orthogonal Polynomials and Their Applications (1991), Scientific Publishing Co), 263-266 · Zbl 0842.46017 Iranzo, V.; Falquès, A., Some spectral approximations for differential equations in unbounded domains, Comput. Methods Appl. Mech. Engrg., 98, 105-126 (1992) · Zbl 0762.76081 Mavriplis, C., Laguerre polynomials for infinite-domain spectral elements, J. Comput. Phys., 80, 480-488 (1989) · Zbl 0665.65066 Funaro, D.; Kavian, O., Approximation of some diffusion evolution equations in unbounded domains by Hermite functions, Math. Comp., 57, 597-619 (1990) · Zbl 0764.35007 Christov, C. I., A complete orthogonal system of functions in $$L^2$$, SIAM J. Appl. Math., 42, 1337-1344 (1982) · Zbl 0562.33009 Boyd, J. P., Spectral method using rational basis functions on an infinite interval, J. Comput. Phys., 69, 112-142 (1987) · Zbl 0615.65090 Adams, R. A., Sobolev Spaces (1975), Academic Press: Academic Press New York · Zbl 0186.19101 Gottlieb, D.; Shu, C. W., On the Gibbs phenomenon. IV. Recovering exponential accuracy in a subinterval from a Gegenbauer partial sum of a piecewise analytic function, Math. Comp., 64, 1081-1095 (1995) · Zbl 0852.42018 Bergh, J.; Löfström, J., Interpolation Spaces, an Introduction (1976), Springer-Verlag: Springer-Verlag Berlin · Zbl 0344.46071 Timan, A. F., Theory of Approximation of Functions of a Real Variable (1963), Pergamon: Pergamon Oxford · Zbl 0117.29001 Canuto, C.; Quarteroni, A., Approximation results for orthogonal polynomials in Sobolev spaces, Math. Comp., 38, 67-86 (1982) · Zbl 0567.41008 Askey, R., Orthogonal Polynomials and Spectral Functions. Orthogonal Polynomials and Spectral Functions, Regional Conference Series in Appl. Math., 21 (1975), SIAM: SIAM Philadelphia · Zbl 0298.26010 Courant, R.; Friedrichs, K. O.; Levy, H., Über die partiellen Differenzengleichungen der mathematischen Physik, Math. Ann., 100, 32-74 (1928) · JFM 54.0486.01 Richtmeyer, R. D.; Morton, K. W., Finite Difference Methods for Initial-Value Problems (1967), Interscience: Interscience New York · Zbl 0155.47502 Ben-yu, Guo, A class of difference schemes of two-dimensional viscous fluid flow, TR SUST (1965) Ben-yu, Guo, Generalized stability of discretization and its applications to numerical solutions of nonlinear differential equations, Contemp. Math., 163, 33-54 (1994) · Zbl 0811.65071 Stetter, H. J., Stability of nonlinear discretization algorithms, (Bramble, J., Numerical Solutions of Partial Differential Equations (1966), Academic Press: Academic Press New York), 111-123 · Zbl 0149.11603\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6375103,"math_prob":0.7554549,"size":4611,"snap":"2023-14-2023-23","text_gpt3_token_len":1337,"char_repetition_ratio":0.12090297,"word_repetition_ratio":0.01540832,"special_character_ratio":0.32162222,"punctuation_ratio":0.27484143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9536429,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-29T19:41:37Z\",\"WARC-Record-ID\":\"<urn:uuid:5bfdefe6-786d-4848-a167-d16bcbd03ad1>\",\"Content-Length\":\"58975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d623b5a3-0551-4957-870b-4cea4c1441f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:14d304ac-b485-4c05-867f-3fd29e6546c0>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/0913.41020\",\"WARC-Payload-Digest\":\"sha1:WKPLBUQYMHD7BQHPWV2Z54RDT6PPLXK5\",\"WARC-Block-Digest\":\"sha1:6XEB7OGE2GCIZRSHYZ5NMAIJFJU57KPJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949025.18_warc_CC-MAIN-20230329182643-20230329212643-00295.warc.gz\"}"} |
https://ashishkumarletslearn.com/chapter-1-sets-class-11-maths-lecture-3/ | [
"“Nobody ever wrote down a plan to be broke, fat, lazy. or stupid. Those things are what happen when you don’t have a plan.” – Larry Winget\n\n NCERT EXERCISE 1.4\n1. Find the union of each of the following pairs of sets:\n• (i) X = {1, 3, 5} Y = {1, 2, 3}\n• A = { a, e, i, o, u} B = {a, b, c}\n• A = {x : x is a natural number and multiple of 3}\nB = {x : x is a natural number less than 6}\n• A = {x : x is a natural number and 1 < x ≤ 6 }\nB = {x : x is a natural number and 6 < x < 10 }\n• (v) A = {1, 2, 3}, B = f\n1. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?\n1. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?\n1. If A = {1, 2, 3, 4},B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find\n• (i) A ∪ B\n• A ∪ C\n• B ∪ C\n• B ∪ D\n• (v) A ∪ B ∪ C\n• A ∪ B ∪ D\n• B ∪ C ∪ D\n1. Find the intersection of each pair of sets of question 1 above.\n1. If A = { 3, 5, 7, 9, 11 },B = {7, 9, 11, 13},C = {11, 13, 15} and D = {15, 17}; find\n• (i) A ∩ B\n• B ∩ C\n• A ∩ C ∩ D\n• A ∩ C\n• (v) B ∩ D\n• A ∩ (B ∪ C)\n• A ∩ D\n• A ∩ (B ∪ D)\n• ( A ∩ B ) ∩ ( B ∪ C )\n• (x) ( A ∪ D) ∩ ( B ∪ C)\n1. If A = {x : x is a natural number},B = {x : x is an even natural number}C = {x : x is an odd natural number} and D = {x : x is a prime number }, find\n• (i) A ∩ B\n• A ∩ C\n• A ∩ D\n• B ∩ C\n• (v) B ∩ D\n• C ∩ D\n1. Which of the following pairs of sets are disjoint\n• (i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }\n• { a, e, i, o, u } and { c, d, e, f }\n• {x : x is an even integer } and {x : x is an odd integer}\n1. If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 },C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find\n• (i) A – B\n• A – C\n• A – D\n• B – A\n• (v) C – A\n• D – A\nB – C\n• B – D\n• C – B\n• (x) D – B\n• C – D\n• D – C\n1. If X= { a, b, c, d } and Y = { f, b, d, g}, find\n• (i) X – Y\n• Y – X\n• X ∩ Y\n1. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?\n1. State whether each of the following statement is true or false. Justify your answer.\n• (i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.\n• {a, e, i, o, u} and {a, b, c, d} are disjoint sets.\n• {2, 6, 10, 14} & {3, 7, 11, 15} are disjoint sets.\n• { 2, 6, 10 } and { 3, 7, 11} are disjoint sets."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85413903,"math_prob":0.9997942,"size":2205,"snap":"2021-31-2021-39","text_gpt3_token_len":1057,"char_repetition_ratio":0.14993185,"word_repetition_ratio":0.13430656,"special_character_ratio":0.54557824,"punctuation_ratio":0.2203125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998597,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T05:18:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f1a4fd34-e577-4dad-b49b-864fcf64deca>\",\"Content-Length\":\"349242\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c5583f3-6361-4cf2-80b3-193739d880b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a4c5a32-d582-48aa-bf42-81987bf78b81>\",\"WARC-IP-Address\":\"192.0.78.147\",\"WARC-Target-URI\":\"https://ashishkumarletslearn.com/chapter-1-sets-class-11-maths-lecture-3/\",\"WARC-Payload-Digest\":\"sha1:4LKWODC6P7DE7MVTNEAVD2NOT4K5OM6Q\",\"WARC-Block-Digest\":\"sha1:2BZHB2ONSMQPGEACKYC3V2RJDPKFTUW5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153931.11_warc_CC-MAIN-20210730025356-20210730055356-00193.warc.gz\"}"} |
https://uk.mathworks.com/help/control/ug/select-inputoutput-pairs-in-mimo-models.html | [
"Documentation\n\n## Select Input/Output Pairs in MIMO Models\n\nThis example shows how to select the response from the first input to the second output of a MIMO model.\n\n1. Create a two-input, one-output transfer function.\n\n```N = {[1 -1],;[1 2],[3 1 4]}; D = [1 1 10]; H = tf(N,D)```\n\n### Note\n\nFor more information about using cell arrays to create MIMO transfer functions, see the `tf` reference page.\n\n2. Select the response from the second input to the output of `H`.\n\nTo do this, use MATLAB® array indexing.\n\n`H12 = H(1,2)`\n\nFor any MIMO system `H`, the index notation `H(i,j)` selects the response from the jth input to the ith output."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69021404,"math_prob":0.9398517,"size":544,"snap":"2019-51-2020-05","text_gpt3_token_len":156,"char_repetition_ratio":0.13148148,"word_repetition_ratio":0.0,"special_character_ratio":0.2959559,"punctuation_ratio":0.144,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95512336,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T04:24:49Z\",\"WARC-Record-ID\":\"<urn:uuid:f3743fc0-c906-4e61-9506-30ebf7ed4221>\",\"Content-Length\":\"63201\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7da4cf84-1d5a-49d4-b384-6e185bd1fd4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bcebd67-26f6-44e5-93d1-93351241d6ad>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://uk.mathworks.com/help/control/ug/select-inputoutput-pairs-in-mimo-models.html\",\"WARC-Payload-Digest\":\"sha1:YCI4JUNTGCXCMJ2344TCMLSBAONY5KOO\",\"WARC-Block-Digest\":\"sha1:R36Z4UMEMU7WJ4SRHCXHXGWKPNWCWEYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594209.12_warc_CC-MAIN-20200119035851-20200119063851-00344.warc.gz\"}"} |
https://web2.0calc.com/questions/algebra-algebraic-manipulation-problem | [
"+0\n\n# Algebra - Algebraic Manipulation Problem\n\n+1\n211\n3\n\nFor a positive integer n , the nth triangular number is $$T(n)=\\dfrac{n(n+1)}{2}.$$\n\nFor example, $$T(3) = \\frac{3(3+1)}{2}= \\frac{3(4)}{2}=6$$ , so the third triangular number is 6.\n\nDetermine the smallest integer $$b>2011$$ such that $$T(b+1)-T(b)=T(x)$$ for some positive integer x .\n\nAug 2, 2022\n\n#2\n+2\n\n$$\\frac{(b+1)(b+2)}{2}-\\frac{b(b+1)}{2}=\\frac{x(x+1)}{2}$$\n\n$$2(b+1)=x(x+1)$$\n\n$$\\frac{x(x+1)}{2}-1=b>2011$$\n\nSo:\n\n$$x(x+1)>4024$$\n\nif x=63, then this satisfies the condition, and is the smallest positive integer x. (As x=62 won't be greater than 4024).\n\nSo:\n\n$$2(b+1)=4032 \\implies b=2015$$ satisfies the condition.\n\nCheck:\n\n$$T(2016)-T(2015)=2033136-2031120=2016=\\frac{63(63+1)}{2}=T(63)$$\n\nAug 2, 2022\n#3\n+2\n\n$$T(b+1) = \\frac{(b+1)(b+2)}{2}=\\frac{b^2+3b+2}{2}$$\n\n$$T(b) = \\frac{b(b+1)}{2}=\\frac{b^2+b}{2}$$\n\n$$T(b+1)-T(b)=b+1$$\n\n$$b+1$$ should be a triangular number.\n\nThe smallest triangular number after 2011 is 2016 (take 2011, double it, square root the result, round that, plug it in as n. It will at the very least get you close to the triangular number wanted and then you will have know that number is the nth triangular number, making finding the next easy).\n\n$$b+1=2016$$\n\n$$b=2015$$\n\nAug 2, 2022"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.65989107,"math_prob":1.0000067,"size":1147,"snap":"2023-40-2023-50","text_gpt3_token_len":449,"char_repetition_ratio":0.16097988,"word_repetition_ratio":0.0,"special_character_ratio":0.4690497,"punctuation_ratio":0.08301887,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T13:13:31Z\",\"WARC-Record-ID\":\"<urn:uuid:7d23dbec-2138-42a5-a2db-21896bbbe49d>\",\"Content-Length\":\"24459\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f256937f-24fb-4fc8-b74e-2e85e73c6ee9>\",\"WARC-Concurrent-To\":\"<urn:uuid:bedcb291-55ce-4cce-9065-97fe95d8f193>\",\"WARC-IP-Address\":\"168.119.251.111\",\"WARC-Target-URI\":\"https://web2.0calc.com/questions/algebra-algebraic-manipulation-problem\",\"WARC-Payload-Digest\":\"sha1:CLAIWE7FHPPI3V5DIOYCEKLPRY7VBGW2\",\"WARC-Block-Digest\":\"sha1:BEJPB3LXW7RUPJDKAQMTGCALNF2IFDWE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00360.warc.gz\"}"} |
http://convert.primail.ch/index.php?page=area | [
"",
null,
"Area Conversions Acres Ares Centiare Hectares Labor (US Survey) Rood Sections Square Square angstrom Square caliber Square centimetres (cm2) Square cubit Square digit Square fathom Square feet Square furlong Square inches (in2) Square kilometres (km2) Square metres (m2) Square miles (Int. Nautical) Square miles (Int.) Square millimetres (mm2) Square rods Square yards Townships\n © 2005 Primail.ch - Alle Rechte vorbehalten. verschiedene Zugriffe bisher\nWebdesign by primail.ch"
] | [
null,
"http://convert.primail.ch/pics/prtitconv.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5519308,"math_prob":0.9975361,"size":381,"snap":"2019-51-2020-05","text_gpt3_token_len":112,"char_repetition_ratio":0.3050398,"word_repetition_ratio":0.0,"special_character_ratio":0.20209974,"punctuation_ratio":0.036363635,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9770817,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T13:55:03Z\",\"WARC-Record-ID\":\"<urn:uuid:96df2ddb-39f7-4a9f-b988-644a536ad059>\",\"Content-Length\":\"21253\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d33df1e1-4b94-4acb-a718-f0849e88bf8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f38e53c4-0657-4627-821c-cd7ee83b4dcb>\",\"WARC-IP-Address\":\"85.10.192.140\",\"WARC-Target-URI\":\"http://convert.primail.ch/index.php?page=area\",\"WARC-Payload-Digest\":\"sha1:JAH7P2PBAN5SCTF2HHB6NJVOZMQQMHL6\",\"WARC-Block-Digest\":\"sha1:WBGVKUH5VGEQQX6D75WG4G4XXCW5CFHE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592636.25_warc_CC-MAIN-20200118135205-20200118163205-00554.warc.gz\"}"} |
https://live.boost.org/doc/libs/1_81_0/libs/multiprecision/doc/html/boost_multiprecision/tut/rational/tommath_rational.html | [
"#",
null,
"Boost C++ Libraries\n\n...one of the most highly regarded and expertly designed C++ library projects in the world.\n\n#### tommath_rational\n\n`#include <boost/multiprecision/tommath.hpp>`\n\n```namespace boost{ namespace multiprecision{\n\ntypedef number<tommath_rational > tom_rational;\n\n}} // namespaces\n```\n\nThe `tommath_rational` back-end is used via the typedef `boost::multiprecision::tom_rational`. It acts as a thin wrapper around `boost::rational<tom_int>` to provide a rational number type that is a drop-in replacement for the native C++ number types, but with unlimited precision.\n\nThe advantage of using this type rather than `boost::rational<tom_int>` directly, is that it is expression-template enabled, greatly reducing the number of temporaries created in complex expressions.\n\nThere are also non-member functions:\n\n```tom_int numerator(const tom_rational&);\ntom_int denominator(const tom_rational&);\n```\n\nwhich return the numerator and denominator of the number.\n\nThings you should know when using this type:\n\n• Default constructed `tom_rational`s have the value zero (this the inherited Boost.Rational behavior).\n• Division by zero results in a `std::overflow_error` being thrown.\n• Conversion from a string results in a `std::runtime_error` being thrown if the string can not be interpreted as a valid rational number.\n• No changes are made to libtommath's global state, so this type can safely coexist with other libtommath code.\n• Performance of this type has been found to be pretty poor - this need further investigation - but it appears that Boost.Rational needs some improvement in this area.\n###### Example:\n```#include <boost/multiprecision/tommath.hpp>\n#include <iostream>\n\nint main()\n{\nusing namespace boost::multiprecision;\n\ntom_rational v = 1;\n\n// Do some arithmetic:\nfor(unsigned i = 1; i <= 1000; ++i)\nv *= i;\nv /= 10;\n\nstd::cout << v << std::endl; // prints 1000! / 10\nstd::cout << numerator(v) << std::endl;\nstd::cout << denominator(v) << std::endl;\n\ntom_rational w(2, 3); // Component wise constructor\nstd::cout << w << std::endl; // prints 2/3\n\nreturn 0;\n}\n```"
] | [
null,
"https://live.boost.org/gfx/space.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5266053,"math_prob":0.85453963,"size":1717,"snap":"2023-14-2023-23","text_gpt3_token_len":432,"char_repetition_ratio":0.14652656,"word_repetition_ratio":0.0,"special_character_ratio":0.2498544,"punctuation_ratio":0.22115384,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867276,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T23:43:02Z\",\"WARC-Record-ID\":\"<urn:uuid:d2653cc4-4850-4588-8637-a7d838429701>\",\"Content-Length\":\"13286\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68a7f906-91a3-4c0b-9946-a40c2095addd>\",\"WARC-Concurrent-To\":\"<urn:uuid:06692fcd-894e-4552-94a0-bfe7306398cd>\",\"WARC-IP-Address\":\"146.20.110.251\",\"WARC-Target-URI\":\"https://live.boost.org/doc/libs/1_81_0/libs/multiprecision/doc/html/boost_multiprecision/tut/rational/tommath_rational.html\",\"WARC-Payload-Digest\":\"sha1:RDDX2SZLJPXWPFRYDCKCGR7R5WFGX6MW\",\"WARC-Block-Digest\":\"sha1:TPX2OKQFWVVZIBSN237VBNRGK2RU6LHV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649348.41_warc_CC-MAIN-20230603233121-20230604023121-00672.warc.gz\"}"} |
https://www.open.edu/openlearn/science-maths-technology/mathematics-and-statistics/mathematics-education/numbers-units-and-arithmetic/content-section-3.11.2 | [
"",
null,
"Numbers, units and arithmetic\n\nThis free course is available to start right now. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation.\n\nFree course\n\n# 3.15.1 Subtraction on the number line\n\nNow what about subtraction? You can think of subtraction as undoing addition: adding 3 to 8 gets you 11, and so subtracting 3 from the answer, 11, gets you back to 8. Therefore, in terms of the number line, subtracting 3 from 11 means starting at 11 and moving 3 units to the left.",
null,
"So 11 − 3 = 8. Subtracting a positive number means moving to the left along the number line. Thinking in terms of the contents of the piggy bank: if Thomas had £11 in his piggy bank and subtracts £3 (to buy a toy), he is left with £8.\n\nSuppose on one occasion Thomas has £2 in his piggy bank. He wants to buy a toy for £5. 2 − 5 means starting at 2 and moving 5 units to the left.",
null,
"This takes you back to 3, so 2 − 5 = 3.\n\nSo Thomas needs to borrow £3 and gets an IOU (from his mother) which he puts in his piggy bank. His piggy bank is worth £3.\n\nMU120_4M1"
] | [
null,
"https://www.open.edu/openlearn/ocw/pluginfile.php/776040/mod_resource/content/0/mu120_4m1_cover_image_1.jpg",
null,
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null,
"https://www.open.edu/openlearn/ocw/pluginfile.php/117429/mod_oucontent/oucontent/2930/758510c5/1b4281f7/mu120_a_i033i.jpg",
null
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http://tibasicdev.wikidot.com/68k:degree | [
"The ° Command",
null,
"Command Summary\n\nConverts an angle to degrees, if necessary.\n\nCommand Syntax\n\nangle°\n\n• Press 2nd MATH to enter the MATH popup menu.\n• Press 2 to enter the Angle submenu.\n• Press 1 to select °.\n\nThis command works on all calculators.\n\n2 bytes\n\nThe ° symbol used after an angle makes sure the angle is interpreted as being in degrees. If the calculator is already in degree mode, x° is equal to x; in radian mode, x° is equal to π*x/180; and in gradian mode, x° is equal to 10*x/9.\n\nIf you're using degree measure extensively in a program, it's a better idea to use setMode() to switch to degree mode and not worry about this. However, there are two reasons you might want to use °:\n\n• If you need an angle in degrees only once or twice, don't bother changing the mode setting.\n• In a function, you're forced to use °, since setMode() isn't valid in a function.\n\n``````:sin(30)\nsin(30)\n:sin(30°)\n1/2\n:180°\nπ```\n```\n\nIn degree mode (no conversion is necessary, so no conversion is done):\n\n``````:sin(30)\n1/2\n:sin(30°)\n1/2\n:180°\n180```\n```\n\nAnother possible use of ° is to write an angle in degrees, minutes, and seconds as x°y'z\" (using the usual apostrophe and quote symbols) — this stands for x degrees, y minutes (equal to 1/60th of a degree) and z seconds (equal to 1/60th of a minute). There's no \"degree/minute/second\" mode setting, so an angle entered in this form will always be simplified: first to (x+y/60+z/3600)2 degrees, and then (if necessary) converted to the correct angle measure. However, you can use ▶DMS to express output in this form."
] | [
null,
"http://tibasicdev.wdfiles.com/local--files/68k:command-index/degree.png",
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https://miro.acadiasi.ro/ns41l9n/d9e8c9-iphone-built-in-apps | [
"# iphone built in apps\n\n1. This allowed us to use Eulerâs theorem and jump to (15.7b), where only a summation with respect to number of moles survived. Find the maximum and minimum values of f(x,) = 2xy - 5x2 - 2y + 4x -4. Let F be a differentiable function of two variables that is homogeneous of some degree. State and prove Euler's theorem for homogeneous function of two variables. Ask Question Asked 5 years, 1 month ago. Indeed, Eulerâs Theorem can be used to show that functions that are homogeneous of degree zero cannot be monotonic when there are two or more variables. Any function f â C1(Rm ++) for m > 1 that is homogeneous of degree zero is not monotonic. Application of Euler Theorem On homogeneous function in two variables. x â
âf(x) = kf(x) Theorem 20.8.1. Then along any given ray from the origin, the slopes of the level curves of F are the same. Proof. Then (2) (3) (4) Let , then (5) This can be generalized to an arbitrary number of variables (6) where Einstein summation has been used. This is normal for such functions. 0. find a numerical solution for partial derivative equations. 1 -1 27 A = 2 0 3. Hiwarekar discussed extension and applications of Eulerâs theorem for finding the values of higher order expression for two variables. Index Termsâ Homogeneous Function, Eulerâs Theorem. Suppose that the function Æ : Rn \\ {0} â R is continuously differentiable. Euler's theorem A function homogeneous of some degree has a property sometimes used in economic theory that was first discovered by Leonhard Euler (1707â1783). Let be a homogeneous function of order so that (1) Then define and . CITE THIS AS: Weisstein, Eric W. \"Euler's Homogeneous Function Theorem.\" (b) State and prove Euler's theorem homogeneous functions of two variables. Functions homogeneous of degree n are characterized by Eulerâs theorem that asserts that if the differential of each independent variable is replaced with the variable itself in the expression for the complete differential INTRODUCTION The Eulerâs theorem on Homogeneous functions is used to solve many problems in engineering, science and finance. But most important, they are intensive variables, homogeneous functions of degree zero in number of moles (and mass). 17 6 -1 ] Solve the system of equations 21 â y +22=4 x + 7y - z = 87, 5x - y - z = 67 by Cramer's rule as well as by matrix method and compare bat results. First, they are convenient variables to work with because we can measure them in the lab. I. 4. Active 5 years, 1 month ago. Euler's Homogeneous Function Theorem. The Euler's theorem on Homogeneous functions is used to solve many problems in engineering, science and finance. 2. Then Æ is positive homogeneous of degree k if and only if. Get the answers you need, now! Question on Euler's Theorem on Homogeneous Functions. Positive homogeneous functions are characterized by Euler's homogeneous function theorem. In mathematics, a homogeneous function is one with multiplicative scaling behaviour: if all its arguments are multiplied by a factor, then its value is multiplied by some power of this factor.. For example, a homogeneous real-valued function of two variables x and y is a real-valued function that satisfies the condition (,) = (,) for some constant k and all real numbers α. Reverse of Euler's Homogeneous Function Theorem. 3 3. In this paper we have extended the result from function of two variables to ⦠Euler theorem on homogeneous function of two variables for two variables C1 ( Rm )! Define and ( and mass ) find the maximum and minimum values higher! Rm ++ ) for m > 1 that is homogeneous of degree k if only... Degree k if and only if Æ is positive homogeneous of some degree of f ( x =! And mass ) application of Euler theorem on homogeneous functions of degree zero in number of moles ( and )... And minimum values of f ( x ) = 2xy - 5x2 - 2y 4x... That is homogeneous of some degree theorem homogeneous functions of degree k if and only if > that. Not monotonic month ago the maximum and minimum values of f are the same let be! For homogeneous function of two variables higher order expression for two variables function â! Æ is positive homogeneous of degree zero is not monotonic and only if, they are intensive variables homogeneous. { 0 } â R is continuously differentiable 5x2 - 2y + 4x -4 discussed and! Given ray from the origin, the slopes of the level curves f! X, ) = kf ( x, ) = kf ( x =. ] discussed extension and applications of Eulerâs theorem on homogeneous function of two variables be a function... Theorem on homogeneous functions is used to solve many problems in engineering, science and finance is homogeneous of k! Function of two variables that is homogeneous of degree zero is not monotonic higher order expression for variables... Euler & # 039 ; s theorem for finding the values of f ( x, ) kf. ] discussed extension and applications of Eulerâs theorem on homogeneous function in two variables prove... Euler 's homogeneous function of two variables, science and finance application of Euler theorem homogeneous! Of Eulerâs theorem on homogeneous functions is used to solve many problems in engineering, and! Is used to solve many problems in engineering, science and finance kf ( x ) 2xy... 4X -4 theorem homogeneous functions is used to solve many problems in engineering, science and finance s! Theorem homogeneous functions is used to solve many problems in engineering, science and finance let be a homogeneous of! Variables, homogeneous functions of degree zero in number of moles ( and mass ) homogeneous of... Of Euler theorem on homogeneous functions is used to solve many problems in engineering, science and.. The values of f are the same then define and 0 } â R is continuously differentiable ) for >! 4X -4 homogeneous of some degree 0 } â R is continuously differentiable f are the.... Euler theorem on homogeneous functions of degree k if and only if ( b ) State and Euler! Introduction the Eulerâs theorem for finding the values of f ( x ) = 2xy - 5x2 - +. This AS: Weisstein, Eric W. `` Euler 's theorem on homogeneous functions of two variables Question Asked years! 'S theorem homogeneous functions is used to solve many problems in engineering, science and.. Curves of f ( x ) = 2xy - 5x2 - 2y + 4x -4 maximum minimum!  R is continuously differentiable s theorem for homogeneous function in two variables ( b ) State prove... Slopes of the level curves of f ( x ) = 2xy - 5x2 - 2y + -4. F be a homogeneous function of two variables hiwarekar [ 1 ] discussed extension applications. Is used to solve many problems in engineering, science and finance of degree k if and only.... ) then define and is used to solve many problems in engineering, science and finance âf ( ). The same any given ray from the origin, the slopes of the level curves of (. This AS: Weisstein, Eric W. `` Euler 's theorem on homogeneous function two. Kf ( x ) = 2xy - 5x2 - 2y + 4x.. C1 ( Rm ++ ) for m > 1 that is homogeneous some. And finance numerical solution for partial derivative equations ) = kf ( x, =... Then define and introduction the Eulerâs theorem on homogeneous function in two.... X, ) = kf ( x, ) = 2xy - 5x2 - 2y + 4x -4 \\. And minimum values of higher order expression for two variables that is homogeneous of degree zero is not.. The origin, the slopes of the level curves of f ( x ) = 2xy - 5x2 2y... In engineering, science and finance given ray from the origin, the slopes of the curves... Is continuously differentiable science and finance but most important, they are intensive variables, homogeneous is! 5 years, 1 month ago used to solve many problems in engineering, science and finance 5,! The Euler 's homogeneous function of order so that ( 1 ) then define and x ) 2xy! Moles ( and mass ) order so that ( 1 ) then define and the Euler 's theorem on functions... Hiwarekar [ 1 ] discussed extension and applications of Eulerâs theorem for function. ( and mass ) 's homogeneous function theorem. mass ) â C1 ( Rm ++ ) m! Theorem on homogeneous function of two variables derivative equations f are the same given. Level curves of f are the same 0. find a numerical solution for partial derivative equations differentiable function of variables... Years, 1 month ago and mass ) 039 ; s theorem for the! K if and only if homogeneous functions of degree k if and only if problems in engineering science... Application of Euler theorem on homogeneous functions of degree zero in number moles... Ask Question Asked 5 years, 1 month ago most important, they are variables! Rm euler's theorem for homogeneous function of two variables ) for m > 1 that is homogeneous of degree zero in number of moles and... F â C1 ( Rm ++ ) for m > 1 that homogeneous! Euler theorem on homogeneous functions is used to solve many problems in engineering, science and.! For homogeneous function in two variables not monotonic years, 1 month ago only if ) then define.... R is continuously differentiable > 1 that is homogeneous of some degree and only.... 2Y + 4x -4: Rn \\ { 0 } â R is differentiable! 0 } â R is continuously differentiable along any given ray from the origin, the slopes of the curves... Be a differentiable function of order so that ( 1 ) then define and zero is not.... Years, 1 month ago State and prove Euler & # 039 ; s theorem for homogeneous function.... The function Æ: Rn \\ { 0 } â R is continuously differentiable function theorem. the! Eric W. `` Euler 's theorem homogeneous functions is used to solve problems. # 039 ; s theorem for finding the values of f are the same Æ: Rn {..., homogeneous functions of two variables for finding the values of higher order expression for variables... Variables that is homogeneous of degree zero is not monotonic x â âf ( x ) = -. That is homogeneous of some degree for homogeneous function of two variables the slopes the. In engineering, science and finance 's homogeneous function of two variables and minimum values of order! K if and only if a differentiable function of order so that 1... Solution for partial derivative equations function of two variables functions is used to solve many problems engineering. Intensive variables, homogeneous functions of degree zero in number of moles ( and )... Homogeneous functions of two variables engineering, science and finance & # 039 ; s for! They are intensive variables, homogeneous functions of two variables a numerical solution for partial derivative.... W. `` Euler 's theorem homogeneous functions is used to solve many problems in engineering, and! Partial derivative equations not monotonic â âf ( x ) = 2xy - 5x2 - +... Find the maximum and minimum values of higher order expression for two variables (! Given ray from the origin, the slopes of the level curves of f ( ). Variables, homogeneous functions of two variables extension and applications of Eulerâs theorem on homogeneous functions is used to many!"
] | [
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https://littlemediastudios.de/balance-ball-charge-grinding-mill-example-calculations.html | [
"Welcome to visit us!\n\nsitemap [email protected]\n1. Home\n2. > Balance Ball Charge Grinding Mill Example Calculations",
null,
"# Balance Ball Charge Grinding Mill Example Calculations\n\nBalance Ball Charge Grinding Mill Example Calculationsri, Balance ball charge grinding mill example calculations balance ball charge grinding mill example calculations the gulin product line consisting of more than 30 machines sets the standard for our industry get price Balance Ball Charge Grinding Mill Example Calculations\n\n[email protected]\n• ### Calculations Mill Charge Plan Brandenburg\n\nCalculate ball mill grinding capacity a total apparent volumetric charge filling including balls and excess slurry on top of the ball charge plus the interstitial voids in between the balls expressed as a percentage of the net internal mill volume inside liners\n\n• ### Balance Ball Charge Grinding Mill Example Calculationsri\n\nBalance ball charge grinding mill example calculations balance ball charge grinding mill example calculations the gulin product line consisting of more than 30 machines sets the standard for our industry get price\n\n• ### Calculation Of Heat Generated From Grinding In Ball Mill\n\nCalculation of heat generated from grinding in ball mill prompt caesar is a famous mining equipment manufacturer wellknown both at home and abroad major in producing stone crushing equipment mineral separation equipment limestone grinding equipment etc\n\n• ### Ballmill Design Calculations Pdf\n\nBallmill design calculations pdf za ball mill power calculation pdf ball mill design calculation xls and pdf file the basic parameters used in ball mill design power calculations rod mill or any tumbling mill sizing are ball mill power calculation pdf power calculation for jaw crusher pdf is a leading global manufacturer of crushing\n\n• ### Mass Balance Grinding Mill\n\nBalance ball charge grinding mill example mass balance infinity for cement hammer crusher heat and mass balance pdf analysis of material flow and consumption in cement mass balance calculations for cement grinding ballmill mass and heat balance calculator ball mill heat balance calculation in process energy consumed in coal\n\n• ### Ball Mill Heat Balance Calculation Worksheet\n\nCement ball mill heat balance calculation posts related to ball mill air balance calculations balance ball charge grinding mill example calculations asking price heat balance of air swept ball mill grinding mill heat balance of air swept ball mill posted at june 5 2013 47 7493 ratings mass balance calculations in vertical more detail\n\n• ### Refusal Of Cement And Grinding Charge Of A Mill Vetura\n\nRefusal of cement and grinding charge of a mill balance ball charge grinding mill example a ball mill grinding circuit is a nonlinear balance ball charge grinding mill example calculations refusal of closed open circuit grinding mill processing differencrefusal of cement and grinding charge of a mill\n\n• ### Amit 135 Lesson 7 Ball Mills Circuits Mining Mill\n\nFor overflow ball mills the charge should not exceed 45 of the mill volume for grate discharge mills the charge should occupy about 50 of the mill volume bond developed a relationship that can be used to determine the percent charge by volume as a function of the vertical height above the charge he and the radius of the mill r ie\n\n• ### Pdf Population Balance Model Approach To Ball Mill\n\nPopulation balance model approach to ball mill optimization in iron ore grinding the model has been applied to describe grinding in a ball mill where the mechanical environment was predicted\n\n• ### Mass Balance Ballmill\n\nBall mill mass balance in steady state 3 replies and 2 comments pitachikondi since it is a close circuit with the sump mass to the ball mill is same as the discharged mass mind that we are not talking about water balance yet question and answer for mass balance calculations in ball mill\n\n• ### Ball Mill Grinding Calculation\n\nBall mill power calculation example 1 a wet grinding ball mill in closed circuit is to be fed 100 tph of a material with a work index of 15 and a size distribution of 80 passing inch 6350 microns the required product size distribution is to be 80 passing 100 mesh 149 microns\n\n• ### Simple Calculation In Ball Mill\n\nOf the bond ball mill grindability test for all model based methods a reliable method to calculate mill power draw for a given mill is required for the calculation of power draw morrell 1996 proposed a mathematical model for autogenous semiautogenous and ball mills which is based on the motion of grinding charge inside the mill get price\n\n• ### Ballmill Design Calculations Pdf\n\nBallmill design calculations pdf za ball mill power calculation pdf ball mill design calculation xls and pdf file the basic parameters used in ball mill design power calculations rod mill or any tumbling mill sizing are ball mill power calculation pdf power calculation for jaw crusher pdf is a leading global manufacturer of crushing\n\n• ### Ball Mill Mass Balance In Steady State Grinding\n\nA ball mill is operated in closed circuit with sieve bend under steady state conditions as shown in the attached solids in each stream are water addition to the sump is 100 cubic metres per hr and to the mill feed is 67cubic metres per hr calculate mass flow of sol\n\n• ### Ball Mill Initial Charge Hemine Mining Machinery Co Ltd\n\nHome ball mill initial charge jaw crushers jaw crushers\n\n• ### Calculations On Mass Balance On Ball Mill Circuit\n\nCalculations on mass balance on ball mill circuit grinding an overview of operation and design cyclone underflow feed to the ball mill determine mass balance of a ball mill grinding circuit balance ball charge grinding mill example calculations mass balance calculations for cement grindingset up population balance circuit charge massand\n\n• ### Calculations Mill Charge\n\nBalance ball charge grinding mill example calculations the ball charge and ore charge volume is a variable subject to what is the target for that operation the type of mill also is a factor as if it is an overflow mill subject to the diameter of the discharge port is usually up to about 4045 if it is a grate discharge you will\n\n• ### Calculations On Mass Balance On Ball Mill Circuit\n\nCalculations on mass balance on ball mill circuit grinding in ball mills modeling and process control grinding in ball mills modeling and the process control in a ball mill grinding circuit faces severe basis of mass balance or energy balance equations get price excel yahoo answers best answer a ball mill is a horizontal\n\n• ### Ball Mill Initial Charge Hemine Mining Machinery Co Ltd\n\nHome ball mill initial charge jaw crushers jaw crushers\n\n• ### Ball Charges Calculators Thecementgrindingoffice Ball\n\nBall charges calculators ball top size bond formula calculation of the top size grinding media balls or cylpebsmodification of the ball charge this calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball\n\n• ### Ball Mill Recirculating Load Calculation\n\n11 crushing and classification the processes of crushing grinding milling micronising etc as ball and rod mills comminution can be derived from a single ordinary differential equation m kl l e classifier is large and the circulating load is defined as the ratio of the\n\n• ### Ball Mill Grinding Calculation\n\nBall mill power calculation example 1 a wet grinding ball mill in closed circuit is to be fed 100 tph of a material with a work index of 15 and a size distribution of 80 passing inch 6350 microns the required product size distribution is to be 80 passing 100 mesh 149 microns\n\n• ### Roller Press With Ball Mill Mass Balance\n\nCaco3 ball mill grinder youtubekeithkirsten ball mill heat balance calculation in process engineering in cement industry roller press with ball mill mass balance how to classify the ball mill charge panies manufacturing jaw crushers feeders screens ball mills in india crusher and ball mill grinding of cement through ball mill kegunaan fungsi mesin ball mill prevent caking in a\n\n• ### Balance Ball Charge Grinding Mill Example Calculations\n\nBalance ball charge grinding mill example calculations mar 17 2017 the basic parameters used in ball mill design power calculations rod mill or any ball mill grinding capacity calculator ball mill motorpower sizing c distance of centre of gravity or charge from centre of mill in feet flotation material balancing excel solver mass balance calculations example\n\nLATEST NEWS\n\n• ### Enquiry For Crusher Iteams",
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https://tentotwelvemath.com/fom-12/1-compound-interest/1-1-numeracy/ | [
"1.1 Numeracy\n\nMultiply by 10, 100, 1000 … 1000000\n\nDivide\n\nDivide by 10, 100, … ,1000000\n\nCalculating percentages using mental arithmetic.\n\n 1% Divide by 100 10% Divide by 10 5% Find 10% then half that. 15% Find 10%, find 5% then add them together. 50% Divide by 2. 25% Find 50% then half that. 95% Find 5% then subtract that from the total. 12% Tax in BC (GST + PST). Find 10%. Find 1%, double it. Add to get 12.\n\nOr, if you happen to be using a calculator, we would perform the following calculations:\n\n 1% Multiply by 1/100 10% Multiply by 10/100 5% Multiply by 5/100 15% Multiply by 15/100 50% Multiply by 50/100 25% Multiply by 25/100 95% Multiply by 95/100 12% Multiply by 12/100\n\nFor example, to calculate BC sales tax (12%) on an item that costs \\$90:\n\n90 times (12/100) = 10.80\n\nThe sales tax is \\$10.80. Therefore the total cost is 90.00 + 10.80 = \\$100.80.\n\nAs another example, suppose we wish to find 11% of 185.\n\nWe type into the calculator: 185 times (11/100) to get 20.35."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8833467,"math_prob":0.9993791,"size":989,"snap":"2019-26-2019-30","text_gpt3_token_len":351,"char_repetition_ratio":0.17766498,"word_repetition_ratio":0.0,"special_character_ratio":0.4651163,"punctuation_ratio":0.15111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99980503,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T18:55:35Z\",\"WARC-Record-ID\":\"<urn:uuid:65439192-359c-4a35-a6a7-9e540da4158a>\",\"Content-Length\":\"64869\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bcc0232a-facb-478f-aeb0-11f8b2944dc4>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a52962c-3d28-4c56-8908-88f9c8b29659>\",\"WARC-IP-Address\":\"37.60.244.209\",\"WARC-Target-URI\":\"https://tentotwelvemath.com/fom-12/1-compound-interest/1-1-numeracy/\",\"WARC-Payload-Digest\":\"sha1:AVGPUFSRQVKOILK35O3IS6ETPYHVA6JF\",\"WARC-Block-Digest\":\"sha1:M7CYDQ7C23BSLTV37CCJP5ZIXD7EKHQC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000414.26_warc_CC-MAIN-20190626174622-20190626200622-00146.warc.gz\"}"} |
https://makeslider.com/slide/slides-for-rosen-5th-edition-academics-wpi-c4hz95 | [
"# Slides for Rosen, 5th edition - Academics | WPI",
null,
"Module #22 - Graphs University of Florida Dept. of Computer & Information Science & Engineering COT 3100 Applications of Discrete Structures Dr. Michael P. Frank Slides for a Course Based on the Text Discrete Mathematics & Its Applications (5th Edition) by Kenneth H. Rosen 02/02/20 (c)2001-2003, Michael P. Frank 1\n\nModule #22 - Graphs Module #22: Graph Theory Rosen 5th ed., chs. 8-9 ~44 slides (more later), ~3 lectures 02/02/20 (c)2001-2003, Michael P. Frank 2 Module #22 - Graphs What are Graphs? Not General meaning in everyday math:\n\nA plot or chart of numerical data using a coordinate system. Technical meaning in discrete mathematics: A particular class of discrete structures (to be defined) that is useful for representing relations and has a convenient webbylooking graphical representation. 02/02/20 (c)2001-2003, Michael P. Frank 3 Module #22 - Graphs Applications of Graphs Potentially anything (graphs can represent relations, relations can describe the extension of any predicate). Apps in networking, scheduling, flow optimization, circuit design, path planning.\n\nMore apps: Geneology analysis, computer game-playing, program compilation, objectoriented design, 02/02/20 (c)2001-2003, Michael P. Frank 4 Module #22 - Graphs Simple Graphs Correspond to symmetric, irreflexive binary relations R. A simple graph G=(V,E) Visual Representation consists of: of a Simple Graph a set V of vertices or nodes (V corresponds to the universe of the relation R), a set E of edges / arcs / links: unordered pairs\n\nof [distinct] elements u,v V, such that uRv. 02/02/20 (c)2001-2003, Michael P. Frank 5 Module #22 - Graphs Example of a Simple Graph Let V be the set of states in the farsoutheastern U.S.: I.e., V={FL, GA, AL, MS, LA, SC, TN, NC} Let E={{u,v}|u adjoins v} ={{FL,GA},{FL,AL},{FL,MS}, {FL,LA},{GA,AL},{AL,MS}, MS {MS,LA},{GA,SC},{GA,TN}, {SC,NC},{NC,TN},{MS,TN}, LA {MS,AL}} 02/02/20\n\n(c)2001-2003, Michael P. Frank TN AL NC SC GA FL 6 Module #22 - Graphs Multigraphs Like simple graphs, but there may be more than one edge connecting two given nodes. A multigraph G=(V, E, f ) consists of a set V of vertices, a set E of edges (as primitive objects), and a function\n\nParallel edges f:E{{u,v}|u,vV uv}. E.g., nodes are cities, edges are segments of major highways. 02/02/20 (c)2001-2003, Michael P. Frank 7 Module #22 - Graphs Pseudographs Like a multigraph, but edges connecting a node to itself are allowed. (R may be reflexive.) A pseudograph G=(V, E, f ) where f:E{{u,v}|u,vV}. Edge eE is a loop if f(e)={u,u}={u}. E.g., nodes are campsites\n\nin a state park, edges are hiking trails through the woods. 02/02/20 (c)2001-2003, Michael P. Frank 8 Module #22 - Graphs Directed Graphs Correspond to arbitrary binary relations R, which need not be symmetric. A directed graph (V,E) consists of a set of vertices V and a binary relation E on V. E.g.: V = set of People, E={(x,y) | x loves y} 02/02/20\n\n(c)2001-2003, Michael P. Frank 9 Module #22 - Graphs Directed Multigraphs Like directed graphs, but there may be more than one arc from a node to another. A directed multigraph G=(V, E, f ) consists of a set V of vertices, a set E of edges, and a function f:EVV. E.g., V=web pages, E=hyperlinks. The WWW is a directed multigraph... 02/02/20 (c)2001-2003, Michael P. Frank\n\n10 Module #22 - Graphs Types of Graphs: Summary Summary of the books definitions. Keep in mind this terminology is not fully standardized across different authors... Term Simple graph Multigraph Pseudograph Directed graph Directed multigraph 02/02/20 Edge type Undir.\n\nUndir. Undir. Directed Directed Multiple edges ok? No Yes Yes No Yes (c)2001-2003, Michael P. Frank Selfloops ok? No No Yes Yes\n\nYes 11 Module #22 - Graphs 8.2: Graph Terminology You need to learn the following terms: Adjacent, connects, endpoints, degree, initial, terminal, in-degree, out-degree, complete, cycles, wheels, n-cubes, bipartite, subgraph, union. 02/02/20 (c)2001-2003, Michael P. Frank 12 Module #22 - Graphs\n\nAdjacency Let G be an undirected graph with edge set E. Let eE be (or map to) the pair {u,v}. Then we say: u, v are adjacent / neighbors / connected. Edge e is incident with vertices u and v. Edge e connects u and v. Vertices u and v are endpoints of edge e. 02/02/20 (c)2001-2003, Michael P. Frank 13 Module #22 - Graphs Degree of a Vertex Let G be an undirected graph, vV a vertex. The degree of v, deg(v), is its number of\n\nincident edges. (Except that any self-loops are counted twice.) A vertex with degree 0 is called isolated. A vertex of degree 1 is called pendant. 02/02/20 (c)2001-2003, Michael P. Frank 14 Module #22 - Graphs Handshaking Theorem Let G be an undirected (simple, multi-, or pseudo-) graph with vertex set V and edge set E. Then deg(v) 2 E vV\n\nCorollary: Any undirected graph has an even number of vertices of odd degree. 02/02/20 (c)2001-2003, Michael P. Frank 15 Module #22 - Graphs Directed Adjacency Let G be a directed (possibly multi-) graph, and let e be an edge of G that is (or maps to) (u,v). Then we say:\n\n02/02/20 u is adjacent to v, v is adjacent from u e comes from u, e goes to v. e connects u to v, e goes from u to v the initial vertex of e is u the terminal vertex of e is v (c)2001-2003, Michael P. Frank 16 Module #22 - Graphs Directed Degree Let G be a directed graph, v a vertex of G. The in-degree of v, deg(v), is the number of edges going to v. The out-degree of v, deg(v), is the number of edges coming from v. The degree of v, deg(v):deg(v)+degdeg(v), is the\n\nsum of vs in-degree and out-degree. 02/02/20 (c)2001-2003, Michael P. Frank 17 Module #22 - Graphs Directed Handshaking Theorem Let G be a directed (possibly multi-) graph with vertex set V and edge set E. Then: 1 deg (v) deg (v) deg(v) E 2 vV vV vV\n\nNote that the degree of a node is unchanged by whether we consider its edges to be directed or undirected. 02/02/20 (c)2001-2003, Michael P. Frank 18 Module #22 - Graphs Special Graph Structures Special cases of undirected graph structures: Complete graphs Kn Cycles Cn Wheels Wn\n\n02/02/20 n-Cubes Qn Bipartite graphs Complete bipartite graphs Km,n (c)2001-2003, Michael P. Frank 19 Module #22 - Graphs Complete Graphs For any nN, a complete graph on n vertices, Kn, is a simple graph with n nodes in which every node is adjacent to every other node: u,vV: uv{u,v}E.\n\nK1 K3 K2 K4 K5 K6 n 1 n(n 1) Note that Kn has i 2 edges. i 1 02/02/20\n\n(c)2001-2003, Michael P. Frank 20 Module #22 - Graphs Cycles For any n3, a cycle on n vertices, Cn, is a simple graph where V={v1,v2, ,vn} and E={{v1,v2},{v2,v3},,{vn1,vn},{vn,v1}}. C3 C4 C5 C6 C7\n\nC8 How many edges are there in Cn? 02/02/20 (c)2001-2003, Michael P. Frank 21 Module #22 - Graphs Wheels For any n3, a wheel Wn, is a simple graph obtained by taking the cycle Cn and adding one extra vertex vhub and n extra edges {{vhub,v1}, {vhub,v2},,{vhub,vn}}. W3 W4\n\nW5 W6 W7 How many edges are there in Wn? 02/02/20 (c)2001-2003, Michael P. Frank W8 22 Module #22 - Graphs n-cubes (hypercubes) For any nN, the hypercube Qn is a simple\n\ngraph consisting of two copies of Qn-1 connected together at corresponding nodes. Q0 has 1 node. Q0 Q1 Q2 Q3 Q4 Number of vertices: 2n. Number of edges:Exercise to try! 02/02/20 (c)2001-2003, Michael P. Frank 23\n\nModule #22 - Graphs n-cubes (hypercubes) For any nN, the hypercube Qn can be defined recursively as follows: Q0={{v0},} (one node and no edges) For any nN, if Qn=(V,E), where V={v1,,va} and E={e1,,eb}, then Qn+deg1=(V{v1,,va}, E{e1,,eb}{{v1,v1},{v2,v2},, {va,va}}) where v1,,va are new vertices, and where if ei={vj,vk} then ei={vj,vk}. 02/02/20 (c)2001-2003, Michael P. Frank 24 Module #22 - Graphs Bipartite Graphs\n\nDefn.: A graph G=(V,E) is bipartite (twopart) iff V = V1V2 where V1V2= and eE: v1V1,v2V2: e={v1,v2}. In English: The graph can be divided into two parts in such a way that all edges go between the two parts. This definition can easily be adapted for the case of multigraphs and directed graphs as well. 02/02/20 (c)2001-2003, Michael P. Frank V1 V2 Can represent with zero-one matrices. 25\n\nModule #22 - Graphs Complete Bipartite Graphs For m,nN, the complete bipartite graph Km,n is a bipartite graph where |V1| = m, |V2| = n, and E = {{v1,v2}|v1V1 v2V2}. That is, there are m nodes in the left part, n nodes in the right part, and every node in the left part is connected to every node in the right part. 02/02/20 K4,3 Km,n has _____ nodes and _____ edges. (c)2001-2003, Michael P. Frank\n\n26 Module #22 - Graphs Subgraphs A subgraph of a graph G=(V,E) is a graph H=(W,F) where WV and FE. G 02/02/20 H (c)2001-2003, Michael P. Frank 27 Module #22 - Graphs\n\nGraph Unions The union G1G2 of two simple graphs G1=(V1, E1) and G2=(V2,E2) is the simple graph (V1V2, E1E2). 02/02/20 a b d e c\n\na d (c)2001-2003, Michael P. Frank b c f 28 Module #22 - Graphs 8.3: Graph Representations & Isomorphism Graph representations: Adjacency lists. Adjacency matrices. Incidence matrices.\n\nGraph isomorphism: Two graphs are isomorphic iff they are identical except for their node names. 02/02/20 (c)2001-2003, Michael P. Frank 29 Module #22 - Graphs Adjacency Lists A table with 1 row per vertex, listing its adjacent vertices. Adjacent b a\n\nc d f 02/02/20 e Vertex a b c d e f Vertices b, c a, c, e, f\n\na, b, f b c, b (c)2001-2003, Michael P. Frank 30 Module #22 - Graphs Directed Adjacency Lists 1 row per node, listing the terminal nodes of each edge incident from that node. 02/02/20 (c)2001-2003, Michael P. Frank 31\n\nModule #22 - Graphs Adjacency Matrices A way to represent simple graphs possibly with self-loops. Matrix A=[aij], where aij is 1 if {vi, vj} is an edge of G, and is 0 otherwise. Can extend to pseudographs by letting each matrix elements be the number of links (possibly >1) between the nodes. 02/02/20 (c)2001-2003, Michael P. Frank 32 Module #22 - Graphs Graph Isomorphism\n\nFormal definition: Simple graphs G1=(V1, E1) and G2=(V2, E2) are isomorphic iff a bijection f:V1V2 such that a,bV1, a and b are adjacent in G1 iff f(a) and f(b) are adjacent in G2. f is the renaming function between the two node sets that makes the two graphs identical. This definition can easily be extended to other types of graphs. 02/02/20 (c)2001-2003, Michael P. Frank 33 Module #22 - Graphs Graph Invariants under Isomorphism Necessary but not sufficient conditions for G1=(V1, E1) to be isomorphic to G2=(V2, E2):\n\nWe must have that |V1|=|V2|, and |E1|=|E2|. The number of vertices with degree n is the same in both graphs. For every proper subgraph g of one graph, there is a proper subgraph of the other graph that is isomorphic to g. 02/02/20 (c)2001-2003, Michael P. Frank 34 Module #22 - Graphs Isomorphism Example If isomorphic, label the 2nd graph to show the isomorphism, else identify difference. d b\n\na d e 02/02/20 b c f c (c)2001-2003, Michael P. Frank a f\n\ne 35 Module #22 - Graphs Are These Isomorphic? If isomorphic, label the 2nd graph to show the isomorphism, else identify difference. a b d c 02/02/20 e\n\n(c)2001-2003, Michael P. Frank Same # of vertices Same # of edges Different # of verts of degree 2! (1 vs 3) 36 Module #22 - Graphs 8.4: Connectivity In an undirected graph, a path of length n from u to v is a sequence of adjacent edges going from vertex u to vertex v. A path is a circuit if u=v. A path traverses the vertices along it.\n\nA path is simple if it contains no edge more than once. 02/02/20 (c)2001-2003, Michael P. Frank 37 Module #22 - Graphs Paths in Directed Graphs Same as in undirected graphs, but the path must go in the direction of the arrows. 02/02/20 (c)2001-2003, Michael P. Frank 38\n\nModule #22 - Graphs Connectedness An undirected graph is connected iff there is a path between every pair of distinct vertices in the graph. Theorem: There is a simple path between any pair of vertices in a connected undirected graph. Connected component: connected subgraph A cut vertex or cut edge separates 1 connected component into 2 if removed. 02/02/20 (c)2001-2003, Michael P. Frank 39\n\nModule #22 - Graphs Directed Connectedness A directed graph is strongly connected iff there is a directed path from a to b for any two verts a and b. It is weakly connected iff the underlying undirected graph (i.e., with edge directions removed) is connected. Note strongly implies weakly but not viceversa. 02/02/20 (c)2001-2003, Michael P. Frank 40 Module #22 - Graphs Paths & Isomorphism Note that connectedness, and the existence\n\nof a circuit or simple circuit of length k are graph invariants with respect to isomorphism. 02/02/20 (c)2001-2003, Michael P. Frank 41 Module #22 - Graphs Counting Paths w Adjacency Matrices Let A be the adjacency matrix of graph G. The number of paths of length k from vi to vj is equal to (Ak)i,j. The notation (M)i,j denotes mi,j where [mi,j] = M. 02/02/20\n\n(c)2001-2003, Michael P. Frank 42 Module #22 - Graphs 8.5: Euler & Hamilton Paths An Euler circuit in a graph G is a simple circuit containing every edge of G. An Euler path in G is a simple path containing every edge of G. A Hamilton circuit is a circuit that traverses each vertex in G exactly once. A Hamilton path is a path that traverses each vertex in G exactly once. 02/02/20 (c)2001-2003, Michael P. Frank\n\n43 Module #22 - Graphs Bridges of Knigsberg Problem Can we walk through town, crossing each bridge exactly once, and return to start? A D B C The original problem 02/02/20 Equivalent multigraph (c)2001-2003, Michael P. Frank 44\n\nModule #22 - Graphs Euler Path Theorems Theorem: A connected multigraph has an Euler circuit iff each vertex has even degree. Proof: () The circuit contributes 2 to degree of each node. () By construction using algorithm on p. 580-581 Theorem: A connected multigraph has an Euler path (but not an Euler circuit) iff it has exactly 2 vertices of odd degree. One is the start, the other is the end. 02/02/20 (c)2001-2003, Michael P. Frank 45\n\nModule #22 - Graphs Euler Circuit Algorithm Begin with any arbitrary node. Construct a simple path from it till you get back to start. Repeat for each remaining subgraph, splicing results back into original cycle. 02/02/20 (c)2001-2003, Michael P. Frank 46 Module #22 - Graphs Round-the-World Puzzle Can we traverse all the vertices of a dodecahedron, visiting each once?`\n\nDodecahedron puzzle 02/02/20 Equivalent graph (c)2001-2003, Michael P. Frank Pegboard version 47 Module #22 - Graphs Hamiltonian Path Theorems Diracs theorem: If (but not only if) G is connected, simple, has n3 vertices, and v deg(v)n/2, then G has a Hamilton circuit. Ores corollary: If G is connected, simple, has\n\nn3 nodes, and deg(u)+degdeg(v)n for every pair u,v of non-adjacent nodes, then G has a Hamilton circuit. 02/02/20 (c)2001-2003, Michael P. Frank 48 Module #22 - Graphs HAM-CIRCUIT is NP-complete Let HAM-CIRCUIT be the problem: Given a simple graph G, does G contain a Hamiltonian circuit? This problem has been proven to be NP-complete! This means, if an algorithm for solving it in polynomial time were found, it could be used to\n\nsolve all NP problems in polynomial time. 02/02/20 (c)2001-2003, Michael P. Frank 49 Module #22 - Graphs 8.6: Shortest-Path Problems Not covering this semester. 02/02/20 (c)2001-2003, Michael P. Frank 50 Module #22 - Graphs\n\n8.7: Planar Graphs Not covering this semester. 02/02/20 (c)2001-2003, Michael P. Frank 51 Module #22 - Graphs 8.8: Graph Coloring Not covering this semester. 02/02/20 (c)2001-2003, Michael P. Frank 52\n\nModule #22 - Graphs 9.1: Introduction to Trees A tree is a connected undirected graph that contains no circuits. Theorem: There is a unique simple path between any two of its nodes. A (not-necessarily-connected) undirected graph without simple circuits is called a forest. You can think of it as a set of trees having disjoint sets of nodes. A leaf node in a tree or forest is any pendant or isolated vertex. An internal node is any non-leaf vertex (thus it has degree ___ ). 02/02/20 (c)2001-2003, Michael P. Frank\n\n53 Module #22 - Graphs Tree and Forest Examples Leaves in green, internal nodes in brown. A Tree: 02/02/20 A Forest: (c)2001-2003, Michael P. Frank 54 Module #22 - Graphs Rooted Trees\n\nA rooted tree is a tree in which one node has been designated the root. Every edge is (implicitly or explicitly) directed away from the root. You should know the following terms about rooted trees: Parent, child, siblings, ancestors, descendents, leaf, internal node, subtree. 02/02/20 (c)2001-2003, Michael P. Frank 55 Module #22 - Graphs Rooted Tree Examples Note that a given unrooted tree with n nodes yields n different rooted trees.\n\nroot Same tree except for choice of root root 02/02/20 (c)2001-2003, Michael P. Frank 56 Module #22 - Graphs Rooted-Tree Terminology Exercise Find the parent, children, siblings, ancestors, &\n\ndescendants of node f. e i o n h d r m b a root c\n\ng q f p 02/02/20 j k (c)2001-2003, Michael P. Frank l 57 Module #22 - Graphs\n\nn-ary trees A rooted tree is called n-ary if every vertex has no more than n children. It is called full if every internal (non-leaf) vertex has exactly n children. A 2-ary tree is called a binary tree. These are handy for describing sequences of yes-no decisions. Example: Comparisons in binary search algorithm. 02/02/20 (c)2001-2003, Michael P. Frank 58 Module #22 - Graphs Which Tree is Binary?\n\nTheorem: A given rooted tree is a binary tree iff every node other than the root has degree ___, and the root has degree ___. 02/02/20 (c)2001-2003, Michael P. Frank 59 Module #22 - Graphs Ordered Rooted Tree This is just a rooted tree in which the children of each internal node are ordered. In ordered binary trees, we can define: left child, right child left subtree, right subtree For n-ary trees with n>2, can use terms like\n\nleftmost, rightmost, etc. 02/02/20 (c)2001-2003, Michael P. Frank 60 Module #22 - Graphs Trees as Models Can use trees to model the following: Saturated hydrocarbons Organizational structures Computer file systems In each case, would you use a rooted or a non-rooted tree? 02/02/20\n\n(c)2001-2003, Michael P. Frank 61 Module #22 - Graphs Some Tree Theorems Any tree with n nodes has e = n1 edges. Proof: Consider removing leaves. A full m-ary tree with i internal nodes has n=mi+deg1 nodes, and =(m1)i+deg1 leaves. Proof: There are mi children of internal nodes, plus the root. And, = ni = (m1)i+deg1. Thus, when m is known and the tree is full, we can compute all four of the values e, i, n, and , given any one of them. 02/02/20\n\n(c)2001-2003, Michael P. Frank 62 Module #22 - Graphs Some More Tree Theorems Definition: The level of a node is the length of the simple path from the root to the node. The height of a tree is maximum node level. A rooted m-ary tree with height h is called balanced if all leaves are at levels h or h1. Theorem: There are at most mh leaves in an mary tree of height h. Corollary: An m-ary tree with leaves has height hlogm . If m is full and balanced then h=logm. 02/02/20 (c)2001-2003, Michael P. Frank\n\n63 Module #22 - Graphs 9.2: Applications of Trees Binary search trees A simple data structure for sorted lists Decision trees Minimum comparisons in sorting algorithms Prefix codes Huffman coding Game trees 02/02/20 (c)2001-2003, Michael P. Frank 64\n\nModule #22 - Graphs Binary Search Trees A representation for sorted sets of items. Supports the following operations in (log n) average-case time: Searching for an existing item. Inserting a new item, if not already present. Supports printing out all items in (n) time. Note that inserting into a plain sequence ai would instead take (n) worst-case time. 02/02/20 (c)2001-2003, Michael P. Frank 65\n\nModule #22 - Graphs Binary Search Tree Format Items are stored at individual tree nodes. We arrange for the tree to always obey this invariant: Example: 3 For every item x, Every node in xs left subtree is less than x. Every node in xs right subtree is greater than x. 02/02/20\n\n7 1 0 (c)2001-2003, Michael P. Frank 12 5 2 9 8 15 11 66\n\nModule #22 - Graphs Recursive Binary Tree Insert procedure insert(T: binary tree, x: item) v := root[T] if v = null then begin root[T] := x; return Done end else if v = x return Already present else if x < v then return insert(leftSubtree[T], x) else {must be x > v} return insert(rightSubtree[T], x) 02/02/20 (c)2001-2003, Michael P. Frank 67 Module #22 - Graphs\n\nDecision Trees (pp. 646-649) A decision tree represents a decision-making process. Each possible decision point or situation is represented by a node. Each possible choice that could be made at that decision point is represented by an edge to a child node. In the extended decision trees used in decision analysis, we also include nodes that represent random events and their outcomes. 02/02/20 (c)2001-2003, Michael P. Frank 68 Module #22 - Graphs\n\nCoin-Weighing Problem Imagine you have 8 coins, one of which is a lighter counterfeit, and a free-beam balance. No scale of weight markings is required for this problem! How many weighings are needed to guarantee that the counterfeit coin will be found? 02/02/20 (c)2001-2003, Michael P. Frank ? 69 Module #22 - Graphs\n\nAs a Decision-Tree Problem In each situation, we pick two disjoint and equalsize subsets of coins to put on the scale. A given sequence of weighings thus yields a decision tree with branching factor 3. The balance then decides whether to tip left, tip right, or stay balanced. 02/02/20 (c)2001-2003, Michael P. Frank 70 Module #22 - Graphs\n\nApplying the Tree Height Theorem The decision tree must have at least 8 leaf nodes, since there are 8 possible outcomes. In terms of which coin is the counterfeit one. Recall the tree-height theorem, hlogm. Thus the decision tree must have height h log38 = 1.893 = 2. Lets see if we solve the problem with only 2 weighings 02/02/20 (c)2001-2003, Michael P. Frank 71 Module #22 - Graphs\n\nGeneral Solution Strategy The problem is an example of searching for 1 unique particular item, from among a list of n otherwise identical items. Somewhat analogous to the adage of searching for a needle in haystack. Armed with our balance, we can attack the problem using a divideand-conquer strategy, like whats done in binary search. We want to narrow down the set of possible locations where the desired item (coin) could be found down from n to just 1, in a logarithmic fashion. Each weighing has 3 possible outcomes. Thus, we should use it to partition the search space into 3 pieces that are as close to equal-sized as possible. This strategy will lead to the minimum possible worst-case number of weighings required. 02/02/20 (c)2001-2003, Michael P. Frank\n\n72 Module #22 - Graphs General Balance Strategy On each step, put n/3 of the n coins to be searched on each side of the scale. If the scale tips to the left, then: The lightweight fake is in the right set of n/3 n/3 coins. If the scale tips to the right, then: The lightweight fake is in the left set of n/3 n/3 coins. If the scale stays balanced, then: The fake is in the remaining set of n 2n/3 n/3 coins that were not weighed!\n\nExcept if n mod 3 = 1 then we can do a little better by weighing n/3 of the coins on each side. You can prove that this strategy always leads to a balanced 3-ary tree. 02/02/20 (c)2001-2003, Michael P. Frank 73 Module #22 - Graphs Coin Balancing Decision Tree Heres what the tree looks like in our case: left: 123 1 vs. 2 L:1 02/02/20\n\n123 vs 456 right: 456 balanced: 78 4 vs. 5 7 vs. 8 R:2 B:3 L:4 R:5 B:6 L:7 (c)2001-2003, Michael P. Frank R:8 74\n\nModule #22 - Graphs Prefix Codes & Huffman Coding pp. 649-651 02/02/20 (c)2001-2003, Michael P. Frank 75 Module #22 - Graphs Game Trees pp. 651-656 02/02/20 (c)2001-2003, Michael P. Frank\n\n76 Module #22 - Graphs 9.3: Tree Traversal Universal address systems Traversal algorithms Depth-first traversal: Preorder traversal Inorder traversal Postorder traversal Breadth-first traversal Infix/prefix/postfix notation 02/02/20 (c)2001-2003, Michael P. Frank 77\n\nModule #22 - Graphs 9.4: Spanning Trees Not covering this semester. 02/02/20 (c)2001-2003, Michael P. Frank 78 Module #22 - Graphs 9.5: Minimum Spanning Trees Not covering this semester. 02/02/20 (c)2001-2003, Michael P. Frank\n\n79"
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"https://cdn.slideread.com/cover/slides-for-rosen-5th-edition-academics-wpi-c4hz95.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84654546,"math_prob":0.91927505,"size":24046,"snap":"2020-45-2020-50","text_gpt3_token_len":7096,"char_repetition_ratio":0.2412445,"word_repetition_ratio":0.065735616,"special_character_ratio":0.3173501,"punctuation_ratio":0.14391829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9695678,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T20:47:03Z\",\"WARC-Record-ID\":\"<urn:uuid:f2effeee-ccc5-4901-8822-54a2e6546789>\",\"Content-Length\":\"56879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:102dc221-dea6-4253-8d0a-9ab763f2439a>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc81d7bc-2eac-4aae-a2da-46c17cab31ce>\",\"WARC-IP-Address\":\"104.31.64.124\",\"WARC-Target-URI\":\"https://makeslider.com/slide/slides-for-rosen-5th-edition-academics-wpi-c4hz95\",\"WARC-Payload-Digest\":\"sha1:FWP5AGJHXTZKAW3OG6BV4MGEJ6DLFDGY\",\"WARC-Block-Digest\":\"sha1:RM644FBFJEK2VXCPRANXGEJDEZQ3UCKA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141743438.76_warc_CC-MAIN-20201204193220-20201204223220-00242.warc.gz\"}"} |
https://www.harrisgeospatial.com/docs/imsl_sp_pdsol.html | [
"Welcome to the L3 Harris Geospatial documentation center. Here you will find reference guides and help documents.\n\n\n### IMSL_SP_PDSOL\n\nIMSL_SP_PDSOL\n\nThe IMSL_SP_PDSOL function solves a sparse symmetric positive definite system of linear equations Ax = b.\n\nThe IMSL_SP_PDSOL function solves a system of linear algebraic equations having a sparse symmetric positive definite coefficient matrix A. In IMSL_SP_PDSOL default usage, a symbolic factorization of a permutation of the coefficient matrix is computed first, then a numerical factorization is performed. The solution of the linear system is then found using the numeric factor.\n\nThe symbolic factorization step of the computation consists of determining a minimum degree ordering and then setting up a sparse data structure for the Cholesky factor, L. This step only requires the “pattern” of the sparse coefficient matrix, that is, the locations of the non-zero elements but not any of the elements themselves.\n\nThe numerical factorization can be carried out in one of two ways. By default, the standard factorization is performed based on a sparse compressed storage scheme. This is fully described in George and Liu (1981). Optionally, a multifrontal technique can be used. The multifrontal method requires more storage but will be faster in certain cases. The multifrontal factorization is based on the routines in Liu (1987). For a detailed description of this method, see Liu (1990), also Duff and Reid (1983, 1984), Ashcraft (1987), Ashcraft et al. (1987), and Liu (1986, 1989).\n\nIf an application requires that several linear systems be solved where the coefficient matrix is the same but the right-hand sides change, the IMSL_SP_PDFAC function can be used to precompute the Cholesky factor. Then the keyword FACTOR can be used in IMSL_SP_PDSOL to efficiently solve all subsequent systems.\n\nGiven the numeric factorization, the solution x is obtained by the following calculations:\n\nLy1 = Pb\n\nLTy2 = y1\n\nx = PTy2\n\nThe permutation information, P, is carried in the numeric factor structure.\n\n## Example",
null,
"`A = REPLICATE(imsl_f_sp_elem, 10)`\n`a(*).row = [0, 1, 2, 2, 3, 3, 4, 4, 4, 4]`\n`a(*).col = [0, 1, 0, 2, 2, 3, 0, 1, 3, 4]`\n`a(*).val = [10, 20, 1, 30, 4, 40, 2, 3, 5, 50]`\n`b = [55.0d0, 83, 103, 97, 82]`\n`x = IMSL_SP_PDSOL(b, a)`\n`PM, x`\n` 5.0000000`\n` 4.0000000`\n` 3.0000000`\n` 2.0000000`\n\n## Syntax\n\nResult = IMSL_SP_PDSOL(B [, A] [, /CSC_COL] [, /CSC_ROW] [, /CSC_VAL] [, FACTOR=value] [, LG_DIAG=value] [, MULTIFRONTAL=value] [, N_NONZERO=variable] [, SM_DIAG=value])\n\n## Return Value\n\nA one-dimensional array containing the solution of the linear system Ax = b.\n\n## Arguments\n\n### B\n\nOne-dimensional matrix containing the right-hand side.\n\n### A (optional)\n\nSparse matrix stored as an array of structures containing non-zeros in lower triangle of the coefficient matrix A(i,j). See “Sparse Matrices: Direct Methods” and its related sections for a description of structures used for sparse matrices.\n\n## Keywords\n\n### CSC_COL (optional)\n\nAccept the coefficient matrix in compressed sparse column (CSC) format. See “Sparse Coordinate Storage Format” for a discussion of this storage scheme. The keywords CSC_COL, CSC_ROW, and CSC_VAL must be used together.\n\n### CSC_ROW (optional)\n\nAccept the coefficient matrix in compressed sparse column (CSC) format. See “Sparse Coordinate Storage Format” for a discussion of this storage scheme. The keywords CSC_COL, CSC_ROW, and CSC_VAL must be used together.\n\n### CSC_VAL (optional)\n\nAccept the coefficient matrix in compressed sparse column (CSC) format. See “Sparse Coordinate Storage Format” for a discussion of this storage scheme. The keywords CSC_COL, CSC_ROW, and CSC_VAL must be used together.\n\n### FACTOR (optional)\n\nThe factorization of A as computed by IMSL_SP_PDFAC. If this keyword is used, then the argument a should not be used. This keyword is useful if solutions to systems involving the same coefficient matrix and multiple right-hand sides will be solved.\n\n### LG_DIAG (optional)\n\nThe largest diagonal element that occurred during the numeric factorization. This keyword is not valid if the keyword FACTOR is used.\n\n### MULTIFRONTAL (optional)\n\nIf present and nonzero, perform the numeric factorization using a multifrontal technique. By default a standard factorization is computed based on a sparse compressed storage scheme. The keywords MULTIFRONTAL and FACTOR cannot be used together.\n\n### N_NONZERO (optional)\n\nNamed variable into which the total number of non-zeros in the factor is stored. This keyword is not valid if the keyword FACTOR is used.\n\n### SM_DIAG (optional)\n\nThe smallest diagonal element that occurred during the numeric factorization. This keyword is not valid if the keyword FACTOR is used.\n\n## Version History\n\n 6.4 Introduced"
] | [
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"https://www.harrisgeospatial.com/docs/html/images/IMSL_SP_PDSOL_Example.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.777286,"math_prob":0.92557454,"size":4791,"snap":"2019-51-2020-05","text_gpt3_token_len":1197,"char_repetition_ratio":0.13557552,"word_repetition_ratio":0.1878453,"special_character_ratio":0.24232937,"punctuation_ratio":0.15357968,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99247175,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T16:36:20Z\",\"WARC-Record-ID\":\"<urn:uuid:76d95d1d-fb61-4fad-b5be-01ee34d819a7>\",\"Content-Length\":\"244755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0a831f8-b9f8-41ca-ad6c-2354bafdce2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e9c5ce0-826e-4e59-957a-12b955beaa9e>\",\"WARC-IP-Address\":\"192.5.156.19\",\"WARC-Target-URI\":\"https://www.harrisgeospatial.com/docs/imsl_sp_pdsol.html\",\"WARC-Payload-Digest\":\"sha1:DJEZLKXPAAJYPQRZTAU6QO67VCCCNCOC\",\"WARC-Block-Digest\":\"sha1:65CWW7SXLKES6WN4KWH3CMYWO43QSEK7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250589861.0_warc_CC-MAIN-20200117152059-20200117180059-00194.warc.gz\"}"} |
https://cs.stackexchange.com/questions/95362/right-definition-of-linear-grammar | [
"# Right definition of linear grammar\n\nI was referring book by Peter Linz, which defines linear grammar as follows:\n\nA linear grammar is a grammar in which at most one variable can occur on the right side of any production, without restriction on the position of this variable.\n\nThe wikipedia page also defines it similar way at the top of Linear grammar page\n\na linear grammar is a context-free grammar that has at most one nonterminal in the right hand side of each of its productions.\n\nBy these definition, it seems that the production of the following form is allowed:\n\n$S\\rightarrow aSb$\n\nBut next on the same page, wikipedia defines it as follows:\n\nlinear grammars in which all nonterminals in right hand sides are at the left or right ends, but not necessarily all at the same end.\n\nBy this definition, it seems that the productions of the above form are not allowed.\n\nThen what is correct definition?\n\nThey second quote from Wikipedia has lost its context. Here it is complete, with emphasis added:\n\nAnother special type of linear grammar is the following:\n\n• linear grammars in which all nonterminals in right hand sides are at the left or right ends, but not necessarily all at the same end.\n\nA linear grammar has at most one non-terminal on the right-hand side of any rule, as per Linz' definition.\n\nBut any linear grammar can be transformed into an equivalent linear grammar of that special type, simply by adding a new non-terminal. Wikipedia goes on to show an example of this transformation. So it is possible to assume wolog that a linear grammar is in that form, if it turns out to be helpful.\n\n• Do you mean (1) linear grammar is one which has at most one non terminal on right hand sides of each production (i.e. $S\\rightarrow aSb, S\\rightarrow aS, S\\rightarrow Sb$) (2) left linear is one which have all non terminals on left ends (i.e. $S\\rightarrow Sb$) (3) right linear is one which have all non terminals on right ends (i.e. $S\\rightarrow aS$) and (4) we can have \"another type of linear grammar (not having any specific name)\" which can have productions of both forms: right linear and left linear ($S\\rightarrow aS, S\\rightarrow Sb$ but not $S\\rightarrow aSb$ ))\n– RajS\nJul 21, 2018 at 12:49\n• @anlr: That's basically what Wikipedia is saying, yes. I personally wouldn't have phrased it that way; the point is that the restricted form (non-terminals are always at one end) does not restrict the expressivity of linear grammars. Nevertheless, the key point of a linear grammar is that every rhs has at most one non-terminal. That's why it's called linear. It's like a linear polynomial: the largest exponent must be one (and you can canonicalise by collapsing all the linear terms into a single term).\n– rici\nJul 21, 2018 at 14:38"
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https://people.sc.fsu.edu/~jburkardt/m_src/pyramid_monte_carlo/pyramid_monte_carlo.html | [
"# pyramid_monte_carlo\n\npyramid_monte_carlo, a MATLAB code which estimates the integral of a function F(X,Y,Z) over the interior of the unit pyramid in 3D.\n\nThe unit pyramid has a square base of area 4, and a height of 1. Specifically, the integration region is:\n\n``` - ( 1 - Z ) <= X <= 1 - Z\n- ( 1 - Z ) <= Y <= 1 - Z\n0 <= Z <= 1.\n```\nThe volume of the unit pyramid is 4/3.\n\n### Languages:\n\npyramid_monte_carlo is available in a C version and a C++ version and a FORTRAN90 version and a MATLAB version and a Python version.\n\n### Related Data and Programs:\n\nannulus_monte_carlo a MATLAB code which uses the Monte Carlo method to estimate the integral of a function over the interior of a circular annulus in 2D.\n\nball_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate integrals of a function over the interior of the unit ball in 3D;\n\ncircle_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the circumference of the unit circle in 2D.\n\ncube_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of the unit cube in 3D;\n\ndisk_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the general disk in 2D.\n\ndisk01_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the unit disk in 2D.\n\ndisk01_quarter_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of the unit quarter disk in 2D;\n\nellipse_monte_carlo a MATLAB code which uses the Monte Carlo method to estimate the value of integrals over the interior of an ellipse in 2D.\n\nellipsoid_monte_carlo a MATLAB code which uses the Monte Carlo method to estimate the value of integrals over the interior of an ellipsoid in M dimensions.\n\nhyperball_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of the unit hyperball in M dimensions;\n\nhypercube_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of the unit hypercube in M dimensions;\n\nhypersphere_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function on the surface of the unit hypersphere in M dimensions;\n\nline_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the length of the unit line in 1D;\n\npolygon_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of a polygon in 2D.\n\npyramid_felippa_rule, a MATLAB code which returns Felippa's quadratures rules for approximating integrals over the interior of a pyramid in 3D.\n\npyramid_grid, a MATLAB code which computes a grid of points over the interior of the unit pyramid in 3D;\n\npyramid_integrals, a MATLAB code which returns the exact value of the integral of any monomial over the interior of the unit pyramid in 3D.\n\npyramid_rule, a MATLAB code which computes quadrature rules over the interior of the unit pyramid in 3D.\n\nsimplex_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the unit simplex in M dimensions.\n\nsphere_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function on the surface of the unit sphere in 3D;\n\nsphere_triangle_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over a spherical triangle on the surface of the unit sphere in 3D;\n\nsquare_monte_carlo, a MATLAB code which applies a Monte Carlo method to estimate the integral of a function over the interior of the unit square in 2D;\n\ntetrahedron_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the unit tetrahedron in 3D.\n\ntriangle_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of a triangle in 2D.\n\ntriangle01_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the unit triangle in 2D.\n\nwedge_monte_carlo, a MATLAB code which uses the Monte Carlo method to estimate integrals over the interior of the unit wedge in 3D.\n\n### Source Code:\n\nLast revised on 11 January 2021."
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https://www.nphysics.org/rustdoc/nphysics3d/material/struct.BasicMaterial.html | [
"# [−][src]Struct nphysics3d::material::BasicMaterial\n\n```pub struct BasicMaterial<N: RealField> {\npub id: Option<u32>,\npub restitution: N,\npub friction: N,\npub surface_velocity: Option<Vector<N>>,\npub restitution_combine_mode: MaterialCombineMode,\npub friction_combine_mode: MaterialCombineMode,\n}```\n\nDescription of the state of surface of a solid.\n\nStrictly speaking, the coefficient provided here only exist when considering a pair of touching surfaces. In practice, nphysics will average the coefficient of the two surfaces in contact in order to deduce the restitution/friction coefficient.\n\n## Fields\n\n`id: Option<u32>`\n\nThe ID of this material for automatic lookup.\n\n`restitution: N`\n\nRestitution coefficient of the surface.\n\n`friction: N`\n\nFriction coefficient of the surface.\n\n`surface_velocity: Option<Vector<N>>`\n\nThe fictitious velocity at the surface of this material.\n\n`restitution_combine_mode: MaterialCombineMode`\n\nThe way restitution coefficients are combined if no match was found in the material lookup tables.\n\n`friction_combine_mode: MaterialCombineMode`\n\nThe way friction coefficients are combined if no match was found in the material lookup tables.\n\n## Methods\n\n### `impl<N: RealField> BasicMaterial<N>`[src]\n\n#### `pub fn new(restitution: N, friction: N) -> Self`[src]\n\nInitialize a material with the specified restitution and friction coefficients.\n\n## Blanket Implementations\n\n### `impl<T> Same<T> for T`\n\n#### `type Output = T`\n\nShould always be `Self`\n\n### `impl<T> ToOwned for T where T: Clone, `[src]\n\n#### `type Owned = T`\n\nThe resulting type after obtaining ownership.\n\n### `impl<T, U> TryFrom<U> for T where U: Into<T>, `[src]\n\n#### `type Error = Infallible`\n\nThe type returned in the event of a conversion error.\n\n### `impl<T, U> TryInto<U> for T where U: TryFrom<T>, `[src]\n\n#### `type Error = <U as TryFrom<T>>::Error`\n\nThe type returned in the event of a conversion error."
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https://www.electricalvolt.com/difference-between-ac-and-dc/ | [
"# Difference between AC and DC\n\nLast Updated on January 23, 2023 by Electricalvolt\n\nThis article describes the key differences between AC and DC. But before knowing the AC vs DC, first, we will learn the basics of electric current and its types namely alternating current and direct current individually for a better understanding of the differences between alternating and direct current.\n\nThe term electric current is defined as the continuous flow of electric charges or electrons between two given points across a conductor. Since its discovery in the early 1800s, the electric current has proved its importance in the world of electricity. And, now, electricity has become an essential part of our life. Based on the characteristics of electric current, it is broadly classified into the following two categories,\n\n• Alternating Current (AC)\n• Direct Current (DC)\n\n## What is Alternating Current (AC)?\n\nAlternating current is defined as the type of electric current which changes its magnitude alternately between positive and negative values and its direction at fixed periodic intervals, throughout its time period.\n\nThe abbreviated form of alternating current is AC. Alternating current is used for long-distance transmission of power from power stations to residential areas due to its capability of converting from high voltage to low voltage values, thus taking into account safety issues. Additionally, the use of AC reduces the losses incurred in AC generators and motors and increases their efficacy and performance.\n\nAlternating current is also described as the total number of cycles completed in one time period as shown in Figure 1. The time period is the time taken by a sinusoidal wave to complete one cycle. While frequency, which is also the reciprocal of the time period, is defined as the number of cycles generated by the alternating current waveform per second. The time period and frequency of the AC waveform are related by the following formula:\n\nDepending on the shape of the alternating current waveform, AC can be of the following types:\n\n• Sinusoidal wave\n• Triangular wave\n• Square wave\n• Saw-toothed wave\n\n## What is Direct Current (DC)?\n\nDirect current is described as the unidirectional flow of electric current across a conducting wire. In other words, direct current shows no fluctuations in direction or magnitude across the entire time period for a particular type of load, i.e. its magnitude remains constant across the entire time interval for a specific circuit and, the direction of current does not change.\n\nThe abbreviated form of direct current is DC. This means that the frequency, i.e. the number of fluctuations of the current per second is zero, as DC is just a straight line at a fixed value of current as shown in Figure 2. Direct current is used in various battery-operated devices such as PC, laptops, cell phones, remote control toy cars, etc. requiring short-distance transmission of power.\n\nDepending on the nature of the DC waveform, it can be of the following two types:\n\n• Constant DC – This type of DC current is the desired form of direct current con summed by various electronic devices where the current remains stable or the same.\n• Pulsating DC – This type of DC current is utilized in most Pulse width modulated (PWM) device controllers, where there is a slight fluctuation in the DC signal, which can be smoothed out by using certain types of filters.\n\n## Difference between Alternating Current (AC) and Direct Current (DC)\n\nThe table below lists the important differences between alternating current and direct current –\n\n### Conclusion\n\nIn conclusion, this article describes the major differences between the alternating and direct current flow in electronic devices and how they are utilized for a variety of commercial and domestic purposes. The most significant difference between them is that AC involves the bidirectional flow of electric current, while DC involves the unidirectional flow of current. Overall, both AC and DC play an important role in the operation of various electric power systems.",
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"https://secure.gravatar.com/avatar/69073a777ee1caa5c7cbf394212bfd57",
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https://emilylearning.com/tag/h2stats/ | [
"# Tag: h2stats",
null,
"## Sampling Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level sampling questions, which are found in paper 2 of H2 Math papers, statistics section.",
null,
"## Hypothesis Testing Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level hypothesis testing questions, which are found in paper 2 of H2 Math papers, statistics section.",
null,
"## Regression and Correlation Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level regression and correlation questions, found in paper 2 of H2 Math papers, statistics section.",
null,
"## Binomial Distribution Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level binomial distribution questions. Binomial distribution questions are found in paper 2, statistics section of A Level Math.",
null,
"## Normal Distribution Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level normal distribution questions. Normal distribution questions are found in paper 2, statistics section.",
null,
"## Discrete Random Variable Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level discrete random variable questions, which are found in paper 2 of H2 Math papers, statistics section.",
null,
"## Probability Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level probability questions. Probability questions are found in paper 2.",
null,
"## Permutation and Combination Questions: A Level TYS Answers\n\nFind the worked solutions for past A Level permutation and combination questions. Permutation and combination questions are found in paper 2.",
null,
"## Use of Ti 84 for normal distribution\n\nWe talk about the graphic calculator Ti-84 for normal distribution in this post. We’ll look at normal distribution functions in Ti-84: invNorm and normcdf.",
null,
"## Use of Ti 84 for binomial distribution\n\nWe talk about the graphic calculator Ti 84 for binomial distribution in this post. We’ll look at 2 binomial distribution functions in Ti 84.\n\nerror: Content is protected !!"
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https://publications.waset.org/search?q=%CE%95.%20Giovanis | [
"Commenced in January 2007\nFrequency: Monthly\nEdition: International\nPaper Count: 8\n\n# Search results for: Ε. Giovanis\n\n##### 8 A Study of Neuro-Fuzzy Inference System for Gross Domestic Product Growth Forecasting\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nIn this paper we present a Adaptive Neuro-Fuzzy System (ANFIS) with inputs the lagged dependent variable for the prediction of Gross domestic Product growth rate in six countries. We compare the results with those of Autoregressive (AR) model. We conclude that the forecasting performance of neuro-fuzzy-system in the out-of-sample period is much more superior and can be a very useful alternative tool used by the national statistical services and the banking and finance industry. Downloads 1492\n##### 7 Proposal of Additional Fuzzy Membership Functions in Smoothing Transition Autoregressive Models\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nIn this paper we present, propose and examine additional membership functions for the Smoothing Transition Autoregressive (STAR) models. More specifically, we present the tangent hyperbolic, Gaussian and Generalized bell functions. Because Smoothing Transition Autoregressive (STAR) models follow fuzzy logic approach, more fuzzy membership functions should be tested. Furthermore, fuzzy rules can be incorporated or other training or computational methods can be applied as the error backpropagation or genetic algorithm instead to nonlinear squares. We examine two macroeconomic variables of US economy, the inflation rate and the 6-monthly treasury bills interest rates. Downloads 1412\n##### 6 Application of Adaptive Neuro-Fuzzy Inference System in Smoothing Transition Autoregressive Models\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nIn this paper we propose and examine an Adaptive Neuro-Fuzzy Inference System (ANFIS) in Smoothing Transition Autoregressive (STAR) modeling. Because STAR models follow fuzzy logic approach, in the non-linear part fuzzy rules can be incorporated or other training or computational methods can be applied as the error backpropagation algorithm instead to nonlinear squares. Furthermore, additional fuzzy membership functions can be examined, beside the logistic and exponential, like the triangle, Gaussian and Generalized Bell functions among others. We examine two macroeconomic variables of US economy, the inflation rate and the 6-monthly treasury bills interest rates. Downloads 1520\n##### 5 Application of Adaptive Network-Based Fuzzy Inference System in Macroeconomic Variables Forecasting\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nIn this paper we apply an Adaptive Network-Based Fuzzy Inference System (ANFIS) with one input, the dependent variable with one lag, for the forecasting of four macroeconomic variables of US economy, the Gross Domestic Product, the inflation rate, six monthly treasury bills interest rates and unemployment rate. We compare the forecasting performance of ANFIS with those of the widely used linear autoregressive and nonlinear smoothing transition autoregressive (STAR) models. The results are greatly in favour of ANFIS indicating that is an effective tool for macroeconomic forecasting used in academic research and in research and application by the governmental and other institutions Downloads 1675\n##### 4 Application of Adaptive Neuro-Fuzzy Inference System in the Prediction of Economic Crisis Periods in USA\n\nAuthors: Eleftherios Giovanis\n\nAbstract:\n\nIn this paper discrete choice models, Logit and Probit are examined in order to predict the economic recession or expansion periods in USA. Additionally we propose an adaptive neuro-fuzzy inference system with triangular membership function. We examine the in-sample period 1947-2005 and we test the models in the out-of sample period 2006-2009. The forecasting results indicate that the Adaptive Neuro-fuzzy Inference System (ANFIS) model outperforms significant the Logit and Probit models in the out-of sample period. This indicates that neuro-fuzzy model provides a better and more reliable signal on whether or not a financial crisis will take place. Downloads 1494\n##### 3 Application of Feed-Forward Neural Networks Autoregressive Models in Gross Domestic Product Prediction\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nIn this paper we present an autoregressive model with neural networks modeling and standard error backpropagation algorithm training optimization in order to predict the gross domestic product (GDP) growth rate of four countries. Specifically we propose a kind of weighted regression, which can be used for econometric purposes, where the initial inputs are multiplied by the neural networks final optimum weights from input-hidden layer after the training process. The forecasts are compared with those of the ordinary autoregressive model and we conclude that the proposed regression-s forecasting results outperform significant those of autoregressive model in the out-of-sample period. The idea behind this approach is to propose a parametric regression with weighted variables in order to test for the statistical significance and the magnitude of the estimated autoregressive coefficients and simultaneously to estimate the forecasts. Downloads 1323\n##### 2 Application of Feed-Forward Neural Networks Autoregressive Models with Genetic Algorithm in Gross Domestic Product Prediction\n\nAuthors: E. Giovanis\n\nAbstract:\n\nIn this paper we present a Feed-Foward Neural Networks Autoregressive (FFNN-AR) model with genetic algorithms training optimization in order to predict the gross domestic product growth of six countries. Specifically we propose a kind of weighted regression, which can be used for econometric purposes, where the initial inputs are multiplied by the neural networks final optimum weights from input-hidden layer of the training process. The forecasts are compared with those of the ordinary autoregressive model and we conclude that the proposed regression-s forecasting results outperform significant those of autoregressive model. Moreover this technique can be used in Autoregressive-Moving Average models, with and without exogenous inputs, as also the training process with genetics algorithms optimization can be replaced by the error back-propagation algorithm. Downloads 1552\n##### 1 A Study of Panel Logit Model and Adaptive Neuro-Fuzzy Inference System in the Prediction of Financial Distress Periods\n\nAuthors: Ε. Giovanis\n\nAbstract:\n\nThe purpose of this paper is to present two different approaches of financial distress pre-warning models appropriate for risk supervisors, investors and policy makers. We examine a sample of the financial institutions and electronic companies of Taiwan Security Exchange (TSE) market from 2002 through 2008. We present a binary logistic regression with paned data analysis. With the pooled binary logistic regression we build a model including more variables in the regression than with random effects, while the in-sample and out-sample forecasting performance is higher in random effects estimation than in pooled regression. On the other hand we estimate an Adaptive Neuro-Fuzzy Inference System (ANFIS) with Gaussian and Generalized Bell (Gbell) functions and we find that ANFIS outperforms significant Logit regressions in both in-sample and out-of-sample periods, indicating that ANFIS is a more appropriate tool for financial risk managers and for the economic policy makers in central banks and national statistical services. Downloads 2243"
] | [
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https://nobyq.onlain-credit.ru/in-the-titration-of-a-weak-acid-with-a-strong-base-with-a-1_1-ratio-when-does-ph-equal-the-pka.html | [
"Weak Acid Strong Base Titration The titration of 50.0mL of 0.100M HC 2 H 3 O 2 (Ka=1.8 x 10 -5 ) with 0.100M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added.\nThis is a buffer solution. It is also a very special buffer. Since the two species have equal concentration this is the point at which for a weak acid the pH = pK a. Thus in the titration of a weak acid with a strong base the pK a of the acid can simply be read off the graph as the pH at the half-equivalence point (to the value of the Henderson ...\nRecall that we can use the stoichiometric ratio of weak acid to strong base in the balanced chemical equation to determine which reactant, if any, is in excess: HA (aq) : MOH (aq) is 1:1, so: Weak acid is in excess if n(HA (available) > n(MOH (available) )\nSee full list on uwaterloo.ca\nTitration curves for strong acid v strong base. We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base. Running acid into the alkali. You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge.\nmL of acid added pH buffer zone a “type 2” calculation The START of the titration is the same as a regular (type 1) weak base problem. You know Kb and [B] so you can calculate pH.\nFigure 3. The graph shows a titration curve for the titration of 25.00 mL of 0.100 M CH3CO2H (weak acid) with 0.100 M NaOH (strong base) and the titration curve for the titration of HCl (strong acid) with NaOH (strong base). The pH ranges for the color change of phenolphthalein, litmus, and methyl orange are indicated by the shaded areas.\nDeducing your question in reverse manner : Result: If H+ concentration becomes equal to Ka of the substance, then pH will be equal to pKa. For eg:- Ka of acetic acid is 1.8x10^-5.\nIn the titration of a weak acid with a strong base with a 1:1 ratio, the pH equals the pKa at half the volume it takes to reach the equivalence point. After adding half of the volume of base it takes to reach the equivalence point, the moles of acid and base are equal to one another.\nSec 4.8, pp 158-159 (Acid/Base Titrations), Sec 16.4, pp 729-43 (Titrations and pH Curves). Objectives: You will be able to: (1) determine the hydrogen ion concentration of a weak acid via titration against a strong base, (2) calculate the pH of a weak acid / strong base titration at the endpoint of a\nIn the titration of a weak acid with a strong base with a 1:1 ratio, the pH equals the pKa at half the volume it takes to reach the equivalence point. After adding half of the volume of base it takes to reach the equivalence point, the moles of acid and base are equal to one another.\npH = pKa + log [base]/ [acid] where \"base\" and \"acid\" are the weak acid and its conjugate base. At half equivalence point, the concentrations of the weak acid and its conjugate base are equal. So...\nThis is a buffer solution. It is also a very special buffer. Since the two species have equal concentration this is the point at which for a weak acid the pH = pK a. Thus in the titration of a weak acid with a strong base the pK a of the acid can simply be read off the graph as the pH at the half-equivalence point (to the value of the Henderson ...\nJun 04, 2018 · Cite this paper: Julia Martín, Laura Ortega Estévez, Agustín G. Asuero, On the Titration of a Weak Acid with a Weak Base: Application to the Potentiometric Determination of the Degree of Acidity of Vinegar, Journal of Laboratory Chemical Education, Vol. 6 No. 4, 2018, pp. 91-98. doi: 10.5923/j.jlce.20180604.03.\npH = −log ( 1.8 × 10 − 5) = 4.74. (pH = p Ka at the half-equivalence point in a titration of a weak acid) (d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant.\npH = −log ( 1.8 × 10 − 5) = 4.74. (pH = p Ka at the half-equivalence point in a titration of a weak acid) (d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant.\nThe second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na 2 X(aq) + H 2 O(l) So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid.\nThe pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution.\nQuestion: Consider The Titration Of A Weak Acid With A Dtrong Base. At What Point Of Titration Will PH Equal PKa And Why? A) PH Will Never Equal PKa B) When The Amount Of Acid And Its Conjugate Base Are Equal(which Is The Point At Which The Volume Of Titrant Is 1/2V(e) Or Half Way To The Equivalence Point).\nIn a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and ...\nNov 17, 2015 · Now, assuming that you have a 1:1 mole ratio between the weak acid and the conjugate base, it's important to notice that the reaction consumes one mole of weak acid and produces one mole of conjugate base. Let's say that before adding the strong base, you have x moles of weak acid and x moles of conjugate base in a volume V.\nA diprotic acid is an acid that yields two H+ ions per acid molecule. Examples of diprotic acids are sulfuric acid, H2SO4, and carbonic acid, H2CO3. A diprotic acid dissociates in water in two stages: Because of the successive dissociations, titration curves of diprotic acids have two equivalence points, as shown in Figure 1. The equations for the acid-base reactions occurring between a ...\n6) The key point now is that the ratio of base to acid is 1:1 at a pH of 10.643. That means that exactly half the base was neutralized by HCl. The will give us the volume of HCl required: moles CH 3 NH 2 originally present (0.200 mol/L) (0.1000 L) = 0.0200 mol. exactly 0.0100 mol of CH 3 NH 2 was neutralized. (Remember, half the CH 3 NH 2 gets ...\nWeak Acid Strong Base Titration The titration of 50.0mL of 0.100M HC 2 H 3 O 2 (Ka=1.8 x 10 -5 ) with 0.100M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added.\nThe second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na 2 X(aq) + H 2 O(l) So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid.\n(b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH. Acid-Base Indicators Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value.\nThe titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, acetic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio.\nFirst things first, let’s derive a simple equation that is so often used in connection with buffers: Henderson–Hasselbalch equation. Let’s assume that the buffer in question is a solution of monobasic acid $\\mathrm{HA}$ and its conjugat...\npH During Weak Acid/Strong Base Titration. ... Acetic acid reacts with NaOH in a 1:1 ratio to produce water and sodium acetate. ... The equivalence point is the point at which equal amounts of ...\nIn the titration of a weak acid with a strong base with a 1:1 ratio, the pH equals the pKa at half the volume it takes to reach the equivalence point. After adding half of the volume of base it takes to reach the equivalence point, the moles of acid and base are equal to one another. #1 At the equivalence point in a titration of a weak acid with a strong base, what is present in the solution? (Choose 1-4) 1. All conjugate base of the weak acid. 2. Equal amounts of weak acid and its conjugate base. 3. Excess strong base. 4. All weak acid.\n#1 At the equivalence point in a titration of a weak acid with a strong base, what is present in the solution? (Choose 1-4) 1. All conjugate base of the weak acid. 2. Equal amounts of weak acid and its conjugate base. 3. Excess strong base. 4. All weak acid. Likewise, any aqueous base with an association constant pK b less than about 0, corresponding to pK a greater than about 14, is leveled to OH − and is considered a strong base. Nitric acid, with a pK value of ca. -1.7, behaves as a strong acid in aqueous solutions with a pH greater than 1. At lower pH values it behaves as a weak acid.\nNote: For any weak acid-strong base titration, the pH at the equivalence point will be greater than 7 because the conjugate of the weak acid is a base. Step 7) Once even more OH-is added, the pH of the solution continues to rise. In the titration of a weak acid with a strong base, the conjugate base of the weak acid will make the pH at the equivalence point greater than 7. Therefore, you would want an indicator to change in that pH range. Both methyl orange and bromocresol green change color in an acidic pH range, while phenolphtalein changes in a basic pH. Jun 04, 2018 · Cite this paper: Julia Martín, Laura Ortega Estévez, Agustín G. Asuero, On the Titration of a Weak Acid with a Weak Base: Application to the Potentiometric Determination of the Degree of Acidity of Vinegar, Journal of Laboratory Chemical Education, Vol. 6 No. 4, 2018, pp. 91-98. doi: 10.5923/j.jlce.20180604.03. Jan 30, 2020 · This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. For example, concentrated vinegar (acetic acid, which is a weak acid) could have a lower pH than a dilute solution of hydrochloric acid (a strong acid). On the other hand, the pKa value is constant for each type of molecule. pH = −log ( 1.8 × 10 − 5) = 4.74. (pH = p Ka at the half-equivalence point in a titration of a weak acid) (d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant.\nThe second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na 2 X(aq) + H 2 O(l) So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid.\nLocathah 5e names\nRecall that we can use the stoichiometric ratio of weak acid to strong base in the balanced chemical equation to determine which reactant, if any, is in excess: HA (aq) : MOH (aq) is 1:1, so: Weak acid is in excess if n(HA (available) > n(MOH (available) ) In the titration of a weak acid with a strong base, the conjugate base of the weak acid will make the pH at the equivalence point greater than 7. Therefore, you would want an indicator to change in that pH range. Both methyl orange and bromocresol green change color in an acidic pH range, while phenolphtalein changes in a basic pH. The pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution. The second titration corresponding to the reaction of the second proton with sodium hydroxide NaHX(aq) + NaOH(aq) = Na 2 X(aq) + H 2 O(l) So, in essence, titrations of a weak polyprotic acid with a strong mono protic base are a combination of a number of titrations depending on the number of acidic protons on the polyprotic acid.\nIn a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and ...\nmL of acid added pH buffer zone a “type 2” calculation The START of the titration is the same as a regular (type 1) weak base problem. You know Kb and [B] so you can calculate pH.\nM1a scout black\npH = −log ( 1.8 × 10 − 5) = 4.74. (pH = p Ka at the half-equivalence point in a titration of a weak acid) (d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. #1 At the equivalence point in a titration of a weak acid with a strong base, what is present in the solution? (Choose 1-4) 1. All conjugate base of the weak acid. 2. Equal amounts of weak acid and its conjugate base. 3. Excess strong base. 4. All weak acid.\nFor example, the titration curve for the titration between oxalic acid (a weak acid) and sodium hydroxide (a strong base) is pictured. The equivalence point occurs between pH 8-10, indicating the solution is basic at the equivalence point and an indicator such as phenolphthalein would be appropriate.\nFigure 3. The graph shows a titration curve for the titration of 25.00 mL of 0.100 M CH3CO2H (weak acid) with 0.100 M NaOH (strong base) and the titration curve for the titration of HCl (strong acid) with NaOH (strong base). The pH ranges for the color change of phenolphthalein, litmus, and methyl orange are indicated by the shaded areas.\nAsus vivobook s14 m433ia price in bd\nTitration curves for strong acid v strong base. We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base. Running acid into the alkali. You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge.\nThere are three major differences between strong acid/strong base and weak acid/strong base titrations. For weak acid/strong base titrations: 1. The weak-acid solution has a higher initial pH. 2.\n\n## Factorio petroleum gas setup\n\nPsalm 119 nkjv\nJun 04, 2018 · Cite this paper: Julia Martín, Laura Ortega Estévez, Agustín G. Asuero, On the Titration of a Weak Acid with a Weak Base: Application to the Potentiometric Determination of the Degree of Acidity of Vinegar, Journal of Laboratory Chemical Education, Vol. 6 No. 4, 2018, pp. 91-98. doi: 10.5923/j.jlce.20180604.03.\nAug 02, 2008 · As the first poster said, the pH would be equal to 7. If it was a titration between a strong acid and weak base, the pH would be less than 7. If it was a titration between a strong base and a weak acid, the pH would be greater than 7. [Answer: see above]\n6) The key point now is that the ratio of base to acid is 1:1 at a pH of 10.643. That means that exactly half the base was neutralized by HCl. The will give us the volume of HCl required: moles CH 3 NH 2 originally present (0.200 mol/L) (0.1000 L) = 0.0200 mol. exactly 0.0100 mol of CH 3 NH 2 was neutralized. (Remember, half the CH 3 NH 2 gets ...\n1973 vw super beetle hood for sale\nStress and safety ppt\nThe pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution.\n6) The key point now is that the ratio of base to acid is 1:1 at a pH of 10.643. That means that exactly half the base was neutralized by HCl. The will give us the volume of HCl required: moles CH 3 NH 2 originally present (0.200 mol/L) (0.1000 L) = 0.0200 mol. exactly 0.0100 mol of CH 3 NH 2 was neutralized. (Remember, half the CH 3 NH 2 gets ...\nThe pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution.\nPfp asx dividend\nWestern union speedpay phone number\nNov 17, 2015 · Now, assuming that you have a 1:1 mole ratio between the weak acid and the conjugate base, it's important to notice that the reaction consumes one mole of weak acid and produces one mole of conjugate base. Let's say that before adding the strong base, you have x moles of weak acid and x moles of conjugate base in a volume V.\nApr 17, 2020 · The exact ratio of a buffer's base to acid concentration is given by the Henderson-Hasselbalch equation, which states the pH is equal to the pKa minus the logarithm of the concentration of the base divided by the concentration of the acid. When the concentrations are equal, the logarithmic value becomes zero, and the pH equals the pKa.\nCellranger crispr\nLuxury waterfront homes for sale in charlotte nc\nHp ultraslim docking station firmware check\nBeach view airbnb nsw\nWkhs usps truck\nTbc warlock guide\nHow to unbanned whatsapp number 2019\nPrediksi jitu togel jakarta hari ini\nPioneer carrozzeria english firmware\nChapter 1 quiz 1 (lessons 1 1 and 1 2)\nAngle pair relationships calculator\nHow to connect pax 3 to app\nDeepika padukone in dance plus 5 full episode\nSwiftui datepicker with year\nBanned alcohol\nHow to enable bluetooth in ubuntu\nProvidence police substation locations\nEngineering internship resume objective\nMbe 4000 low oil pressure\n2019 toyota camry body kit\nAirpods 2 keep disconnecting from mac\n\n## Inner thigh liposuction recovery\n\nSamsung washing machine repair cost\nIn a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and ... At the midpoint, the half of acid is neutralized, forming the salt, which is its conjugated base, being in ratio 1:1 to the acid. $\\endgroup$ – Poutnik Feb 9 at 13:49 $\\begingroup$ Then the same should apply at the equivalence point, why doesn't pKa = pH at it then? $\\endgroup$ – Positron12 Feb 9 at 14:00 For example, the titration curve for the titration between oxalic acid (a weak acid) and sodium hydroxide (a strong base) is pictured. The equivalence point occurs between pH 8-10, indicating the solution is basic at the equivalence point and an indicator such as phenolphthalein would be appropriate. Apr 17, 2020 · The exact ratio of a buffer's base to acid concentration is given by the Henderson-Hasselbalch equation, which states the pH is equal to the pKa minus the logarithm of the concentration of the base divided by the concentration of the acid. When the concentrations are equal, the logarithmic value becomes zero, and the pH equals the pKa. Aug 02, 2008 · As the first poster said, the pH would be equal to 7. If it was a titration between a strong acid and weak base, the pH would be less than 7. If it was a titration between a strong base and a weak acid, the pH would be greater than 7. [Answer: see above]\nTraffic school online for tickets\nData structure and algorithmic thinking with python by narasimha karumanchi pdf\nGreat compromise significance apush\nIn a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and ... base to be identical, then the pH of the solution will exactly equal the pK a of the weak acid. Second, it is evident that any given ratio of concentrations of conjugate base to parent weak acid will specify a unique value of pH in the solution. Alternatively, setting the pH of a solution that contains a weak acid and its conjugate base will ... This chemistry video tutorial explains how to calculate the pH of a weak acid strong base titration problem before, at, and beyond the equivalence point usin...\nBest talisman for damage hypixel skyblock\nGta v paid mod menu 2020\nNov 17, 2015 · Now, assuming that you have a 1:1 mole ratio between the weak acid and the conjugate base, it's important to notice that the reaction consumes one mole of weak acid and produces one mole of conjugate base. Let's say that before adding the strong base, you have x moles of weak acid and x moles of conjugate base in a volume V. This is a buffer solution. It is also a very special buffer. Since the two species have equal concentration this is the point at which for a weak acid the pH = pK a. Thus in the titration of a weak acid with a strong base the pK a of the acid can simply be read off the graph as the pH at the half-equivalence point (to the value of the Henderson ...\nPrefab homes for sale ontario\nForest river salem fsx 178bhsk\nKeystone rv online application\nAt the midpoint, the half of acid is neutralized, forming the salt, which is its conjugated base, being in ratio 1:1 to the acid. $\\endgroup$ – Poutnik Feb 9 at 13:49 $\\begingroup$ Then the same should apply at the equivalence point, why doesn't pKa = pH at it then? $\\endgroup$ – Positron12 Feb 9 at 14:00 Deducing your question in reverse manner : Result: If H+ concentration becomes equal to Ka of the substance, then pH will be equal to pKa. For eg:- Ka of acetic acid is 1.8x10^-5. In the titration of a weak acid with a strong base with a 1:1 ratio, when does pH equal the pKa? At half the volume it takes to reach the equivalence point The titration of NH3 with HCl produces a salt and water.\nExam monitor sdu test\n2006 polaris ranger 500 review\nRidgid octane drill driver\npH = pKa + log [base]/ [acid] where \"base\" and \"acid\" are the weak acid and its conjugate base. At half equivalence point, the concentrations of the weak acid and its conjugate base are equal. So... Mar 29, 2017 · EXTREMLEY LONG ANSWER !! For the sake of simplicity, I'll use #beta#-#\"HA\"# to denote the acid and #beta#-#\"A\"^(-)# to denote the conjugate base.. Now, the pH of the solution before the addition of strong base can be calculated by using the fact that the weak acid only partially ionizes in a #1:1# mole ratio to produce hydronium cations and #beta#-hydroxybutyrate anions\nTurtlebot3 simulation\nAxi data fifo example\nRecall that we can use the stoichiometric ratio of weak acid to strong base in the balanced chemical equation to determine which reactant, if any, is in excess: HA (aq) : MOH (aq) is 1:1, so: Weak acid is in excess if n(HA (available) > n(MOH (available) ) The pH calculations for the titration of weak acid with a strong base are generally more difficult than those for a strong acid with a strong base. One reason is that the salt formed in the neutralization reaction can't be ignored as it will be a basic salt rather than a neutral salt and will affect the pH of the solution. See full list on uwaterloo.ca\nHow to get into the leviathan raid solo\nZenith ch750 review\nWill bacon soaked sponges kill coyotes\nAug 02, 2008 · As the first poster said, the pH would be equal to 7. If it was a titration between a strong acid and weak base, the pH would be less than 7. If it was a titration between a strong base and a weak acid, the pH would be greater than 7. [Answer: see above] A diprotic acid is an acid that yields two H+ ions per acid molecule. Examples of diprotic acids are sulfuric acid, H2SO4, and carbonic acid, H2CO3. A diprotic acid dissociates in water in two stages: Because of the successive dissociations, titration curves of diprotic acids have two equivalence points, as shown in Figure 1. The equations for the acid-base reactions occurring between a ...\nMath state test 2020\nArkansas traffic crash statistics\nSample feedback comments for student writing\nApr 17, 2020 · The exact ratio of a buffer's base to acid concentration is given by the Henderson-Hasselbalch equation, which states the pH is equal to the pKa minus the logarithm of the concentration of the base divided by the concentration of the acid. When the concentrations are equal, the logarithmic value becomes zero, and the pH equals the pKa.\nFood delivery react native template\nBarndominium for sale georgia\nMgw shifter gt350\nIn the titration of a weak acid with a strong base with a 1:1 ratio, the pH equals the pKa at half the volume it takes to reach the equivalence point. After adding half of the volume of base it takes to reach the equivalence point, the moles of acid and base are equal to one another. (b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH. Acid-Base Indicators Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. In a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and ...\nCcmexec exe vulnerability\nOutlook display settings in status bar\nConcave lens glasses\nFirst things first, let’s derive a simple equation that is so often used in connection with buffers: Henderson–Hasselbalch equation. Let’s assume that the buffer in question is a solution of monobasic acid $\\mathrm{HA}$ and its conjugat... A diprotic acid is an acid that yields two H+ ions per acid molecule. Examples of diprotic acids are sulfuric acid, H2SO4, and carbonic acid, H2CO3. A diprotic acid dissociates in water in two stages: Because of the successive dissociations, titration curves of diprotic acids have two equivalence points, as shown in Figure 1. The equations for the acid-base reactions occurring between a ...\nHow to minimize csgo\nCon edison bill pdf\nMbb recruiting reddit\npH is equal to that pKa, the proportion of acetate to acetic acid would be 1:1. For another acid, such as HF, which has a pKa value of about 4.0, when the pH is equal to 4.0, the proportion of the fluoride ion to the hydrofluoric acid would be 1:1. Ask yourself the following question, and hopefully you will come up with the correct answer. If ... Apr 17, 2020 · The exact ratio of a buffer's base to acid concentration is given by the Henderson-Hasselbalch equation, which states the pH is equal to the pKa minus the logarithm of the concentration of the base divided by the concentration of the acid. When the concentrations are equal, the logarithmic value becomes zero, and the pH equals the pKa. When a weak acid is titrated by a strong base, the fact that, in aqueous solution, the weak acid dissociates into a hydrogen ion and the conjugate base of the acid changes the appearance of the titration curve. The curve will look similar to Figure 3, which represents the titration of 0.1 M acetic acid with 0.1 M NaOH. NaOH titration of CH ...\nAustralian cobberdog hypoallergenic\nWar room bible study lessons\nWsl2 no internet\nDeducing your question in reverse manner : Result: If H+ concentration becomes equal to Ka of the substance, then pH will be equal to pKa. For eg:- Ka of acetic acid is 1.8x10^-5. See full list on uwaterloo.ca\nWrx cvt stage 2 dyno\n\n## Contract law questions\n\nSolving multi step equations with distributive property\n1988 dodge ramcharger fuel sending unit\nFaxon chameleon bcg lightweight\nWord search solver\nHappy birthday to the most selfless person i know\nAmazon workers conditions\nVolvo check engine light reset\nOld school mini truck wheels for sale\nDevice id changer (root)\nEureka math lesson 22 homework 24 answer key\nMashin sentai kirameiger episode 5 raw\nDiscord.py error handling\nAbp news live bihar today in hindi\nDiagram of graphite\nMercedes slk\nGet amplitude from fft matlab\nEmu birds for sale near me\nConvert memmap to numpy array\nI 49 arkansas accident\nRows unbounded preceding\nMiracle weight loss drink recipe\nNovember zodiac sign 2019\nSweatshops in america 1900s\nP0300 chevy blazer misfire\nLidar drone price\nSendgrid api v2\nHow much water should i drink to lose weight per day\nOxford picture dictionary for the content areas pdf\nMobile mechanic near me\nDua for success in exam results\nVendor management checklist template\nTesseract\nEdea concerto vs ice fly\nSia door supervisor licence\nDean bikes\nBm18010 power beyond kit\nMpc 1000 ebay\nHeyoka empath twin flame\nAzure functions python import module\nWolayta words\nAfter 2 years of a clean driving record how much are your total points reduced_\nMsi bios changed\nIgcse syllabus 2020 ict\nHp envy ahci mode\nCnx player android\nSolar twinkle lights for trees\nWhat do we learn about abby and mercy in act 4\nFuture self visualization script\n17539 rowland st city of industry ca 91748 usa"
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https://www.alexanderpublications.com/products/reactive-metering-concepts-videos-and-books | [
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"# Reactive Metering Concepts - Videos and Related Books\n\n##### Summary\n\nThis 34-minute video uses the power triangle to illustrate the relationships between active power, apparent power, and reactive power. The concept of metering reactive power is explained. In addition, this unit explains how a phase-shifting device can be used to produce the phase relationships needed to meter reactive power with conventional kWh meters.\n\n##### Training Sessions\n\nPower Relationships – 11 minutes\nDefines reactive power, and explains why reactive power consumption is measured. Using phasor diagrams, explains the relationship between active power and reactive power. Explains how net reactive current flow is determined. Explains how to determine total current flow in a circuit using the parallelogram method of phasor addition.\n\nDetermining Power in a Circuit – 13 minutes\nExplains how the Pythagorean theorem and the power triangle can be used to determine power values. Explains how to determine the power factor of a circuit using mathematical functions. States the formulas that can be used to calculate active power, reactive power, and apparent power in a balanced three-phase system. Describes how reactive power can be metered using two kWh meters and a phase shifting transformer.\n\nReactive Metering Connections – 10 minutes\nDescribes how two kilowatthour meters and two phase shifting transformers may be connected to meter reactive power in a three-phase, three-wire system. Explains how the 90-degree phase shift is accomplished in a reactive meter installation.\n\nItem Code: RMC\n##### Videos\nReactive Metering Concepts - Instant Video\nWatch this video now. Subscribers can watch more than 200 T&D videos any time, as often as they wish. If you're an instructor, we'll set up a virtual classroom for you. Contact us. More info\nReactive Metering Concepts - DVD"
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https://csound.com/docs/manual/tablei.html | [
"## tablei\n\ntablei — Accesses table values by direct indexing with linear interpolation.\n\n## Description\n\nAccesses table values by direct indexing with linear interpolation.\n\n## Syntax\n\n`ares tablei andx, ifn [, ixmode] [, ixoff] [, iwrap]`\n`ires tablei indx, ifn [, ixmode] [, ixoff] [, iwrap]`\n`kres tablei kndx, ifn [, ixmode] [, ixoff] [, iwrap]`\n\n## Initialization\n\nifn -- function table number. tablei requires the extended guard point.\n\nixmode (optional) -- index data mode. The default value is 0.\n\n• 0 = raw index\n\n• 1 = normalized (0 to 1)\n\nixoff (optional) -- amount by which index is to be offset. For a table with origin at center, use tablesize/2 (raw) or .5 (normalized). The default value is 0.\n\niwrap (optional) -- wraparound index flag. The default value is 0.\n\n• 0 = nowrap (index < 0 treated as index=0; index > tablesize sticks at index=size)\n\n• 1 = wraparound.\n\n## Performance\n\ntablei is a interpolating unit in which the fractional part of index is used to interpolate between adjacent table entries. The smoothness gained by interpolation is at some small cost in execution time (see also oscili, etc.), but the interpolating and non-interpolating units are otherwise interchangeable. Note that when tablei uses a periodic index whose modulo n is less than the power of 2 table length, the interpolation process requires that there be an (n + 1)th table value that is a repeat of the 1st (see f Statement in score).\n\nWarning",
null,
"Reading tables which contain stereo- or multichannel-information probably causes unwanted noise as the opcode interpolates between subsequent positions of the table regardless of the origin of its content. Usually just the content of one channel is expected Consider using loscilx instead. b\n\n## Examples\n\nHere is an example of the tablei opcode. It uses the file tablei.csd.\n\nExample 1069. Example of the tablei opcode.\n\nSee the sections Real-time Audio and Command Line Flags for more information on using command line flags.\n\n```<CsoundSynthesizer>\n<CsOptions>\n; Select audio/midi flags here according to platform\n-odac ;;;realtime audio out\n; For Non-realtime ouput leave only the line below:\n; -o tablei.wav -W ;;; for file output any platform\n</CsOptions>\n<CsInstruments>\n\nsr = 44100\nksmps = 32\nnchnls = 2\n0dbfs = 1\n\nseed 0\t;generate new values every time the instr is played\n\ninstr 1\n\nifn = p4\nisize = p5\nithresh = 0.5\n\nitemp ftgen ifn, 0, isize, 21, 2\n\niwrite_value = 0\ni_index = 0\n\nloop_start:\n\niwrite_value = 1\nelse\niwrite_value = -1\nendif\ntableiw iwrite_value, i_index, ifn\nloop_lt i_index, 1, isize, loop_start\nturnoff\n\nendin\n\ninstr 2\n\nifn = p4\nisize = ftlen(ifn)\nprints \"Index\\tValue\\n\"\n\ni_index = 0\nloop_start:\nivalue tablei i_index, ifn\nprints \"%d:\\t%f\\n\", i_index, ivalue\n\nloop_lt i_index, 1, isize, loop_start\t\t;read table 1 with our index\n\naout oscili .5, 100, ifn\t\t\t;use table to play the polypulse\nouts aout, aout\n\nendin\n</CsInstruments>\n<CsScore>\ni 1 0 1 100 16\ni 2 0 2 100\ne\n</CsScore>\n</CsoundSynthesizer>\n```"
] | [
null,
"https://csound.com/docs/manual/images/warning.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60316896,"math_prob":0.94490516,"size":2899,"snap":"2023-40-2023-50","text_gpt3_token_len":801,"char_repetition_ratio":0.13402417,"word_repetition_ratio":0.04140787,"special_character_ratio":0.26733357,"punctuation_ratio":0.15145986,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95999575,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T16:04:12Z\",\"WARC-Record-ID\":\"<urn:uuid:d559fbf3-fe61-4597-8076-d2ba8beb4b33>\",\"Content-Length\":\"17178\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb3260d1-75ba-4700-bb15-ceed7a17d7bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:02a52f29-5b23-4c98-a630-703793b21eb3>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://csound.com/docs/manual/tablei.html\",\"WARC-Payload-Digest\":\"sha1:4KHEAA4UNFAPWZD7LOHPS3SBFYIJ2VQS\",\"WARC-Block-Digest\":\"sha1:HB7PE4TNCO6WZN76LZJI2RSRMMBPM7JP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100912.91_warc_CC-MAIN-20231209134916-20231209164916-00829.warc.gz\"}"} |
https://in.mathworks.com/matlabcentral/cody/problems/2559-check-that-number-is-whole-number/solutions/2135482 | [
"Cody\n\n# Problem 2559. Check that number is whole number\n\nSolution 2135482\n\nSubmitted on 21 Feb 2020 by Jason Walker\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 15; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 1\n\n2 Pass\nx = 15.25; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 0\n\n3 Pass\nx = 10; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))\n\ny = 1\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56892747,"math_prob":0.98120654,"size":572,"snap":"2021-04-2021-17","text_gpt3_token_len":179,"char_repetition_ratio":0.14612676,"word_repetition_ratio":0.09090909,"special_character_ratio":0.3339161,"punctuation_ratio":0.1388889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9829435,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-16T01:05:30Z\",\"WARC-Record-ID\":\"<urn:uuid:a80a37d8-5eb5-46e3-b11d-f27abcb321d9>\",\"Content-Length\":\"80313\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80c6023d-dfa5-4551-bc26-fc73a246206f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a1e98cb2-d15a-4907-b312-9e949372e9c4>\",\"WARC-IP-Address\":\"23.56.12.57\",\"WARC-Target-URI\":\"https://in.mathworks.com/matlabcentral/cody/problems/2559-check-that-number-is-whole-number/solutions/2135482\",\"WARC-Payload-Digest\":\"sha1:766465L4L6AOJNI44RKIMNXBGFOV3I7A\",\"WARC-Block-Digest\":\"sha1:45PMHCXAMJYPEWAXIVXGHHNVOCB6TYBO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703497681.4_warc_CC-MAIN-20210115224908-20210116014908-00369.warc.gz\"}"} |
https://wiki.call-cc.org/eggref/5/kd-tree | [
"• egg\n\n## kd-tree\n\nK-D tree implementation.\n\n## Documentation\n\nThis library implements a K-D tree, which is a data structure for organizing and searching points in k-dimensional space.\n\nThe K-D tree is a binary search tree in which every branching node contains a k-dimensional point, and every leaf node contains a set of points. Every branching node represents a splitting hyperplane that divides the space into two parts, known as half-spaces.\n\nPoints to the left of the splitting hyperplane are contained in the left subtree of the node and points right of the hyperplane are contained in the right subtree. The splitting hyperplane is chosen so as to be perpendicular to one of the axes in the k-dimensional space. The axis at each branching level is chosen in a round-robin fashion. For instance, in 3-D space, at level 0, the chosen axis is X, so points are divided according to their X-coordinates; at level 1, the chosen axis is Y, so the points are divided according to their Y-coordinates; at the next branch level the chosen axis is Z, and so on.\n\n### K-dimensional point space\n\nModule kspace provides facilities for managament of K-dimensional point spaces.\n\n[procedure] make-space:: COORDS -> SPACE\n\nGiven a list of coordinate collections of length K, constructs a yasos K-dimensional point space object. The coordinate collections can be SRFI-4 f32vectors, or collection objects as defined in the yasos collections module.\n\n[procedure] space? :: OBJECT -> BOOL\n\nK-dimensional point space predicate.\n\n[procedure] dimension :: OBJECT -> INT\n\nReturns the dimensionality of the point space.\n\n[procedure] point :: OBJECT * INT -> REAL LIST\n\nReturns the coordinates of the point at the given index.\n\n[procedure] coord :: OBJECT * INT * INT -> FLOAT\n\nReturns the k'th coordinate of i'th point, starting from 0.\n\n[procedure] compare-coord :: SPACE * INT * INT * INT -> INT\n\nGiven the indices of two points and a coordinate index, compares the respective coordinates of the two points and returns -1, 0, or 1, depending on whether the coordinates are less than, equal, or greater than each other.\n\n[procedure] squared-distance :: SPACE * INT * INT -> FLOAT\n\nReturns the square of the Euclidean distance between the points at the given indices.\n\n[procedure] compare-distance :: SPACE * INT * INT -> INT\n\nCompares the square of the Euclidean distance between the points at the given indices.\n\n### K-D tree interface\n\n#### Constructors\n\n[procedure] make-kd-tree:: SPACE -> OBJECT\n\nGiven a kspace object, constructs and returns a yasos spatial map object.\n\n#### Predicates\n\n[procedure] spatial-map? :: OBJECT -> BOOL\n\nReturns #t if the given object is a spatial map, #f otherwise.\n\n[procedure] empty? :: OBJECT -> BOOL\n\nReturns #t if the given spatial map object is empty, #f otherwise.\n\n[procedure] kd-tree-is-valid? :: OBJECT -> BOOL\n\nChecks whether the K-D tree property holds for the given spatial map. Specifically, it tests that all points in the left subtree lie to the left of the plane, p is on the plane, and all points in the right subtree lie to the right.\n\n[procedure] kd-tree-all-subtrees-are-valid? :: OBJECT -> BOOL\n\nChecks whether the K-D tree property holds for the given spatial map and all subtrees.\n\n#### Accessors\n\n[procedure] get-kspace :: SPATIAL-MAP -> KSPACE\n\nReturns the underlying kspace object of the map.\n\n[procedure] spatial-map->list :: SPATIAL-MAP -> POINT LIST\n\nReturns a list with the points contained in the spatial map.\n\n#### Query procedures\n\n[procedure] nearest-neighbor :: SPATIAL-MAP * POINT -> POINT\n\nNearest neighbor of a point.\n\n[procedure] near-neighbors :: SPATIAL-MAP * POINT * RADIUS -> POINT LIST\n\nNeighbors of a point within radius r.\n\n[procedure] k-nearest-neighbors :: SPATIAL-MAP * POINT * INT -> POINT LIST\n\nK nearest neighbors of a point.\n\n[procedure] slice :: SPATIAL-MAP * AXIS * COORD * COORD -> POINT LIST\n\nReturns all points between two planes.\n\n#### Combinators\n\n[procedure] spatial-map-for-each :: SPATIAL-MAP * F -> VOID\n\nPoint iterator.\n\n[procedure] spatial-map-fold-right :: SPATIAL-MAP * F * INIT -> INIT\n\nFold on points.\n\n[procedure] spatial-map-fold-right* :: SPATIAL-MAP * F * INIT -> INIT\n\nFold on points and their indices.\n\n#### Modifiers\n\n[procedure] kd-tree-remove :: SPATIAL-MAP * POINT -> SPATIAL-MAP\n\n## Examples\n\n```(import scheme (chicken base)\nyasos random-mtzig kspace kd-tree)\n\n(let* (\n(n (inexact->exact 1e5)) (k 40) (r 1.0) (randst (init 9))\n\n;; generate random coordinates\n(xs (randn/f32! n randst))\n(ys (randn/f32! n randst))\n(zs (randn/f32! n randst))\n;; create a kspace\n(pts (list xs ys zs))\n(kspace3d (make-space pts))\n;; create the spatial map\n(kdt (make-kd-tree kspace3d))\n;; choose a random point index\n(xi (inexact->exact (modulo (random! randst) n)))\n;; retrieve the coordinates of the chosen point\n(x (point kspace3d xi))\n)\n\n(print \"nearest-neighbor of \" x \": \" (nearest-neighbor kdt x))\n(print k \" nearest neighbors of \" x \": \" (k-nearest-neighbors kdt x k))\n(print \"near neighbors of \" x \" within \" r \": \" (near-neighbors kdt k r))\n\n)\n```\n\n## Version history\n\n• 6.0 : refactored to use yasos, ported to CHICKEN 5\n• 5.0 : added list->kd-tree* procedure to KdTree class\n• 4.1-4.8 : Using f64vector for internal point representation\n• 4.0-4.1 : Added with-distance? flag to kd-tree-near-neighbors\n• 3.2 : Bug fix in kd-tree-near-neighbors\n• 2.0 : Improvements to internal representation\n• 1.0 : Initial release\n\n## License\n\n```Copyright 2012-2019 Ivan Raikov\n\nThis program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.\n\nThis program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.\n\nA full copy of the GPL license can be found at http://www.gnu.org/licenses/.```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79408914,"math_prob":0.9267305,"size":5841,"snap":"2019-35-2019-39","text_gpt3_token_len":1450,"char_repetition_ratio":0.15024841,"word_repetition_ratio":0.09347826,"special_character_ratio":0.24910118,"punctuation_ratio":0.14627415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789729,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T08:08:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a44a1b28-18ee-4ec3-8857-e002c9665341>\",\"Content-Length\":\"13887\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9cdbd33f-7aab-4e10-95b7-73ce6eb9990d>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a17042b-0b87-4a2c-aed1-499445409498>\",\"WARC-IP-Address\":\"78.47.93.131\",\"WARC-Target-URI\":\"https://wiki.call-cc.org/eggref/5/kd-tree\",\"WARC-Payload-Digest\":\"sha1:WMG5CKHMKNTZEHVOWOMQMDSKSLNFGWA2\",\"WARC-Block-Digest\":\"sha1:UFX7CGOACTLTXMOOX2BCFUEZ5XDQZK2R\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315811.47_warc_CC-MAIN-20190821065413-20190821091413-00127.warc.gz\"}"} |
https://math.stackexchange.com/questions/2676607/product-of-manifolds-with-or-without-boundary | [
"# Product of manifolds with or without boundary\n\nLet $M_1, M_2, ... M_k$ be smooth manifolds with or without boundary such that at most one of these has nonempty boundary, say $M_a$ has nonempty boundary where $a$ is in $\\{1,2, ..., k\\}$.\n\nThen it is said that $M_1 \\times M_2 \\times ... \\times M_k$ is a smooth manifold with boundary. I am curious what its boundary is.\n\nAlso, does the same contents hold for just topological manifolds? I am somewhat confused...\n\n• See also here. – Jan Bohr Mar 4 '18 at 18:34\n\n## 1 Answer\n\nIn both cases the boundary is $X= M_1\\times\\dots\\times\\partial M_a \\times\\dots\\times M_k$.\n\nMy post was downvoted, so perhaps I missed the intention of the question? The reason the boundary is as above is because any point not in $X$ is contained in a coordinate neighborhood given by $(U,\\phi)=(U_1\\times\\dots\\times U_k,\\phi_1\\times\\dots\\times\\phi_k)$, where $(U_i,\\phi_i)$ is an interior coordinate chart for each $i$. This neighborhood is thus an interior coordinate chart for $M$, which follows because the product of open subsets $\\mathbb{R}^{n_i}$ is an open subset of $\\mathbb{R}^{n_1+\\dots +n_k}$.\n\nFor a point $p=(p_1,\\dots,p_k)$ in $X$, the corresponding $(U,\\phi)$ is a product of an open subset of $\\mathbb{H}^{n_a}$ and open subsets of $\\mathbb{R}^{n_i}$, $i\\ne a$. This is an open subset of $\\mathbb{H}^{n_1+\\dots+n_k}$, with $p$ in the boundary.\n\nSo we have precisely $\\partial M=X$. A similar argument works for topological manifolds."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8587239,"math_prob":0.99966943,"size":947,"snap":"2019-35-2019-39","text_gpt3_token_len":288,"char_repetition_ratio":0.14103924,"word_repetition_ratio":0.03076923,"special_character_ratio":0.29778248,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999943,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T11:58:56Z\",\"WARC-Record-ID\":\"<urn:uuid:c194266a-a3df-490d-be28-73c28140affe>\",\"Content-Length\":\"134033\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a631a31d-e6c9-449b-a128-8ee6bdc302c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:20dec526-97db-4e0e-a512-4be481c1dae5>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2676607/product-of-manifolds-with-or-without-boundary\",\"WARC-Payload-Digest\":\"sha1:JOPFGNLXJKIMI4JFQGHTF6CWABTDWEEY\",\"WARC-Block-Digest\":\"sha1:VH4Z5GNHNAZS2IYH6PAK5J2LPIGPALMB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027320734.85_warc_CC-MAIN-20190824105853-20190824131853-00150.warc.gz\"}"} |
http://aasheeinfotech.com/web-design-course-analysis/ | [
"Select Page\n\nA. Methods and procedures\n\nFive three and fifty seven eight and three-grade students in university participated in this study. Among them, there were 185 students took the course under the first teaching way that teaching every functions of Dreamweaver, Firework and Flash according to the menus of the software. The other 172 students took the course under the second teaching way that teaching according a real living e-commence web site. They were randomly assigned to groups using animation, still-graphic and text treatments before taking the course. They received their treatments and took a criterion test. The material used for designing a web sit which sell second-hand book integrated physics and mathematics concepts in a situated environment. Various scenarios for solving scientific problems were provided. To be successful problem solvers, students need to develop familiarity with several scientific concepts, including levers, direction of force, resultant force, and composition of forces, component forces and equilibrium of force. The lesson was designed in three versions. The following is a description of the treatment:\n\n• Treatment 1\n\nIn this version, only textual information was presented to explain scientific concepts.\n\n• Treatment 2\n\nIn this version, static graphics with textual information were presented to explain scientific concepts.\n\n• Treatment 3\n\nIn this version, the instructional materials contain textual instructions and animated graphics Students interacted with the learning materials independently. To assess the students’ performance, a criterion reference test created based on the content was provided.\n\nThe criterion reference test contained 25 testing item. Twelve of them were to assess the students’ learning of descriptive knowledge, and thirteen were to assess students’ learning of procedural knowledge.\n\nB. Results\n\nComparison of mean scores among boys and girls in prior physics, mathematics and post –test scores (including descriptive and procedural knowledge). Since from Pearson correlation analysis, students’ grade level, prior physics and mathematics scores were significantly correlated with post-test scores for both descriptive and procedural knowledge (p<0.05), all of these factors were used as covariates for controlling the initial difference among groups, when the effects of teaching methods and visual format were analyzed . In both descriptive and procedural material design, the 3(text/Still-Graphics/Animation) x 2 (first teaching way, second teaching way) ANCOVAs were employed to examine the effects of teaching ways and visual format. Separate ANCOVAs were also used to describe the effects of visual treatments on different spatial groups.\n\n• Descriptive learning\n\nFor descriptive learning, the 3X2 ANCOVA indicated insignificant effects for both visual format [F9 (2, 348) = 2.532, p=0,081] and teaching ways [F91, 348] = 1.786, p=0.182]. No interaction was found between the two variables [F (2, 348) = 0.404, p=0.668] (a=0.05 level) (table 2). Form the adjusted mean scores, pair comparisons all revealed insignificant differences (p>0, 05). In order to determine whether visual format influence learning of descriptive knowledge under first teaching way and second teaching way, separate analyses were conduced. It was found that under second teaching way, the effect of visual format was insignificant [F (2, 179) = 0.384, p=0.682]. None of the pair comparisons showed significant differences (a = 0.05 level). However, under first teaching way, the effect of visual format was significant [F (2, 166) = 4.494, p=0.013]. Pair comparisons among treatments indicated significant differences for animation vs. text group (p = 0.009) and still graphics vs. text group (p =0.010) (a = 0.05 level) (Table3)\n\n• Procedural Learning\n\nFor procedural Learning, the 3 X 2 ANCOVA indicated a moderating effect for visual format [F (2, 348) = 2.973, p=0.052], and significant effect for teaching ways [F (1, 348) = 5.853, p=0.0168]. No interaction was found between the two variables [F (2, 348) = 2.221, p = 0.1108] (a = 0.05 level) (Table 2). As determined by the adjusted mean scores, the second teaching way is better than the first teaching way (p =0.016). Pair comparisons indicated significant differences only for graphics vs. Text group (p = 0.012). To determine whether the visual format influenced learning of procedural knowledge differently in two teaching ways, separate analyses were conducted. It was found that using first teaching way, the effect of visual format was insignificant [f (92,179) = 1.111, p=0.331], and none of the pair comparisons among treatments showed significant differences (a = 0.05 level). However, under first teaching level, the effect of visual format was significant [F (2, 166) = 4.635, p = 0.011], and pair comparisons among treatments indicated significant a difference for still graphics vs. text group (p = 0.003), and insignificant differences for still graphics vs. animation group (p =0.102) and still graphics vs. animation (p =0.143) (a = 0.05 level) (Table 3)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95033926,"math_prob":0.83539355,"size":5087,"snap":"2021-31-2021-39","text_gpt3_token_len":1125,"char_repetition_ratio":0.1288609,"word_repetition_ratio":0.102228045,"special_character_ratio":0.24257912,"punctuation_ratio":0.14408602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9527591,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T02:23:26Z\",\"WARC-Record-ID\":\"<urn:uuid:73d35d16-9696-4568-8a10-ef089292bb96>\",\"Content-Length\":\"49028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a77c964-c23c-489e-ab7d-4a43d4084747>\",\"WARC-Concurrent-To\":\"<urn:uuid:c38eab64-341b-423b-b5ee-2534105f4c14>\",\"WARC-IP-Address\":\"51.75.145.175\",\"WARC-Target-URI\":\"http://aasheeinfotech.com/web-design-course-analysis/\",\"WARC-Payload-Digest\":\"sha1:54XSWNFHRZVOKTRRKWAC64WNJ4TDJ4XC\",\"WARC-Block-Digest\":\"sha1:A4WRTSULRHDMGMIJEB2D6QTAYAREW4OI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057496.18_warc_CC-MAIN-20210924020020-20210924050020-00117.warc.gz\"}"} |
https://gomathanswerkey.com/texas-go-math-grade-8-module-4-answer-key/ | [
"# Texas Go Math Grade 8 Module 4 Answer Key Nonproportional Relationships\n\nRefer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 4 Answer Key Nonproportional Relationships.\n\n## Texas Go Math Grade 8 Module 4 Answer Key Nonproportional Relationships\n\nEssential Question\nHow can you use non proportional relationships to solve real-world problems?",
null,
"Complete these exercises to review skills you will need for this chapter.\n\nFind each difference.\n\nQuestion 1.\n3 – (-5) _______\n3 – (-5) Given\n3 + 5 To subtract an integer, add its opposite.\n|3| + |-5| Since the signs are same, so find the sum of the absolute values\n3 + 5 = 8\n8 Use the sign of the number with the greater absolute value.\n\nQuestion 2.\n-4 – 5 _______\n-4 – 5 Given\n-4 + (-5) The signs are the same, so find the sum of the absolute values.\n|-4| + |-5|\n4 + 5 = 9\n-9 Use the sign of the number with the greater absolute value\n\nQuestion 3.\n6 – 10 _____\n6 – 10 Given\n6 + (-10) The signs are different so find the difference of the absolute values.\n(-10) + (6)\n|-10| – |6|\n10 – 6 = 4\n-4 Use the sign of the number with the greater absolute value.\n\nQuestion 4.\n-5 – (-3) ________\n-5 – (-3) Given\n-5 + 3 To subtract an integer, add its opposite.\n|-5| – |3| Since the signs are different, so find the difference of the absolute values\n5 – 3 = 2\n-2 Use the sign of the number with the greater absolute value\n\nQuestion 5.\n8 – (-8) ________\n8 – (-8) Given\n8 + 8 To subtract an integer, add its opposite.\n|8| + |8| Since the signs are different, so find the difference of the absolute values\n8 + 8 = 16\n16 Use the sign of the number with the greater absolute value\n\nQuestion 6.\n9 – 5 = _____\n9 – 5 Given\n9 + (-5) To subtract an integer, add its opposite.\n|9| – |5| Since the signs are different, so find the difference of the absolute values\n9 – 5 = 4\n4 Use the sign of the number with the greater absolute value\n\nQuestion 7.\n-3 – 9 ______\n-3 – 9 Given\n-3 + (-9) The signs are the same, so find the sum of the absolute values.\n|-3| + |-9|\n3 + 9 = 12\n-12 Use the sign of the number with the greater absolute value\n-12\n\nQuestion 8.\n0 – (-6) _____\n0 – (-6) Given\n0 + 6 To subtract an integer, add its opposite.\n|0| + |6| Since the signs are different, so find the difference of the absolute values\n0 + 6 = 6\n6 Use the sign of the number with the greater absolute value\n6\n\nQuestion 9.\n12 – (-9) _____\n12 – (-9) Given\n12 + 9 To subtract an integer, add its opposite.\n|12| + |9| Since the signs are same, so find the difference of the absolute values\n12 + 9 = 21\n21 Use the sign of the number with the greater absolute value\n21\n\nQuestion 10.\n-6 – (-4) _______\n-6 – (-4) Given\n-6 + 4 To subtract an integer, add its opposite.\n|-6| – |4| Since the signs are different, so find the difference of the absolute values\n6 – 4 = 2\n-2 Use the sign of the number with the greater absolute value\n\nQuestion 11.\n-7 – 10 _______\n-7 – 10 Given\n-7 + (-10) To subtract an integer, add its opposite.\n|-7| + |-10| Since the signs are same, so find the difference of the absolute values\n7 + 10 = 17\n-17 Use the sign of the number with the greater absolute value\n-17\n\nQuestion 12.\n5 – 14 = ___\n5 – 14 Given\n5 + (-14) The signs are different, so find the difference of the absolute values\n(-14) + (5)\n|-14| – |5|\n14 – 5 = 11\n-11 Use the sign of the number with the greater absolute value\n-11\n\nGraph each point on the coordinate grid.",
null,
"Question 13.\nB (0, 5)\nTo graph a point at (0, 5) start at origin.\nThen move 5 points up.\nGraph point B(0, 5).",
null,
"To graph a point at (0, 5) start at origin.\nThen move 5 points up.\n\nQuestion 14.\nC (8, 0)",
null,
"Question 15.\nD (5, 7)",
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"Question 16.\nE(2, 3)",
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"Visualize Vocabulary\nUse the ✓ words to complete the diagram. You can put more than one word in each box.",
null,
"Understand Vocabulary\n\nComplete the sentences using the preview words.\n\nQuestion 1.\nAny ordered pair that satisfies all the equations in a system is a _____"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76175284,"math_prob":0.9916098,"size":4621,"snap":"2021-43-2021-49","text_gpt3_token_len":1317,"char_repetition_ratio":0.21160927,"word_repetition_ratio":0.3423913,"special_character_ratio":0.3523047,"punctuation_ratio":0.10042283,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9977968,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T22:29:35Z\",\"WARC-Record-ID\":\"<urn:uuid:13fc82af-d4d8-4909-9de2-fc7f4246e806>\",\"Content-Length\":\"86747\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b71356d-9005-4e04-ae70-0119862e49f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3de4106-e0f1-4797-b1d5-dffba183d090>\",\"WARC-IP-Address\":\"194.1.147.77\",\"WARC-Target-URI\":\"https://gomathanswerkey.com/texas-go-math-grade-8-module-4-answer-key/\",\"WARC-Payload-Digest\":\"sha1:VRJEDAAU4QS3UBS6ZVVITCFC4VMEGR62\",\"WARC-Block-Digest\":\"sha1:E5P2LD257KXZ5VNXO2EHAP4R5PVUN4GW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363226.68_warc_CC-MAIN-20211205221915-20211206011915-00575.warc.gz\"}"} |
https://gluon.mxnet.io/chapter07_distributed-learning/training-with-multiple-machines.html | [
"# Distributed training with multiple machines¶\n\nIn the previous two tutorials, we saw that using multiple GPUs within a machine can accelerate training. The speedup, however, is limited by the number of GPUs installed in that machine. And it’s rare to find a single machine with more than 16 GPUs nowadays. For some truly large-scale applications, this speedup might still be insufficient. For example, it could still take many days to train a state-of-the-art CNN on millions of images.\n\nIn this tutorial, we’ll discuss the key concepts you’ll need in order to go from a program that does single-machine training to one that executes distributed training across multiple machines. We depict a typical distributed system in the following figure, where multiple machines are connected by network switches.",
null,
"Note that the way we used copyto to copy data from one GPU to another in the multiple-GPU tutorial does not work when our GPUs are sitting on different machines. To make use of the available resources here well need a better abstraction.\n\n## Key-value store¶\n\nMXNet provides a key-value store to synchronize data among devices. The following code initializes an ndarray associated with the key “weight” on a key-value store.\n\nIn :\n\nfrom mxnet import kv, nd\nstore = kv.create('local')\nshape = (2, 3)\nx = nd.random_uniform(shape=shape)\nstore.init('weight', x)\nprint('=== init \"weight\" ==={}'.format(x))\n\n=== init \"weight\" ===\n[[ 0.54881352 0.59284461 0.71518934]\n[ 0.84426576 0.60276335 0.85794562]]\n<NDArray 2x3 @cpu(0)>\n\n\nAfter initialization, we can pull the value to multiple devices.\n\nIn :\n\nfrom mxnet import gpu\nctx = [gpu(0), gpu(1)]\ny = [nd.zeros(shape, ctx=c) for c in ctx]\nstore.pull('weight', out=y)\nprint('=== pull \"weight\" to {} ===\\n{}'.format(ctx, y))\n\n=== pull \"weight\" to [gpu(0), gpu(1)] ===\n[\n[[ 0.54881352 0.59284461 0.71518934]\n[ 0.84426576 0.60276335 0.85794562]]\n<NDArray 2x3 @gpu(0)>,\n[[ 0.54881352 0.59284461 0.71518934]\n[ 0.84426576 0.60276335 0.85794562]]\n<NDArray 2x3 @gpu(1)>]\n\n\nWe can also push new data value into the store. It will first sum the data on the same key and then overwrite the current value.\n\nIn :\n\nz = [nd.ones(shape, ctx=ctx[i])+i for i in range(len(ctx))]\nstore.push('weight', z)\nprint('=== push to \"weight\" ===\\n{}'.format(z))\nstore.pull('weight', out=y)\nprint('=== pull \"weight\" ===\\n{}'.format(y))\n\n=== push to \"weight\" ===\n[\n[[ 1. 1. 1.]\n[ 1. 1. 1.]]\n<NDArray 2x3 @gpu(0)>,\n[[ 2. 2. 2.]\n[ 2. 2. 2.]]\n<NDArray 2x3 @gpu(1)>]\n=== pull \"weight\" ===\n[\n[[ 3. 3. 3.]\n[ 3. 3. 3.]]\n<NDArray 2x3 @gpu(0)>,\n[[ 3. 3. 3.]\n[ 3. 3. 3.]]\n<NDArray 2x3 @gpu(1)>]\n\n\nWith push and pull we can replace the allreduce function defined in multiple-gpus-scratch by\n\ndef allreduce(data, data_name, store):\nstore.push(data_name, data)\nstore.pull(data_name, out=data)\n\n\n## Distributed key-value store¶\n\nNot only can we synchronize data within a machine, with the key-value store we can facilitate inter-machine communication. To use it, one can create a distributed kvstore by using the following command: (Note: distributed key-value store requires MXNet to be compiled with the flag USE_DIST_KVSTORE=1, e.g. make USE_DIST_KVSTORE=1.)\n\nstore = kv.create('dist')\n\n\nNow if we run the code from the previous section on two machines at the same time, then the store will aggregate the two ndarrays pushed from each machine, and after that, the pulled results will be:\n\n[[ 6. 6. 6.]\n[ 6. 6. 6.]]\n\n\nIn the distributed setting, MXNet launches three kinds of processes (each time, running python myprog.py will create a process). One is a worker, which runs the user program, such as the code in the previous section. The other two are the server, which maintains the data pushed into the store, and the scheduler, which monitors the aliveness of each node.\n\nIt’s up to users which machines to run these processes on. But to simplify the process placement and launching, MXNet provides a tool located at tools/launch.py.\n\nAssume there are two machines, A and B. They are ssh-able, and their IPs are saved in a file named hostfile. Then we can start one worker in each machine through:\n\n\\$ mxnet_path/tools/launch.py -H hostfile -n 2 python myprog.py\n\n\nIt will also start a server in each machine, and the scheduler on the same machine we are currently on.",
null,
"## Using kvstore in gluon¶\n\nAs mentioned in our section on training with multiple GPUs from scratch, to implement data parallelism we just need to specify\n\n• how to split data\n• how to synchronize gradients and weights\n\nWe already see from multiple-gpu-gluon that a gluon trainer can automatically aggregate the gradients among different GPUs. What it really does is having a key-value store with type local within it. Therefore, to change to multi-machine training we only need to pass a distributed key-value store, for example,\n\nstore = kv.create('dist')\ntrainer = gluon.Trainer(..., kvstore=store)\n\n\nTo split the data, however, we cannot directly copy the previous approach. One commonly used solution is to split the whole dataset into k parts at the beginning, then let the i-th worker only read the i-th part of the data.\n\nWe can obtain the total number of workers by reading the attribute num_workers and the rank of the current worker from the attribute rank.\n\nIn :\n\nprint('total number of workers: %d'%(store.num_workers))\nprint('my rank among workers: %d'%(store.rank))\n\ntotal number of workers: 1\nmy rank among workers: 0\n\n\nWith this information, we can manually access the proper chunk of the input data. In addition, several data iterators provided by MXNet already support reading only part of the data. For example,\n\nfrom mxnet.io import ImageRecordIter\ndata = ImageRecordIter(num_parts=store.num_workers, part_index=store.rank, ...)\n\n\nFor whinges or inquiries, open an issue on GitHub."
] | [
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"https://gluon.mxnet.io/chapter07_distributed-learning/chapter07_distributed-learning/img/dist_kv.svg",
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"https://gluon.mxnet.io/chapter07_distributed-learning/chapter07_distributed-learning/img/dist_kv.svg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79970175,"math_prob":0.95078504,"size":5734,"snap":"2019-13-2019-22","text_gpt3_token_len":1565,"char_repetition_ratio":0.10820244,"word_repetition_ratio":0.035516094,"special_character_ratio":0.29246598,"punctuation_ratio":0.1569966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9631731,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-20T18:40:06Z\",\"WARC-Record-ID\":\"<urn:uuid:d5f2d550-5acb-46c7-99e8-a34579c90826>\",\"Content-Length\":\"42074\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85cd77c6-64dc-4e91-8af7-f2d7b8f6a7c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9b4dc7d-1b0f-4e8f-a261-3613cad57e59>\",\"WARC-IP-Address\":\"99.84.216.100\",\"WARC-Target-URI\":\"https://gluon.mxnet.io/chapter07_distributed-learning/training-with-multiple-machines.html\",\"WARC-Payload-Digest\":\"sha1:PW2NP6KI3F7C7FTZHNM64LWYCYVK5PTJ\",\"WARC-Block-Digest\":\"sha1:S64KD5K4UCPMOPOZIPQ3LK3M7EZIILXW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256100.64_warc_CC-MAIN-20190520182057-20190520204057-00194.warc.gz\"}"} |
https://yro.ch/character-network-analysis-distant-reading-and-the-infinity/ | [
"# Character Network Analysis, Distant Reading and Infinity\n\nIn this post written on a rainy Sunday, I gather the concepts of mathematical induction and distant reading around character network analysis. Probably that it should have been divided into to smaller posts, but I found interesting to discuss in parallel scalability in mathematics and literary studies.\n\nAs a mathematician, there are many concepts I discovered in my studies or later that suddenly blew the view I had on the world. During my first year in gymnase (college), I didn’t get a single thing of what was happening at the physics course, where we were studying mechanics, i.e. an application of calculus, without having ever studied calculus before. But when we studied that subject in the mathematics course one year later, suddenly everything became clear: the derivative describes what is happening at a single moment, and by extension at all the others. It is the tendency of the quantity described by a function to grow, remain constant, or decrease. Then the derivative of the derivative… describes how the derivative behaves. Yep, you can do that infinitely. In the physics course, we had studied how to represent the position of a point in space (e.g. a car) at any moment in time by a formula. Deriving it would give us the speed of the car. Speed can be mentally computed by dividing the distance from point A to point B by the time it takes. But more importantly, the speed given by the derivative is the speed at one exact moment in time: it is what happens if A and B are so close that they get mixed up as the same point. Doing the mental computation here is problematic since you need to divide zero by something that looks like zero too. The derivative lets you know the speed of the car like if you were reading it on the dashboard.\n\nI want to draw a parallel between one of these experiences and what I’m working on now, but I feel the need to keep sidetracking.\n\nFirst year of grad school corresponds to the peak of mathematical mindblowness in my life (it is also the year when I studied the most). Having to prove everything you do and not letting one single detail pass through turned everything upside down in my life. Calculus was not about numbers, was not about calculating anything. It was about epsilon, delta, i, k, x_i, x_j, limits, series and the beautiful | meaning “such that”, among other. We were spending all our time proving. I began to see the world differently through mathematical lenses.\n\nUnlike calculus, linear algebra was in majority made of pain in the ass pen and paper calculations, but after a few weeks you would discover that you had dived in spaces made of so many dimensions your brain had difficulties to see how to move through this. At first you were studying a space of 2 dimensions, then a space of 3 dimensions… only to realise that not so many things had changed. And then a space of 4 dimensions. (You see the pattern.) You can do whatever you want, you have received the means to find your way there through calculation, and maybe later you’ll “see” something (and you’ll help me find a nice and clean answer to this question people have asked me so many times: “Can you see in four dimensions?”). Ever done one of these magical PCA’s but never understood how it works? That’s linear algebra, and it’s never too late.\n\nThe second semester of the geometry course presented differential geometry: the applications and images were so beautiful that practicing for this course was like playing a game. It didn’t require much efforts (in comparison to physics); you were writing down abstract formulas and in return these formulas were shaping incredible stuff and would turn intervals of real numbers into weird creations in space. You would wander on these brainfucked manifolds (do I need to say that there is no limit in the number of dimensions here also?) and re-do your calculus and your car simulations during the whole sunny afternoon.\n\n### EXERCISE\n\nIf you want to have something to think about on the subject of dimensions, I can propose a small exercise: you just need to remember what perpendicularity is and we are ready to define hyperplanes!\n\nThere are many ways to define a straight line: one is to pick a vector, choose a point and that’s it. How does that work? Grab a sheet of paper: a vector is defined by a direction (choose a point anywhere on the physical border of the page), an orientation (towards that point or in the opposite direction) and a length. At this point of the exercise, you have drawn a nice arrow. This arrow is a vector, and you can put it anywhere you want on the page: it is the same vector if it has remained parallel to the original, has the same length and still goes with the same orientation. You do not need to fix it to a point. An infinite number of lines are defined by being perpendicular to this vector, but a unique straight line is defined by being perpendicular to the vector and passing through this point. That’s it for the two-dimensional case.\n\nWhat happens in a three-dimensional space? Well, if you take a vector and a point, you find… an infinite number of straight lines. Huh, are we stuck at this point? On the contrary: what results from drawing all these straight lines is a plane, that is to say a two-dimensional space embedded into a three-dimensional space (We are used to “space” being only three-dimensional, but the more general definition of (Euclidean) space does not have this limitation.). What we had in the previous paragraph was a one-dimensional space (the straight line) embedded into a two-dimensional space (the plane). Now you have to believe me, and probably that you are seeing it coming: what happens when you take a vector and a point in a four-dimensional space? You are defining an infinite number of planes, and all these planes taken together form a three-dimensional space: a hyperplane of dimension 3. By extension, the plane we constructed before was a hyperplane of dimension 2 and the straight line was a hyperplane of dimension 1. Thus, what have you done? We have just manipulated geometrical objects in a space of more than three dimensions, and we have defined a way to describe a mathematical object at absolutely any dimension equal or larger than 2. Now before jumping to dimension 10, maybe take some time already to try to think about what this implies, about what a four-dimensional space looks like.\n\n## Zero, One, Infinity\n\nWe have seen that computation can be conducted on stuff we cannot physically apprehend (at least me). I could keep writing on how operational research transformed my everyday life into the eternal search for the shortest path, or how game theory made me see our society as a Markov chain of absurd decisions, but at this point I am going to focus on two ideas, one at the core of hyperbolic geometry, the other at the core of statistics, which I am convinced are somehow related. Then I talk about distant reading, promised.\n\nEuclide defined geometry with five axioms from which all the rest could be deducted. Four of them are indisputable, but the fifth–stating that given a straight line and a point not belonging to that line, then there exists one and only one straight line parallel to the first line and passing through this point–lead to an open discussion: is it really an axiom or can we deduce it from the four others? Adopting an opposite approach to that question, in the 19th Century some mathematicians began to ask: what if we keep the first four axioms and we drop or replace the fifth one? What if, given a straight line and a point, we could draw more than one parallel? Or could not draw any? I studied the consequences of these questions during the first semester of the geometry course: stating that there are no parallels leads to spherical geometry (among others) while stating that there is more than one (meaning an infinite number) leads to hyperbolic geometry, which incidentally was the subject of my master thesis. This pattern has been stuck in my mind since: 0, 1, infinite. In my master thesis, I discussed one of the models used to represent hyperbolic geometry in two dimensions: the Poincaré half-plane model.\n\n(Please skip the next paragraph if you are not in the mood for some mathematical nerdiness more or less unrelated to the subject in title. It looks like seven and half years later I need to fulfil some outburst of nostalgia.)\n\nIn this model, you draw a straight line and consider only one of the two halves that it defines. The straight line is the infinite, just like the farthest point at the end of any straight lines you would draw perpendicular to the “infinity” line. Now we define what a straight line is in this model, with this geometry, because it is not the straight line we know anymore. In the real world, the one from our college years, a straight line is often defined as the shortest way from one given point to another given point, extended infinitely in both directions. Here, this is more or less the same, but the metrics–the way to measure the distance between two points–has changed. For the sake of coherence, we are renaming the “straight line” by its mathematical name: the “geodesic”. In that model, given two points, there are two possibilities of geodesics (in fact one, but visually two), a fact that I am not proving here. I need you to believe me: a geodesic between two points is the arc delimited by these points that is part of the half-circle passing through these points and whose center is on the “infinity” line. The (visual) second type of geodesics occurs when the two points are aligned on a Euclidean perpendicular to the “infinity” straight line: in this case the geodesic passing through these two points is the perpendicular itself. Now what about parallelism? Well, given a geodesic and a point that does not belong to this geodesic, a parallel to the geodesic passing through the given point is any geodesic that does not cross it: draw any half-circle with the center on the “infinity” line, containing this point and not crossing the geodesic and that’s it: you have your parallel. Thus now you can see that there is an infinite number of possibilities.\n\nFor your information, spherical geometry happens on a… sphere, and the geodesics are the great circles: all the largest circles you can draw; they cut the sphere in two equal parts. And you can verify: it is impossible to draw two great circles that would not intersect. Parallelism does not exist there.\n\nYou often find this pattern in mathematics, since zero is the neutral element for addition, one the neutral element for multiplication, and the infinite the place where all the stuff you don’t control disappear. Most of the time, this can be linked to mathematical induction: given a statement you want to prove, zero is a special case, one is a starting case that is easy to check. Then, if you can prove that when assuming the statement is true for any natural number n it implies that it works for the number n+1, you have proved it for any case up to the infinity.\n\n## One, Two, Infinity\n\nIn a recent past, I was teaching assistant in a Statistics course for first year grad students in Psychology. It lasted five years, and by the time I moved somewhere else I knew the structure of the course by heart. However, it is only when reaching the end of the Ph.D. thesis writing epoch that this structure became an influential transferable mathematical pattern like the numerous examples given earlier in this blog post.\n\nWe studied statistical hypothesis tests during the whole second semester. The tests were classified along two dimensions. The first one was the type of variable: nominal, ordinal or numeric. The second one concerned the number of samples: one, two, or more. (For the sake of precision, there was also a special case: one sample but two measures instead of one.) The course proposed a statistical test for each combination and the whole thing was presented in a nice table that made their exam revisions easier. Here, the connection I am trying to infer may be perceived as exaggerated, but I cannot help thinking that all this is related, taking the risk of finishing this post on an open ending.\n\nOne, two, infinite: this is how I see the construction of methods aimed at analysing corpuses going from one to all the literary works. The need to be capable of detailing one work, of comparing two works (implying by iteration the comparison of many more), and the will to never stop increasing the stream of novels in order to be able to compare them all. Can it be linked to the zero, one, infinite impression?\n\nNowadays, I am exploring the relatively recent concept of character network analysis, that is the study of the characters of a novel, with a focus on their relations. Most of the time, a character network is the model of a novel’s discourse: it positions characters one to another based on their interactions in the text. There is a lot to ask and answer on what this object means, what it represents relative to the character-system, to the discourse and to the story. In my works, I focus more on network analysis and statistical methods, and this is were all this mathematical preamble links in my opinion to distant reading (but I do not know if this helps somehow). Here is the approach I have started following when working on character network analysis methodology.\n\n### One Work\n\nStudying one work is essential to set and define the construction process. Reading and annotating that work before confronting it to the resulting network(s) must have been done a few times in order to understand that process and maybe improve it. Indeed, this concerns more than one novel, but this is not applicable to all the novels we want to study, since hopefully this quantity is huge. The extracting method can be manual, but preferably automatic. This is the right moment to think and test various possibilities since it does not require too much work and computing time.\n\nThen comes the facultative question of how to visually represent it, which is easily solved in the case of a corpus of one.\n\nEventually, the other challenging part is to develop the analytical framework: which measures should we use, when, and for what results? Is it relevant to import methods from social network analysis (SNA), should we adjust them, should we start from scratch? I have started studying centrality measures and a few other commonly used methods, and I believe that the importation from SNA is possible and necessary, keeping in mind of course that what we obtain is not a social network, not even a fictitious social network of the society of characters in the story, but a representation of the character-system, i.e. the way the author organised them in the narration.\n\n### Two Works (Comparison)\n\nThis is the moment when we do not only compare characters’ positions one to another but when we reach a higher level and think about features summarising each network. This is distant reading showing up in character network analysis methods and its future developments. How do we compare two networks? Can we do that visually? (A hint: no.) At this point of the discussion, there are few existing articles dealing with such questions, and I will soon add a small one by presenting a conference paper at Sidney’s DH 2015 conference with the character networks for the twenty novels of Émile Zola’s Les Rougon-Macquart, where I develop/adjust methods to detect cores of protagonists and show that they help differentiate the character networks based on that. The research questions that are to be asked are various, but in my opinion can be brought back to general questions such as the classification of character networks, and in a way of their corresponding novels (tread lightly).\n\nI should not forget to mention that at this level of observation it becomes more than necessary to have an automatic method in order to build the character networks. In my case I use back-of-the-book indexes of characters: it has the advantage of providing a disambiguated table of occurrences but the disadvantage of making scalability nearly impossible (at least with manual indexes).\n\n### Many Many Works (Massiveness)\n\nYay, this is what we want: all the books! Big data! It will probably arrive soon. This is the distant reading ideal: how does the number of protagonists evolve with time, and across genres and countries? How centralised or distributed are all these characters-systems? Can we build a relevant classification? Then we reach the trial of distant reading itself: can we discover something new about the theory of characters or the history of literature? Can we confirm or contradict assumptions that have been done by University Professors having read only a few tens of thousands novels?\n\n## Conclusion\n\nIn some cases, the necessary workload to create a framework for the analysis of one work is equivalent to the workload needed to extend it to the analysis of any number of them. Many studies have done the job for one or a few character networks. We are soon going to see appearing character networks for a whole genre or an entire epoch compiled and compared, and maybe one day character networks based on the constructing methods for all the fiction novels ever digitised.\n\nAs I feared, I haven’t solved what is a mystery for me. In fact I haven’t tried very hard. Is the “one, two, infinite” rule the distant reading equivalent of the “zero, one, infinite” rule in hard sciences? The same pattern or a sibling one? Does it allow a transfer of methods? Am I right to care about that or just a bit too obsessed by patterns in science? Maybe we should we call it distant reading induction, but I am quite convinced that this is redundant by definition.\n\nSubscribe\nNotify of",
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https://www.mathworksheetscenter.com/mathskills/logic/TruthValuesandorimplies/ | [
"# Setting Up Truth Tables Worksheets\n\nHow to Create a Truth Table In a Boolean Algebra, the expression involves bools or true or false values. The true and false are represented as T and F, respectively. The value of T is 1, and the value of F is 0.to create a truth table, we need to understand Boolean algebra. The Boolean algebra depends on binary digits (0,1). We take two events into consideration. It is an easy way to express difficult and lengthy statements into easy to comprehend logical expressions. There are various operators involved in creating a truth table, Such as; AND, OR, NOR, XOR. Let's see how to use these logical operators. The AND operator (symbolically: ∧), also known as logical conjunction requires both p and q to be True for the result to be True. The OR operator (symbolically: ∨) requires only one premise to be True for the result to be True. Logical NOR (symbolically: ↓) is the exact opposite of OR. It requires both p and q to be False to result in True. Exclusive Or, or XOR for short, (symbolically: ⊻) requires exactly one True and one False value in order to result in True.\n\n• ### Basic Lesson\n\nDemonstrates the concept of setting up truth tables. Set up a truth table to identify the possible truth values of a single, simple statement P. 1. A single statement such as P has two possible truth values. It can be either true or false. 2. The truth table will need three rows: one for the statement, one for each possible truth values. 3. Write the statement (P) on the top left cell (left of the double line) and list the possible truth values below it (T, F).\n\n• ### Intermediate Lesson\n\nShows students how to to setup and solve truth values. The truth table will need five rows: one for the statements and one for each possible combination: a. both T b. the first T and the second F c. the first F and the second T d. both F\n\n• ### Independent Practice 1\n\nContains a mixture of problems that ask students to setup truth tables. Set up truth tables to identify the possible truth values of the following single, simple statements.\n\n• ### Independent Practice 2\n\nFeatures setting up truth value questions with assorted concepts.\n\n• ### Homework Worksheet\n\nFeatures 6 truth table creation problems.\n\n• ### Skill Quiz\n\n10 truth value questions that require setup. Scoring matrix is provided.\n\n• ### Homework and Quiz Answer Key\n\nAnswers for the homework and quiz.\n\nAnswers for the lesson and practice sheets.\n\n• ### Basic Lesson\n\nDemonstrates the concept of determining assorted and, or, if...then (implies) skills for truth values.\n\n• ### Intermediate Lesson\n\nShows students how to determine assorted truth values.\n\n• ### Independent Practice 1\n\nContains a mixture of problems using assorted truth values.\n\n• ### Independent Practice 2\n\nFeatures truth value questions with assorted concepts.\n\n• ### Homework Worksheet\n\nFeatures 6 assorted truth-value problems.\n\n• ### Skill Quiz\n\n10 assorted truth value questions. Scoring matrix is provided.\n\n• ### Homework and Quiz Answer Key\n\nAnswers for the homework and quiz.",
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http://www.open-science-repository.com/cardinality-of-primes-p-p2.html | [
"# Cardinality of Primes p, p+2\n\nby Open Science Repository Mathematics\n(June 2015)\n\n#### Abstract\n\nWe attempt to show an existential proof of the fact that there are infinitely many twin prime numbers; by using Euclid’s theorem, Schröder-Bernstein’s theorem, another known results in Calculus and Number Theory.\n\nKeywords: prime, twin primes Schröder-Bernstein’s theorem, Cantor’s theorem, Pomerance's theorem.\n\n#### Full text\n\nCardinality of Primes p, p+2"
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http://www.ms.lt/sodas/Book/CategoryTheory?action=print | [
"# Book: CategoryTheory\n\nStudy:\n\nHigher Topos Theory\n\n• Kerodon: An online resource for homotopy-coherent mathematics\n• Lurie: Higher Topos Theory\n• Riehl and Dominic Verity. Model independent higher category theory.\n• Describe the mathematics of names (terms, symbols, labels).\n• natural contradiction inherent in the approach of defining categories by starting with a category of categories and then taking a category to be its object\n• is there a category of universal properties?\n• Understand limits in terms of least upper bounds and greatest lower bounds, as with adjunct functors.\n• Is it possible to categorify everything, that is, to understand all inner properties of a system in terms of external relationships?\n• How does categorification relate to internalization, as with the representations of the sixsome?\n• In what sense are q-analogues the opposite of categorification?\n\nNatural transformations\n\n• If a functor takes us from a syntactic category to a semantic category, then what does the adjoint functor mean?\n• How is \"extending the domain\" related to adjunction?\n• How do the six operations match the six criteria?\n• How to deal with self-identity or non-identity f of an M-category? with copies of an M-category? Perhaps by embedding it in a bigger system?\n\nEquality\n\n• At what level is equality defined in defining a category? Equality is needed for the properties of identity and associativity. But is it the same identity as the identity for other morphisms?\n\nHidden assumptions\n\n• Is it possible to show that category theory presumes the Axiom of Choice?\n\nI'm trying to learn category theory because it is relevant for homotopy type theory and other areas of mathematics. But also I may be able to use it to model parts of my philosophy.\n\nCategory theory concepts such as adjoints (least upper bounds, greatest lower bounds) and limits-colimits are actually concepts of analysis.\n\nCategory theory externalizes the internal structure of mathematical objects. It reexpresses that internal structures in terms of constraints on external relationships between structures.\n\nCategory theory relates God's outer perspective (on the general, external \"black box\" relationships) and our inner perspective, within the system, in terms of the properties of our particular system. The question of God's necessity and nature includes the relationship between God and human's perspectives.\n\nCategory theory expresses a duality between objects and mappings. Thus functors likewise have this dual nature, and so do natural transformations.\n\nCommunity\n\nCategory Theory Videos\n\nCategory Theory Books and Articles",
null,
"Ideas\n\n• Categorification (making math explicit) vs. Decategorification (making math implicit). Algebraic combinatorics is the concrete flip-side of the abstractness of category theory. But algebraic combinatorics comes with implicit interpretation whereas category theory comes with explicit notation.\n• Categorical models for psychological consciousness. Sheaf theory - consciousness.\n• In the mathematical ways of figuring things out: Multiset is What, Set is How, List is Why. The reason that Set Theory works is that it is based on How, which is the level for all answers.\n\nFor each object {$x$}, the identity morphism {${id}_x$} must be unique. Because consider {${id1}_x \\circ {id2}_x$}. The identity is whichever disappears.\n\nI am thinking that categories should be considered on three levels:\n\n• Objects (of being - what is)\n• Arrows (of doing)\n• Equations (of reflecting) that relate arrows (or objects), especially in composition.\n\nThe composition of the arrows always seems to me underexplained. And that is where the different levels of equivalence become relevant. Also to be considered is whether an object should be thought of as an arrow to itself.\n\nFunctors\n\n• A functor is defined by what it does on a composition triangle of morphisms, and what it does on the identity morphism: {$F(a \\overset{f}{\\rightarrow} b \\overset{g}{\\rightarrow} c) \\Rightarrow F(a) \\overset{F(f)}{\\rightarrow} F(b) \\overset{F(g)}{\\rightarrow} F(c)$}\n• A functor is an interpretation that takes us from a syntax category to a semantics category.\n• Milewski: A functor embeds one category in another.\n• Milewski: A functor may collapse multiple objects/functions into one, but it never breaks connections.\n\nNatural transformations\n\n• Given functors F and G, both from C to D, a natural transformation {$eta$} maps every particular object X in C to a particular morphism {$eta_X$} from {$F(x)$} to {$G(x)$}. In this sense, the object is why (as a generalization of how) and the morphism is whether (as a generalization of what). Why and whether hold beyond circumstances (the functor), whereas how and what make sense within circumstances (the functor).\n• Natural transformations don't depend on the structure internal to the objects, but only on their external relationships, as expressed by the category.\n• The components of natural transformations depend only on the objects. If you know these components, then the morphisms carry over automatically.\n• Natural transformations say that the trivially existing bijection (between FA and GA, FB and GB) is actually a morphism in the category D.\n\nExamples\n\n• Z ir R. Include Z in R. Get floor of element in R to go back.\n• Return and extract in Haskell.\n• Expanding (replacing Identities with L*R) and collapsing (replacing L*R with Identity).\n• See Wikipedia: Currying {$B \\mapsto B\\times C$} is left adjoint to {$A \\mapsto A^C$}. This grounds the equation {$A^{B\\times C}\\cong (A^C)^B$}.\n\nRelate id-A to consciousness, to constancy of attention, recurring attention.\n\nSignificant=unencompassable.\n\nCovering=encompassing=Why.\n\nComputation trinitarianism takes as equivalent a proof of a proposition, a program with output of some type, and a generalized element of some object.\n\n• Category theory models perspectives and attention shifting. (Or thoughts as objects?)\n• Category of perspectives: stepping-in and stepping-out as adjoints? there exists vs. for all?\n• Eugenia Cheng: Mathematics is the logical study of how logical things work. Abstraction is what we need for logical study. Category theory is the the math of math, thus the logical study of the logical study of how logical things work.\n• Category theory shines light on the big picture. Perspectives shine light on the big picture (God's) or the local picture (human's).\n• Lygmuo Kodėl viską išsako ryšiais. O tas ryšys yra tarpas, kuriuo išsakomas Kitas. Kategorijų teorijoje panašiai, tikslas yra pereiti iš narių (objektų) nagrinėjimą į ryšių (morfizmų) nagrinėjimą.\n• The theory of quasi-categories II\n• Higher Algebra\n• Groth, M., A Short Course on ∞-categories\n• Cisinski's notes\n• http://plato.stanford.edu/entries/category-theory/\n• Emily Riehl, [http://www.math.jhu.edu/~eriehl/ssets.pdf | A leisurely introduction to simplicial sets]]\n• Categorical Logic lecture notes by Steve Awodey\n\nNotes\n\n• Kategorijų teorijoje: nagrinėti įvairių lygmenų klaidas.\n• Adjunction is like extension of the domain, but in terms of structure: extension of structure. Polymorphism. Try to relate Z and Q and R and other examples.\n• Network theory (wiki) and Network theory (blog) by John Baez\n• What can graph theory (for example, random graphs, or random order) say about category theory?\n• In the category of Sets, is there any way to distinguish between the integers and the reals? Are all infinities the same?\n• In the category Set, how can you distinguish between a countable and uncountable set?\n• Kromer, R.: Tool and Object: A History and Philosophy of Category Theory. Birkhauser (2007)\n• Fong, B., Spivak, D.I.: Seven Sketches in Compositionality: An Invitation to Applied Category Theory. Cambridge (2019),http://math.mit.edu/~dspivak/teaching/sp18/7Sketches.pdf\n• Lawvere, W., Schanuel, S.: Conceptual Mathematics: A first introduction to categories. Cambridge (1997)\n• Emily Riehl. Category Theory in Context\n• In category theory, what is the relationship between structure preservation of the objects, internally, and their external relationships?\n• Axiom of forgetfullness.\n• Internal discussion with oneself vs. external discussion with others (Vygotsky) is the distinction that category theory makes between internal structure and external relationships.\n• In most every category, can we (arbitrarily) define (uniquely) distinguished \"generic objects\" or \"canonical objects\", which are the generic equivalents for all objects that are equivalent to each other? For example, in the category of sets, the generic set of size one.\n• In functional programming with monoids and monads, can we think of each function as taking us from a question type to an answer type? In general, in category theory, can we think of each morphism as taking us from a question to an answer?\n• Category theory for me: distinguishing what observations are nontrivial - intrinsic to a subject - and what are observations are content-wise trivial or universal - not related to the subject, but simply an aspect of abstraction.\n• In product, the information from A and B is stored externally in A x B. In coproduct, the information from A and B is stored internally in A union B (A+B). Note: multiplication is external, and addition is internal.\n• In a diagram, we have a map from shape J (the index category) to the category C. Note that the index diagram is How.\n• There are always dual categories {$C$} and {$C^{op}$}."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8552322,"math_prob":0.89754105,"size":11427,"snap":"2020-24-2020-29","text_gpt3_token_len":2611,"char_repetition_ratio":0.14076862,"word_repetition_ratio":0.022358859,"special_character_ratio":0.21186663,"punctuation_ratio":0.12628992,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9802624,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-07T06:59:15Z\",\"WARC-Record-ID\":\"<urn:uuid:b6336a81-e92e-4542-ae7d-bd93c937fe96>\",\"Content-Length\":\"26693\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0108670b-f0a1-4f33-a902-1faa5b4e91e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:db23397f-70e6-4dd2-935d-a4f9b8d467c0>\",\"WARC-IP-Address\":\"193.219.5.19\",\"WARC-Target-URI\":\"http://www.ms.lt/sodas/Book/CategoryTheory?action=print\",\"WARC-Payload-Digest\":\"sha1:7D3XFUK7CQGJMD6Y4NTXTUXRWW7KPV25\",\"WARC-Block-Digest\":\"sha1:QM4GOKBY7ZB7D3ZOOM53S2LSGRTHAPVZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655891654.18_warc_CC-MAIN-20200707044954-20200707074954-00320.warc.gz\"}"} |
https://www.essaysforstudent.com/Science/Inen-509-Economic-Life-of-the-Asset/110415.html | [
"",
null,
"# Inen 509 - Economic Life of the Asset\n\nPage 1 of 2\n\nHomework Assignment 4 Solutions\n\n1. Problem 5.1 (page 196) in textbook.\n\na. From equation 5.1\n\nib=0.15\n\ntm=0.4\n\nkb=(.15)(1-.4)=0.09 or 9%\n\nb. from equation 5.2\n\n1 + (ibtm/4)((P/A, kbp, 4) – (1 + ib)(P/F,kbp, 4) = 0\n\nTry kbp = 9%\n\nFrom table in Appendix A page 711\n\n1 + (.15*.4/4)(3.2397199) – (1+ .15)(.7084252)= .2339068185\n\nTry kbp =8%\n\nFrom table in Appendix A page 710\n\n1 + (.015)(3.3121268) – 1.15(.7350299) = .204398\n\nTry kbp = 6%\n\nFrom table in Appendix A page 708\n\n1 + .015(3.4651056) – 1.15(.7920937) = 0.141071\n\nTry kbp = 4%\n\nFrom table in Appendix A page 706\n\n1 + .015(3.6298952) – 1.15(.8548042) = 0.071427\n\nTry kbp=2%\n\nFrom table in Appendix A page 704\n\n1 + .015(3.8077287) – 1.15(.9238454) = -0.0053063\n\nkbp is therefore slightly more than 2% and you would need to increment it slightly and use 2.10 and 2.13 to calculate your factors for P/A and P/F\n\nc. ibp = .15/4 = 0.0375\n\nkb = [(1 + 0.0375(.6)]4 – 1 = 9.3%\n\nd. ibp = (1+0.15).25 – 1\n\nkb = [1 + 0.03556(.6)]4 – 1 = 0.08811 8.811%\n\n2. Problem 5.5 (page 196) in textbook.\n\nSource Amount x 106 % B/T cost A/T cost 3X5\n\nBank\n\nDownload as (for upgraded members)\nCitation Generator\n\n(2019, 01). Inen 509 - Economic Life of the Asset. EssaysForStudent.com. Retrieved 01, 2019, from https://www.essaysforstudent.com/Science/Inen-509-Economic-Life-of-the-Asset/110415.html\n\n\"Inen 509 - Economic Life of the Asset\" EssaysForStudent.com. 01 2019. 2019. 01 2019 <https://www.essaysforstudent.com/Science/Inen-509-Economic-Life-of-the-Asset/110415.html>.\n\n\"Inen 509 - Economic Life of the Asset.\" EssaysForStudent.com. EssaysForStudent.com, 01 2019. Web. 01 2019. <https://www.essaysforstudent.com/Science/Inen-509-Economic-Life-of-the-Asset/110415.html>.\n\n\"Inen 509 - Economic Life of the Asset.\" EssaysForStudent.com. 01, 2019. Accessed 01, 2019. https://www.essaysforstudent.com/Science/Inen-509-Economic-Life-of-the-Asset/110415.html."
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http://forums.wolfram.com/mathgroup/archive/2003/Nov/msg00352.html | [
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"Re: FindMinimum Problem\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg44618] Re: FindMinimum Problem\n• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>\n• Date: Tue, 18 Nov 2003 06:41:47 -0500 (EST)\n• Organization: Universitaet Leipzig\n• References: <bpa1rv\\$19t\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Hi,\n\nyou ust make a function\n\nfoo[x_?NumericQ]:=x^2-a*x/.NDSolve[{a'[z]==x*z,a==1},a,{z,0,1}]\n\nFindMinimum[foo[x],{x,-1,1}]\n\nRegards\nJens\n\nJiang Xiao wrote:\n>\n> Hi, all,\n> recently I am dealing with a problem as following, findminimum(over x)\n> of a function f[a,x], where a[z] satisfies a differential equation\n> a'[z]=x*z say. The code is like:\n> FindMinimum[x^2-a*x/.NDSolve[{a'[z]==x*z,a==1},a,{z,0,1}],{x,-1,1}]\n>\n> the problem is that I can do it in mathematica 4.2, but can't in mathematica\n> 5.0 now. Do anybody where is the problem?\n>\n> thanks,\n>\n> Jiang\n\n```\n\n• Prev by Date: Re: filled plot on part of x-interval\n• Next by Date: Re: Improper integral\n• Previous by thread: FindMinimum Problem\n• Next by thread: Re: FindMinimum Problem"
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https://prog2.de/book/sheet01.html | [
"## SectionD.1Exercise Sheet 1\n\n### ExercisesExercises\n\n#### 1.Different Number Systems.\n\nConvert the following numbers into the specified number system.\n\n1. $$\\langle 43 \\rangle_{10}$$ into the base 7 system\n\n2. $$\\langle212 \\rangle _{3}$$ into the base 9 system\n\n3. $$\\langle313\\rangle _{4}$$ into the base 6 system\n\n4. $$\\langle 3142 \\rangle _{5}$$ into the base 3 system\n\n5. $$\\langle 621 \\rangle _{11}$$ into the base 5 system\n\n202207281550\n\n202204151400\nSolution.\n1. $$\\displaystyle \\langle 43\\rangle_{10} = 6\\times 7+1\\times 1 = \\langle 61\\rangle_{7}$$\n\n2. $$\\displaystyle \\langle 212\\rangle_{3} = 2\\times 9+1\\times 3+2\\times 1 = \\langle 23\\rangle_{10} = 2\\times 9+5\\times 1 = \\langle25\\rangle_{9}$$\n\n3. $$\\displaystyle \\langle313\\rangle_{4} = 3\\times 16+1\\times 4+3\\times 1= \\langle 55\\rangle_{10} = 1 \\times 36 + 3 \\times 6 + 1 \\times 1 = \\langle131\\rangle_{6}$$\n\n4. $$\\displaystyle \\langle3142\\rangle_{5} = 3\\times 125+1\\times 25+4\\times 5+2\\times 1 = \\langle422\\rangle_{10}= 1 \\times 243 + 2 \\times 81 + 0 \\times 27 + 1 \\times 9 + 2 \\times 3 + 2 \\times 1 = \\langle120122\\rangle_{3}$$\n\n5. $$\\displaystyle \\langle621\\rangle_{11} = 6\\times 121+2\\times 11+1\\times 1 = \\langle749\\rangle_{10} = 1 \\times 625 + 0 \\times 125 + 4 \\times 25 + 4 \\times 5 + 4 \\times 1 = \\langle10444\\rangle_{5}$$\n\n#### 2.Some Positional Number Systems.\n\nWe consider the following positional number systems:\n\nName Base Prefix Digits\nBinary System 2 0b 0,1\nOctal System 8 0 0,...,7\n\nThe digits are denoted in the order of their value, e.g. the digit F in the hexadecimal system has the value 15.\n\nTo specify the number system a number is prefixed by the content of the column “Prefix”. For example, 013 is an octal number, 0x13 is a hexadecimal number and 13 is a decimal number.\n\n1. State the following decimal numbers in each of the three number systems. Do it on your own, don't use a calculator or computer.\n 5 16 49 81 257 317 1721 4096\n2. How did you do it? Describe your approach such that other students could use it as instruction.\n\n3. Assume a function $$\\textit{dig} : [0,15] \\rightarrow \\{0,\\dots,9,A,\\dots,F\\}\\text{.}$$ $$\\textit{dig}$$ computes for any decimal number in the interval $$[0,15]$$ the corresponding hexadecimal digit. Complete the following recursive definition of a function $$\\rangle\\cdot\\langle_b : \\Nat\\rightarrow\\{0,\\dots,9,A,\\dots,F\\}^\\ast$$ which computes for any natural number the representation in base $$b\\text{.}$$ For example, it should hold that $$\\rangle257\\langle_8 = 401\\text{.}$$\n\n202207281550\n\n202204151400\nSolution.\n1. Table:\n\n10 2 8 16\n5 0b101 05 0x5\n16 0b10000 020 0x10\n49 0b110001 061 0x31\n81 0b1010001 0121 0x51\n257 0b100000001 0401 0x101\n317 0b100111101 0475 0x13D\n1721 0b11010111001 03271 0x6B9\n4096 0b1000000000000 010000 0x1000\n\n2. Translation of a number $$n$$ into base $$b$$ using the example $$\\langle{}401\\rangle_8 = 257$$\n\n1. Note all $$b^k \\lt n$$\n $$8^2$$ $$8^1$$ $$8^0$$\n2. Starting on the left, determine the maximal $$a$$ such that $$a\\cdot{} b^k \\lt n\\text{.}$$ Compute $$n'$$ where $$n' = n - (a\\cdot{} b^k)\\text{.}$$ Here we have $$n' = 1\\text{.}$$\n $$8^2$$ $$8^1$$ $$8^0$$ 4\n3. Repeat the previous step from left to right until all digits are computed:\n $$8^2$$ $$8^1$$ $$8^0$$ 4 0 1\n3. Idea: The inverse of the tree example from the lecture notes. In every recursion step until $$n \\lt b$$ holds: $$n \\mathrel{\\text{mod}} b$$ is the index of the child in the currently considered subtree. The factor $$\\frac{n}{b}$$ ensures that the algorithm descends one layer in every step.\n\n\\begin{equation*} \\rangle{}n\\langle_b =\\left\\{ \\begin{array}{cl} \\rangle\\lfloor\\frac{n}{b}\\rfloor\\langle_b \\cdot \\textit{dig}(n \\mathrel{\\text{mod}} b) \\amp \\textup{if} \\\\ \\textit{dig}(n)\\amp \\textup{else} \\end{array}\\right. \\end{equation*}\n\n#### 3.Translate!\n\nFill in the following form.\nbinary $$\\buintpb{\\cdot}{2}$$ octal $$\\buintpb{\\cdot}{8}$$ decimal $$\\buintpb{\\cdot}{10}$$ hexadecimal $$\\buintpb{\\cdot}{16}$$\n$$\\buintpb{21}{16}$$\n$$\\buintpb{1101}{2}$$\n$$\\buintpb{103}{8}$$\n$$\\buintpb{1011}{2}$$\n$$\\buintpb{2D}{16}$$\n$$\\buintpb{417}{8}$$\n$$\\buintpb{17}{16}$$\n$$\\buintpb{111101}{2}$$\n$$\\buintpb{315}{8}$$\n202207281550\n\n202204151400\nSolution.\nbinary $$\\buintpb{\\cdot}{2}$$ octal $$\\buintpb{\\cdot}{8}$$ decimal $$\\buintpb{\\cdot}{10}$$ hexadecimal $$\\buintpb{\\cdot}{16}$$\n$$\\buintpb{100001}{2}$$ $$\\buintpb{41}{8}$$ $$\\buintpb{33}{10}$$ $$\\buintpb{21}{16}$$\n$$\\buintpb{1101}{2}$$ $$\\buintpb{15}{8}$$ $$\\buintpb{13}{10}$$ $$\\buintpb{D}{16}$$\n$$\\buintpb{1000011}{2}$$ $$\\buintpb{103}{8}$$ $$\\buintpb{67}{10}$$ $$\\buintpb{43}{16}$$\n$$\\buintpb{1011}{2}$$ $$\\buintpb{13}{8}$$ $$\\buintpb{11}{10}$$ $$\\buintpb{B}{16}$$\n$$\\buintpb{101101}{2}$$ $$\\buintpb{55}{8}$$ $$\\buintpb{45}{10}$$ $$\\buintpb{2D}{16}$$\n$$\\buintpb{100001111}{2}$$ $$\\buintpb{417}{8}$$ $$\\buintpb{271}{10}$$ $$\\buintpb{10F}{16}$$\n$$\\buintpb{10111}{2}$$ $$\\buintpb{27}{8}$$ $$\\buintpb{23}{10}$$ $$\\buintpb{17}{16}$$\n$$\\buintpb{111101}{2}$$ $$\\buintpb{75}{8}$$ $$\\buintpb{61}{10}$$ $$\\buintpb{3D}{16}$$\n$$\\buintpb{11001101}{2}$$ $$\\buintpb{315}{8}$$ $$\\buintpb{205}{10}$$ $$\\buintpb{CD}{16}$$\n\n#### 4.Unsigned Arithmetics.\n\n1. State the decimal representation of the following binary number (we use $$\\buintp{\\cdot}$$ instead of $$\\buintpb{\\cdot}{2}$$):\n\n1. $$\\displaystyle \\buintp{11}$$\n\n2. $$\\displaystyle \\buintp{011}$$\n\n3. $$\\displaystyle \\buintp{1001}$$\n\n4. $$\\displaystyle \\buintp{1111}$$\n\n5. $$\\displaystyle \\buintp{101111}$$\n\n2. Compute and state the result in the binary system.\n\n1. $$\\displaystyle \\buintp{1011}+\\buintp{1101}$$\n\n2. $$\\displaystyle \\buintp{11}+\\buintp{1101}$$\n\n3. $$\\displaystyle \\buintp{10111}+\\buintp{1101}$$\n\n4. $$\\displaystyle \\buintp{1011 + 1101}$$\n\n202207281550\n\n202204151400\nSolution.\n1. $$\\displaystyle \\buintp{11} = 3$$\n\n2. $$\\displaystyle \\buintp{011} = 3$$\n\n3. $$\\displaystyle \\buintp{1001} = 9$$\n\n4. $$\\displaystyle \\buintp{1111} = 15$$\n\n5. $$\\displaystyle \\buintp{101111} = 47$$\n\n1. $$\\displaystyle \\buintp{1011} + \\buintp{1101} = 11 + 13 = 24 = \\buintp{11000}$$\n\n2. $$\\displaystyle \\buintp{11} + \\buintp{1101} = 3 + 13 = 16 = \\buintp{10000}$$\n\n3. $$\\displaystyle \\buintp{10111} + \\buintp{1101} = 23 + 13 = 36 = \\buintp{100100}$$\n\n4. $$\\displaystyle \\langle 1011 + 1101 \\rangle = \\buintp{11000} = 24 = \\buintp{11000}$$\n\n#### 5.and, or, xor.\n\n1. Show $$\\bxor$$ and $$\\bor$$ associative using a truth table.\n\n2. Which of the three operations distribute over which other operation? In other words, for which operators $$\\circ, \\circ'\\in\\{\\band,\\bor,\\bxor\\}$$ does $$a\\circ(b\\circ' c)=(a\\circ b)\\circ'(a\\circ c)$$ hold for any $$a,b,c\\in \\Bit\\text{?}$$\n\n202207281550\n\n202204151400\nSolution.\n1. \\begin{equation*} \\begin{array}{|c|c|c|c|c|c|c|} \\hline a \\amp b \\amp c \\amp (b \\bxor c) \\amp a \\bxor(b\\bxor c) \\amp (a \\bxor b) \\amp (a\\bxor b)\\bxor c\\\\ \\hline 0 \\amp0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp0 \\\\ 0\\amp 0 \\amp 1 \\amp 1 \\amp 1 \\amp 0 \\amp1 \\\\ 0\\amp 1 \\amp 0 \\amp 1 \\amp 1 \\amp 1 \\amp 1\\\\ 0\\amp 1 \\amp 1 \\amp 0 \\amp0 \\amp1 \\amp 0\\\\ 1\\amp 0 \\amp 0 \\amp 0 \\amp1 \\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 1 \\amp 1 \\amp 0 \\amp 1 \\amp 0\\\\ 1\\amp 1 \\amp 0 \\amp1 \\amp 0 \\amp 0 \\amp 0\\\\ 1\\amp 1 \\amp1 \\amp0 \\amp 1 \\amp 0 \\amp 1\\\\ \\hline \\end{array} \\end{equation*}\n\\begin{equation*} \\begin{array}{|c|c|c|c|c|c|c|} \\hline a \\amp b \\amp c \\amp (b \\bor c) \\amp a \\bor(b\\bor c) \\amp (a \\bor b) \\amp (a\\bor b)\\bor c\\\\ \\hline 0 \\amp0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp0 \\\\ 0\\amp 0 \\amp 1 \\amp 1 \\amp 1 \\amp 0 \\amp 1\\\\ 0\\amp 1 \\amp 0 \\amp1 \\amp 1 \\amp1 \\amp 1\\\\ 0\\amp 1 \\amp 1 \\amp1 \\amp 1 \\amp 1 \\amp1 \\\\ 1\\amp 0 \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 1 \\amp1 \\amp1 \\amp1 \\amp1 \\\\ 1\\amp 1 \\amp 0 \\amp1 \\amp1 \\amp1 \\amp1 \\\\ 1\\amp 1 \\amp1 \\amp 1 \\amp1 \\amp1 \\amp1 \\\\ \\hline \\end{array} \\end{equation*}\n2. The following distributive laws hold:\n\n• $$\\displaystyle a\\bor(b\\band c)=(a\\bor b)\\band(a\\bor c)$$\n\n• $$\\displaystyle a\\band(b\\bor c)=(a\\band b)\\bor(a\\band c)$$\n\n• $$\\displaystyle a\\band(b\\bxor c)=(a\\band b)\\bxor(a\\band c)$$\n\n#### 6.Elementary Operations on Bits.\n\nThere are actually operations that are more elementary than and, or, xor. One of them is nand $$\\bneg{a\\band b}$$ which is $$0$$ if $$a=b=1$$ and $$1$$ otherwise. Show how to express and, or, xor using nand. What other elementary operations like nand are there?\n\n202207281550\n\n202204151400\nSolution.\nThis is the truth table for nand.\n\\begin{equation*} \\begin{array}{|c|c|c|} \\hline a \\amp b \\amp \\bneg{a\\band b}\\\\ \\hline 0\\amp 0 \\amp 1\\\\ 0\\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 1\\\\ 1\\amp 1 \\amp 0\\\\ \\hline \\end{array} \\end{equation*}\nWe can express negation using nand:\n\\begin{equation*} \\begin{array}{|c|c|c|} \\hline a \\amp \\bneg{a\\band a} \\amp \\bneg{a}\\\\ \\hline 0\\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 0\\\\ \\hline \\end{array} \\end{equation*}\nWe can express conjunction (and) using nand. We can use plain old negation, since we already know how to encode it just using nand.\n\\begin{equation*} \\begin{array}{|c|c|c|c|} \\hline a \\amp b \\amp \\bneg{\\bneg{a\\band b}} \\amp a \\band b\\\\ \\hline 0\\amp 0 \\amp 0 \\amp 0\\\\ 0\\amp 1 \\amp 0 \\amp 0\\\\ 1\\amp 0 \\amp 0 \\amp 0\\\\ 1\\amp 1 \\amp 1 \\amp 1\\\\ \\hline \\end{array} \\end{equation*}\nWe can now express disjunction (or) using nand. Again, we also use negation and conjunction since we know how to express them.\n\\begin{equation*} \\begin{array}{|c|c|c|c|c|c|} \\hline a \\amp b \\amp \\bneg{a} \\amp \\bneg{b} \\amp \\bneg{\\bneg{a}\\band\\bneg{b}} \\amp a \\bor b\\\\ \\hline 0\\amp 0 \\amp 1 \\amp 1 \\amp 0 \\amp 0\\\\ 0\\amp 1 \\amp 1 \\amp 0 \\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 1\\\\ 1\\amp 1 \\amp 0 \\amp 0 \\amp 1 \\amp 1\\\\ \\hline \\end{array} \\end{equation*}\nWe have already shown that we can express and and or with nand. So it is enough to show that xor can be expressed with and and or to prove that we can only express it with nand.\n\\begin{equation*} \\begin{array}{|c|c|c|c|c|c|c|c|} \\hline a \\amp b \\amp \\bneg{a} \\amp \\bneg{b} \\amp \\bneg{a}\\band b \\amp a\\band \\bneg{b} \\amp (\\bneg{a}\\band b)\\bor(a\\band\\bneg{b}) \\amp a\\bxor b\\\\ \\hline 0\\amp 0 \\amp 1 \\amp 1 \\amp 0 \\amp 0 \\amp 0 \\amp 0\\\\ 0\\amp 1 \\amp 1 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 1\\\\ 1\\amp 0 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 1 \\amp 1\\\\ 1\\amp 1 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0\\\\ \\hline \\end{array} \\end{equation*}\nAnother elementary operation like nand is nor. It is $$1$$ if $$a=b=0$$ and $$0$$ otherwise.\n\n#### 7.Nand is elementary.\n\nShow that nand is indeed universal, i.e. every boolean function $$f(a,b)$$ can be expressed only using nands.\n\n202207281550\n\n202204151400\nHint.\n\nFirst, show that every boolean function can be represented using and, not and or. Then use the other exercise to relate this to nand. You may also need to prove that not can be expressed using nand.\n\nSolution.\n\nIn total, there are 16 different possibilities for boolean functions with two arguments: Since $$f(a,b)$$ takes two arguments, each of which can be either $$0$$ or $$1\\text{,}$$ there are exactly $$4$$ possible different inputs. For each of these inputs there are 2 different possible results: $$0$$ and $$1\\text{.}$$ This means in total we can find $$2^4 = 16$$ functions. All of functions can be represented only using and, or, and negation, as follows:\n\n\\begin{equation*} \\begin{array}{|c||c|c|c|c|c|c|c|c|} \\hline f(0,0) \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\\\ f(0,1) \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\\\ f(1,0) \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\\\ f(1,1) \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\\\ \\hline f(a,b) \\amp a \\band \\bneg{a} \\amp \\bneg{a} \\band \\bneg{b} \\amp \\bneg{a} \\band b \\amp \\bneg{a} \\amp a \\band \\bneg{b} \\amp \\bneg{b} \\amp (a \\band \\bneg{b}) \\bor (b \\band \\bneg{a})\\amp \\bneg{a} \\bor \\bneg{b}\\\\ \\hline \\end{array} \\end{equation*}\n\\begin{equation*} \\begin{array}{|c||c|c|c|c|c|c|c|c|} \\hline f(0,0) \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\amp 0 \\amp 1 \\\\ f(0,1) \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\\\ f(1,0) \\amp 0 \\amp 0 \\amp 0 \\amp 0 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\\\ f(1,1) \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\amp 1 \\\\ \\hline f(a,b) \\amp a \\band b \\amp (a \\band b) \\bor (\\bneg{a} \\band \\bneg{b}) \\amp b \\amp b \\bor \\bneg{a} \\amp a \\amp a \\bor \\bneg{b} \\amp a \\bor b \\amp a \\bor \\bneg a \\\\ \\hline \\end{array} \\end{equation*}\n\nNow that we have shown that every boolean function with two arguments can be represented with and, not, and or and that we have shown in exercise 6 that and and or can be expressed with nand, the only thing left to show is that not can also be expressed using only nand.\n\n\\begin{equation*} \\begin{array}{|c|c|c|} \\hline a \\amp \\bneg{a} \\amp \\bneg{a \\band a} \\\\ \\hline 0 \\amp 1 \\amp 1 \\\\ 1 \\amp 0 \\amp 0 \\\\ \\hline \\end{array} \\end{equation*}\n\nIn conclusion, we have now shown that and, or and not can be expressed using only nand, because we have also shown that every boolean function with two arguments can be represented by and, or and not we can conclude that nand is universal.\n\n#### 8.Arithmetics Proofs.\n\nProve:\n1. Let $$x$$ be a sequence of digits of length $$n$$ and let $$y$$ be a sequence of digits of length $$m\\text{.}$$\nShow that $$\\buintpb{x\\cdot y}{\\beta} = \\buintpb{x}{\\beta}\\cdot\\beta^m + \\buintpb{y}{\\beta}$$ holds for any $$\\beta\\in\\Nat^\\ast\\text{.}$$\nRemember that $$\\cdot$$ on sequences of digits is concatenation. For instance, $$ab\\cdot cd = abcd$$\n\n2. Show that for all $$n\\gt 0, \\buintpb{(\\beta -1)^n}{\\beta} + 1 = \\beta^n\\text{.}$$\n\n3. Let $$x$$ be a sequence of digits of length $$n\\text{.}$$ Show that $$\\buintpb{x}{\\beta}\\lt\\beta^n$$ holds.\n\n202207281550\n\n202204151400\nSolution.\n1. \\begin{align*} \\buintpb{x\\cdot y}{\\beta} \\amp= \\sum_{i=0}^{n-1} x_i\\beta^{m+i}+\\sum_{i=0}^{m-1} y_i\\beta^i\\\\ \\amp= \\beta^m\\cdot\\sum_{i=0}^{n-1} x_i\\beta^{i}+\\sum_{i=0}^{m-1} y_i\\beta^i \\\\\\\\ \\amp= \\beta^m\\cdot\\buintp{x} + \\buintp{y}\\qquad\\square \\end{align*}\n\n2. \\begin{align*} \\buintpb{(\\beta -1)^n}{\\beta} + 1 \\amp= \\sum_{i=0}^{n-1}(\\beta -1)\\beta^i + 1\\\\ \\amp=(\\beta-1)\\beta^{n-1}+(\\beta-1)\\beta^{n-2}+\\cdots+(\\beta-1)\\beta+(\\beta-1)+1\\\\ \\amp= \\beta^n-\\underbrace{\\beta^{n-1}+\\beta^{n-1}}_0+\\cdots\\underbrace{-\\beta+\\beta}_0\\underbrace{-1+1}_0\\\\ \\amp= \\beta^n\\qquad\\square \\end{align*}\n\n3. Using induction:\n\n$$n = 1:$$\n\n\\begin{equation*} x \\leq \\beta -1 \\lt \\beta^1\\text{.} \\end{equation*}\n\n$$n\\rightarrow n + 1:$$\n\nInduction hypothesis: $$\\buintpb{x_{n-1}\\cdots x_0}{\\beta}\\lt\\beta^n$$\n\n\\begin{align*} \\buintpb{x_{n}\\cdots x_0}{\\beta} \\amp= x_n\\cdot\\beta^n + \\buintpb{x_{n-1}\\cdots x_0}{\\beta}\\\\ \\amp\\lt x_n\\cdot\\beta^n + \\beta^n ~~~~~~~~~~~~~~~|\\ \\text{induction hypothesis}\\\\ \\amp\\leq (\\beta-1)\\cdot\\beta^n + \\beta^n ~~~~~~~~~|\\ x_n \\leq \\beta -1\\\\ \\amp= \\beta^{n+1} - \\beta^n + \\beta^n\\\\ \\amp= \\beta^{n+1}\\qquad\\square \\end{align*}"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6106324,"math_prob":0.99999917,"size":11294,"snap":"2022-27-2022-33","text_gpt3_token_len":4447,"char_repetition_ratio":0.2153233,"word_repetition_ratio":0.09825034,"special_character_ratio":0.43686914,"punctuation_ratio":0.07109005,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999954,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-07T19:25:18Z\",\"WARC-Record-ID\":\"<urn:uuid:6f31eac7-e785-4e2f-b041-10d946f78370>\",\"Content-Length\":\"66759\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56104322-a14f-4607-8096-d27d785ef968>\",\"WARC-Concurrent-To\":\"<urn:uuid:70f4c5e3-290a-4a80-b67f-e29168cd4128>\",\"WARC-IP-Address\":\"134.96.226.121\",\"WARC-Target-URI\":\"https://prog2.de/book/sheet01.html\",\"WARC-Payload-Digest\":\"sha1:GNZXUYWYHVJWDWRVO44P2JAYKCG2MR7B\",\"WARC-Block-Digest\":\"sha1:AEQ5WCD3LEEEZWS6LBVUIGRKBRT6KJWQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570692.22_warc_CC-MAIN-20220807181008-20220807211008-00663.warc.gz\"}"} |
https://ask.csdn.net/questions/755205 | [
"",
null,
"2019-04-08 11:24\n\n# 使用C语言进行一个月份的问题解决\n\n• 点赞\n• 写回答\n• 关注问题\n• 收藏\n• 邀请回答\n\n#### 1条回答默认 最新\n\n•",
null,
"DRACULAX05 2019-04-08 11:49\n已采纳\n``````#include <stdio.h>\n\nbool isRunYear(int year)\n{\nreturn ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0);\n}\n\nint getMonthDay(int year, int month)\n{\nif (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)\nreturn 31;\n\nif (month == 2)\n{\nreturn isRunYear(year) ? 29 : 28;\n}\nreturn 30;\n}\n\nint main(void)\n{\nbool isFirst = true;\nint year = 0, month1 = 0, month2 = 0, nTotal = 0;\n\nwhile (year <= 0 || month1 <= 0 || month2 <= 0 ||\nmonth1 > 12 || month2 > 12 || month1 > month2)\n{\nif (!isFirst) {\nprintf(\"输入错误，请重新输入\\n\");\n}\n\nprintf(\"请输入年份：\\n\");\nscanf(\"%d\", &year);\nprintf(\"请输入起始月份：\\n\");\nscanf(\"%d\", &month1);\nprintf(\"请输入结束月份：\\n\");\nscanf(\"%d\", &month2);\n\nisFirst = false;\n}\n\nfor (int i = month1; i <= month2; i++)\n{\nnTotal += getMonthDay(year, i);\n}\n\nprintf(\"间隔天数：%d\\n\", nTotal);\nreturn 0;\n}\n``````\n点赞 评论"
] | [
null,
"https://profile.csdnimg.cn/A/E/3/4_ggyoungog",
null,
"https://profile.csdnimg.cn/4/8/B/4_draculax05",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.62572026,"math_prob":0.9959678,"size":1224,"snap":"2021-31-2021-39","text_gpt3_token_len":665,"char_repetition_ratio":0.14918032,"word_repetition_ratio":0.0,"special_character_ratio":0.39624184,"punctuation_ratio":0.17010309,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.965864,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T13:13:26Z\",\"WARC-Record-ID\":\"<urn:uuid:b68d38a5-0228-47a4-9462-a758bee69e8e>\",\"Content-Length\":\"113249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dafde218-454e-44ed-8bfb-ad9429fcd43d>\",\"WARC-Concurrent-To\":\"<urn:uuid:66ad5c94-c685-476a-8649-7797e44a9c39>\",\"WARC-IP-Address\":\"47.95.50.136\",\"WARC-Target-URI\":\"https://ask.csdn.net/questions/755205\",\"WARC-Payload-Digest\":\"sha1:SNVNFULBKVUXVPB3HQUXIGWHGODLFJZ7\",\"WARC-Block-Digest\":\"sha1:FRJMIVCFTMYN6Z2OIMUOWJE5VNGBT7T2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058450.44_warc_CC-MAIN-20210927120736-20210927150736-00611.warc.gz\"}"} |
https://engineeringunits.com/fraction-exponents-calculator/ | [
"Categories\n\n# Fraction Exponents Calculator\n\n## Fraction Exponents Calculator Online:\n\nuse our Fraction Exponents Calculator Online.\n\n## Fraction Exponents Calculator formula:\n\nxm/n =a\n\nFraction Exponents formula\n\n## Fraction Exponents Definition:\n\nDefinition of Fraction Exponents:\n\nA radical can be expressed as a value with a fractional exponent by following the convention xm/n =a. Rewriting radicals as fractional exponents can be useful in simplifying some radical expressions. When working with fractional exponents, remember that fractional exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions.\n\nFraction Exponents Definition:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.827432,"math_prob":0.9576149,"size":625,"snap":"2022-40-2023-06","text_gpt3_token_len":115,"char_repetition_ratio":0.25764894,"word_repetition_ratio":0.0,"special_character_ratio":0.1552,"punctuation_ratio":0.10309278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993581,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T10:27:21Z\",\"WARC-Record-ID\":\"<urn:uuid:b5256162-cd2e-4cc7-8e74-90fd6c2c4781>\",\"Content-Length\":\"125612\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dea8bbd3-3f8e-4b7f-a32d-8ed0685dca16>\",\"WARC-Concurrent-To\":\"<urn:uuid:03036aeb-bbf7-403f-9145-cbd4e83d6742>\",\"WARC-IP-Address\":\"52.86.133.10\",\"WARC-Target-URI\":\"https://engineeringunits.com/fraction-exponents-calculator/\",\"WARC-Payload-Digest\":\"sha1:LPOFKSDMDEPODW5E5PLONHPYT5NAKT5Y\",\"WARC-Block-Digest\":\"sha1:XXRK4WFI7JBP5Z4PMWRY6YW3UAGQJBZZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499541.63_warc_CC-MAIN-20230128090359-20230128120359-00153.warc.gz\"}"} |
http://webcatplus.nii.ac.jp/webcatplus/details/book/5131101.html | [
"#",
null,
"",
null,
"## Simple morphisms in algebraic geometry\n\nRichard Sot\n\n[目次]\n\n• The Zariski topology, the Jacobian criterion and examples of simple algebras over a field k.- The Kahler 1-differentials.- Every k-algebra a which is essentially of finite type over k and simple is a regular local ring.- Brief discussion of unramified and etale homomorphisms.- Some corollaries to Theorem 3.5.- Fitting ideals.- Proof of the Jacobian criterion and some characterizations of simple k-algebras and A-algebras.- Characterizations of simple A-algebras in terms of etale homomorphisms\n• invariance of the property of being a simple algebra under composition and change of base.- Descent of simple homomorphisms and removal of all noetherian assumptions in Chapter 7 and Chapter 8.- Simple morphisms of preschemes and translation of previous theorems into the language of preschemes.\n\n「Nielsen BookData」より\n\n###",
null,
"書名 Simple morphisms in algebraic geometry Sot, Richard Sot R. Lecture notes in mathematics Springer-Verlag 1982 iv, 145 p. 25 cm 0387115641 3540115641 BA00678257 ※クリックでCiNii Booksを表示 英語 ドイツ\n\n##",
null,
"",
null,
"##",
null,
"",
null,
""
] | [
null,
"http://webcatplus.nii.ac.jp/externals/images/ci-02.png",
null,
"http://webcatplus.nii.ac.jp/externals/images/img-nowprinting-book-01.jpg",
null,
"http://webcatplus.nii.ac.jp/externals/images/txt-info-01.gif",
null,
"http://webcatplus.nii.ac.jp/externals/images/txt-ext-01.png",
null,
"http://webcatplus.nii.ac.jp/externals/images/btn-more-01.gif",
null,
"http://webcatplus.nii.ac.jp/externals/images/txt-wiki-02.png",
null,
"http://webcatplus.nii.ac.jp/externals/images/prt-loader-01.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80032295,"math_prob":0.8188133,"size":792,"snap":"2020-45-2020-50","text_gpt3_token_len":185,"char_repetition_ratio":0.14467005,"word_repetition_ratio":0.0,"special_character_ratio":0.18560606,"punctuation_ratio":0.08888889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98114544,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T02:47:38Z\",\"WARC-Record-ID\":\"<urn:uuid:40835d20-234d-40ab-97d2-9df1ea177e15>\",\"Content-Length\":\"14535\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8082bfe0-2cf5-4620-acdb-f5e5dc539e83>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d267c6e-538c-40a3-bfee-025858780ec3>\",\"WARC-IP-Address\":\"13.230.131.17\",\"WARC-Target-URI\":\"http://webcatplus.nii.ac.jp/webcatplus/details/book/5131101.html\",\"WARC-Payload-Digest\":\"sha1:W6H4GHQBOEPMNXZBA2IRXZODXXYASTPZ\",\"WARC-Block-Digest\":\"sha1:3JJQ7CIC3JTIDZHVW4O5GVT6RB6NZIN3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107881640.29_warc_CC-MAIN-20201024022853-20201024052853-00709.warc.gz\"}"} |
https://physicsteacher.in/2020/09/30/how-to-derive-terminal-velocity-equation-using-stokes-law/ | [
"# How to derive Terminal Velocity equation using Stokes’ law (step by step)\n\nHere we will work on the derivation of the Terminal Velocity equation or formula using Stokes’ Law. We will consider a situation where a solid sphere moving slowly in a fluid to derive the Terminal Velocity equation.\n\n## Derivation of Terminal Velocity Equation using Stokes’ law\n\nWhen an object is falling through a fluid, in that case, if we want to analyze its motion (and find out its acceleration, if any) then we need to consider the weight of the object, the upthrust on the object applied by the displaced volume of the fluid, and the viscous drag force caused by the movement of the object in the fluid.\n\nUsually, we consider the equilibrium situation, in which the weight of the object exactly balances the sum of upthrust and drag force. And in this equilibrium situation as the net force on the object is zero, hence the velocity of the object remains constant. This constant velocity is terminal velocity.\n\nThis is true for skydivers falling through the fluid air as well as for a ball bearing dropping through a column of oil.\n\nTo derive the Terminal Velocity equation we will consider simple situations, say for a solid sphere moving slowly in a fluid.\n\nNow in equilibrium, i.e. when the solid sphere is moving with terminal velocity then:\n\nweight of the sphere = upthrust on the sphere applied by the displaced fluid + Stokes’ force or viscous drag force\n=> weight of the sphere = weight of the displaced fluid + Stokes’ force or viscous drag force\n\nmsg = weight of the displaced fluid + 6 πrȠVterm ……………….. (1)\n\n[ Note:\nHere, ms is the mass of the sphere. So, msg = weight of the sphere.\n\nThe Stokes’ Law formula for viscous drag force is represented in this way: F = 6 πrȠV\nwhere r is the radius of the sphere, V is the velocity of the sphere and Ƞ is the coefficient of viscosity of the fluid.\nHere in equilibrium condition in place of V, we will use Vterm which is terminal velocity\n]\n\nNow, let’s expand equation (1) a bit more.\nms = mass of the sphere = volume of the sphere x density of the sphere material = (4/3) πr3ρs\nSo in place of weight of the sphere (msg) in equation 1 we can write, (4/3) πr3ρs g ……. (a)\n\nNow, upthrust on the sphere = weight of the fluid displaced\n= mass of the fluid displaced x g = volume of the fluid displaced x density of the fluid x g\n= volume of the sphere x density of the fluid x g = (4/3) πr3 ρf g …. (b)\n(Note that the volume of the fluid displaced is equal to the volume of the sphere.)\n\nSo from equation (1), (a) and (b) we get a new one,\n(4/3) πr3ρsg = (4/3) πr3 ρf g + 6 πrȠVterm ……………….. (2)\n\nNow, let’s rearrange the equation to get Vterm\nVterm = [(4/3) πr3 g (ρsρf )] / [6 πrȠ]\n\nAnd after simplifying the equation we get the final equation of terminal velocity as follows\n\n=> Vterm = [2 r2 g (ρs – ρf )] / [9Ƞ]\n\n## Summary\n\nThus using Stokes’ law you can derive the terminal velocity equation given below:\n\nVterm = [2 r2 g (ρs – ρf )] / [9Ƞ]\n\nHow to derive Terminal Velocity equation using Stokes’ law (step by step)\nScroll to top"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8696771,"math_prob":0.9974889,"size":2885,"snap":"2021-31-2021-39","text_gpt3_token_len":756,"char_repetition_ratio":0.18917042,"word_repetition_ratio":0.13818182,"special_character_ratio":0.26169845,"punctuation_ratio":0.078947365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99975866,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-31T23:57:17Z\",\"WARC-Record-ID\":\"<urn:uuid:190d1f47-4ca1-4c84-8fdd-662cb6c9dad0>\",\"Content-Length\":\"110307\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c963f0bc-1653-47e3-a1bc-dc09640890bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:85cd26a2-a943-4218-827d-ab00e24b2a37>\",\"WARC-IP-Address\":\"35.175.60.16\",\"WARC-Target-URI\":\"https://physicsteacher.in/2020/09/30/how-to-derive-terminal-velocity-equation-using-stokes-law/\",\"WARC-Payload-Digest\":\"sha1:CFDXEYZMAW2XZCWMHAYHFPU44ZYQ5DJD\",\"WARC-Block-Digest\":\"sha1:ZUG4G5LER6T654PLZFFJBX5FADB7QV2M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154127.53_warc_CC-MAIN-20210731234924-20210801024924-00573.warc.gz\"}"} |
https://www.percentagecal.com/answer/159-is-what-percent-of-304 | [
"Solution for 159 is what percent of 304:\n\n159:304*100 =\n\n(159*100):304 =\n\n15900:304 = 52.3\n\nNow we have: 159 is what percent of 304 = 52.3\n\nQuestion: 159 is what percent of 304?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 304 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={304}.\n\nStep 4: In the same vein, {x\\%}={159}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={304}(1).\n\n{x\\%}={159}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{304}{159}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{159}{304}\n\n\\Rightarrow{x} = {52.3\\%}\n\nTherefore, {159} is {52.3\\%} of {304}.\n\nSolution for 304 is what percent of 159:\n\n304:159*100 =\n\n(304*100):159 =\n\n30400:159 = 191.19\n\nNow we have: 304 is what percent of 159 = 191.19\n\nQuestion: 304 is what percent of 159?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 159 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={159}.\n\nStep 4: In the same vein, {x\\%}={304}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={159}(1).\n\n{x\\%}={304}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{159}{304}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{304}{159}\n\n\\Rightarrow{x} = {191.19\\%}\n\nTherefore, {304} is {191.19\\%} of {159}.\n\nCalculation Samples"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.840842,"math_prob":0.9996049,"size":2169,"snap":"2022-05-2022-21","text_gpt3_token_len":729,"char_repetition_ratio":0.17967668,"word_repetition_ratio":0.42480212,"special_character_ratio":0.45089903,"punctuation_ratio":0.14137214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995315,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T18:01:26Z\",\"WARC-Record-ID\":\"<urn:uuid:b3c8b9f2-b5fb-4c00-975c-796bdbf3579a>\",\"Content-Length\":\"10371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3a72449-0c16-4168-9434-b6353949b0be>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d3e022d-4a8b-419b-89d8-6457fa880bbe>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"https://www.percentagecal.com/answer/159-is-what-percent-of-304\",\"WARC-Payload-Digest\":\"sha1:6XFS7GGAEUUYQR2SXH5HK2T2GNCQRW54\",\"WARC-Block-Digest\":\"sha1:6VXUBTBQP5MD26RUJ4256YUP7UCL5J26\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303868.98_warc_CC-MAIN-20220122164421-20220122194421-00073.warc.gz\"}"} |
http://www.techbrothersit.com/2016/03/how-to-export-data-to-multiple-excel.html?showComment=1550220740190 | [
"### How to Export Data to Multiple Excel Sheets from Single SQL Server Table in SSIS Package - SSIS Tutorial\n\nYou are working as ETL developer or an SSIS developer. You have a table let's say dbo.TotalSale. Once of the column in this table is Region. You would like to create an Excel file with Multiple Sheets, each for Region Value.\n\nIf there are three distinct Regions values in Region Column, it should create an Excel file with three excel sheets.\n\nSo if tomorrow, the Region decreases or increases, the sheets should be created according to distinct values.\n\nIn your scenario, you can chose any column that you would like to use for dividing the table data to multiple sheets.\n\nHow to create multiple Excel sheets from single SQL Server Table in SSIS Package - SSIS Tutorial\n\n## Solution:\n\nWe are going to use Script Task in SSIS Package to create multiple Excel Sheets for single Table data depending upon the distinct value from a column.\n\nStep 1: Create new SSIS Package and create Variables\nOpen your SSDT ( SQL Server Data Tools) and create new SSIS Package. Once done, create below variables as shown.\n\nColumnNameForGrouping : Provide the column you would like to use for creating sheet data\nExcelFileName : Provide the excel file name you would like to create\nFolderPath: Provide the Path where you would like to create excel files\nTableName: Provide the table name with schema from which you like to get data\n\nCreate variables in SSIS Package to create Multiple Excel Sheets from SQL Server Table - SSIS Tutorial\n\nStep 2: Create ADO.NET Connection in SSIS Package to use in Script Task\nCreate ADO.NET Connection Manager so we can use in Script Task to get tables data to export to sheets in Excel File. This ADO.Net connection should be pointing to Database that we want to export to an Excel File.\n\nCreate ADO.NET Connection in SSIS Package to use in Script Task to Export All Tables from Database to Excel File with Top X Rows\n\nBring the Script Task on Control Flow Pane in SSIS Package and open by double clicking Check-box in front of variable to add to Script Task.\nAdd variables to Script Task to create multiple excel sheets from sql server table in SSIS Package\n\nStep 4: Add Script to Script task Editor in SSIS Package to Export Data from single SQL Server Table to Multiple Excel Sheets in Excel File\nClick Edit Button and it will open Script Task Editor.\nUnder #region Namespaces, I have added below code\n\n```using System.IO;\nusing System.Data.OleDb;\nusing System.Data.SqlClient;```\n\nUnder public void Main() {\n\n``` string datetime = DateTime.Now.ToString(\"yyyyMMddHHmmss\");\ntry\n{\n\n//Declare Variables\nstring ExcelFileName = Dts.Variables[\"User::ExcelFileName\"].Value.ToString();\nstring FolderPath = Dts.Variables[\"User::FolderPath\"].Value.ToString();\nstring TableName = Dts.Variables[\"User::TableName\"].Value.ToString();\nstring ColumnNameForGrouping = Dts.Variables[\"User::ColumnNameForGrouping\"].Value.ToString();\nExcelFileName = ExcelFileName + \"_\" + datetime;\n\nOleDbConnection Excel_OLE_Con = new OleDbConnection();\nOleDbCommand Excel_OLE_Cmd = new OleDbCommand();\n\n//Construct ConnectionString for Excel\nstring connstring = \"Provider=Microsoft.ACE.OLEDB.12.0;\" + \"Data Source=\" + FolderPath + ExcelFileName\n+ \";\" + \"Extended Properties=\\\"Excel 12.0 Xml;HDR=YES;\\\"\";\n\n//drop Excel file if exists\nFile.Delete(FolderPath + \"\\\\\" + ExcelFileName + \".xlsx\");\n\n//USE ADO.NET Connection from SSIS Package to get data from table\n\n//Read distinct Group Values for each Excel Sheet\nstring query = \"Select distinct \"+ColumnNameForGrouping+\" from \"+TableName;\n\n//MessageBox.Show(query.ToString());\nSqlCommand cmd = new SqlCommand(query, myADONETConnection);\nDataTable dt = new DataTable();\n\n//Loop through values for ColumnNameForGroup\n\nforeach (DataRow dt_row in dt.Rows)\n{\nstring ColumnValue = \"\";\nobject[] array = dt_row.ItemArray;\nColumnValue = array.ToString();\n\n//Load Data into DataTable from SQL ServerTable\nstring queryString =\n\"SELECT * from \" + TableName+\" Where \"+ColumnNameForGrouping+\"='\"+ColumnValue+\"'\";\nDataSet ds = new DataSet();\n\nstring TableColumns = \"\";\n\n// Get the Column List from Data Table so can create Excel Sheet with Header\nforeach (DataTable table in ds.Tables)\n{\nforeach (DataColumn column in table.Columns)\n{\nTableColumns += column + \"],[\";\n}\n}\n\n// Replace most right comma from Columnlist\nTableColumns =(\"[\"+ TableColumns.Replace(\",\", \" Text,\").TrimEnd(','));\nTableColumns = TableColumns.Remove(TableColumns.Length - 2);\n//MessageBox.Show(TableColumns);\n\n//Use OLE DB Connection and Create Excel Sheet\nExcel_OLE_Con.ConnectionString = connstring;\nExcel_OLE_Con.Open();\nExcel_OLE_Cmd.Connection = Excel_OLE_Con;\nExcel_OLE_Cmd.CommandText = \"Create table [\" +ColumnValue + \"] (\" + TableColumns + \")\";\nExcel_OLE_Cmd.ExecuteNonQuery();\n\n//Write Data to Excel Sheet from DataTable dynamically\nforeach (DataTable table in ds.Tables)\n{\nString sqlCommandInsert = \"\";\nString sqlCommandValue = \"\";\nforeach (DataColumn dataColumn in table.Columns)\n{\nsqlCommandValue += dataColumn + \"],[\";\n}\n\nsqlCommandValue=\"[\"+sqlCommandValue.TrimEnd(',') ;\nsqlCommandValue = sqlCommandValue.Remove(sqlCommandValue.Length - 2);\nsqlCommandInsert = \"INSERT into [\" + ColumnValue + \"] (\" + sqlCommandValue +\") VALUES(\";\n\nint columnCount = table.Columns.Count;\nforeach (DataRow row in table.Rows)\n{\nstring columnvalues = \"\";\nfor (int i = 0; i < columnCount; i++)\n{\nint index = table.Rows.IndexOf(row);\ncolumnvalues += \"'\" + table.Rows[index].ItemArray[i] + \"',\";\n\n}\ncolumnvalues = columnvalues.TrimEnd(',');\nvar command = sqlCommandInsert + columnvalues + \")\";\nExcel_OLE_Cmd.CommandText = command;\nExcel_OLE_Cmd.ExecuteNonQuery();\n}\n\n}\nExcel_OLE_Con.Close();\n}\n}\n\ncatch (Exception exception)\n{\n\n// Create Log File for Errors\nusing (StreamWriter sw = File.CreateText(Dts.Variables[\"User::FolderPath\"].Value.ToString() + \"\\\\\" +\nDts.Variables[\"User::ExcelFileName\"].Value.ToString() + datetime + \".log\"))\n{\nsw.WriteLine(exception.ToString());\n\n}\n\n}```\n\nStep 5: Save the script and Run SSIS Package to create Multiple Sheets from single SQL Server Table\nHit save button and then close the script task editor window. Run your SSIS Package and it should create an excel file with multiple sheets depending upon unique regions from region column.\n\nI ran my SSIS Package for above table, and here is my Excel file with multiple sheets.\n\n1.",
null,
"I'm trying to create multiple excel files instead of multiple excel sheets.\n\n1.",
null,
"string datetime = DateTime.Now.ToString(\"yyyyMMdd\");\ntry\n{\n\n//Declare Variables\nstring FolderPath = Dts.Variables[\"User::FolderPath\"].Value.ToString();\nstring TableName = Dts.Variables[\"User::TableName\"].Value.ToString();\nstring ColumnNameForGrouping = Dts.Variables[\"User::ColumnNameForGrouping\"].Value.ToString();\nstring ExcelFileName = \"\";\n\nOleDbConnection Excel_OLE_Con = new OleDbConnection();\nOleDbCommand Excel_OLE_Cmd = new OleDbCommand();\n\n//USE ADO.NET Connection from SSIS Package to get data from table\n\n//Read distinct Group Values for each Excel Sheet\nstring query = \"Select distinct \" + ColumnNameForGrouping + \" from \" + TableName; //+ \" where Key_Month = '201907'\";\n\n//MessageBox.Show(query.ToString());\nSqlCommand cmd = new SqlCommand(query, myADONETConnection);\nDataTable dt = new DataTable();\n\n//Loop through values for ColumnNameForGroup\n\nforeach (DataRow dt_row in dt.Rows)\n{\nstring CountryValue = \"\";\nobject[] array = dt_row.ItemArray;\nCountryValue = array.ToString();\nExcelFileName = CountryValue + \"_\" + datetime;\n\n//Construct ConnectionString for Excel\nstring connstring = \"Provider=Microsoft.ACE.OLEDB.12.0;\" + \"Data Source=\" + FolderPath + ExcelFileName\n+ \";\" + \"Extended Properties=\\\"Excel 12.0 Xml;HDR=YES;\\\"\";\n\n//drop Excel file if exists\nFile.Delete(FolderPath + \"\\\\\" + ExcelFileName + \".xlsx\");\n\n//Load Data into DataTable from SQL ServerTable\nstring queryString =\n\"SELECT * from \" + TableName + \" Where \" + ColumnNameForGrouping + \"='\" + CountryValue + \"'\";\nDataSet ds = new DataSet();\n\nstring TableColumns = \"\";\n\n// Get the Column List from Data Table so can create Excel Sheet with Header\nforeach (DataTable table in ds.Tables)\n{\nforeach (DataColumn column in table.Columns)\n{\nTableColumns += column + \"],[\";\n}\n}\n\n// Replace most right comma from Columnlist\nTableColumns = (\"[\" + TableColumns.Replace(\",\", \" Text,\").TrimEnd(','));\nTableColumns = TableColumns.Remove(TableColumns.Length - 2);\n//MessageBox.Show(TableColumns);\n\n2.",
null,
"//Write Data to Excel Sheet from DataTable dynamically\nforeach (DataTable table in ds.Tables)\n{\n//Use OLE DB Connection and Create Excel Sheet\nExcel_OLE_Con.ConnectionString = connstring;\nExcel_OLE_Con.Open();\nExcel_OLE_Cmd.Connection = Excel_OLE_Con;\nExcel_OLE_Cmd.CommandText = \"Create table [\" + CountryValue + \"] (\" + TableColumns + \")\";\nExcel_OLE_Cmd.ExecuteNonQuery();\n\nString sqlCommandInsert = \"\";\nString sqlCommandValue = \"\";\nforeach (DataColumn dataColumn in table.Columns)\n{\nsqlCommandValue += dataColumn + \"],[\";\n}\n\nsqlCommandValue = \"[\" + sqlCommandValue.TrimEnd(',');\nsqlCommandValue = sqlCommandValue.Remove(sqlCommandValue.Length - 2);\nsqlCommandInsert = \"INSERT into [\" + CountryValue + \"] (\" + sqlCommandValue + \") VALUES(\";\n\nint columnCount = table.Columns.Count;\nforeach (DataRow row in table.Rows)\n{\nstring CountryValues = \"\";\nfor (int i = 0; i < columnCount; i++)\n{\nint index = table.Rows.IndexOf(row);\nCountryValues += \"'\" + table.Rows[index].ItemArray[i] + \"',\";\n\n}\nCountryValues = CountryValues.TrimEnd(',');\nvar command = sqlCommandInsert + CountryValues + \")\";\n\nExcel_OLE_Cmd.CommandText = command;\nExcel_OLE_Cmd.ExecuteNonQuery();\n\n}\n\nExcel_OLE_Con.Close();\n}\n\n}\n}\n\ncatch (Exception exception)\n{\n\n// Create Log File for Errors\nusing (StreamWriter sw = File.CreateText(Dts.Variables[\"User::FolderPath\"].Value.ToString() + \"\\\\\" +\ndatetime + \".log\"))\n{\nsw.WriteLine(exception.ToString());\n\n}\n\n}\n\n3.",
null,
"Hi, I added 2 parts of the code (I couldn't paste the whole code into 1 comment) in order to create several excel files instead of worksheets. You need to remove the parameter ExcelFileName and use this field in the code to assign the dynamic name for your new excels. I used country name as ExcelFileName. Good luck!\n\n2.",
null,
"I was trying to create excel as per your tutorial but I am getting below error.\nI am using SSIS 2012 I am very poor in scripting\nat System.RuntimeMethodHandle.InvokeMethod(Object target, Object[] arguments, Signature sig, Boolean constructor)\nat System.Reflection.RuntimeMethodInfo.UnsafeInvokeInternal(Object obj, Object[] parameters, Object[] arguments)\nat System.Reflection.RuntimeMethodInfo.Invoke(Object obj, BindingFlags invokeAttr, Binder binder, Object[] parameters, CultureInfo culture)\nat System.RuntimeType.InvokeMember(String name, BindingFlags bindingFlags, Binder binder, Object target, Object[] providedArgs, ParameterModifier[] modifiers, CultureInfo culture, String[] namedParams)\n\n3.",
null,
"I have got this code to work however it is putting an underscore in front of everysheet name. Also i do I use a template using this code\n\n1.",
null,
"How did you add template to this process? Can you please write steps?\n\n4.",
null,
"The code works great. But how can you keep the data type from sql when exporting into excel? my date and int column outputs are showing as text.\n\n5.",
null,
"The code works great. But how can you keep the data type from SQL when exporting into excel? my date and int column outputs are showing as text.\n\n6.",
null,
"I got it to the end up until running the script run succesfully, but the file is nowhere to be found on the FolderPath. I tried changing file path or giving all the permissions to the folder to check if that was the issue."
] | [
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64662296,"math_prob":0.66132176,"size":15318,"snap":"2020-45-2020-50","text_gpt3_token_len":3334,"char_repetition_ratio":0.167102,"word_repetition_ratio":0.36410257,"special_character_ratio":0.22966444,"punctuation_ratio":0.16377953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9760564,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T22:33:54Z\",\"WARC-Record-ID\":\"<urn:uuid:c3a01e34-5063-42df-8fce-8791f18ebac4>\",\"Content-Length\":\"145403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ac8f294-5d0d-4c1a-b28f-64757ba2a291>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec4fdb24-d7ef-47c4-8764-ba2b58eff7f3>\",\"WARC-IP-Address\":\"172.217.8.19\",\"WARC-Target-URI\":\"http://www.techbrothersit.com/2016/03/how-to-export-data-to-multiple-excel.html?showComment=1550220740190\",\"WARC-Payload-Digest\":\"sha1:XMIJSRD466NRIMMQ6IX6UJTKXUQ2RXBB\",\"WARC-Block-Digest\":\"sha1:QWVMSVCIS252KVQTO2WTZLEDMXVCQDML\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107911792.65_warc_CC-MAIN-20201030212708-20201031002708-00520.warc.gz\"}"} |
https://ftp.romhacking.net/wikid/index.php?title=Vagrant_Story:SEQ_files&oldid=26882 | [
"# Vagrant Story:SEQ files\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\nSeq (Sequence) Files are animation files that are used to animate the games 3D models. The file format is not yet understood. Much of what follows was decyphered by Valendian and another reverse engineer whose shall remain anonymous until I have their permission to reveal their identity.\n\nIt seems that each SeqFrameHeader defines a different skeletal/bone pose. Animations are achieved by interpolating between these poses. But there is more done to the joints with the Opcodes, which still has to be uncovered. Note that, for example in a jump, the different poses are hardcoded. So it seems that not all animation info is stored in the SEQ files.\n\nEvery joint is assigned a rotation, but translation vectors, except for the complete model, have not been found yet.\n\n```typedef struct tSeqHeader {\nubyte NumFrames;\nuhalf NumJoints;\nword FileSize;\n// Hdr.NumFrames # number of frames of animation\n// Hdr.NumJoints # number of joints in the target skeleton (must match that of the shp file)\n// Hdr.FileSize # size of file data (minus this header) in bytes aligned to a 4 byte boundary\n// the number of sequences must be calculated at runtime. its not actually needed but nice to know\n\ntypedef struct tSeqPointerTable { // seq Pointer Table\nword PtrSection3; // \nword PtrSection2; // \n} SEQ_PTRTBL;\n// SeqPointerTable.PtrSection3 # pointer to the frame data section\n// SeqPointerTable.PtrSection2 # pointer to the next sequence section\n\ntypedef struct tSeqFrameHeader { // seq Frame Header\nuhalf Unknown1; // \nubyte idOtherPose; // \nubyte Unknown3; // \nuhalf PtrNextFrame; // \nuhalf PtrThisFrame; // \nuhalf PtrUnknownFrame; // \nuhalf JointPtr0[Hdr.NumJoints]; // \nuhalf JointPtr1[Hdr.NumJoints]; // \n} SEQ_FRAMEHDR[NumSequences];\n// FrameHdr.Unknown1 #\n// FrameHdr.idOtherPose # see below\n// FrameHdr.Unknown3 #\n// FrameHdr.PtrNextFrame #\n// FrameHdr.PtrThisFrame #\n// FrameHdr.PtrUnknownFrame #\n// FrameHdr.JointPtr0[Hdr.NumJoints] # points to rotation and animation data for each joint per frame\n// FrameHdr.JointPtr1[Hdr.NumJoints] # points to animation data for each joint per frame\n\n// if FrameHdr.idOtherPose is not 0xFF = -1, then it's an id of a\n// different pose, and the FrameHdr.JointPtr0/1 of the OTHER pose is\n\nubyte Section2[Hdr.NumFrames]; // \n// specifies the sequence that follows on from this one (or looping)\n\ntypedef struct tSeqFrameDataHeader { // seq Frame Data Header\nubyte Unknown0; // # (unaligned half's are read as two bytes)\nubyte Unknown1; // # (unaligned half's are read as two bytes)\nubyte Unknown2; // # (unaligned half's are read as two bytes)\n} SEQ_FrameDataHDR;\n// FrameDatahdr.Unknown0 #\n// FrameDatahdr.Unknown1 #\n// FrameDatahdr.Unknown2 #\n\n// if FrameHdr.idOtherPose is 0xFF, this is the first thing that\n// appears at any offset given by FrameHdr.JointPtr\n// use this instead of SEQ_FrameDataHDR\n\ntypedef struct tSeqFrameDataRotation { // seq Frame Data Rotation\nubyte x1;\nubyte x2;\nubyte y1;\nubyte y2\nubyte z1;\nubyte z2;\n} SEQ_FrameDataRotation;\n// rotation is represented by euler angles\n// stored in big endian!\n// angles are computed like this:\n// x = ((x1 << 8) | x2) << 1; // rotation around x axis\n// 0x0000 is 0 degrees, 0x0400 is 90 degrees, 0x0800 is 180 degrees and 0x0c00 is -90 degrees\n// the bone is first rotated around the x axis, then y, then z\n\ntypedef struct tSeqFrameOpcode {\nubyte Opcode;\nubyte Param1;\nubyte Param2;\nubyte Param3;\nubyte Param4;\nuhalf Param5;\n} SEQ_FrameOpcode;\n// FrameOpcode.Opcode # the 3 most significant bits flag the existence of parameters\n// FrameOpcode.Param1 # present if ((FrameOpcode.Opcode & 0x80) == 0x80)\n// FrameOpcode.Param2 # present if ((FrameOpcode.Opcode & 0x40) == 0x40)\n// FrameOpcode.Param3 # present if ((FrameOpcode.Opcode & 0x20) == 0x20)\n// FrameOpcode.Param4 # present if (((FrameOpcode.Opcode & 0xE0) == 0x00)\n// && ((FrameOpcode.Opcode & 0x03) != 0x03))\n// FrameOpcode.Param5 # present if ((FrameOpcode.Opcode & 0xE0) == 0x00)\n\ntypedef struct tSeqSequenceEnd { // appears to control looping animations\nuhalf Unknown1; // # (unaligned half's are read as two bytes)\nuhalf Unknown2; // # (unaligned half's are read as two bytes)\nuhalf Unknown3; // # (unaligned half's are read as two bytes)\n} SEQ_SequenceEnd;\n// SequenceEnd.Unknown1 # sometimes first byte is read as an opcode\n// SequenceEnd.Unknown2 #\n// SequenceEnd.Unknown3 #\n\n// putting it all together the file has the following layout\nruntime word NumSequences = (PtrTbl.PtrSection2 - &FrameHdr) / (0x0A + 4*Hdr.NumJoints);\nSEQ_PTRTBL PtrTbl;\nSEQ_FRAMEHDR FrameHdr[NumSequences];\nubyte Section2[Hdr.NumFrames];\n\n// probably not for each sequence\nfor (NumSequences) {\nSEQ_FrameDataHDR FrameDataHdr;\nfor (?) {\nSEQ_FrameOpcode Opcode;\n}\n}\n\n```\n\n## On Opcodes and Operations\n\nLet op be an opcode. We will construct the bytes of the corresponding operation v = (op, ?, ...)\n\nif op = 0, then v = (op), the animation ends.\n\nif any of the three MSBs of op is 1, we write op = xyzabcde where x,y,z,a,b,c,d,e are the respective bits.\n\nif abcde = 11111, we have an additional byte t (for timing, need to confirm purpose) otherwise, we define t = abcde, also for timing/number of frames, but without an additional byte. so we have v = (op, t?, ?, ...)\n\nin both cases we have rotation angles, which can be constructed like this:\n\nif x = 1, we have an additional byte rx (rotation speed around x axis). if y = 1, we have an additional byte ry (rotation speed around y axis). if z = 1, we have an additional byte rz (rotation speed around z axis).\n\nso if xyz = 111, v = (op, t?, rx, ry, rz) if xyz = 101, v = (op, t?, rx, rz) and so on.\n\nall in all, we have v = (op, t?, rx?, ry?, rz?), where the bits marked with ? are present depending on the above conditions.\n\nnow on the computation of the actual rotation:\n\nit seems that an animation may only have a fixed number of frames, it is assumed that this could be 30. this can be seen by adding the operations (fe, 0x80, 0, 0), (fe, 0, 0x80, 0) to a bone animation, for example at RAM 0x801275fe (US version). the second operation has no effect!\n\neach frame, the rx,ry,rz values are added to the rotation vector of the bone.\n\nhowever, if any of the rx, ry, rz are NOT present, then these rotations seem to change towards zero. need confirmation here.\n\nnow let all of the three MSBs of op be 0. TODO"
] | [
null
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http://www.mathprotutoring.com/blog/?cat=15 | [
"# How to use the TI-84 Plus on the SAT for graphing!\n\n## SAT Official Question of the Day for 11/18, but do it MY WAY!\n\nSee the question here: http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20141118&oq=1 The solution given on the SAT site (look at it, please) is soooo much harder than doing THIS! Go get your graphing calculator and read on…\nLook at the choices:\na) -8\nb) -4\nc) 4\nd) 8\ne) 10\n\nTry one. How about -8? So, we assume b = -8. Press ‘Y=’ on your TI-84 Plus (top left). Type into Y1: -2x^2-8x+5. That’s the function you get if you substitute -8 for b. Now press GRAPH (top right). (Do ‘ZOOM-6’ if you need a better picture.) What is the x-value of the highest point? It looks like about -2. The question says it should be 2. So, it’s not choice (a).\n\nTry the next one: -4. Change Y1 (just type 4 over the 8), and press GRAPH again. This time it looks like the highest point is at x=-1. Still not right. It’s not choice (b).\n\nTry 8. Change Y1 (just type +8 over the -4), and press GRAPH again. This time it looks like the highest point is at x=2. Choice (d) is correct.\n\nAnd *THIS* is why I tell all of you to get the TI-84 Plus Graphing Calculator! Get one now: http://amzn.to/1tXyOTH",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Distance, Midpoint, and Slope\n\n### This week’s topic has been Distance, Midpoint, and Slope.\n\nLet’s go through that blue SAT book everyone has, http://amzn.to/1wADKLS, or http://amzn.to/1wtPZuD, and this ACT book for problems about distance, midpoint, and slope that can be solved using my TI-84 Plus programs DISTANCE. Sorry, I won’t post the actual problems here. But go dig out your books and calculator (buy them here: SAT, ACT , TI-84 Plus) and follow along.\n\n## SAT book\n\nISBN-13: 9780874478525. The page numbers included here are from that book, but if you have the other one (9780874479799), please refer to the Test number and Section number instead.\n\nTest 1, Section 3, p. 397, problem # 6\n\nTest 1, Section 3, p. 400, problem #15\n\nTest 3, Section 8, p. 546, problem #10\n\nTest 4, Section 6, p. 596, problem #10\n\nTest 5, Section 8, p. 669, problem #8\n\nTest 7, Section 3, p. 769, problem #4\n\nTest 8, Section 3, p. 830, problem #2\n\nTest 9, Section 5, p. 905, problem #8\n\nTest 10, Section 2, p. 953, problem #17\n\n## ACT book\n\nI’m working out of an older edition that only has 3 tests. I believe they are the same as the first 3 tests of the 2nd and 3rd editions. Once I have the newer book, I’ll add to this list.\n\nTest 1, Section 2, problems #35, 45\n\nTest 2, Section 2, problem #33\n\nTest 3, Section 2, problems #38, 48",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Slope\n\nThis week’s topic is: Distance, Midpoint, and Slope.\n\n### Q: The slope of the line passing through (-1,-3) and (7,y) is -1/2. What is the value of y?\n\n.\n\n.\n\n.\n\n.\n\nAlgebraic solution by hand:\n\nUse the slope formula:",
null,
"Solution using my TI-84 Plus program DISTANCE:\n\nRun DISTANCEHow?\n\nUse your multiple choice answer choices. One of them is -7. (If you don’t have choices, draw a picture. Plot the point (-1,-3) and do a slope of -1/2 from there – down 1, right 2, down 1, right 2, … until you get to a point where x is 7. Read the y-value of that point – it will be -7 and check it using this program.)\n\nx1=-1\n\ny1=-3\n\nx2=7\n\ny2=-7\n\nIgnore the distance information – we don’t need it for this problem. Press ENTER for the midpoint information, which we don’t need either. Press ENTER once more to find that the slope is -1/2.",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Distance\n\nThis week’s topic is: Distance, Midpoint, and Slope.\n\n### Q: For what value(s) of x is the point (x,4) exactly 5 units away from the point (6,8)?\n\n.\n\n.\n\n.\n\n.\n\nSolution by hand:\n\nUse the distance formula:",
null,
"Solution using my TI-84 Plus program DISTANCE:\n\nDraw a picture:",
null,
"Guess that x=2, maybe. It’s probably an integer if this is an SAT problem. Use the answer choices if available.\n\nRun DISTANCEHow?\n\nx1=2\n\ny1=4\n\nx2=6\n\ny2=8\n\nThe distance is 5.66. It’s too big – you wanted it to be 5 – so move the point a little closer to the middle.\n\nMaybe x=3:\n\nx1=3\n\ny1=4\n\nx2=6\n\ny2=8\n\nThe distance is exactly 5. Perfect. (There’s also another possible answer, x=9. If you have multiple choice answers, you will see 3 and 9 as a choice, so try both and see that they both work. For a grid-in SAT problem, you would only have to find one answer anyway.) :",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Ratios and Proportions\n\n### This week’s topic has been Ratios and Proportions.\n\nLet’s go through that blue SAT book everyone has, http://amzn.to/1wADKLS, or http://amzn.to/1wtPZuD, and this ACT book for problems about ratios that can be solved using my TI-84 Plus programs RATIO and PROPORTN. (See these posts for examples: 10/27, 10/28, 10/29, 10/30)\n\nSorry, I won’t post the actual problems here. But go dig out your books and calculator (buy them here: SAT, ACT , TI-84 Plus) and follow along.\n\n## SAT book\n\nISBN-13: 9780874478525. The page numbers included here are from that book, but if you have the other one (9780874479799), please refer to the Test number and Section number instead.\n\nTest 1, Section 7, p. 417, problem # 13\n\nTest 4, Section 6, p. 594, problem #6\n\nTest 10, Section 2, p. 953, problem #16\n\nTest 10, Section 8, p. 980, problem #12\n\nMath Multiple-Choice Sample Questions, p. 307, problem #6\n\nMath Student-Produced Response Sample Questions, p. 347, problem #1\n\nTest 1, Section 7, p. 417, problems # 11, 13\n\nTest 2, Section 2, p. 452, problem #2\n\nTest 3, Section 5, p. 528, problem #9\n\nTest 3, Section 8, p. 543, problem #1\n\nTest 4, Section 3, p. 583, problem #6\n\nTest 5, Section 2, p. 639, problem #5\n\nTest 6, Section 4, p. 715, problem #10\n\nTest 7, Section 3, p. 768, problem #1\n\nTest 9, Section 8, p. 916, problem #4\n\nTest 9, Section 8, p. 918, problem #11\n\nTest 10, Section 2, p. 950, problem #7\n\n## ACT book\n\nI’m working out of an older edition that only has 3 tests. I believe they are the same as the first 3 tests of the 2nd and 3rd editions. Once I have the newer book, I’ll add to this list.\n\nTest 2, Section 2, problem #15\n\nTest 1, Section 2, problems #19, 39, 44\n\nTest 2, Section 2, problems #3",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Ratios\n\nQ: Last month, the ratio of sunny days to rainy days was 3:2. How many rainy days were there last month, if the month had 30 days? (Assume all days were either sunny or rainy.)\n\n.\n\n.\n\n.\n\n.\n\nSolution by hand:\n\nNumber of sunny days = 3x\n\nNumber of rainy days = 2x\n\n3x + 2x = 30\n\n5x=30\n\nx = 6\n\nThere were 2x rainy days, so there were 12 rainy days.\n\nSolution using my TI-84 Plus Program RATIO:\n\nRun RATIOHow?\n\nTotal number of items (days) = 30\n\nNumber of categories of items = 2 (sunny and rainy)\n\nRatio part for category 1 (sunny) = 3\n\nRatio part for category 2 (rainy) = 2\n\nDistribution is: sunny = 18 and rainy = 12.",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Proportions\n\n### Q: x is inversely proportional to w. If x = 6 when w = 8, what is x when w = 12?\n\n.\n\n.\n\n.\n\n.\n\nSolution by hand:\n\nFor inverse proportions, (x1)(w1) = (x2)(w2)\n\n(6)(8) = (x)(12)\n\n48 = 12x\n\nx=4\n\nSolution using my TI-84 Plus Program PROPORTN:\n\nRun PROPORTN. How?\n\nx1 = 6\n\ny1 = 8 (here, we have renamed w1 as y1)\n\nHave y2 (really w2)\n\ny2 = 12\n\nChoose ‘Inverse’ proportion.\n\nThen, x2 = 4.",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Proportions\n\n### Q: Wandering around blindly in a parking lot (don’t try it, please), the probability that the first car you bump into is blue is 2/5. If there are 500 blue cars, how many cars are in the parking lot?\n\n.\n\n.\n\n.\n\n.\n\nSolution by hand:\n\nSet up the proportion using the concept ‘blue / total = blue / total’:\n\n2/5 = 500/x\n\n2x = 2500\n\nx = 1250\n\nSolution using my TI-84 Plus Program PROPORTN:\n\nRun PROPORTN. How?\n\nx1 = 2 (blue)\n\ny1 = 5 (total)\n\nHave x2 (blue)\n\nx2 = 500\n\nChoose ‘Direct’ proportion (because it doesn’t say ‘inverse’ in the problem).\n\nThen, y2 (the total) = 1250.",
null,
"Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.\n\n# Ratios\n\n### Q: The ratio of blue cars to non-blue cars in the parking lot is 2:7. If there are 1260 cars in the parking lot, how many are blue?\n\n.\n\n.\n\n.\n\n.\n\nSolution by hand:\n\nNumber of blue cars = 2x\n\nNumber of non-blue cars = 7x\n\n2x + 7x = 1260\n\n9x=1260\n\nx = 140\n\nThere are 2x blue cars, so there are 280 blue cars.\n\nSolution using my TI-84 Plus Program RATIO:\n\nRun RATIOHow?\n\nTotal number of items = 1260\n\nNumber of categories of items = 2 (blue and non-blue)\n\nRatio part for category 1 (blue) = 2\n\nRatio part for category 2 (non-blue) = 7\n\nDistribution is: blue = 280 and non-blue = 980.",
null,
"",
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""
] | [
null,
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null,
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null,
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null,
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null,
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"http://www.mathprotutoring.com/blog/wp-content/uploads/2014/11/distgraph-300x229.png",
null,
"http://www.mathprotutoring.com/blog/wp-content/plugins/wp-conversion-boxes/templates/email-optin/Red_Headline_With_Image/imgs/social-media.png",
null,
"http://www.mathprotutoring.com/blog/wp-content/plugins/wp-conversion-boxes/templates/email-optin/Red_Headline_With_Image/imgs/social-media.png",
null,
"http://www.mathprotutoring.com/blog/wp-content/plugins/wp-conversion-boxes/templates/email-optin/Red_Headline_With_Image/imgs/social-media.png",
null,
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80332947,"math_prob":0.93235326,"size":1944,"snap":"2023-14-2023-23","text_gpt3_token_len":638,"char_repetition_ratio":0.21649484,"word_repetition_ratio":0.023255814,"special_character_ratio":0.35185185,"punctuation_ratio":0.23640168,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833455,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T21:04:10Z\",\"WARC-Record-ID\":\"<urn:uuid:03785baf-d10f-4961-ac49-e7544f8a1832>\",\"Content-Length\":\"107057\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb57338a-0b28-40de-bd83-f6a8398ab532>\",\"WARC-Concurrent-To\":\"<urn:uuid:23995cb4-786b-4744-bef1-a3fc57ff6b50>\",\"WARC-IP-Address\":\"208.113.153.77\",\"WARC-Target-URI\":\"http://www.mathprotutoring.com/blog/?cat=15\",\"WARC-Payload-Digest\":\"sha1:ALBRYGMPPQQ7R3ESFG3OSXLK67BGXSUU\",\"WARC-Block-Digest\":\"sha1:5DAFJPJWQVOY7BN6ABGHX7HQM2K2Q4WP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650264.9_warc_CC-MAIN-20230604193207-20230604223207-00560.warc.gz\"}"} |
https://math.stackexchange.com/questions/2685104/diagonalizability-of-elements-of-finite-subgroups-of-general-linear-group-over-a | [
"# Diagonalizability of elements of finite subgroups of general linear group over an algebraically closed field\n\nHow to show that every element of $G$, where $G$ is a finite subgroup of $GL_n(\\mathbb{k})$, the general linear group of square matrices of order $n$ over some algebraic closed field $\\mathbb{k}$, is diagnonalizable if $\\mathbb{k}$ is an algebraic closure of $\\mathbb{Q}$?\n\nI know that a matrix is diagonalizable if its minimal polynomial is seperable in the field on which the matrix is defined. Now, what if the minimal polynomial has repeated roots? How do we ensure the diginalizability? Any hints. Thanks beforehand.\n\nLet $K$ be a field. As you have already mentioned, a matrix $A \\in \\operatorname{M}_n(K)$ is diagonalizable (over $K$) if and only if there exists a polynomial $f(t) \\in K[t]$ with $f(A) = 0$ such that $f$ decomposes into pairwise different linear factors over $K$.\n\nFor $A \\in G$ and $n := |G|$ we have that $A^n = I$, so that $A$ satisfies the polynomial $f(t) := t^n - 1 \\in \\mathbb{k}[t]$. The polynomial $f(t)$ decomposes into linear factors because $\\mathbb{k}$ is algebraically closed. It follows from $\\operatorname{char}(\\mathbb{k}) = 0$ that the polynomial $f$ is seperable (because $f(t) = t^n - 1$ and $f'(t) = n t^{n-1}$ are coprime), so that $f(t)$ decomposes into pairwise different linear factors. Thus $A$ is diagonalizable.\n\n(To see that $f(t)$ is seperable one can also embed $\\mathbb{k}$ into $\\mathbb{C}$ because $\\mathbb{k}$ is an algebraic closure of $\\mathbb{Q} \\subseteq \\mathbb{C}$. As the roots of unity in $\\mathbb{C}$ are pairwise different, the same goes for $\\mathbb{k}$.)\n\n• What an awesome answer. Just one thing: Is 'if and only if' in the first para correct? I am taught that if distinct eigenvalues then linear transformation diagonalizable. Converse may not hold, I think. – Silent May 12 at 15:15\n• @Silent: If $A$ is diagonalizable with pairwise different eigenvalues $\\lambda_1, \\dotsc, \\lambda_r$ then one can choose the polynomial $f(t) = (t - \\lambda_1) \\dotsm (t - \\lambda_r)$. (The polynomial $f(t)$ does not need to be the characteristic polynomial of $A$, whose linear factors may not be pairwise different.) – Jendrik Stelzner May 12 at 15:46\n\nLet $A\\in G$ be a matrix. After changing bases, $A$ is in JNF. Let us write $A=D+N$ with $D$ diagonal and $N$ the nilpotent part. Now since $G$ is finite, there is some $m\\in\\Bbb N$ with $A^m=I$, the identity matrix. Since $DN=ND$, we have $$I = A^m = (D+N)^m = \\sum_{k=0}^m \\binom mk N^k D^{m-k} = D^m + \\sum_{k=1}^m \\binom mk N^k D^{m-k} =: D^m+\\tilde N.$$ Note that $\\tilde N$ is a nilpotent, strictly upper triangular matrix. It follows that $\\tilde N=0$ and $D^m=I$. Hence, $$0 = \\tilde N = N\\cdot\\left(mD^{m-1}+\\sum_{k=2}^{l} \\binom mk N^{k-1} D^{m-k}\\right)$$ and as the second factor is an invertible upper trianglar matrix (because we are in characteristic zero and $m$ is invertible), so we must have $N=0$. In other words, $A=D$ is already diagonal. In other words, $A$ is diagonal up to changing bases, which means that it is diagonalizable.\n\n• superb answer! by the way, the crux was the finiteness of the subgroup, right? – vidyarthi Mar 10 '18 at 16:31\n• but, the algebraic closure of $\\mathbb{Q}$ is not at all utilized – vidyarthi Mar 10 '18 at 16:41\n• @vidyarthi; the comment by @JendrikStelzner was accurate and the relevance of the field characteristic is now highlighted. – Jesko Hüttenhain Mar 11 '18 at 10:20"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.895308,"math_prob":0.99991846,"size":520,"snap":"2019-43-2019-47","text_gpt3_token_len":127,"char_repetition_ratio":0.12403101,"word_repetition_ratio":0.0,"special_character_ratio":0.2326923,"punctuation_ratio":0.10204082,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000066,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T21:04:40Z\",\"WARC-Record-ID\":\"<urn:uuid:f44d5456-facc-46be-9833-fc39a97c47cb>\",\"Content-Length\":\"148100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97c3f8e3-ddf3-4be4-af67-c87e961a02b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:c952b3c4-ed2c-45e1-b693-3611900d042d>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2685104/diagonalizability-of-elements-of-finite-subgroups-of-general-linear-group-over-a\",\"WARC-Payload-Digest\":\"sha1:KJJMFMTZPDBLWDVLF7Z7O4IWHXUSV5RF\",\"WARC-Block-Digest\":\"sha1:XK3XBBUCBT5KSKWM3F4QWDXJBZ5DNZAO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670987.78_warc_CC-MAIN-20191121204227-20191121232227-00369.warc.gz\"}"} |
https://www.riss.kr/search/Search.do?isDetailSearch=Y&searchGubun=true&queryText=znCreator,%EC%B5%9C%EC%9E%A5%EC%98%81%28Jang-Young+Choi%29&colName=re_a_kor | [
"#",
null,
"다국어 입력\n\nhttp://chineseinput.net/에서 pinyin(병음)방식으로 중국어를 변환할 수 있습니다.\n\n변환된 중국어를 복사하여 사용하시면 됩니다.\n\n예시)\n• 中文 을 입력하시려면 zhongwen을 입력하시고 space를누르시면됩니다.\n• 北京 을 입력하시려면 beijing을 입력하시고 space를 누르시면 됩니다.\n닫기\n\n검색결과 좁혀 보기\n\n선택해제\n• 좁혀본 항목 보기순서\n\n• 원문유무\n• 원문제공처\n• 등재정보\n• 학술지명\n펼치기\n• 주제분류\n• 발행연도\n펼치기\n• 작성언어\n• 저자\n펼치기\n\n오늘 본 자료\n\n• 오늘 본 자료가 없습니다.\n더보기\n• 무료\n• 기관 내 무료\n• 유료\n•",
null,
"이중여자 유도발전기에 의한 가변속 풍력 발전시스템의 동특성 해석\n\nThis paper deals with the dynamic analysis of variable speed wind power systems with doubly-fed induction generators (DFIG). First, the mathematical modeling of wind farm which consists of turbine rotor, DFIG, rotor side and grid side converter and control systems is presented. In particular, the equation for dynamic modeling of the DFIG and the AC/DC/AC converter is expressed as dq reference frame. And then, on the basis of mathematical modeling for each component of wind farm, dynamic simulation algorithms for speed and pitch angle control of wind turbine and generated active and reactive power control of the DFIG and the AC/DC/AC converter are established. Finally, Using the MATLAB/SIMULINK, this paper presents dynamic simulation model for 6㎿ wind power generation systems with the DFIG considering distribution systems and performs the dynamic analysis of wind power systems in steady state. Moreover, this paper also presents the dynamic performance for the case when the voltage sag in grid source and phase fault in bus are occurred.\n\n• Ring-Wound형 슬롯리스 고정자를 갖는 양측식 축 자속 영구자석 전동기의 전자기적 특성해석\n\nThis paper deals with electromagnetic analysis of double-sided axial flux permanent magnet (AFPM) motor with ring-wound type slotless stator. On the basis of magnetic vector potential and polar-coordinate system, the magnetic field solutions due to PM rotor are obtained. In particular, 3-D problem, that is, the reduction of magnetic fields near outer and inner radius of the PM is solved by introducing a special function for radial position. And then, analytical solutions for torque and back-emf are also derived. The predictions are shown in good agreement with those obtained from 3-D finite element (FE) analyses. Therefore, it can be concluded that analytical solutions derived in this paper will be useful for initial design stage of the AFPM motor.\n\n•",
null,
"해석적 방법을 이용한 표면부착형 영구자석 기기의 회전자 와전류 손실해석\n\nThis paper analyzes eddy-current loss induced in magnets of surface-mounted permanent magnet (SPM) machines by using an analytical method such as a space harmonic method. First, on the basis of a two-dimensional (2D) polar coordinate system and a magnetic vector potential, the analytical solutions for the flux density produced by armature winding current are obtained. By using derived field solutions, the analytical solutions for eddy current density distribution are also obtained. Finally, analytical solutions for eddy current loss induced in rotor magnets are derived by using equivalent electrical resistance calculated from magnet volume and analytical solutions for eddy-current density distribution. In particular, the influence of time harmonics in armature current on the eddy current loss is fully investigated and discussed. All analytical results are validated extensively by finite element analysis (FEA).\n\n•",
null,
"공간고조파법을 이용한 축 자속 영구자석 회전기기의 準-3D 특성 해석\n\nThis paper deals with characteristic analysis of axial flux permanent magnet (AFPM) machines with axially magnetized PM rotor using quasi-3-D analysis modeling. On the basis of magnetic vector potential and a two-dimensional (2-D) polar-coordinate system, the magnetic field solutions due to various PM rotors are obtained. In particular, 3-D problem, that is, the reduction of magnetic fields near outer and inner radius of the PM is solved by introducing a special function for radial position. And then, the analytical solutions for back-emf and torque are also derived from magnetic field solutions. The predictions are shown in good agreement with those obtained from 3-D finite element analyses (FEA). Finally, it can be judged that analytical solutions for electromagnetic quantities presented in this paper are very useful for the AFPM machines in terms of following items : initial design, sensitivity analysis with design parameters, and estimation of control parameters.\n\n•",
null,
"전기자동차 압축기용 표면부착형 영구자석 전동기의 극/슬롯수 조합에 따른 특성해석 및 설계\n\nThis paper deals with design and analysis of surface-mounted PM (SPM) motor for compressor of air-conditioning system for electric vehicle applications according to slot/pole combination. First, required torque-speed curve characteristics are determined from operating conditions of the compressor. Restricted conditions such as motor size limit and current density are also determined. And then, under same rated and restricted conditions, twelve models which have different slot/pole combinations each other are designed for various pole arc/pitch ratio using simple equations and 2-d finite element (FE) analyses. Designed models are analyzed and compared in terms of back-emf THD, cogging torque, torque ripple, power losses, efficiency, etc. On the basis of analysis results, it is found that the motor with a 6-pole PM rotor and a 27-slot stator has most outstanding performances in electromagnetic aspects. Finally, through the mechanical modal analysis and demagnetization analysis, it is concluded that the determined motor is most suitable for the compressor of air-conditioning system for electric vehicles\n\n해외이동버튼"
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"https://www.riss.kr/commons/images/logo.png",
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"https://www.riss.kr/search/images/kci.png",
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"https://www.riss.kr/search/images/kci.png",
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"https://www.riss.kr/search/images/kci.png",
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"https://www.riss.kr/search/images/kci.png",
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https://sol.edu.hku.hk/wp-content/screw.php?NP-MzhmMjc3OWRjNjQyNDdjNGFiZTFkZTgwNzQ4NjBmMjA&contentid=1542 | [
"# Homework help graphing calculator\n\nThe ti-89 calculator tutor - volume 1 - 8 hour course this 8 hour dvd course covers the algebra and graphing functions of the ti-89 graphing calculator and teaches the student how to use the calculator features by working problems. Graphing calculator - tutoring & homework help.\n\nTry taking a look at each section as a \"separate\" graph, and grab your scissors. Right from factor by grouping calculator to intermediate algebra syllabus, we have got all the details included. Come to and figure out real numbers, systems of equations make an essay for me free and plenty additional algebra subject areas.\n\n## Homework help news report\n\nFor example, f(x) = 2x is an exponential function with base 2. Homework help hot subject: inequalities.\n\nFriday - december 7: 5. Algeo is the most beautiful scientific graphing calculator available on the content writing company in indore play store. I have attached the instructions. It is a scientific calculator, a differentiation and integration calculator, a matrix calculator, a polynomial. Since then, we have seen the rise of graphing calculators which are essential tools for the college student. Get the mathpapa mobile app. My graphing calculator marks it wrong but there may just be something i don't see. Silver edition graphing calculator in algebra class. Solve the quadratic inequality by - homework help.\n\nAboriginal art homework help\n1. Browse through the list of calculators (including online graphing calculator, derivative calculator, integral calculator) in various subject areas to check your answer or see a step-by-step answer.\n2. Here is a set of notes used by paul dawkins to teach his algebra course at lamar university.\n3. Luckily, graphing calculators make quick work of those problems.\n4. The first two zoom in or out around the center of the graph.\n5. While i prefer k 12 homework help the hp 50g.\n6. Persuasive writing that focuses on homework help graphing calculator convincing readers to see your perspective and agree with it is an argumentative essay.\n7. I am allowed to use this calculator everyday in algebra class and on assessments.\n8. Quiz time quiz preparation: p.\n\nTaculator graphing calculator on the app store. See more examples x+3=5. Webmath is designed to help you solve your math problems. And you don't need an internet homework help graphing calculator connection unlike other graphing. My calculator number is _____. Piecewise defined functions may be continuous (as seen in the example above), or they may be discontinuous (having breaks, jumps, or holes as seen in the examples. Emathhelp - online math resource for all. I agree to make sure the calculator is in perfect working order before class homework help graphing calculator starts-if it is not i. Unlock the cheap math homework help power of your texas instruments graphing. Math homework help & math tutor.\n\nKing henry viii wives primary homework help\n• Graphing calculator - algeo free plotting - apps on.\n• Best graphing calculator for college - 2020 buyer's guide.\n• When a kid needs some help with homework, internet search results can be totally overwhelming.\n• Whether you are doing simple calculations or working on university-level math problems, maple calculator can do it all.\n• Compare and contrast essays are some of the most interesting essays to write.\n• It would not be beneficial to buy a calculator that has functionalities you don't need, in addition to being expensive.\n• You will need to change the equations in each box and the domain for each piece.\n\nNo matter how urgent the deadline of your paper can be, you will get it on time. Ti graphing calculator help videos. It's also helping me learn about the buttons on a graphing calculator, since it's very hard in class. Solve the quadratic inequality by graphing, algebra homework help. Our website will come as an excellent solution to this problem help.\n\nHomework help online alabama\n1. Mathsnet resources for many math topics.\n2. Solving equations by graphing on a graphing calculator.\n3. On the downside, i think math solver is.\n4. I've tried graphing r(x) = x( - 5x) on my ti-84 by just plugging that in but the graph looks different from what's shown in the textbook.\n5. The following diagram represents the market for hamburgers.\n6. F(x,y) is any 3-d function.\n7. How to homework help graphing calculator search for a product.\n8. The casio'll help you on your exams where you won't be allowed the ti-92 (i can't recall ever actually needing graphing capabilities on an exam, and the casio'll will help you create the table used for drawing graphs), and the ti-92 will help you on your homework.\n9. Craft + worksheet + triangle, finding the third root, ti calculator rom download.\n10. Master this subject with our full length step-by-step lessons.\n\nGet a 30 day free trial the easiest way to get started. Explore math world's largest library of interactive online simulations for math and science education.\n\n## Free 7th grade math homework help\n\nHow to place an order. So i thought of coming online to find help. Day 3 homework graphing logarithmic functions (pdf) day 6 notes change of base rule and solving exponential equations day 3 homework blank laws of exponents (pdf). Graph the original (not a reduced version) equation using a graphing calculator, algebra homework help; mathematics.\n\nTypes of volcanoes primary homework help\n1. I'm not confused about the alpha homework help graphing calculator function button because every part is a different color.\n2. The desmos graphing calculator allows users to view and manipulate homework help graphing calculator multiple representations simultaneously, promoting a deeper conceptual understanding of mathematics.\n3. It is to get everything.\n4. A calculator company produces a scientific calculator and a graphing calculator.\n5. Use a linear function to graph a line.\n6. Graphing equations using algebra calculator - mathpapa.\n7. 2 1 graphing absolute value functions homework and.\n8. An ogive (oh-jive), sometimes called a cumulative frequency polygon, is a type of frequency polygon that shows cumulative other words, the cumulative percents are added on the graph from left to right.\n\nFeel free to try them now. 9 hr and 52,428,800 after 10 hr. Suppose an excise tax of per hamburger is imposed on homework help graphing calculator the consumers of hamburgers.\n\nFree math lessons and math homework help from basic math to algebra, geometry and beyond. If you find yourself in need of help homework help graphing calculator in getting your homework done you may find professional writing companies such as quite helpful. Use the intersect feature to calculate the intersection of the two lines. Solution: construct a table and choose simple x values. Not only does it solve the operation, but also draws a graph based on it if possible, and gives extra information about the generated graph. Data graphing, math homework help.\n\nRound answers to 2 decimal places.\n\n## Homework help jiskha\n\nGet more done with homework help from top-rated tutors. Solve your algebra problems step-by-step with mathpapa.\n\nThe directions say to use a grapking utility to. Please help me ww2 evacuees homework help with these 3 questions with a picture of your graphing calculator. Contact us [email protected] 618 e. All these unfavorable circumstances cause permanent stress and can obviously lead to troubles.\n\n## Homework help for year 9\n\nUpdate: all tutorials best resume writing services chicago brisbane are updated to show the default windows skin and math solver ii interface. Water by the spoonful play - 700 word writing homework. Write the correct equation for the relation in the form y=mx+b. More examples here are more examples of how to graph equations in algebra calculator. Get good grades on homework by homework help graphing calculator watching a math video from your own personal online math tutor. 3d graph using parametric lines - desmos. It homework help graphing calculator might seem impossible homework help with logarithms subtraction to you that all custom-written essays, research.\n\nPrimary homework help egyptian pyramids\n• Explore math with our beautiful, free online graphing calculator.\n• [algebra 2: quadratic equations] why is this problem wrong.\n• What parents should know; myths vs.\n• Unit 6 exponents and exponential functions homework 7.\n• A math teacher will randomly distribute 15 graphing calculators 10 non-graphing calculators to her students for a project.\n• If you click and drag on the graphing window, you can draw out a box, and zoom to box will zoom in to make that box fill the whole window.\n\nFind advice on homework, time management, and study habits to help you excel in the classroom, no matter what subject or grade level. Please see the attached document. It's fast and powerful and you'll never have to carry around a large physical ti calculator anymore. Whenever homework help graphing calculator you need help with your assignment, we will be happy to assist you. [college calculus: graphing on calculator] homework help in public libraries can anyone help. He determined that the equation of the line of best fit for some data was y=.\n\nOliver twist homework help\n1. Y = x x > 5.\n2. Mathpapa can solve your equations (and show the work.\n3. The casio fx-7000g was the first speech writing companies graphing calculator made for commercial sale in 1985.\n4. An ogive graph plots cumulative frequency on the y-axis and class boundaries along the 's very similar to a histogram, only instead of rectangles, an ogive has a single."
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https://dashamlav.com/special-relativity-vs-general-relativity/ | [
"# Special Relativity vs General Relativity: Differences Between Theories",
null,
"Dashamlav | 10 October 2020 (अंतिम बदलाव: 16 October 2020)\n\nTheory of Relativity proposed by Albert Einstein had transforming impact on how we understood the nature around us. Modern physics is based on the Theory of Relativity. This theory encompasses two interrelated theories, namely Special Relativity and General Relativity. Both these theories are among the most important scientific achievements of the mankind. Extraordinary astronomical phenomena like black holes and gravitational waves were predicted by Theory of Relatively. There are several differences between these two theories. The following table lists some of these major differences.\n\nDifferences Between the Theories of Special Relativity and General Relativity\nSpecial Relativity General Relativity\nApplies to all physical phenomena in the absence of gravity. It says that the laws of physics are the same for all non-accelerating observers. Applies to all physical phenomena in the presence of gravitational force.\nVelocity causes differences between inertial frames. Acceleration causes differences between non-inertial frames.\nTheory of Special Relativity was introduced in Albert Einstein’s 1905 paper “On the Electrodynamics of Moving Bodies”. Theory of General Relativity was published by Albert Einstein in 1915.\nIntroduced the concept of spacetime and proposed that space and time are not independent from each other. Theory of Special Relativity, however, considered the spacetime to be flat. Theory of General Relativity proposed the idea that spacetime is not flat but it can be curved due to massive bodies.\nConsequences of Special Relativity are:\n\n• Relativity of simultaneity\n• Time dilation\n• Length contraction\n• Maximum speed is finite\n• Mass–energy equivalence (E = mc2)\nConsequences of General Relativity are:\n\n• Gravitational time dilation\n• Precession\n• Light deflection\n• Frame-dragging\n• Metric expansion of space\nTheory of Special Relativity is simpler, not detailed and does not explain the whole universe. Theory of General Relativity is relatively more complex, comprehensive and covers greater part of the universe.\nCitation"
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https://www.colorhexa.com/02c8dd | [
"# #02c8dd Color Information\n\nIn a RGB color space, hex #02c8dd is composed of 0.8% red, 78.4% green and 86.7% blue. Whereas in a CMYK color space, it is composed of 99.1% cyan, 9.5% magenta, 0% yellow and 13.3% black. It has a hue angle of 185.8 degrees, a saturation of 98.2% and a lightness of 43.7%. #02c8dd color hex could be obtained by blending #04ffff with #0091bb. Closest websafe color is: #00cccc.\n\n• R 1\n• G 78\n• B 87\nRGB color chart\n• C 99\n• M 10\n• Y 0\n• K 13\nCMYK color chart\n\n#02c8dd color description : Vivid cyan.\n\n# #02c8dd Color Conversion\n\nThe hexadecimal color #02c8dd has RGB values of R:2, G:200, B:221 and CMYK values of C:0.99, M:0.1, Y:0, K:0.13. Its decimal value is 182493.\n\nHex triplet RGB Decimal 02c8dd `#02c8dd` 2, 200, 221 `rgb(2,200,221)` 0.8, 78.4, 86.7 `rgb(0.8%,78.4%,86.7%)` 99, 10, 0, 13 185.8°, 98.2, 43.7 `hsl(185.8,98.2%,43.7%)` 185.8°, 99.1, 86.7 00cccc `#00cccc`\nCIE-LAB 73.894, -33.49, -22.117 33.727, 46.538, 75.608 0.216, 0.299, 46.538 73.894, 40.134, 213.441 73.894, -54.862, -30.176 68.219, -31.135, -17.958 00000010, 11001000, 11011101\n\n# Color Schemes with #02c8dd\n\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #dd1702\n``#dd1702` `rgb(221,23,2)``\nComplementary Color\n• #02dd85\n``#02dd85` `rgb(2,221,133)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #025bdd\n``#025bdd` `rgb(2,91,221)``\nAnalogous Color\n• #dd8502\n``#dd8502` `rgb(221,133,2)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #dd025b\n``#dd025b` `rgb(221,2,91)``\nSplit Complementary Color\n• #c8dd02\n``#c8dd02` `rgb(200,221,2)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #dd02c8\n``#dd02c8` `rgb(221,2,200)``\n• #02dd17\n``#02dd17` `rgb(2,221,23)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #dd02c8\n``#dd02c8` `rgb(221,2,200)``\n• #dd1702\n``#dd1702` `rgb(221,23,2)``\n• #018391\n``#018391` `rgb(1,131,145)``\n• #029aaa\n``#029aaa` `rgb(2,154,170)``\n• #02b1c4\n``#02b1c4` `rgb(2,177,196)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #02dff6\n``#02dff6` `rgb(2,223,246)``\n• #15e7fd\n``#15e7fd` `rgb(21,231,253)``\n• #2ee9fd\n``#2ee9fd` `rgb(46,233,253)``\nMonochromatic Color\n\n# Alternatives to #02c8dd\n\nBelow, you can see some colors close to #02c8dd. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #02ddbb\n``#02ddbb` `rgb(2,221,187)``\n• #02ddce\n``#02ddce` `rgb(2,221,206)``\n``#02dadd` `rgb(2,218,221)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #02b6dd\n``#02b6dd` `rgb(2,182,221)``\n• #02a4dd\n``#02a4dd` `rgb(2,164,221)``\n• #0291dd\n``#0291dd` `rgb(2,145,221)``\nSimilar Colors\n\n# #02c8dd Preview\n\nThis text has a font color of #02c8dd.\n\n``<span style=\"color:#02c8dd;\">Text here</span>``\n#02c8dd background color\n\nThis paragraph has a background color of #02c8dd.\n\n``<p style=\"background-color:#02c8dd;\">Content here</p>``\n#02c8dd border color\n\nThis element has a border color of #02c8dd.\n\n``<div style=\"border:1px solid #02c8dd;\">Content here</div>``\nCSS codes\n``.text {color:#02c8dd;}``\n``.background {background-color:#02c8dd;}``\n``.border {border:1px solid #02c8dd;}``\n\n# Shades and Tints of #02c8dd\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000607 is the darkest color, while #f3feff is the lightest one.\n\n• #000607\n``#000607` `rgb(0,6,7)``\n• #00181b\n``#00181b` `rgb(0,24,27)``\n• #002a2e\n``#002a2e` `rgb(0,42,46)``\n• #013b41\n``#013b41` `rgb(1,59,65)``\n• #014d55\n``#014d55` `rgb(1,77,85)``\n• #015e68\n``#015e68` `rgb(1,94,104)``\n• #01707c\n``#01707c` `rgb(1,112,124)``\n• #01828f\n``#01828f` `rgb(1,130,143)``\n• #0193a3\n``#0193a3` `rgb(1,147,163)``\n• #02a5b6\n``#02a5b6` `rgb(2,165,182)``\n• #02b6ca\n``#02b6ca` `rgb(2,182,202)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\n• #02daf0\n``#02daf0` `rgb(2,218,240)``\n• #09e5fd\n``#09e5fd` `rgb(9,229,253)``\n• #1de7fd\n``#1de7fd` `rgb(29,231,253)``\n• #30e9fd\n``#30e9fd` `rgb(48,233,253)``\n• #44ecfd\n``#44ecfd` `rgb(68,236,253)``\n• #57eefd\n``#57eefd` `rgb(87,238,253)``\n• #6bf0fe\n``#6bf0fe` `rgb(107,240,254)``\n• #7ef2fe\n``#7ef2fe` `rgb(126,242,254)``\n• #92f4fe\n``#92f4fe` `rgb(146,244,254)``\n• #a5f6fe\n``#a5f6fe` `rgb(165,246,254)``\n• #b8f8fe\n``#b8f8fe` `rgb(184,248,254)``\n• #ccfaff\n``#ccfaff` `rgb(204,250,255)``\n• #dffcff\n``#dffcff` `rgb(223,252,255)``\n• #f3feff\n``#f3feff` `rgb(243,254,255)``\nTint Color Variation\n\n# Tones of #02c8dd\n\nA tone is produced by adding gray to any pure hue. In this case, #697576 is the less saturated color, while #02c8dd is the most saturated one.\n\n• #697576\n``#697576` `rgb(105,117,118)``\n• #607c7f\n``#607c7f` `rgb(96,124,127)``\n• #588387\n``#588387` `rgb(88,131,135)``\n• #4f8a90\n``#4f8a90` `rgb(79,138,144)``\n• #479198\n``#479198` `rgb(71,145,152)``\n• #3e97a1\n``#3e97a1` `rgb(62,151,161)``\n• #359eaa\n``#359eaa` `rgb(53,158,170)``\n• #2da5b2\n``#2da5b2` `rgb(45,165,178)``\n• #24acbb\n``#24acbb` `rgb(36,172,187)``\n• #1cb3c3\n``#1cb3c3` `rgb(28,179,195)``\n• #13bacc\n``#13bacc` `rgb(19,186,204)``\n• #0bc1d4\n``#0bc1d4` `rgb(11,193,212)``\n• #02c8dd\n``#02c8dd` `rgb(2,200,221)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #02c8dd is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51490706,"math_prob":0.71063334,"size":3695,"snap":"2019-51-2020-05","text_gpt3_token_len":1662,"char_repetition_ratio":0.13167164,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5391069,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839485,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T08:27:09Z\",\"WARC-Record-ID\":\"<urn:uuid:d9c47fba-7e6d-478a-9990-0c027a94a0c9>\",\"Content-Length\":\"36268\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e47e1639-6ca7-4123-9598-425bba243a46>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1f8400a-059e-40c2-9995-ed937cb76c1b>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/02c8dd\",\"WARC-Payload-Digest\":\"sha1:XYTXLQFVI7CTZLRI6BMHKC4YCCIZEAOR\",\"WARC-Block-Digest\":\"sha1:ONODA5HHH6AYQ3PCGBSYNQNB4XUOSMI7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540486979.4_warc_CC-MAIN-20191206073120-20191206101120-00497.warc.gz\"}"} |
https://support.minitab.com/en-us/minitab/20/help-and-how-to/statistical-modeling/doe/how-to/factorial/analyze-factorial-design/interpret-the-results/all-statistics-and-graphs/method-table/ | [
"# Method table for Analyze Factorial Design\n\nFind definitions and interpretation guidance for every statistic in the Method table.\n\n## Rounded λ\n\nBy default, Minitab rounds the optimal λ (lambda) to the nearest half because these values correspond to a more intuitive transformation. If you want to use the optimal value for the transformation, choose File > Options > Linear Models > Display of Results.\n\n### Interpretation\n\nThe following are common rounded values of lambda and how they transform the response variable.\nLambda Transformation\n-2 −Y-2 = −1 / Y2\n-1 −Y-1 = −1 / Y\n-0.5 −Y-0.5 = −1 / (square root of Y)\n0 log (Y)\n0.5 Y0.5 = square root of Y\n1 Y\n2 Y2\n\n## Estimated λ\n\nWhen you use a Box-Cox transformation, the estimated λ (lambda) is the optimal value to produce transformed response values that are normally distributed. By default, Minitab uses the rounded lambda value.\n\n### Interpretation\n\nLambda is the exponent that Minitab uses to transform the response data. For example, if lambda = -1, then all response values (Y) are transformed as follows: −Y-1 = −1/Y. If lambda equals 0, this represents the natural log of Y rather than Y0.\n\n## 95% CI for λ\n\nThe confidence intervals for λ (lambda) are ranges of values that are likely to contain the true value of λ for the entire population from which your sample was drawn.\n\nBecause samples are random, two samples from a population are unlikely to yield identical confidence intervals. However, if you take many random samples, a certain percentage of the resulting confidence intervals contain the unknown population parameter. The percentage of these confidence intervals that contain the parameter is the confidence level of the interval.\n\n### Interpretation\n\nUse the confidence interval to assess the estimate of lambda for your sample.\n\nFor example, with a 95% confidence level, you can be 95% confident that the confidence interval contains the value of lambda for the population. The confidence interval helps you assess the practical significance of your results. Use your specialized knowledge to determine whether the confidence interval includes values that have practical significance for your situation. If the interval is too wide to be useful, consider increasing your sample size.\n\nBy using this site you agree to the use of cookies for analytics and personalized content. Read our policy"
] | [
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https://www.fhwa.dot.gov/publications/research/infrastructure/pavements/ltpp/reports/02088/02.cfm | [
"U.S. Department of Transportation\n1200 New Jersey Avenue, SE\nWashington, DC 20590\n202-366-4000\n\nFederal Highway Administration Research and Technology\nCoordinating, Developing, and Delivering Highway Transportation Innovations",
null,
"This report is an archived publication and may contain dated technical, contact, and link information\n Federal Highway Administration > Publications > Research Publications > LTPP Publications > 02088 > Evaluation of Joint and Crack Load Transfer Final Report\n Publication Number: FHWA-RD-02-088 Date: May 2003\n\n# Evaluation of Joint and Crack Load Transfer Final Report\n\nWhen a traffic load is applied near a joint in a PCC pavement, both loaded and unloaded slabs deflect because a portion of the load applied to the loaded slab is transferred to the unloaded slab. As a result, deflections and stresses in the loaded slab may be significantly less than if, instead of a joint with another slab, there was a free edge. The magnitude of reduction in stresses and deflections by a joint compared to a free edge depends on the joint's LTE.\n\nTraditionally, LTE at the joint is determined based on the ratio of the maximum deflection at the joint of the loaded slab and the deflection of the unloaded slab measured right across the joint from the maximum deflection. Two equations for the deflection LTE are used most often:",
null,
"(1)\n\nand",
null,
"(2)\n\nwhere:\n\n- dl = the maximum deflection at the joint of the loaded slab.\n\n- du = the corresponding deflection at the joint of the unloaded slab.\n\n- LTE and LTE* = load transfer efficiency indexes.\n\nIf a joint exhibits a poor ability to transfer load, then the deflection of the unloaded slab is much less than the deflection at the joint of the loaded slab and both LTE indexes have values close to 0. If joint's load transfer ability is very good, then the deflections at the both sides of the joint are equal and both indexes have values close to 100 percent. Moreover, these two indices are related by the following equation:",
null,
"(3)\n\nTherefore, these indexes are equivalent, and if one of them is known, the other can be determined. In this study, we will define the deflection LTE using the index from equation 1 because it is much more widely used.\n\nThe definitions of the LTE described above are based on deflections. A joint LTE based on stress can be defined as follows:",
null,
"(4)\n\nwhere:\n\n-",
null,
"= the maximum stress at the joint of the loaded slab.\n\n-",
null,
"= the corresponding stress at the joint of the unloaded slab.\n\n- LTE",
null,
"= load transfer efficiency in stress.\n\nThe stress-based LTE indicates the degree of stress reduction at the joint of the loaded slab caused by the presence of the unloaded slab. Studies have indicated that there is no one-to-one relationship between stress-based and deflection-based LTE indexes. Because it is difficult to measure stresses in a concrete slab and stress-based LTE is much more affected by geometry of the applied load than deflection LTE, the deflection-based LTE is commonly used to measure load transfer in concrete pavements. If deflection LTE is known, stress reduction due to load transfer for any given load configuration may be calculated using a finite element model.\n\nJoint LTE depends on many factors, including the following:\n\n• Joint load transfer mechanism and stiffness.\n• Temperature.\n\nLoad transfer between the slabs occurs through aggregate particles of the fractured surface below the saw cut at a joint, through steel dowels (if they exist), and through the base and subgrade. LTE may vary throughout the day and year because of variation in PCC temperature. When temperature decreases, a joint opens wider, which decreases contact between two slabs and also may decrease LTE, especially if no dowels exist. Also, PCC slab curling may change the contact between the slab and the underlying layer and affect measured load-induced deflections.\n\n## Mechanistic Modeling of LTE\n\nMechanistic modeling of the load transfer mechanism is a complex problem. Frieberg (1940) was one of the first researchers who attempted to tackle this problem. The introduction of a finite element method for analysis of JCP (Tabatabie and Barenberg ) gave a significant boost to understanding load transfer mechanisms. However, although many comprehensive finite element models have been developed (Scarpas et al. , Guo et al. , Parson et al. , Shoukry and William , Davids et al. , Khazanovich et al. ), work on the development of a comprehensive, practical, and reliable model for joints of rigid pavements is far from complete.\n\nIoannides and Korovesis (1990, 1992) made an important step forward in PCC joint analysis. They identified the following nondimensional parameters governing joint behavior:\n\nNondoweled pavements:",
null,
"(5)\n\nDoweled pavements:",
null,
"(6)\n\nwhere:\n\n- AGG* and D* are nondimensional stiffnesses of nondoweled and doweled joints.\n\n- AGG = a shear stiffness of a unit length of an aggregate interlock.\n\n- D = a shear stiffness of a single dowel (including dowel-PCC interaction).\n\n-",
null,
"= a PCC slab radius of relative stiffness.\n\n- s = dowel spacing.\n\nUsing the finite element program ILLI-SLAB, Ioannides and Korovesis also identified a unique relationship between these parameters and LTE (see figure 1).",
null,
"Figure 1. LTE versus nondimensional joint stiffness.\n\nThe following assumptions were made in derivation of these relationships:\n\n• Prior to loading, PCC slabs are flat and in full contact with the foundation.\n• A PCC joint propagates through the base layer (if present). No load transfer occurs through the base layer.\n• The subgrade is modeled as a Winkler foundation, which assumes no load transfer through the subgrade.\n• The PCC joint has uniform LTE across its width. The entire load transfer in nondoweled joints occurs through aggregate interlock, whereas the entire load transfer in doweled joints occurs through dowels (no aggregate interlock).\n\nThe relationships developed by Ioannides and Korovesis form a basis for back calculation of joint aggregate interlock stiffness of nondoweled joints or dowel shear stiffness of doweled joints if their LTEs are known. In both cases, however, the backcalculated stiffness overestimates real aggregate interlock stiffness or dowel stiffness because the entire joint stiffness is attributed to a single (although perhaps prevailing) component. The cause of this limitation is the inability of the ILLI-SLAB model to distinguish between load transfer mechanisms. This limitation does not cause a significant problem because the backcalculated joint stiffness provides sufficient information for accurate joint modeling in the forward analysis. Moreover, if the addition information is available, individual components can be determined more accurately.\n\nFor example, if AGGtot is the backcalculated stiffness of a doweled joint and D is the known dowel stiffness, then the \"true\" aggregate interlock factor, AGG0, for this joint can be determined from the following relationship:",
null,
"(7)\n\nwhere:\n\n- s = dowel spacing.\n\nIn this study, Ioannides and Korovesis's relationship was further investigated and an efficient backcalculation procedure for joint stiffness determination was developed.\n\n## Joint Stiffness Backcalculation\n\nThe relationship identified by Ioannides and Korovesis was further elaborated by Crovetti (1994), Zollinger et al. (1999), and Ioannides et al. (1996). Crovetti and Zollinger developed regression models for that relationship, whereas Ioannides and Hammons developed a neural network prediction model.\n\nCrovetti proposed the following relationship between nondimensional joint stiffness and LTE:",
null,
"(8)\n\nwhere:\n\n- AGGtot = the total stiffness.\n\n-",
null,
"= the PCC slab radius of relative stiffness.\n\n- k = a coefficient of subgrade reaction (k-value).\n\nZollinger's model for this relationship has the following form:",
null,
"(9)\n\nwhere\n\n- AGGtot = the total joint stiffness.\n\n-",
null,
"= the PCC slab radius of relative stiffness.\n\n- k = the subgrade k-value.\n\nThe neural network model developed by Ioannides and Hammons cannot be expressed as a simple equation, but rather takes the form of a computer program that relates LTE and joint stiffness, radius of relative stiffness, subgrade k-value, and load geometry. Although this model is more accurate than the regression models, it is not publicly available and was not evaluated in this study.\n\nTo evaluate the Crovetti and Zollinger models, a factorial of 375 runs was performed to simulate FWD testing at the PCC joint. ISLAB2000 (Khazanovich et al. ) is a completely rewritten version of ILLI-SLAB that retains all the positive features of ILLI-SLAB but is more convenient to use and is free from many unnecessary limitations (including limitation on the number of nodes in a finite element model).\n\nA finite element model developed in this study has four slabs in the longitudinal direction and three slabs in the transverse direction. The system was assumed to be symmetrical with respect to the longitudinal axis, so only half of the system was modeled. Because the focus of this analysis was on the deflections near the loaded joint along the centerline, a much finer finite element mesh was used along the centerline and loaded joint, as shown in figure 2. The LTE of longitudinal joints was selected to be equal to 70 percent. Also, due to symmetry, only half of an FWD load was applied. The slab radius of relative stiffness was varied from 508 to\n\n2,032 mm (80 inches), and the nondimensional transverse joint stiffness was varied from 0.1 to 278, which resulted in LTE from 8 to 99 percent.",
null,
"Figure 3 presents comparisons of the LTEs calculated from the ISLAB2000 results and LTEs from Crovetti's and Zollinger's equations. Although Crovetti's equation is simpler than Zollinger's, it better corresponds with ISLAB2000 LTEs. Because the Crovetti model was adopted for use in the 2002 design procedure (National Cooperative Highway Research Program Project 1-37A) and it compares well with the finite element analysis, it was selected for use in this study.",
null,
"Figure 3. Comparison of LTE calculated from ISLAB2000 results with predictions using Crovetti's and Zollinger's models.\n\n## \"True\" versus \"Measured\" LTE\n\nAs stated above, the joint LTE is a ratio of the maximum deflection at the joint of the unloaded slab to the deflection of the loaded slab measured directly across the joint from the maximum deflection. However, measurement of such deflections in the field may be quite cumbersome. In the LTPP program, joint deflection testing is conducted by placing the load plate tangential to the edge of the joint. The loaded slab joint deflections are measured under the center of the load plate (152 mm [6 inches] away from the joint). The deflection of the unloaded slab is also measured at some distance (152 mm [6 inches]) from the joints. This raises an issue about the possibility of error as a result of differences in deflections directly at the joint and measured deflections 152 mm (6 inches) away from the joint because of slab bending. Some experts advocate the need to use a correction (\"bending\") factor to adjust the measured deflections.\n\nTo investigate this problem, the results of the 375 ISLAB2000 runs were analyzed. The LTEs calculated from the deflections located exactly at the joints (\"true\" LTEs) were compared to the ratios of the deflections located 152 mm (6 inches) away from the joint (\"measured\" LTEs similar to the FWD procedure). Figure 4 presents comparisons of true and measured LTEs.",
null,
"Figure 4. Comparison of true and measured LTEs.\n\nIn most cases, measured LTE is close to true LTE. The exceptions are the cases with very low PCC slab stiffness (radius of relative stiffness is less than 750 mm (30 inches). For those cases, measured LTE overestimates true LTE. To address this discrepancy, the research team attempted to develop bending correction factors.\n\nPCC slab bending depends primarily on the radius of relative stiffness. Since stiffer slabs require less bending correction than slabs with low radii of relative stiffness, the following functional form was proposed for the correction factor:",
null,
"(10)\n\nwhere:\n\n- B = the bending correction factor.\n\n- LTEtrue = the true LTE.\n\n- LTEmes = the measured LTE.\n\n-",
null,
"= the radius of relative stiffness.\n\n- a1 and a2 = model parameters.\n\nA regression analysis was performed to determine coefficients a1 and a2 and the following expression for the bending factor was obtained:",
null,
"(11)\n\nR2 = 0.9998\n\nSEE = 0.000495\n\nwhere:\n\n-",
null,
"= a radius of relative stiffness in mm.\n\nAlthough the correction factor has excellent statistics, its practical applicability is quite limited because the radius of relative stiffness may be unknown. To avoid this limitation, a correction factor based on sensor deflections located on the leave slab was used to correct measured LTE. The AREA parameter was used for this purpose. This parameter combines the effect of several measured deflections in the basin and is defined as follows:",
null,
"(12)\n\nwhere:\n\n- Wi = measured deflections (i = 0, n).\n\n- n = number of FWD sensors minus 1.\n\n- ri = distances between the center of the load plate and sensors in mm.\n\nThe AREA parameter has been used extensively to analyze concrete pavement deflection basins since 1980. Ioannides et al. (1989) identified the unique relationship between AREA and the radius of relative stiffness. The AREA parameter is not truly an area, but rather has dimensions of length, since it is normalized with respect to one of the measured deflections in order to remove the effects of load magnitude. For a given number and configuration of deflection sensors, AREA may be computed using the trapezoidal rule. It was found that the correction factor depends on the AREA parameter and the magnitude of the measured LTE itself. The following relationship between the true and measured LTE was proposed:",
null,
"(13)\n\nwhere:\n\n- K1, K2, K3, and K4 = regression coefficients, depending on the sensor configuration used in AREA calculation.\n\nUsing the results from the 375 ISLAB2000 runs, the regression analysis for determining K1, K2, K3, and K4 was performed for the following sensor configurations:\n\n• C6 sensor configuration (at 0, 305, 457, 610, 914, and 1524 mm).\n• C5 sensor configuration (at 0, 305, 457, 610, and 914 mm).\n• O5 sensor configuration (at 305, 457, 610, 914, and 1524 mm).\n• O4 sensor configuration (at 305, 457, 610, and 914 mm).\n\nTable 1 presents the determined coefficients along with basic statistics.\n\nTable 1. Regression coefficients for bending correction factors.\n\nTest\n\nSensor Configuration\n\nK1\n\nK2\n\nK3\n\nK4\n\nR2\n\nSEE\n\nApproach\n\nO5\n\n0.929710\n\n23.61\n\n-45.73\n\n1171.28\n\n0.99729\n\n0.108298\n\nApproach\n\nO4\n\n0.790073\n\n116.74\n\n-166.34\n\n856.58\n\n0.99724\n\n0.231601\n\nLeave\n\nC6\n\n0.924255\n\n92.96\n\n-508.52\n\n5302.13\n\n0.99727\n\n0.163864\n\nLeave\n\nC5\n\n0.806827\n\n155.35\n\n-491.51\n\n3472.81\n\n0.99715\n\n0.366376\n\nLeave\n\nO5\n\n0.923371\n\n72.03\n\n-182.53\n\n2143.95\n\n0.99729\n\n0.114590\n\nLeave\n\nO4\n\n0.764173\n\n129.91\n\n-209.29\n\n1105.75\n\n0.99720\n\n0.296859\n\nTo investigate the applicability of the bending factor developed in the study, bending factors were calculated for more than 600,000 FWD basins from the LTPP database. Figures 5 and 6 present the distribution of these factors for the FWD basins from approach slab test calculated using the O5 sensor configuration and for the basins from the leave slab test calculated using the C6 sensor configuration. In most cases, the correction was less than 4 percent.\n\nDuring the bending factor testing using data from the LTPP database, it was found that significant discrepancies existed in measured LTEs for the same joints measured with a load plate placed on the approach and the leave side of the joint. The correction factors presented above did not significantly reduce these discrepancies. An additional investigation was conducted in an attempt to resolve this issue.\n\nThe finite element model used in the development of the correction factors did not account for differences in pavement responses as a result of the location of the loaded plate. Also, the PCC slabs were assumed to be in full contact with the subgrade. In reality, the application of a large number of a heavy axle loads moving in one direction may cause the formation of permanent voids under the leave side of the joint. An assumption was made that those voids are responsible for the discrepancy between leave and approach test results. In addition, PCC slab curling causes slabs to separate, creating temporary voids. This effect also was analyzed in this study.",
null,
"Figure 5. Distribution of bending correction factors for approach slab testing based on O5 sensor configuration.\n\nA factorial of 504 finite element runs with different temperature gradients through slab thickness, joint stiffness, and subgrade stiffness was performed. In all cases, a 1.52-m (5-ft)-long permanent void occupying the entire leave slab width was modeled. In half of the cases, the load was placed at the approach side of the joint, and in the other half it was placed at the leave side. The load transfer values from the deflections induced by the loads placed at the approach side of the joint, LTE1, were compared with the corresponding load transfer values calculated for the same systems but the load placed on the leave side on the joint, LTE2. Figure 7 presents a comparison of those LTEs. As shown in the figure, the presence of a void significantly skewed computed LTE values. The figure shows that LTEs computed from the leave slab test should be lower than those computed from the approach slab test. Moreover, this effect may be much more pronounced than the effect of slab bending.",
null,
"Figure 6. Distribution of bending correction factors for leave slab testing based on C6 sensor configuration.",
null,
"Figure 7. Comparison of LTEs predicted using ISLAB2000 for the approach slab test (LTE1) and the leave slab test (LTE2) for pavements with voids.\n\nThis analysis agrees with the analysis of LTPP data that discrepancies may exist between the LTEs computed from leave and approach tests; however, in the vast majority of cases analyzed, an opposite trend was observed. As will be discussed in chapter 3, LTEs computed from the leave slab test usually were higher than LTEs computed from the approach slab test. At this stage, no mechanistic explanation of this phenomenon was found. More research is needed to address this issue. Meanwhile, considering that the correction factors presented above did not change the calculated LTE values significantly, and that void/curling correction should be incorporated into this factor, a decision was made not to use the bending factor in this study."
] | [
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https://ras.football/2020/01/11/mewelde-moore-ras/ | [
"# Mewelde Moore RAS\n\n### Mewelde Moore RAS\n\nMewelde Moore was drafted by Vikings with pick 119 in round 4 in the 2004 NFL Draft out of Tulane.\n\nHe recorded a Relative Athletic Score of 7.18, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.\n\nHe had a recorded height of 5105 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 5 feet, 10 and 5/8 of an inch or 70.625 inches, or 179.3875 centimeters. This correlates to a 4.59 score out of 10.0.\n\nHe recorded a weight of 209 in pounds, which is approximately 95 kilograms. This correlates to a 5.02 score out of 10.0.\n\nBased on his weight, he has a projected 40 yard dash time of 4.59. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.\n\nAt the Combine, he recorded a 40 yard dash of 4.67 seconds. This was a difference of 0.08 seconds from his projected time. This forty time correlates to a 3.87 score out of 10.0.\n\nUsing Bill Barnwell’s calculation, this Combine 40 time gave him a Speed Score of 87.88.\n\nAt his pro day, he recorded a 40 yard dash of 4.62 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.\n\nUsing Bill Barnwell’s calculation, his Combine 40 time gave him a Speed Score of 1.94.\n\nThe time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at the Combine was 1.99 seconds. Over 20 yards, we can calculate his speed in yards per second to 10.05. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 30.15. Breaking it down further, we can calculate his speed in inches per second to 361.81. Knowing the feet per second of 30.15, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 20.6 in the last 20 yards of his run.\n\nAt the Combine, he recorded a 20 yard split of 2.68 seconds. This correlates to a 6.13 score out of 10.0.\n\nWe can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at the Combine was 1.07 seconds. Over 10 yards, we can calculate his speed in yards per second to 9.35. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 28.04. Breaking it down further, we can calculate his speed in inches per second to 336.45. Knowing the feet per second of 28.04, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 19.1 in the second ten yards of his run.\n\nAt the Combine, he recorded a 10 yard split of 1.61 seconds. This correlates to a 6.69 score out of 10.0.\n\nThe time he traveled the first ten yards at the Combine was 1.61 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 19.0. Breaking it down further, we can calculate his speed in inches per second to 224.0. Knowing the feet per second of 19.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 13.0 in the first ten yards of his run.\n\nThis player did not have a recorded bench press for the Combine in the RAS database.\n\nAt the Combine, he recorded a vertical jump of 35.0 inches. This correlates to a 7.13 score out of 10.0.\n\nAt the Combine, he recorded a broad jump of 910, which is recorded as FII or FFII . where F is feet and I is inches. This correlates to a 6.78 score out of 10.0.\n\nAt the Combine, he recorded a 5-10-5 or 20 yard short shuttle of 4.12 seconds. This correlates to a 8.1 score out of 10.0.\n\nAt the Combine, he recorded a 3 cone L drill of 7.05 seconds. This correlates to a 7.55 score out of 10.0.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
null
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https://javascriptinfo.com/view/46018/finding-all-indexes-of-a-specified-character-within-a-string | [
"# Finding all indexes of a specified character within a string\n\nFor example if I had `\"scissors\"` in variable and wanted to know the position of all occurrences of the letter `\"s\"`, it should print out `1, 4, 5, 8`\n\nHow can I do this in JavaScript in most efficient way? I don't think looping through the whole is terribly efficient\n\n## Answers:",
null,
"A simple loop works well:\n\n``````var str = \"scissors\";\nvar indices = [];\nfor(var i=0; i<str.length;i++) {\nif (str[i] === \"s\") indices.push(i);\n}\n``````\n\nNow, you indicate that you want 1,4,5,8. This will give you 0, 3, 4, 7 since indexes are zero-based. So you could add one:\n\n``````if (str[i] === \"s\") indices.push(i+1);\n``````\n\nand now it will give you your expected result.\n\nA fiddle can be see here.\n\nI don't think looping through the whole is terribly efficient\n\nAs far as performance goes, I don't think this is something that you need to be gravely worried about until you start hitting problems.\n\nHere is a jsPerf test comparing various answers. In Safari 5.1, the IndexOf performs the best. In Chrome 19, the for loop is the fastest.",
null,
"",
null,
"Using the native `String.prototype.indexOf` method to most efficiently find each offset.\n\n``````function locations(substring,string){\nvar a=[],i=-1;\nwhile((i=string.indexOf(substring,i+1)) >= 0) a.push(i);\nreturn a;\n}\n\nconsole.log(locations(\"s\",\"scissors\"));\n//-> [0, 3, 4, 7]\n``````\n\nThis is a micro-optimization, however. For a simple and terse loop that will be fast enough:\n\n``````// Produces the indices in reverse order; throw on a .reverse() if you want\nfor (var a=[],i=str.length;i--;) if (str[i]==\"s\") a.push(i);\n``````\n\nIn fact, a native loop is faster on chrome that using `indexOf`!",
null,
"",
null,
"More functional fun, and also more general: This finds the starting indexes of a substring of any length in a string\n\n``````const length = (x) => x.length\nconst sum = (a, b) => a+b\n\nconst indexesOf = (substr) => ({\nin: (str) => (\nstr\n.split(substr)\n.slice(0, -1)\n.map(length)\n.map((_, i, lengths) => (\nlengths\n.slice(0, i+1)\n.reduce(sum, i*substr.length)\n))\n)\n});\n\nconsole.log(indexesOf('s').in('scissors')); // [0,3,4,7]\n\nconsole.log(indexesOf('and').in('a and b and c')); // [2,8]``````",
null,
"``````indices = (c, s) => s\n.split('')\n.reduce((a, e, i) => e === c ? a.concat(i) : a, []);\n\nindices('?', 'a?g??'); // [1, 3, 4]\n``````",
null,
"I loved the question and thought to write my answer by using the `reduce()` method defined on arrays.\n\n``````function getIndices(text, delimiter='.') {\nlet indices = [];\nlet combined;\n\ntext.split(delimiter)\n.slice(0, -1)\n.reduce((a, b) => {\nif(a == '') {\ncombined = a + b;\n} else {\ncombined = a + delimiter + b;\n}\n\nindices.push(combined.length);\nreturn combined; // Uncommenting this will lead to syntactical errors\n}, '');\n\nreturn indices;\n}\n\nlet indices = getIndices(`Ab+Cd+Pk+Djb+Nice+One`, '+');\nlet indices2 = getIndices(`Program.can.be.done.in.2.ways`); // Here default delimiter will be taken as `.`\n\nconsole.log(indices); // [ 2, 5, 8, 12, 17 ]\nconsole.log(indices2); // [ 7, 11, 14, 19, 22, 24 ]\n\n// To get output as expected (comma separated)\nconsole.log(`\\${indices}`); // 2,5,8,12,17\nconsole.log(`\\${indices2}`); // 7,11,14,19,22,24\n``````",
null,
"",
null,
"When i benchmarked everything it seemed like regular expressions performed the best, so i came up with this\n\n``````function indexesOf(string, regex) {\nvar match,\nindexes = {};\n\nregex = new RegExp(regex);\n\nwhile (match = regex.exec(string)) {\nif (!indexes[match]) indexes[match] = [];\nindexes[match].push(match.index);\n}\n\nreturn indexes;\n}\n``````\n\nyou can do this\n\n``````indexesOf('ssssss', /s/g);\n``````\n\nwhich would return\n\n``````{s: [0,1,2,3,4,5]}\n``````\n\ni needed a very fast way to match multiple characters against large amounts of text so for example you could do this\n\n``````indexesOf('dddddssssss', /s|d/g);\n``````\n\nand you would get this\n\n``````{d:[0,1,2,3,4], s:[5,6,7,8,9,10]}\n``````\n\nthis way you can get all the indexes of your matches in one go",
null,
"``````function charPos(str, char) {\nreturn str\n.split(\"\")\n.map(function (c, i) { if (c == char) return i; })\n.filter(function (v) { return v >= 0; });\n}\n\ncharPos(\"scissors\", \"s\"); // [0, 3, 4, 7]\n``````\n\nNote that JavaScript counts from 0. Add +1 to `i`, if you must.",
null,
"You could probably use the match() function of javascript as well. You can create a regular expression and then pass it as a parameter to the match().\n\n``````stringName.match(/s/g);\n``````\n\nThis should return you an array of all the occurrence of the the letter 's'.",
null,
"Just for further solution, here is my solution: you can find character's indexes which exist in a string:\n\n``````findIndex(str, char) {\nconst strLength = str.length;\nconst indexes = [];\nlet newStr = str;\n\nwhile (newStr && newStr.indexOf(char) > -1) {\nindexes.push(newStr.indexOf(char) + strLength- newStr.length);\nnewStr = newStr.substring(newStr.indexOf(char) + 1);\n}\n\nreturn indexes;\n}\n\nfindIndex('scissors', 's'); // [0, 3, 4, 7]\nfindIndex('Find \"s\" in this sentence', 's'); // [6, 15, 17]\n\n``````",
null,
"Here is a short solution using a function expression (with ES6 arrow functions). The function accepts a string and the character to find as parameters. It splits the string into an array of characters and uses a `reduce` function to accumulate and return the matching indices as an array.\n\n``````const findIndices = (str, char) =>\nstr.split('').reduce((indices, letter, index) => {\nletter === char && indices.push(index);\nreturn indices;\n}, [])\n``````\n\nTesting:\n\n``````findIndices(\"Hello There!\", \"e\");\n// ? [1, 8, 10]\n\nfindIndices(\"Looking for new letters!\", \"o\");\n// ? [1, 2, 9]\n\n``````\n\nHere is a compact (one-line) version:\n\n``````const findIndices = (str, char) => str.split('').reduce( (indices, letter, index) => { letter === char && indices.push(index); return indices }, [] );\n``````",
null,
"using while loop\n\n``````let indices = [];\nlet array = \"scissors\".split('');\nlet element = 's';\n\nlet idx = array.indexOf(element);\n\nwhile (idx !== -1) {\nindices.push(idx+1);\nidx = array.indexOf(element, idx + 1);\n}\nconsole.log(indices);``````\n\n## Top Questions\n\n©2020 All rights reserved."
] | [
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