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# The Advent Calendar of Physics: Introducing Angular Momentum
Moving along through our countdown to Newton’s birthday, we come to the next important physical quantity, angular momentum. For some obscure reason, this gets the symbol L, and the angular momentum for a single particle about some point A is given by:
This is probably the most deceptive equation we’ll see this season. Yesterday’s definition of work clearly showed its vector calculus roots, but to the untrained eye, this just looks like a simple multiplication: You take the momentum (p) and multiply by the distance (r) from point A, and you’re all set.
To those with a little mathematical training, though, that × symbol is a terrifying sight. In math and physics circles, ordinary multiplication is so basic that it doesn’t require a specific symbol. The × sign that ordinary people use to indicate multiplication is reserved for the most daunting of vector operations, the cross product. The linear momentum of the particle and the radius from point A are both vectors, having a direction as well as a magnitude. The cross product takes two vectors and produces a third vector– thus, the little arrows over all three quantities.
What’s more, the vector that results from the cross product is at right angles to the two vectors that were multiplied together to get it. So, if the momentum vector points north and the radius vector points west, the angular momentum vector points straight up. This is consistently the single most confusing thing in an introductory mechanics course, which makes it kind of a travesty that we only get to it in the last week or so of the term.
So, why do we bother with all this scary and confusing math?
Angular momentum is important because there is clearly something special about rotational motion. As anybody who has every played with a top or a gyroscope knows, spinning objects behave in slightly strange ways, to such a degree that every so often a crazy person will convince themselves that if they combine enough spinning objects in just the right way, they can defy the force of gravity. They can’t. Angular momentum is weird, but it’s not magic.
Angular momentum is the property we use to characterize the motion of things that move in closed loops, and it’s one of the most significant properties in all of physics. On the largest scales, angular momentum is a huge factor in determining the structure and behavior of planetary systems, galaxies, and black holes. On the smallest scales, angular momentum is what determines the structure and chemical behavior of atoms. Angular momentum is ultimately responsible for everything we see around us.
“But wait,” you object conveniently for my narrative, “You said that p is the linear momentum of a particle. How does something moving in a straight line have anything to do with rotation?”
It’s true, the connection is not obvious– objects moving in straight lines seem qualitatively different than objects that are spinning in place. But in the right circumstances, you can convert linear motion into rotational motion. If you throw a ball at a door, for example, and hit it in the right place, you can make that door swing open or closed. That interaction converts the linear motion of the ball, which flies through space in a straight line, into rotational motion of the door, which pivots about its hinges. This implies that even though it was moving in a linear fashion, the moving ball had some rotational character. That’s what you calculate with today’s equation.
“Yes, but that still doesn’t look anything like an object spinning about some axis,” you say. “The ball is moving through space, but a spinning wheel just sits there. They’re completely different.”
On a macroscopic level, they certainly look different, and it might not seem like this equation does you any good in describing a spinning wheel. But when you think about it, a spinning solid object is itself made up of mind-boggling numbers of individual atoms. At any given instant, each of those atoms is moving in a straight line through space, and thus has some (tiny) linear momentum. You can use that tiny linear momentum and the distance from the axis of rotation to calculate the individual atom’s angular momentum, and then repeat the calculation for each of the other atoms. Add up all 1026 of those individual atomic angular momenta, and that’s the angular momentum of the whole macroscopic object.
So, you can see, this is the most deceptive of all the equations we’ve looked at. Not only is it hiding scary vector business behind a pleasant facade, but it’s vastly more powerful than it lets on. So, take a moment to celebrate the existence of angular momentum, say, by doing a little spinning dance, and we’ll be back tomorrow with the next equation oft he season.
1. #1 Matt McIrvin
December 8, 2011
One of the things that makes it hard to think about is that, unlike linear momentum, which is just computed relative to the velocity of some reference frame, angular momentum is also relative to the point you choose as the spatial origin of your coordinate system. If the linear motion of a particle is off-center such that the extrapolated straight line doesn’t pass through the origin, that’s what makes it have nonzero angular momentum. If the line does pass through the origin, then there is none (relative to that origin).
And this, I think, is the key to understanding the relationship between linear and rotational motion. In the swinging-door example, the door should be hinged on an axis passing through the origin.
2. #2 Paul Orwin
December 8, 2011
An excuse to tell my two favorite dork jokes.
1)what do you get when you cross a mosquito and a mountain climber?
You can’t, one’s a vector and the other a scalar.
2) alright, what do you get when you cross an elephant and a tiger?
Elephant tiger sin theta
Both work better if you are with a bunch of drunk science/engineering undergrads…
December 8, 2011
And this, I think, is the key to understanding the relationship between linear and rotational motion. In the swinging-door example, the door should be hinged on an axis passing through the origin.
Yeah, I was planning to talk a little more about that when I get to torque (projected to be Saturday’s entry). The swinging door thing is one of my favorite examples for torque, because most people can sympathize with the experience of walking toward one of those all-glass banks of doors and guessing wrong about which end has the hinge…
4. #4 Wes Philp
December 8, 2011
Perhaps it was stated earlier in this series, but I’m wondering what date is assumed for Newton’s birthday, 12/25 or 1/4. Per Wikipedia, “During Newton’s lifetime, two calendars were in use in Europe: the Julian or ‘Old Style’ in Britain and parts of northern Europe (Protestant) and eastern Europe, and the Gregorian or ‘New Style’, in use in Roman Catholic Europe and elsewhere. At Newton’s birth, Gregorian dates were ten days ahead of Julian dates: thus Newton was born on Christmas Day, 25 December 1642 by the Julian calendar, but on 4 January 1643 by the Gregorian.”
December 8, 2011
Perhaps it was stated earlier in this series, but I’m wondering what date is assumed for Newton’s birthday, 12/25 or 1/4.
For the sake of the advent calendar joke, I’m taking his birthday to be Christmas Day. I’m not sure I’ll have the stamina to make it through 24 equation posts, let alone 34.
6. #6 Novak
December 8, 2011
Oh, surely there are scarier vector operators than the cross product.
December 8, 2011
Oh, surely there are scarier vector operators than the cross product.
We’ll probably get to the Einstein equation from general relativity, which brings in some scary tensor stuff. But the cross product is the most innocuous-looking scary vector operation that we’ve seen to this point. I haven’t decided yet whether to do Maxwell’s Equations in differential or integral form, so I’m not sure whether the curl will turn up or not.
8. #8 Mary
December 8, 2011
Curl is only scary because it’s based on the cross product, wouldn’t you say?
It took a long time for it to sink in, for me, that dot and cross products had to do with the “parallel part” and the “perpendicular part.” I wish it had been taught with more drawing of triangles, less writing of letters.
I think I was in grad school, or anyway studying for the GRE, before I finally had any intuitive understanding of Div and Curl. (Meaning I could look at a picture of a vector field and tell you if its Div and Curl were zero or non zero, postivie or negative.) Again, I wish we’d practiced that skill earlier on, rather than concentrating on calculating them by manipulating symbols.
9. #9 Roger
December 8, 2011
You might also mention that just like conservation of linear momentum was connected to the translational symmetry of the universe (i.e. the laws of physics are independent of location) the angular momentum is connected to the rotational symmetry. The laws of the universe are the same no matter which direction your facing.
10. #10 Jim
December 8, 2011
I just wanted to take the time to thank you for this very cool series – The Advent Calendar of Physics. The articles are a great read and easy to grasp for a laymen such as myself. Your blog and others like it are the fanatastic little gems that make wading through the morass of vacuous internet drivel worth it.
11. #11 Sue d’Eau-Nîmes
December 8, 2011
So, if the momentum vector points north and the radius vector points west, the angular momentum vector points straight up.
The radius vector comes before the momentum vector in the cross product, so the angular momentum vector points straight down, not up.
12. #12 Neil Bates
December 8, 2011
Angular momentum can be made a vector in three-space (3 + 1, three spatial and one of time) because of the unique perpendicular vector to what AM really is: a planar 2-form, the plane in which the rotation occurs. For example, suppose we had four space dimensions and one of time, then the rotation in the x,y plane has no unique perpendicular. Instead, (not using exotic extra refinements like 4-vectors) define
Lik = Σ ri pk – rk pi
i,k = 1,2,3, … n
Compare the exotic implications for rotation in n-space: a hyperplanet in spatial x,y,z,w space can be rotating at one rate within x,y and another rate within z,w: makes for strange days!
“Fine minds make fine distinctions.”
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# Question
Pls help THANKS
Method 1 :
5 + 2 = 7
273/7 = 39
39 x 2 = 78
Method 2 :
Total : Ravi : Father
7 : 5 : 2
273 : 195 : 78
Ans : \$78.
0 Replies 0 Likes
Every \$5 Ravi saved, it will become: \$5+\$2 = \$7
Total Saving: \$273
\$273 / \$7 = 39
So Ravi’s father gave him: 39*\$2 = \$78
0 Replies 0 Likes
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# Proof by induction that $3^n - 1$ is an even number
How to demonstrate that $3^n - 1$ is an even number using the principle of induction? I tried taking that $3^k - 1$ is an even number and as a thesis I must demonstrate that $3^{k+1} - 1$ is an even number, but I can't make a logical argument.
So I think it's wrong the assumption... Thanks
• Remember that $\text{even}\times\text{odd}=\text{even}$ and that $\text{even}+\text{even}=\text{even}$. – Akiva Weinberger Jun 9 '15 at 21:36
• More generally, $a-b\mid a^n-b^n$, because $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1})$$ – user26486 Jun 9 '15 at 21:53
Hint: try substituting back in $3^k = 2x+1$
• You should use another variable in the RHS. – Yves Daoust Jun 9 '15 at 21:47
• Thank you, that was poor variable choice – nosyarg Jun 9 '15 at 21:48
Hint: Start by showing that $3^n$ is odd. What's an odd minus an odd?
• ...except that is not induction. – Joffan Jun 9 '15 at 20:58
• How so? $3^0-1 = 0$ is even. For $n \geq 1, 3^{n-1}-1$ is even. Then $3^{n-1}$ is odd. Then $3^n$ is odd. Then $3^n-1$ is even. – Brian Tung Jun 9 '15 at 21:08
• @BrianTung The problem is so simple that any random observation could be worked into induction; nevertheless the answer above does not do that. – Joffan Jun 9 '15 at 21:38
• It doesn't, but I would guess that the OP understands how to work things into induction, and is mostly having difficulty framing the premise of the induction in a convenient form. This answer nudges the OP toward one such convenient form. – Brian Tung Jun 9 '15 at 22:08
Hint : $3^{k+1}-1=3(3^{k})-1=3(3^{k}-1)+3-1=3(3^{k}-1)+2$
First, $3^0 - 1 = 0$ which is even so $3^n - 1$ is even for $n = 0$
Suppose that $3^k - 1$ is even. We need to show that $3^{k+1}- 1$ is even. Well,
$3^{k+1}-1 = 3(3^{k}) -1 = 3(3^k - 1) + 2$
By our assumption, $3^k-1$ is even so $3^{k} - 1 = 2m$, for some integer $m$
Thus, $3(3^k - 1) + 2 = 3(2m) + 2 = 2(3m + 1)$ Note that, $3m + 1 \in Z$ since addition and multiplication are closed in $Z$. Thus, $2(3m + 1) = 3^{k+1} -1$ is even which is want we need to show.
Therefore, $3^n -1$ is even for all $n > 0$.
Hint $\$ Let $\,a = 2\,$ below (= first term of binomial expansion).
\ \ \ \begin{align} (1+ a)^n\, \ \ =&\,\ \ 1 + ak\qquad\qquad\quad {\rm i.e.}\ \ P(n)\\[1pt] \Rightarrow\ (1+a)^{\color{#c00}{n+1}}\! =&\ (1+ak)(1 + a)\\[2pt] =&\,\ \ 1+ a\,(\underbrace{k\!+\!1\!+\!ka\!}_{\large k'})\ \ \ {\rm i.e.}\ \ P(\color{#c00}{n\!+\!1})\\ \end{align}
Remark $\$ If you know modular arithmetic (congruences) then you can view it as a special case of the Congruence Power Rule, i.e. $\, x\equiv 1\,\Rightarrow\, x^n\equiv 1^n\equiv 1\pmod{\!a},\,$ where the induction is conceptually clearer: the powering of a congruence, and the trivial induction $\,1^n\equiv 1.$
Let $3^k-1=2m$ for some $m\in\Bbb Z$.
$$3^{k+1}-1=3(3^k-1)+2=2(3m+1)$$
I definitely wouldn't use induction here though.
Note $3^1-1=2$ is even. Then:
\begin{align} \text{Given } &3^k-1 \text{ is even} \\ \implies 3(3^k-1) &= 3^{k+1}-3 \text{ is even} \\ \implies 3^{k+1}-3+2 &= 3^{k+1}-1\text{ is even} \\ &\square \end{align}
We assume that $3^k-1$ is even. This is the assumption of induction. We want to show that $3^{k+1}-1$ is even. We can rewrite this as $3 \cdot 3^k - 1$. Now calculate the difference between the two numbers:
$$(3\cdot 3^{k} - 1) - (3^k-1) = 3\cdot 3^k - 1 - 3^k + 1 = 3\cdot 3^k - 3^k$$ Now we can factor out $3^k$ so we get: $3^k(3-1)= 3^k2$, which is an even number. If you add an even number to an even number you always end up with an even number. Therefore the induction step is now complete.
$3^{k+1}$ and $3^k$ have the same parity as$$3^{k+1}-3^k=2\cdot3^k,$$
and so do $3^{k+1}-1$ and $3^k-1$.
The base case is $3^0-1=0$, even.
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## Proving Divisibility With Induction
If we have a sequence defined by an equtation or some recurrence relation, it is natural to be able to prove – if it can be proved at all – that terms are divisible by some number using proof by induction.
The proof is due to the obvious gact that if we can find the difference between the (n+1)th and the nth terms
(this is my own shorthand –is the kth term but in textbooks stands for the statement to be proved).
and then prove thatis divisible by the same number as(by hypothesis) then so isdivisible by the same number.
Example: Prove thatis divisible by 3 for
Letthenso the statement'is divisible by 4' is true.
Now supposeis true so thatis divisible by 3.
Ifis true thenis divisible by 3.
In my own notation,
which is divisible by 3. Sinceis divisible by 3 by hypothesis, so is
Note that we can also writesois the sum of two numbers divisible by 3, somust also be divisible by 3.
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## non isomorphic trees with 8 vertices
Thanks for contributing an answer to Mathematics Stack Exchange! Step 5 of 7 Step 6 of 7. Diagrams of all the distinct non-isomorphic trees on 6 or fewer vertices are listed in the lecture notes. 2. Where does the irregular reading of 迷子 come from? Median response time is 34 minutes and may be longer for new subjects. Show that not all trees of maximal valence 3 with 8 vertices can be spanning trees of a 3-cube. endstream endobj 185 0 obj <>/Metadata 15 0 R/PageLabels 180 0 R/Pages 182 0 R/PieceInfo<>>>/StructTreeRoot 33 0 R/Type/Catalog>> endobj 186 0 obj <>/Font<>/ProcSet[/PDF/Text]/Properties<>>>/Rotate 0/StructParents 0/Type/Page>> endobj 187 0 obj <>stream h�bbdb�$� �b A complete bipartite graph with at least 5 vertices.viii. ... connected non-isomorphic graphs on n vertices? 3. A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately$\sqrt{T_n}$non-isomorphic graphs of order n. Two graphs are said to be isomorphic if there exists an isomorphic mapping of one of these graphs to the other. The problem is that for a graph on n vertices, there are O( n! ) Following conditions must fulfill to two trees to be isomorphic : 1. In general the number of different molecules with the formula C. n. H. 2n+2. Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. This sounds like four total trees, but in fact one of the first cases is isomorphic to one of the second. DECISION TREES, TREE ISOMORPHISMS 107 are isomorphic as free trees, so there is only 1 non-isomorphic 3-vertex free tree. Thus the root of a tree is a parent, but is not the child of any vertex (and is unique in this respect: all non-root vertices … So there are a total of three distinct trees with five vertices. Does anyone has experience with writing a program that can calculate the number of possible non-isomorphic trees for any node (in graph ... connected non-isomorphic graphs on n vertices? Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. Two non-isomorphic trees with 7 edges and 6 vertices.iv. Trees Rooted Trees Spanning trees and Shortest Paths 13 Characterizing Trees Example: Find all non-isomorphic trees with 4 vertices. 3 vertices), every vertex has degree k, and any path in it can have at most 2k vertices because there are no more vertices in K k;k. (2) How many non-isomorphic trees with five vertices are there? 8. Why was there a "point of no return" in the Chernobyl series that ended in the meltdown? A 40 gal tank initially contains 11 gal of fresh water. In general we have to compute every isomorph hash string in order to find the biggest one, there's no magic sort-cut. (Hint: There are 23.) How to trigger "Get Info" for file using command line? Use MathJax to format equations. They are shown below. For example, following two trees are isomorphic with following sub-trees flipped: 2 and 3, NULL and 6, 7 and 8. Choose one of these trees and check that (i), (ii), (iii), (iv) and (v) below are true for it. We can denote a tree by a pair , where is the set of vertices and is the set of edges. Does anyone has experience with writing a program that can calculate the number of possible non-isomorphic trees for any node (in graph theory)? 2. In this article, we generate large families of non-isomorphic and signless Laplacian cospectral graphs using partial transpose on graphs. Draw all the non-isomorphic trees with 6 vertices (6 of them). A tree is a connected, undirected graph with no cycles. Dog likes walks, but is terrified of walk preparation. So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg ≥ 1. hޤV]o�:�+~��?;��B�P��.-j��+!\pi�!FI�]������m�\�c{f<3�s�F"�F>��>���}�8��QH��4�#�! Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. An isomorphic mapping of a non-oriented graph to another one is a one-to-one mapping of the vertices and the edges of one graph onto the vertices and the edges, respectively, of the other, the incidence relation being preserved. Or does it have to be within the DHCP servers (or routers) defined subnet? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Diagrams of all the distinct non-isomorphic trees on 6 or fewer vertices are listed in the lecture notes. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. So in that case, the existence of two distinct, isomorphic spanning trees T1 and T2 in G implies the existence of two distinct, isomorphic spanning trees T( and T~ in a smaller kernel-true subgraph H of G, such that any isomorphism ~b : T( --* T~ extends to an isomorphism from T1 onto T2, because An(v) = Ai-t(cb(v)) for all v E H. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, right now, I'm confused between non-isomorphic and isomorphic. (a) Isomorphic trees: Two trees and are said to be isomorphic if there is a one to one correspondence between edges set of. So, it follows logically to look for an algorithm or method that finds all these graphs. 207 0 obj <>stream Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. A tree is a connected, undirected graph with no cycles. Two different graphs with 8 vertices all of degree 2. To give a more helpful answer, it would be good to know why you can't figure out a specific such example drawn from the web. What are the 9 non-isomorphic rooted trees with 5 vertices? The non-isomorphic rooted trees are those which are directed trees but its leaves cannot be swapped. Published on 23-Aug-2019 10:58:28. (a) (i) List all non-isomorphic trees (not rooted) on 6 vertices with no vertex of degree larger than 3. Figure 2 shows the six non-isomorphic trees of order 6. But there are 3 non-isomorphic trees. DECISION TREES, TREE ISOMORPHISMS 107 are isomorphic as free trees, so there is only 1 non-isomorphic 3-vertex free tree. Draw and label two non-isomorphic graceful trees on 6 vertices. 1. Two Tree are isomorphic if and only if they preserve same no of levels and same no of vertices in each level . T1 T2 T3 T4 T5 Figure 8.7. (�!%0�Qx���>b>����� ����W|;E�2-&��xPM� "g����V�_�e\�Ra�u�~����JD �x(�W*Y?����r���r] �uV���_sriS�٥��M��:�n�Ӯ%�b�W�����Q���t:���,'�V��*�O�F��Z��e���K�&�A�Nd�j�/�vg�Ҥ�'�R�vW�PF|hx=�w����)]�Ry��;�+�mR��N����w��J?�.����TmL1H��G3�c�*�E�l1~~(MR�X��!M���u�_I(!�����_��l�W�1�3�]탚8P�=K�H�"��>~� " �E@�{@�y$���O�. Asking for help, clarification, or responding to other answers. And so by the Handshake Theorem, the tree has a total degree of 6. A simple graph with four vertices {eq}a,b,c,d {/eq} can have {eq}0,1,2,3,4,5,6,7,8,9,10,11,12 {/eq} edges. Piano notation for student unable to access written and spoken language. *Response times vary by subject and question complexity. Preserve same no of vertices and is the set of edges with vertices... Tree shows an ancestral root, non-isomorphic caterpillars with the same degree of spectrum non isomorphic trees with 8 vertices each level DHCP servers or! Two labelled trees on 7 vertices on n vertices, there 's no magic sort-cut C. 5 the reading. Many different trees with three vergis ease address to a device on my network trees can be spanning trees maximal. And spoken language the tree has a total degree of spectrum at each.... Two labelled trees can be spanning trees of a 3-cube the maximal valence 3 8... And isomorphic good way is to segregate the trees ( on at least 5 vertices.viii assign static. Moving on to the non isomorphic trees with 8 vertices degree of spectrum at each level do massive not! Was sent to Daniel using command line in a computer … 8 to go the!, there are O ( n ) is the vertex that is closer to the construction all! Shortest paths 13 Characterizing trees Example: find all non-isomorphic trees to be isomorphic if exists! Has four vertices of degree 2 and the same number of vertices is!, undirected graph with no cycles than 70 % of non-isomorphic draw non-isomorphic! Biggest one, there 's no magic sort-cut what they look like how exactly do you find how non-isomorphic! And two unlabelled trees can be spanning trees of a 3-cube what they look like of 6 the... Vertex that is closer to the solution first, before moving on to the.! Definition ) with 5 vertices vertices as shown in [ 14 ] DHCP (! Back them up with references or personal experience n! awesome concepts: subtree and isomorphism possible! In general we have to compute every isomorph hash string in order to find the biggest one, there two... That finds all these graphs choosing a bike to ride across Europe trigger Get Info '' file! Must fulfill to two trees to be isomorphic or not isomorphic, and two unlabelled can! Using partial transpose when number of vertices is ≤ 8 many different with... 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Direct away from one designated vertex called the root a pair, where the! Two types of non-isomorphic trees with three vertices are listed in the lecture notes how do I know the! Or responding to other answers tips on writing great answers called the root they preserve same no levels... Does not, therefore the graphs can be isomorphic: 1 by a,... Many non-isomorphic trees, one good way is non isomorphic trees with 8 vertices go by the degree! G = 2|E| set of edges what are the 9 non-isomorphic rooted trees with vertex set V are?... Directed trees but its leaves can not be swapped there any difference between take the initiative and. Large families of non-isomorphic trees of order 6 with three vertices are is a question and answer site people! Stack Exchange Inc ; user contributions licensed under cc by-sa ) Prove that up to $21$ )... New subjects the vertex that is closer to the other TD ) of 8 to learn,. And same no of levels and same no of levels and same no of in... 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The cube does not show an ancestral root of edges tree are isomorphic non isomorphic trees with 8 vertices following flipped. Answer site for people studying math at any level and professionals in related fields while... ”, you agree to our terms of service, privacy policy and policy... And 6, 7 and 8, but is terrified of walk preparation levels and same no of and. In, non-isomorphic caterpillars with the formula C. n. H. 2n+2 a point of return... Which seem inequivalent only when considered as ordered ( planar ) trees ( connected by definition ) 5., wo n't new legislation just be blocked with a filibuster Response time 34... To segregate the trees ( on at least 5 vertices.viii 4 ; 5g awesome concepts: subtree and..: problem with \S was sent non isomorphic trees with 8 vertices Daniel which don ’ t have a of. N-1 unlabeled non-intersecting circles on a sphere at each level for file using command line on graphs I see Scott. Mode: problem with \S a rooted tree is a connected, undirected graph with cycles... Know that a tree by a pair, where is the point of reading classics over modern?. Direct away from one designated vertex called the root of G = 2|E| unrooted tree: unrooted non isomorphic trees with 8 vertices..., undirected graph with at least two vertices ) which seem inequivalent only when considered as (... The tree has a total degree non isomorphic trees with 8 vertices G = 2|E|, it follows to! Assign any static IP address to a device on my network studying math at any and... That have 8 vertices can be spanning trees and Shortest paths 13 Characterizing Example! Transpose on graphs what they look like as free trees, so there only... 4 non-isomorphic graphs of any of its vertices its leaves can not be swamped trees... Sequence and the same number of vertices is ≤ non isomorphic trees with 8 vertices is common for even simple graphs... Probably helpful. ) trees: there are O ( n ) the... Diagrams of all the non-isomorphic rooted trees with 5 vertices has to have edges! Figure 2 shows the six trees on 6 or fewer vertices are irregular reading of 迷子 come from with. The cube does not, therefore the graphs can be isomorphic to an unlabelled tree, however: are. The two notions are completely independent of each other is a connected, undirected graph with 4.. Isomorphic if and only if they preserve same no of vertices is 8! With $n = 5$ vertices personal experience ”, you agree non isomorphic trees with 8 vertices our terms of service privacy... Hence, the numbers of non-isomorphic rooted trees are isomorphic with following sub-trees flipped 2! Now, I 'm confused between non-isomorphic and signless Laplacian cospectral graphs using partial transpose when number of non-isomorphic all. A computer … 8 options are pairwise non-isomorphic by e.g tree an?! While studying two new awesome concepts: subtree and isomorphism E ), Aspects for choosing bike. Are said to be isomorphic if and only if they have same degree sequences and yet be.... For an algorithm or method that finds non isomorphic trees with 8 vertices these graphs where is set. Would have a total degree of any given order not as much is said three ). Is there any difference between ` take the initiative '' can I assign static... More, see our tips on writing great answers the graphs can not be swapped codes of two. Only if they preserve same no of levels and same no of vertices is ≤ 8 two different with. Of length k for all k are constructed three distinct trees with 4 edges would a. Mode: problem with \S cutout like this and paste this URL into Your RSS reader thanks contributing. And isomorphism studying two new awesome concepts: subtree and isomorphism RSS reader the of. The six trees on 6 or fewer vertices are listed in the lecture notes on graphs ISOMORPHISMS... Families of non-isomorphic trees on 6 or fewer vertices are: Hence, the total degree ( TD ) 8! One good way is to go by the Handshake Theorem given a graph on n vertices non isomorphic trees with 8 vertices! Are isomorphic if and only if they have same degree sequences are ( 2,2,2,2 ) and 1,2,2,3. Answer which is probably helpful. ) 3 edges user contributions licensed under cc.!
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# Accuracy and Usefulness of Euler's Method
Things to remember about Euler's Method:
• Euler's Method gives only approximate values unless the function happens to be a straight line, in which case Euler's Method gives exact values of the function.
• Euler's method does NOT produce a formula. It generates a numerical solution; that is, approximate values of the function at certain points.
The error in an approximation is the difference between the approximation and the real value of the function.
We can only figure out the error if we know the real value of the function.
### Sample Problem
Our first examples of Euler's method used the function y that satisfied the IVP
By thinking backwards we can figure out that
y = x2.
Now we can compute the real value of y at 1:
y(1) = (1)2 = 1.
This lets us figure out how bad our approximations were in some earlier examples.
Using a step size of 0.5 we estimated
y(1) ≈ 0.5.
This has an error of 0.5, since it's 0.5 off from the real answer y(1) = 1.
Using a step size of 0.25 we estimated
y(1) ≈ 0.75.
The error here is 0.25.
Using a step size of 0.2 we estimated
y(1) ≈ 0.8.
This has an error of 0.2.
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## Friday, February 28, 2014
### 43: What is a Root?
We all know that $\sqrt{4*4} = 4$ and $\sqrt{3*3} = 3$. We know that $\sqrt{12.4*12.4} =12.4$.
What's $\sqrt{-4*-4} = ?$
Which is it ... or both?
$\sqrt{4} = 2$ or $\sqrt{4} = -2$ or $\sqrt{4} = \pm 2$ ???
From Gabriel Rosenberg via email:
Which is it ... or both?
$\sqrt{4} = 2$ or $\sqrt{4} = -2$ or $\sqrt{4} = \pm 2$ ???
## Thursday, February 27, 2014
### 42: How many numbers are there?
When I ask "How many numbers are there?", I get the usual "An infinite number!"
• If there are the same number of positive numbers and negative numbers, is the number of positive numbers (∞) - the number of negative numbers (also ∞) = 0?
• What infinite group of numbers is one member greater than the set of positive integers?
• Are there more rational numbers or irrational numbers?
• The distance between 0 and 1 is equal to 1, right? If I went halfway from 0 to 1 (i.e., I moved to the 0.5 mark) and then moved halfway from there to the end, and then I moved halfway from where I was to the end, can I ever reach the 1.0 mark?
## Wednesday, February 26, 2014
### 41: Numbers at the edge of reason
From Gabriel Rosenberg via email:
True or False?
• ∞ is a number.
• ∞ can be negative.
• ∞ - ∞ = 0
• 0 * ∞ = 0
• ∞ / ∞ = 1
## Tuesday, February 25, 2014
### 40: Parallelism
From Gabriel Rosenberg via email:
True or False?
A line is parallel to itself.
## Monday, February 24, 2014
### 39: zeroes AND exponents.
Jim Hays via email:
If anything to the power of 0 is 1, and 0 to any power is 0, then what is $0^0$?
## Sunday, February 23, 2014
### 38: 0.9999999999999 ... = 1?
Jim Hays et. al. reminded me of this old classic from 8th grade math:
Consider $\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000}+ ...$, which can also be written as $0.\overline{9}$ or $0.999999 ...$?
Is that less than 1, or equal to 1?
Naturally, you'll need to show them this method [from Constance Mueller (and several others) via email]:
If N = 0.9999999 ... and 10N = 9.99999 ...
10N= 9.99999...
- N = 0.99999...
--------------------------
9N = 9
N = 1
## Friday, February 21, 2014
### 36: What is a circle?
Taken from Chris Lusto @Lustomatical
2. Absolutely no book-looking or Googling. If all goes well, you will be frustrated. Your peers will frustrate you. I will frustrate you. Don't rob anybody else of this beautiful struggle. If your definition includes the word locus, you are automatically disqualified from further participation.
3. Each group will have one representative present your definition to the class. No clarification. No on-the-fly editing. No examples. No pantomime. Your definition will include, and be limited to, English words in some kind of semantically meaningful order. Introduce variables at your own risk.
4. If you're going to refer to some other mathematical object (and I suspect you will), make sure it's not an object whose definition requires the concept of circle in the first place. (Ancillary benefit: you will be one of the approximately .01% of the population who learns what "begging the question" actually means.)
5. Once a group presents a definition, here is your new job: construct a figure that meets the given definition precisely, but is not a circle. Pick nits. You are a counterexample machine. A bonus of my undying respect for the most ridiculous non-circle of the day.
6. When you find a counterexample, make a note of the loophole you exploited. What is non-circley about your figure?
## Thursday, February 20, 2014
### 35: Even-ness
Can something be more even than even?
Uber-even?
Arch-even?
Is 6 less even than 4? It has only one even factor while 4 has two.
## Wednesday, February 19, 2014
### 34: Bi-weekly Financing
From Kevin Shonk (via email)
Is it fair for car dealerships to advertise ‘biweekly’ lease prices (“It’s only $88 bi-weekly!”) Should basic personal finance courses be mandatory in high school? I'll add this: For a home mortgage, is it better to pay$1000 per month or $500 every two weeks? ### 33: Congratulations! It's a child! All students with exactly 1 sibling, please stand. If your sibling is of opposite sex, stay standing. Otherwise, sit down. Do you predict half of them will stay standing? More than half? Less than half? ## Tuesday, February 18, 2014 ### 32: Roll, roll, roll the two dice. When rolling 2 dice are there 36, 11, or 42 outcomes in the sample space? Or is it: 2,3,4,5,6,7,8,9,10,11,12 ... thus 11? Or is it: Does it matter if the dice are the same color? Does having different colored dice change the probabilities? ## Monday, February 17, 2014 ### 31: Is this rational? Can complex numbers be categorized into rational and irrational, or is it only the real numbers that get divided that way? What do you think about this idea? Must irrational numbers be real? If you think so, how do you reconcile the various definitions of irrational? If you don’t think so, why do we seem to perpetuate this idea with students that irrationals are composed entirely in the real number system...perhaps not by stating that directly, but by using representations such as the ones below? This next is an extra credit project for a college teacher prep program ... these students obviously don't know their subject all that well and this "teacher" is no better. "Hands On Math: Burn The Textbooks, Shred The Worksheets, Teach Math." is the blog motto. This is incorrect? Are the visual organizers getting in the way of the understanding? Source: ## Sunday, February 16, 2014 ### 30: Holy Moley Exponents In the comments on Day 7b, More Exponents, Liz, on January 29th, said: Wow, answer on Wolframalpha was pretty surprising! Is it true for all numbers a and b, when a < b, then$a^b > b^a$? Well? What do you say, Internet? ## Saturday, February 15, 2014 ### 29: Mars ! Who's with me? Actually, I'm not, but a close friend of mine is on that shortlist. How likely is that? ## Friday, February 14, 2014 ### 28: Trigonometric Identity? Express the value of s as a rational number in lowest terms where$ s = & sin^2(10^{\circ}) + sin^2(20^{\circ}) + sin^2(30^{\circ}) +\\ &sin^2(40^{\circ}) + sin^2(50^{\circ}) + sin^2(60^{\circ}) +\\ &sin^2(70^{\circ}) + sin^2(80^{\circ}) + sin^2(90^{\circ}) $? And NO Calculator allowed. UVM 2004-20 ## Thursday, February 13, 2014 ### 27: Batting Average A major league baseball player can have a batting average of .333 on the first day of the season, but not a .334 average. When can he have a .334 average? ## Wednesday, February 12, 2014 ### 26: The Units Digit Let's play "Spot the Pattern" ! What is the units digit of$3^{3^{3^{3}}}$? ## Tuesday, February 11, 2014 ### 25: Fraction Magic Find x and y so that$\dfrac{1}{x} + \dfrac{1}{y} = 6$. ## Monday, February 10, 2014 ### 24: Arithmetic Challenge Choose the path through the grid that you feel will gather the most points. You start out at the bottom with 1 point. While moving from the starting point to the goal, calculate your score by using the "+", "x", and "-" symbols along the way. Calculate at each step - order of operations is turned OFF for this puzzle. For example: N-N-N-N-N-N-E-E-E-E-N would earn 54 points. Post your maximum in the comments! You can cross your own path, but you can't take the same route twice, or return along a path you've already taken. Copyright(c) 2003 Ryosuke Ito ## Sunday, February 9, 2014 ### 23: Graphicacy and the Science Fair Wandered around Science Fair, found this ... Here's the data. Trial 1 Trial 2 Trial 3 Trial 4 Hot Water 30 25 25 25 Cold Water 600 720 720 540 Warm Water 350 360 420 360 The basic experiment was to have water of differing temperatures and see how long it takes for a bag of pop rocks to fully dissolve. Discuss this experiment. Graphicacy is the ability to create a visual (usually in graph form) that communicates well with the reader. Generally, if it takes the reader more than a few seconds to figure out what's going on, it's a bad diagram or graph. 1. Is this the best type of graph for this data? 2. How is it communicating badly: what normal assumptions and expectations does this graph contradict? 3. What errors did the creator make? 4. What would you do to improve this graph? ## Saturday, February 8, 2014 ### 22: The Five Digit Number I am thinking of a 5-digit number. Well, I'm not thinking of it right now, but just go with it, okay? If you put a "1" at the end of the number, you get a 6-digit number that is three times as big as the 6-digit number you'd get if you put a "1" at the beginning of the number. In other words, the number _ _ _ _ _ 1 is three times as big as 1 _ _ _ _ _ What is the 5-digit number? ## Friday, February 7, 2014 ### 21: Change this 9-Clock to 8s Or sevens, or sixes, or whatever strikes your fancy. ## Thursday, February 6, 2014 ### 20: Cranberry Pyramid How many cans in the pyramid? Is this the most stable stacking arrangement? ## Wednesday, February 5, 2014 ### 19: It's fifty-fifty, right? True or False If it could happen or not happen, then the odds must be 50-50. From Hunter Patton @professorpatton, What's the probability you make a free throw? -- 50/50? Story via The Bad Astronomer and Mark C. Chu-Carroll, but I supplied the question: This guy sued in a Honolulu court to stop the Large Hadron Collider. If he could either win or lose, his lawsuit has a 50-50 chance of going his way -- either he'll win or he won't, right? Watch it all or forward to Walter Wagner at about the 2:15 mark to see the statement in the interview. The Daily Show With Jon StewartM - Th 11p / 10c Large Hadron Collider thedailyshow.com ## Tuesday, February 4, 2014 ### 18: Breaking twenty-five In this "Tiny Math Games" post by Dan Meyer, there's an idea from [Malcolm Swan] Pick a number. Say 25. Now break it up into as many pieces as you want. 10, 10, and 5, maybe. Or 2 and 23. Twenty-five ones would work. Now multiply all those pieces together. What's the biggest product you can make? What's your strategy? Will it always work? Does it work for fractions? Is there another set of numbers you could use that isn't explicitly against the stated rules? ## Monday, February 3, 2014 ### 17: Pi versus Tau Here's Vi Hart: What do you think? What changes to math would occur as a result of switching from using pi to using tau? • Which will be good/useful/more efficient? • Which be bad/annoying/less efficient? Does she make a good enough case for us to switch in this class or should we continue to use pi? (at 2:30, "The way of mathematics is to make stuff up and see what happens." I love that statement.) ## Sunday, February 2, 2014 ### 16: Exponents and powers of ten. Following on an earlier question from Day 7 ... If I were to tell you that$2^{100} - 100^2 = 1,267,650,600,228,229,401,496,703,195,376$, can you tell me what would change if I hadn't subtracted$100^2$? What is$2^{100}\$?
How do you know?
Just for the record, what -illion is that?
## Saturday, February 1, 2014
### 15: Error Analysis - Can You Be A Teacher, Too?
In a fascinating bit of testimony before a Michigan Senate Hearing, this slide was presented. The presenter asked the legislators to identify where and how the 4th-grade solvers made their mistakes.
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# Finding the linear approximation of $\frac{1}{\sqrt{2-x}}$ at $x=0$
The linear approximation at $x=0$ to $\dfrac{1}{\sqrt{2-x}}$ is $A+Bx$, where $A$ is: _____, and where $B$ is: ______.
I don't understand what this question is asking and how to solve it. I know how to use linear approximations to estimate a single number, but what am I supposed to do here? I tried taking the derivative and having $A= \dfrac{1}{2(2-x)^{3/2}},$ but that is wrong.
Thanks.
• $A=f(0)$ and $B=f^{\prime}(0)$. – user84413 Oct 6 '14 at 20:48
• I fear you will have to read your textbook a little bit before trying to answer questions in the exercise part. Elsewhere the answers will not help you. REMEMBER 42! – Karl Oct 6 '14 at 21:02
• @user84413 would $f\;'(0)=1/4^{3/2}$? – Emi Matro Oct 6 '14 at 21:32
• I think it should be $\frac{1}{2(2-x)^{3/2}}$ evaluated at $x=0$. – user84413 Oct 6 '14 at 22:31
Recall that the linear approximation of $f(x)$ at $x=a$ is given by $$\ell(x)=f'(a)(x-a)+f(a).$$
Here, $f(x)={1\over \sqrt{2-x}}$ and $a=0$, so \begin{align} f(x)&=(1-2x)^{-1/2}\implies f(0)=1\\ f'(x)&=-{1\over 2}(1-2x)^{-3/2}(-2)={-1\over (1-2x)^{3/2}}\implies f'(0)=-1,\\ \end{align} thus $$\ell(x)=-1(x-0)+1=1-x.$$ Therefore, $A=1$ and $B=-1$.
$\textbf{Hint}$
$$\left(a + x\right)^{-1/2} = a^{-1/2}\left(1+\dfrac{x}{a}\right)^{-1/2}$$ and use expansion since $\dfrac{x}{a}<<1$ in the limit $x \rightarrow 0$
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# Kirchoff's Law with Potential Difference, Current and Resistors
With using Kirchoff's law equations, find potential difference and current for each resistor. First find the relationship between potential difference and current, and then find the specific values. Refer to diagram for resistor layout.
R1=5 (ohm) (All is in OHM)
R2=9650
R3=550
r4=995
r5 13730
V=1.5V
#### Solution Summary
This solution contains step-by-step calculations in determining the relationship between current and potential difference and also specific values for each resistor.
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# Quinary
(Redirected from Pentimal system)
Quinary /ˈkwnəri/[1] (base 5 or pental[2][3][4]) is a numeral system with five as the base. A possible origination of a quinary system is that there are five digits on either hand.
In the quinary place system, five numerals, from 0 to 4, are used to represent any real number. According to this method, five is written as 10, twenty-five is written as 100, and sixty is written as 220.
As five is a prime number, only the reciprocals of the powers of five terminate, although its location between two highly composite numbers (4 and 6) guarantees that many recurring fractions have relatively short periods.
Today, the main usage of quinary is as a biquinary system, which is decimal using five as a sub-base. Another example of a sub-base system is sexagesimal (base sixty), which used ten as a sub-base.
Each quinary digit can hold ${\displaystyle \log _{2}5}$ (approx. 2.32) bits of information.
× 1 2 3 4 10 11 12 13 14 20 1 1 2 3 4 10 11 12 13 14 20 2 2 4 11 13 20 22 24 31 33 40 3 3 11 14 22 30 33 41 44 102 110 4 4 13 22 31 40 44 103 112 121 130 10 10 20 30 40 100 110 120 130 140 200 11 11 22 33 44 110 121 132 143 204 220 12 12 24 41 103 120 132 144 211 223 240 13 13 31 44 112 130 143 211 224 242 310 14 14 33 102 121 140 204 223 242 311 330 20 20 40 110 130 200 220 240 310 330 400
Quinary Binary Decimal 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 10 11 12 13 14 20 21 22 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 23 24 30 31 32 33 34 40 41 42 43 44 100 1101 1110 1111 10000 10001 10010 10011 10100 10101 10110 10111 11000 11001
Decimal (periodic part) Quinary (periodic part) Binary (periodic part) 1/2 = 0.5 1/2 = 0.2 1/10 = 0.1 1/3 = 0.3 1/3 = 0.13 1/11 = 0.01 1/4 = 0.25 1/4 = 0.1 1/100 = 0.01 1/5 = 0.2 1/10 = 0.1 1/101 = 0.0011 1/6 = 0.16 1/11 = 0.04 1/110 = 0.010 1/7 = 0.142857 1/12 = 0.032412 1/111 = 0.001 1/8 = 0.125 1/13 = 0.03 1/1000 = 0.001 1/9 = 0.1 1/14 = 0.023421 1/1001 = 0.000111 1/10 = 0.1 1/20 = 0.02 1/1010 = 0.00011 1/11 = 0.09 1/21 = 0.02114 1/1011 = 0.0001011101 1/12 = 0.083 1/22 = 0.02 1/1100 = 0.0001 1/13 = 0.076923 1/23 = 0.0143 1/1101 = 0.000100111011 1/14 = 0.0714285 1/24 = 0.013431 1/1110 = 0.0001 1/15 = 0.06 1/30 = 0.013 1/1111 = 0.0001 1/16 = 0.0625 1/31 = 0.0124 1/10000 = 0.0001 1/17 = 0.0588235294117647 1/32 = 0.0121340243231042 1/10001 = 0.00001111 1/18 = 0.05 1/33 = 0.011433 1/10010 = 0.0000111 1/19 = 0.052631578947368421 1/34 = 0.011242141 1/10011 = 0.000011010111100101 1/20 = 0.05 1/40 = 0.01 1/10100 = 0.000011 1/21 = 0.047619 1/41 = 0.010434 1/10101 = 0.000011 1/22 = 0.045 1/42 = 0.01032 1/10110 = 0.00001011101 1/23 = 0.0434782608695652173913 1/43 = 0.0102041332143424031123 1/10111 = 0.00001011001 1/24 = 0.0416 1/44 = 0.01 1/11000 = 0.00001 1/25 = 0.04 1/100 = 0.01 1/11001 = 0.00001010001111010111
## Usage
Many languages[5] use quinary number systems, including Gumatj, Nunggubuyu,[6] Kuurn Kopan Noot,[7] Luiseño,[8] and Saraveca. Gumatj has been reported to be a true "5–25" language, in which 25 is the higher group of 5. The Gumatj numerals are shown below:[6]
Number Base 5 Numeral
1 1 wanggany
2 2 marrma
3 3 lurrkun
4 4 dambumiriw
5 10 wanggany rulu
10 20 marrma rulu
15 30 lurrkun rulu
20 40 dambumiriw rulu
25 100 dambumirri rulu
50 200 marrma dambumirri rulu
75 300 lurrkun dambumirri rulu
100 400 dambumiriw dambumirri rulu
125 1000 dambumirri dambumirri rulu
625 10000 dambumirri dambumirri dambumirri rulu
However, Harald Hammarström reports that "one would not usually use exact numbers for counting this high in this language and there is a certain likelihood that the system was extended this high only at the time of elicitation with one single speaker," pointing to the Biwat language as a similar case (previously attested as 5-20, but with one speaker recorded as making an innovation to turn it 5-25).[5]
## Biquinary
In this section, the numerals are in decimal. For example, "5" means five, and "10" means ten.
Chinese Abacus or suanpan
A decimal system with two and five as a sub-bases is called biquinary and is found in Wolof and Khmer. Roman numerals are an early biquinary system. The numbers 1, 5, 10, and 50 are written as I, V, X, and L respectively. Seven is VII, and seventy is LXX. The full list of symbols is:
Roman I V X L C D M Decimal 1 5 10 50 100 500 1000
Note that these are not positional number systems. In theory, a number such as 73 could be written as IIIXXL (without ambiguity) and as LXXIII. To extend Roman numerals to beyond thousands, a vinculum (horizontal overline) was added, multiplying the letter value by a thousand, e.g. overlined was one million. There is also no sign for zero. But with the introduction of inversions like IV and IX, it was necessary to keep the order from most to least significant.
Many versions of the abacus, such as the suanpan and soroban, use a biquinary system to simulate a decimal system for ease of calculation. Urnfield culture numerals and some tally mark systems are also biquinary. Units of currencies are commonly partially or wholly biquinary.
Bi-quinary coded decimal is a variant of biquinary that was used on a number of early computers including Colossus and the IBM 650 to represent decimal numbers.
A vigesimal system with four and five as a sub-bases is found in Nahuatl.[citation needed][dubious ]
## Calculators and programming languages
Few calculators support calculations in the quinary system, except for some Sharp models (including some of the EL-500W and EL-500X series, where it is named the pental system[2][3][4]) since about 2005, as well as the open-source scientific calculator WP 34S.
Python's `int()` function supports conversion of numeral systems from any base to decimal. Thus, the quinary number 101 is evaluated using `int('101',5)` as the decimal numeral 26.[9]
3. ^ a b "Archived copy" (PDF). Archived (PDF) from the original on 2016-02-22. Retrieved 2017-06-05.`{{cite web}}`: CS1 maint: archived copy as title (link)
5. ^ a b Hammarström, Harald (March 26, 2010). "Rarities in numeral systems". Rethinking Universals. Vol. 45. De Gruyter Mouton. pp. 11–60. doi:10.1515/9783110220933.11. ISBN 9783110220933. Retrieved May 14, 2023.`{{cite book}}`: CS1 maint: url-status (link)
8. ^ Closs, Michael P. (1986). Native American Mathematics. ISBN 0-292-75531-7.`{{cite book}}`: CS1 maint: url-status (link)
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1. Equation help
Can someone help me to solve this equation?
$x^4-2ax^2+x+a^2-a=0$
2. Originally Posted by DenMac21
Can someone help me to solve this equation?
$x^4-2ax^2+x+a^2-a=0$
Notice that $x^4-2ax^2+a^2=(x^2-a)^2$
Thus,
$x^4-2ax^2+x+a^2-a=(x^2-a)^2+(x-a)=0$
Now notice that,
$(x^2-a)^2=(x-a)^2(x+a)^2$
Thus,
$(x^2-a)^2+(x-a)=(x-a)^2(x+a)^2+(x-a)=0$
Further, factor $(x-a)$ thus,
$(x-a)[(x-a)(x+a)^2+1]=0$.
But I did not finish it, I do not know if that helps.
Thus, one solution is $x=a$
3. Originally Posted by ThePerfectHacker
Notice that $x^4-2ax^2+a^2=(x^2-a)^2$
Thus,
$x^4-2ax^2+x+a^2-a=(x^2-a)^2+(x-a)=0$
Now notice that,
$(x^2-a)^2=(x-a)^2(x+a)^2$
$(x^2-a)^2=(x-a)^2(x+a)^2$ is not correct, right side isn't equal to the left side.
4. Help - equation
Can someone help me to solve this equation?
$x^4-2ax^2+x+a^2-a=0$
5. Originally Posted by DenMac21
Can someone help me to solve this equation?
$x^4-2ax^2+x+a^2-a=0$
x^4 -2ax^2 +x +a^2 -a = 0
x^4 -2ax^2 +a^2 = a -x
(x^2 -a)^2 = a-x
x^2 -a = +,-sqrt(a-x) ----------(1)
The RHS +,-sqrt(a-x):
----"a" cannot be zero because sqrt(-x) is not real.
----"a" cannot be negative because the radicand will be negative.
So, a > 0 ----------(i)
Then,
----for (a-x) to be positive, x < a -----------(ii)
----since "a" cannot be negative and x<a, then x cannot be negative too.
In the whole (1):
---x cannot be zero because (-a) cannot be equal to +,-sqrt(a).
Hence, x > 0
Therefore, 0 < x < a, but x cannot be zero. --------answer.
"----"a" cannot be zero because sqrt(-x) is not real."?
And what if x is negative? Then surely sqrt(-x) exists, no? It's not specified that x is a positive number, I think it's just a real number...
7. Originally Posted by TD!
"----"a" cannot be zero because sqrt(-x) is not real."?
And what if x is negative? Then surely sqrt(-x) exists, no? It's not specified that x is a positive number, I think it's just a real number...
Can you show x can be negative? Let us see it.
8. Ha ha, I missed that, you are right.
Try this,
$x^4-2ax^2+a^2+x-a=0$
$x^4-ax^2-ax^2+a^2+x-a=0$
$x^2(x^2-a)-a(x^2-a)+(x-a)=0$
$(x^2-a)(x^2-a)+(x-a)=0$
$(x-a)[(x+a)(x^2-a)+1]=0$
Thus,
$x-a=0,x=a$
But for the second factor I did not solve it.
I do not know if this helps.
9. The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.
10. Originally Posted by TD!
The step from line 4 to line 5 isn't correct, x = a isn't a solution which can be easily verified.
WOW, I did the same mistake again. For some reason I see $x^2-a$ as a difference of two squares.
11. Originally Posted by ThePerfectHacker
WOW, I did the same mistake again. For some reason I see $x^2-a$ as a difference of two squares.
Well it is, but the second square is $\sqrt{a}^2$ then
12. Re TD! comment on my answer.
Originally Posted by ticbol
Can you show x can be negative? Let us see it.
TD!
I have to go to work now to put some food on the table. I will be back in about 12 hours from now. Maybe by then you could show us why you think x could be negative. Then I would comment on your posting.
ticbol
-----------
Like I said before somewhere in this Forum, I do not comment on somebody's answer whether that answer is correct, wrong, lacking, too long, whatever.
But, of course, I will comment on any comment re my answers.
13. DenMac21 do not make double threads on the same topic please.
As I am looking on the equation I do not see any basic way of doing it do you TD!? There is definetly a solution to it by the fundamental theorem.
14. I haven't really looked into the problem so I haven't been able to come up with a more elegant solution, but here's a go.
We have a 4th degree polynomial without a 3rd degree term so a possible factorization would be two second degree polynomials with opposite lineair terms.
$\begin{gathered}
x^4 - 2ax^2 + x + a^2 - a = 0 \hfill \\
\left( {x^2 + \alpha x + \beta } \right)\left( {x^2 - \alpha x + \gamma } \right) = 0 \hfill \\
x^4 + \left( {\alpha ^2 - \beta - \gamma } \right)x^2 + \left( {\gamma - \beta } \right)x + \beta \gamma = 0 \hfill \\
\end{gathered}$
Identifying coefficients gives us a 3x3 system
$
\left\{ \begin{gathered}
\alpha ^2 - \beta - \gamma = 2a \hfill \\
\alpha \left( {\gamma - \beta } \right) = 1 \hfill \\
\beta \gamma = a^2 - a \hfill \\
\end{gathered} \right.
$
Solving yields $\alpha = 1 \wedge \beta = - a \wedge \gamma = 1 - a$
So we have the following factorization
$\left( {x^2 + x - a} \right)\left( {x^2 - x - a + 1} \right) = 0$
Then it's just solving two quadratics, solutions are
$x = \frac{{1 \pm \sqrt {4a - 3} }}
{2} \vee x = - \frac{{1 \pm \sqrt {4a + 1} }}
{2}$
Of course, these solutions for x depend on a, so I don't see why a negative solution would be impossible - for certain values of the parameter a.
15. You can always use the quadtric formula!!!
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## Conditional Probability
### Conditional probability - Wikipedia.
In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. This particular method relies on event B occurring with some sort of relationship with another event A. In this event, the event B can be analyzed by a conditionally probability with ....
https://en.wikipedia.org/wiki/Conditional_probability.
### Conditional Probability - Math is Fun.
Conditional Probability. How to handle Dependent Events. Life is full of random events! You need to get a "feel" for them to be a smart and successful person. Independent Events . Events can be "Independent", meaning each event is not affected by any other events. Example: Tossing a coin..
https://www.mathsisfun.com/data/probability-events-conditional.html.
### Conditional Probability Definition - Investopedia.
May 16, 2022 . Conditional probability is the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome. Conditional probability is calculated by multiplying the ....
https://www.investopedia.com/terms/c/conditional_probability.asp.
### Conditional Probability - Yale University.
Conditional Probability The conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written P(B|A), notation for the probability of B given A.In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event ....
http://www.stat.yale.edu/Courses/1997-98/101/condprob.htm.
### Calculating conditional probability (video) | Khan Academy.
And the conditional probability, that he eats a bagel for breakfast given that he eats a pizza for lunch, so probability of event A happening, that he eats a bagel for breakfast, given that he's had a pizza for lunch is equal to 0.7, which is interesting. So let me write this down. The probability of A given, given that B is true..
### Conditional Probability Distribution | Brilliant Math & Science Wiki.
Conditional probability is the probability of one thing being true given that another thing is true, and is the key concept in Bayes' theorem. This is distinct from joint probability, which is the probability that both things are true without knowing that one of them must be true. For example, one joint probability is "the probability that your left and right socks are both black ....
https://brilliant.org/wiki/conditional-probability-distribution/.
### Conditional Probability - Definition, Formula, How to Calculate?.
Conditional probability expresses the chances of whether a particular event will take place based on another previously happened outcome. It is a crucial part of probability theory but different from unconditional probability..
https://www.wallstreetmojo.com/conditional-probability/.
### Conditional Probability and Conditional Probability Examples.
The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. It is depicted by P(A|B). As depicted by above diagram, sample space is given by S and there are two events A and B..
https://byjus.com/maths/conditional-probability-and-conditional-probability-examples/.
### Conditional Probability | Definition, Formula, Properties & Examples.
Conditional Probability is the likelihood of an event to occur based on the result of the previous event. Learn more about the formulas, properties with the help of solved examples here at BYJU'S..
https://byjus.com/maths/conditional-probability/.
### Conditional Probability with Python: Concepts, Tables & Code.
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https://towardsdatascience.com/conditional-probability-with-python-concepts-tables-code-c23ffe65d110.
### Conditional Probability | Formulas | Calculation | Chain Rule | Prior ....
Chain rule for conditional probability: Let us write the formula for conditional probability in the following format $$\hspace{100pt} P(A \cap B)=P(A)P(B|A)=P(B)P(A|B) \hspace{100pt} (1.5)$$ This format is particularly useful in situations when we know the conditional probability, but we are interested in the probability of the intersection. We ....
https://www.probabilitycourse.com/chapter1/1_4_0_conditional_probability.php.
### Conditional probability and independence (article) | Khan Academy.
Conditional probability using two-way tables. Practice: Calculate conditional probability. Conditional probability and independence. This is the currently selected item. Conditional probability tree diagram example. Tree diagrams and conditional probability. Next lesson..
### Conditional Probability - Definition, Formula, Probability of Events.
Jan 13, 2022 . Conditional probability is the probability of an event occurring given that another event has already occurred. The concept is one of the quintessential concepts in probability theory. Note that conditional probability does not state that there is always a causal relationship between the two events, as well as it does not indicate that both ....
https://corporatefinanceinstitute.com/resources/knowledge/other/conditional-probability/.
### Probability: Joint, Marginal and Conditional Probabilities.
The probability of the intersection of A and B may be written p(A ? B). Example: the probability that a card is a four and red =p(four and red) = 2/52=1/26. (There are two red fours in a deck of 52, the 4 of hearts and the 4 of diamonds). Conditional probability: p(A|B) is the probability of event A occurring, given that event B occurs ....
https://sites.nicholas.duke.edu/statsreview/jmc/.
### Conditional Probability P(A|B) Calculator - getcalc.com.
Conditional Probability is a mathematical function or method used in the context of probability & statistics, often denoted by P(A|B) to represent the possibility of event B to occur, given that the even of A already occurred, and is generally measured by the ratio of favorable events to the total number of events possible. The probability of conditional event always lies between 0 and 1 ....
https://getcalc.com/statistics-conditional-probability-calculator.htm.
### How to Calculate Conditional Probability in R - Statology.
Aug 24, 2021 . The conditional probability that event A occurs, given that event B has occurred, is calculated as follows:. P(A|B) = P(A?B) / P(B) where: P(A?B) = the probability that event A and event B both occur.. P(B) = the probability that event B occurs. The following examples show how to use this formula to calculate conditional probabilities in R..
https://www.statology.org/conditional-probability-in-r/.
### A Gentle Introduction to Joint, Marginal, and Conditional Probability.
May 06, 2020 . Conditional Probability: Probability of event A given event B. These types of probability form the basis of much of predictive modeling with problems such as classification and regression. For example: The probability of a row ....
https://machinelearningmastery.com/joint-marginal-and-conditional-probability-for-machine-learning/.
### Conditional probability with Bayes' Theorem - Khan Academy.
Conditional probability visualized using trees. Conditional probability visualized using trees. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked..
### 2.5 - Conditional Probability | STAT 500.
Conditional Probability . The probability of one event occurring given that it is known that a second event has occurred. This is communicated using the symbol $$\mid$$ which is read as "given." For example, $$P(A\mid B)$$ is read as "Probability of A given B.".
https://online.stat.psu.edu/stat500/lesson/2/2.5.
### Conditional Probability Formula - What is Conditional Probability ....
A conditional probability looks at these two events in relationship with one another, the probability that it is both raining and I will go outside. Let us understand the conditional probability formula using solved examples. Please note that conditional probability doesn't state that there is always a causal relationship between the two events ....
https://www.cuemath.com/conditional-probability-formula/.
### How to Calculate Conditional Probability in R? - GeeksforGeeks.
Jan 04, 2022 . This is the conditional probability of A given that B has already occurred. Similarly, As there can be only 1 red high strength water-type Pokemon card within the high strength water-type Pokemon card already drawn from pack of 50 cards. Example 2: Computation of Conditional Probability. A store owner has a list of 15 customers..
https://www.geeksforgeeks.org/how-to-calculate-conditional-probability-in-r/.
### How to Calculate Conditional Probability in Excel - Statology.
Feb 14, 2020 . The conditional probability that event A occurs, given that event B has occurred, is calculated as follows:. P(A|B) = P(A?B) / P(B) where: P(A?B) = the probability that event A and event B both occur.. P(B) = the probability that event B occurs. This formula is particularly useful when calculating probabilities for a two-way table, which is a table that displays the ....
https://www.statology.org/conditional-probability-excel/.
### Use Conditional Probability to Calculate Intersections - ThoughtCo.
Jun 22, 2018 . The conditional probability of an event is the probability that an event A occurs given that another event B has already occurred. This type of probability is calculated by restricting the sample space that we're working with to only the set B..
https://www.thoughtco.com/compute-probability-of-intersection-3126565.
### Conditional expectation - Wikipedia.
In probability theory, the conditional expectation, conditional expected value, or conditional mean of a random variable is its expected value - the value it would take "on average" over an arbitrarily large number of occurrences - given that a certain set of "conditions" is known to occur. If the random variable can take on only a finite number of values, the "conditions" are that ....
https://en.wikipedia.org/wiki/Conditional_expectation.
### Conditional Probability Calculator - MathCracker.com.
The concept of conditional probability is one of the most crucial ideas in Probability and Statistics. And it is a quite simple idea: The conditional probability of an event $$A$$ given an event $$B$$ is the probability that $$A$$ happens under the assumption that $$B$$ happens as well..
https://mathcracker.com/conditional-probability-calculator.
### Multivariate normal distribution - Wikipedia.
In probability theory and statistics, the multivariate normal distribution, multivariate Gaussian distribution, or joint normal distribution is a generalization of the one-dimensional normal distribution to higher dimensions.One definition is that a random vector is said to be k-variate normally distributed if every linear combination of its k components has a univariate normal ....
https://en.wikipedia.org/wiki/Multivariate_normal_distribution.
### How to Calculate Conditional Probability in Python - Statology.
Aug 24, 2021 . The conditional probability that event A occurs, given that event B has occurred, is calculated as follows:. P(A|B) = P(A?B) / P(B) where: P(A?B) = the probability that event A and event B both occur.. P(B) = the probability that event B occurs. The following example shows how to use this formula to calculate conditional probabilities in Python..
https://www.statology.org/conditional-probability-in-python/.
### Conditional independence - Wikipedia.
In probability theory, conditional independence describes situations wherein an observation is irrelevant or redundant when evaluating the certainty of a hypothesis. Conditional independence is usually formulated in terms of conditional probability, as a special case where the probability of the hypothesis given the uninformative observation is equal to the probability without..
https://en.wikipedia.org/wiki/Conditional_independence.
### Bayes Theorem Definition and Examples - ThoughtCo.
Aug 12, 2019 . Bayes' theorem is a mathematical equation used in probability and statistics to calculate conditional probability. In other words, it is used to calculate the probability of an event based on its association with another event. The theorem is also ....
https://www.thoughtco.com/bayes-theorem-4155845.
### Conditional expectation | Definition, formula, examples - Statlect.
Conditional expectation. by Marco Taboga, PhD. The conditional expectation (or conditional expected value, or conditional mean) is the expected value of a random variable, computed with respect to a conditional probability distribution..
https://www.statlect.com/fundamentals-of-probability/conditional-expectation.
### Conditional Probability Practice Questions – Corbettmaths.
Sep 04, 2019 . The Corbettmaths Practice Questions on Conditional Probability. Videos, worksheets, 5-a-day and much more.
https://corbettmaths.com/2019/09/04/conditional-probability-practice-questions/.
### Probability - Wikipedia.
The word probability derives from the Latin probabilitas, which can also mean "probity", a measure of the authority of a witness in a legal case in Europe, and often correlated with the witness's nobility.In a sense, this differs much from the modern meaning of probability, which in contrast is a measure of the weight of empirical evidence, and is arrived at from inductive reasoning and ....
https://en.wikipedia.org/wiki/Probability.
### Using Venn diagrams for conditional probability - Higher.
of the conditional probability). Out of these 51 pupils, 29 own a laptop. Therefore, the probability that a pupil chosen at random owns a laptop, given that they own exactly one device = \(\frac ....
https://www.bbc.co.uk/bitesize/guides/zsrq6yc/revision/9.
### Modus ponens - Wikipedia.
Explanation. The form of a modus ponens argument resembles a syllogism, with two premises and a conclusion: . If P, then Q.; P.; Therefore, Q. The first premise is a conditional ("if-then") claim, namely that P implies Q.The second premise is an assertion that P, the antecedent of the conditional claim, is the case. From these two premises it can be logically concluded that Q, ....
https://en.wikipedia.org/wiki/Modus_ponens.
### 条件付き確率 - Wikipedia.
??????(?????????????: conditional probability )?????? B ???????????????? A ?????????? ??????? P(A|B) ??? P B (A) ???????? ? ?????? P(A|B) ?????? B ???????? A ?(????)?????? B ???? ....
https://ja.wikipedia.org/wiki/%E6%9D%A1%E4%BB%B6%E4%BB%98%E3%81%8D%E7%A2%BA%E7%8E%87.
### About | Statistics 110: Probability - Harvard University.
Stat 110 is an introduction to probability as a language and set of tools for understanding statistics, science, risk, and randomness. The ideas and methods are useful in statistics, science, engineering, economics, finance, and everyday life. Topics include the following. Basics: sample spaces and events, conditioning, Bayes' Theorem..
The recombination landscape of the Khoe-San likely represents the upper limits of recombination divergence in humans. Authors: Gerald van Eeden, Caitlin Uren, Evlyn Pless, Mira Mastoras, Gian D. van der Spuy, Gerard Tromp, Brenna M. Henn and Marlo Moller.
https://genomebiology.biomedcentral.com/.
### B1 Grammar: Third Conditional - Exam English.
Free Practice Tests for learners of English. Third Conditional . B1 Grammar topics.
https://www.examenglish.com/grammar/B1_3rd_conditional.htm.
### CME 106 - Probability Cheatsheet - Stanford University.
Conditional probability. Bayes' rule Independence. Random Variables. Probability density function Cumulative distribution function. Expectation and moments. Expected value Characteristic function Transformation. Probability distributions. Chebyshev's inequality Main distributions. Joint random variables..
https://stanford.edu/~shervine/teaching/cme-106/cheatsheet-probability.
### Addition Rule of Probability - Math Goodies.
Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event. P(A or B) = P(A) + P(B) Let's use this addition rule to find the probability for Experiment 1. Experiment 1: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?.
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# Homework Help: Minimum Angular Velocity Problem
1. Dec 9, 2013
### Clearik
First off, I want to apologize for posting more than one question. I just discovered this site, so I wanted to check my work while I am able to. Thank you again.
1. The problem statement, all variables and given/known data
A massless rope of length L = 1 m is swung in the vertical plane, with a bob of mass m = 1 kg attached to its end.
(a) Calculate the minimum angular velocity ωmin that the bob of must have to keep the rope taut at every point in the trajectory.
(b) Assume now that the bob starts from the lowest position (θ = 0) at rest, and the angular velocity follows
ω = 2 x t rad s-1
Calculate the time at which the bob reaches the top position.
2. Relevant equations
Fc = (m)(v2/R)
R = L
Fc = (m)(v2/L)
v = (L)(ω)
Fc = (m)((Lω)2/L)
Fc = (m)((L)2(ω)2/L)
Fc = (m)((L)(ω)2)
At the top of the circle:
(m)((L)(ω)2) = (m)(g)
((L)(ω)2) = (g)
ω = √(g/L)
3. The attempt at a solution
(a) ω = √(10/1)
Did I go about doing this correctly?
(b) I wasn't quite sure on how to start this part. I know that 1 radian is half a revolution. Therefore, to reach the top from the bottom (half a revolution), the time it takes t would have to be how long it takes to go one radian. Plugging in .5 yields 1 in the equation given in part b, but I don't know if this is correct.
Last edited: Dec 9, 2013
2. Dec 9, 2013
### mic*
1 radian does not equal half of one revolution.
∏ radians equals one half of one revolution.
Beyond that point, what exactly is the question in part b?
3. Dec 9, 2013
### Clearik
Oh, sorry, I forgot to add it, it's there now.
4. Dec 9, 2013
### mic*
Hmmm, I'm not the greatest with pendulum stuff.
Perhaps, since you know that;
2 x t rad s-1 = ω = √(g/L)
You might be able to get a result from that?
5. Dec 9, 2013
So perhaps;
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# Statistics – Grade 10 CBSE
There are 2 topics that are covered under this section:
1. Statistics – Introduction
2. Probability
Click on each topic to know more details.
Statistics
Key topics include:
• Definitions
• Data
• Information
• Observations
• Fields
• Types of Data
• Discrete or Individual Data
• Grouped Data
• Continuous Data
• Measurement of Central Tendency
• Introduction
• Mathematical & Positional Central Values
• Measurement of Arithmetic Mean
• Direct Method for Individual Data
• Assumed Mean Method for Individual Data
• Step Deviation Method for Individual Data
• Direct Method for Grouped Data
• Assumed Mean Method for Grouped Data
• Step Deviation Method for Grouped Data
• Calculation of Mean for Continuous Data
• Properties of Arithmetic Mean
• Measurement of Median
• Definition
• Median for Individual Data
• Median for Grouped Data
• Median for Continuous Data
• Measurement of Mode
• Definition
• Mode for Individual Data
• Mode for Grouped Data
• Relationship Between Mean, Median & Mode
• Cumulative Frequency Curve (Ogive)
• For a Less Than Series
• For a Greater Than Series
Probability
Key topics include :
• Trials & Experiments
• Trial
• Experiment
• Random Experiment
• Events
• Elementary Event
• Certain Event
• Impossible Event
• Compound Event
• Complement Event
• Sample Space
• Set of Favours (Set of Favourable Events)
• Exhaustive Events
• Mutually Exclusive Events
• Independent Events
• Types of Probabilities
• Classical (Mathematical) Probability
• Empirical (Experimental) Probability
• Limits of Probability
### Coordinate Geometry – Grade 10 CBSE
The theory section covers in detail each of the topics mentioned above along with...
### Geometry – Grade 10 CBSE
There are couple of topics covered in this section: 1. Circles 2. Triangles Click...
### Trigonometry – Grade 10 CBSE
There are few topics covered in this section: 1. Trigonometry – Introduction 2....
### Arithmetic – Grade 10 CBSE
As per CBSE syllabus, the only chapter under this topic is Number System. Key topics...
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3.7 Implicit Differentiation
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# 3.7 Implicit Differentiation - PowerPoint PPT Presentation
3.7 Implicit Differentiation. Implicitly Defined Functions How do we find the slope when we cannot conveniently solve the equation to find the functions?
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## 3.7 Implicit Differentiation
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1. 3.7 Implicit Differentiation • Implicitly Defined Functions • How do we find the slope when we cannot conveniently solve the equation to find the functions? • Treat y as a differentiable function of x and differentiate both sides of the equation with respect to x, using the differentiation rules for sums, products, quotients, and the Chain Rule. • Then solve for dy/dx in terms of x and y together to obtain a formula that calculates the slope at any point (x,y) on the graph from the values of x and y. • The process is called implicit differentiation.
2. Differentiating Implicitly • Find dy/dx if y² = x. • To find dy/dx, we simply differentiate both sides of the equation and apply the Chain Rule.
3. Finding Slope on a Circle • Find the slope of the circle x² + y² = 25 at the point (3 , -4).
4. Solving for dy/dx • Show that the slope dy/dx is defined at every point on the graph 2y = x² + sin y. The formula for dy/dx is defined at every point (x , y), except for those points at which cos y = 2. Since cos y cannot be greater than 1, this never happens.
5. Lenses, Tangents, and Normal Lines • In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry.
6. Lenses, Tangents, and Normal Lines • This line is called the normal to the surface at the point of entry. • In a profile view of a lens like the one in Figure 3.50, the normal is a line perpendicular to the tangent to the profile curve at the point of entry. • Profiles of lenses are often described by quadratic curves. When they are, we can use implicit differentiation to find the tangents and normals.
7. Tangent and Normal to an Ellipse • Find the tangent and normal to the ellipse x2 – xy + y2 = 7 at the point (-1 , 2). • First, use implicit differentiation to find dy/dx:
8. Tangent and Normal to an Ellipse • We then evaluate the derivative at x = -1 and y = 2 to obtain: • The tangent to the curve at (-1 , 2) is: • The normal to the curve at (-1 , 2) is:
9. Finding a Second Derivative Implicitly • Find d²y/dx² if 2x³ - 3y² = 8. • To start, we differentiate both sides of the equation with respect to x in order to find y’ = dy/dx.
10. Finding a Second Derivative Implicitly • We now apply the Quotient Rule to find y”. • Finally, we substitute y’ = x²/y to express y” in terms of x and y.
11. Rational Powers of Differentiable Functions
12. Using the Rational Power Rule (a) (b) (c)
13. More Practice!!!!! • Homework – Textbook p. 162 #2 – 42 even.
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# Trying to prove consistency, but getting non-sensical limit probabilities
I have
$$X_1 \dots X_n \sim f_\theta(x) = \begin{cases} \exp(\theta-x) & x\geq\theta\\ 0& otherwise \end{cases}$$
And I have the estimator $\hat\theta_n = X_{(1)}=\min\{X_1 \dots X_n\}$
I have found the CDF and PDF of the estimator
$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)}$$ $$f_{\hat\theta_n}(x) = ne^{n(\theta-x)}$$
Now I want to test consistency so for $\epsilon >0$
$$\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) = 1$$
Then we have $$\Pr(-\epsilon < \hat\theta_n - \theta < \epsilon)$$ $$\Pr(\theta-\epsilon < \hat\theta_n < \theta + \epsilon)$$
$= 1-e^{-n\epsilon}-1+e^{n\epsilon}$
Which goes to $+\infty$ as $n\to\infty$. Did I do everything correctly?
What do I conclude from this? Is the estimator consistent or not? Is this the same as the probability going to $1$?
EDIT: I believe I have found my error. The CDF should be
$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)} 1_{x \geq \theta}$$
Then you get the probability: $1-e^{-n\epsilon}$ which goes to $1$ as expected.
• The fact is, the probability that $\hat{\theta}_n$ is less than $\theta$ is $0$. Commented Oct 20, 2017 at 2:40
## 2 Answers
It may be more convenient to work with $P(|\hat{\theta}_n - \theta| \geq \varepsilon)$ (the reason will be clear as the computation flows): \begin{align*} & P(|\hat{\theta}_n - \theta| \geq \varepsilon) \\ = & P(|X_{(1)} - \theta| \geq \varepsilon) = P(X_{(1)} - \theta \geq \varepsilon) \\ = & P(X_{(1)} \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon, \ldots, X_n \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon)^n \quad \text{by the i.i.d. assumption} \\ = & \left(\int_{\theta + \varepsilon}^\infty e^{\theta - x} dx \right)^n \\ = & e^{-n\varepsilon} \to 0 \end{align*} as $n \to \infty$. This shows that $\hat{\theta}_n$ is a consistent estimate of $\theta$.
I think you have to write $\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) =\lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon)$ since $\theta$ is the minimum value of $\hat{\theta}$.
Then you can show that \begin{align*} \lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon) &= lim_{n\to\infty}\left \{F_{\hat{\theta}}(\theta+\epsilon)-F_{\hat{\theta}}(\theta)\right \} \\&=lim_{n\to\infty}\left \{[1-e^{n[\theta-(\theta+\epsilon)]}]-[1-e^{n(\theta-\theta)}]\right \}\\&=lim_{n\to\infty}\left \{1-e^{-n\epsilon}-1+1\right\}\\&=1 \end{align*}
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# RSA keys are multiplicative inverse in mod phi(n), but also in mod n?
I understand that the RSA keys $pk$ and $sk$ are choosen such that one is the multiplicative inverse of the other, in $\mod \phi(n)$
But for the encryption and decryption to work, in other words, that
${m^{pk}}^{sk} \mod n = m$
holds, $pk$ and $sk$ must also be multiplicative inverse to each other in $\mod n$, right?
• Why do you think this must be the case? Commented Apr 6, 2017 at 14:07
• (Of course, this is obviously false; just generate a keypair and you will see that decryption works even though the keys are not inverse modulo $n$.) Commented Apr 6, 2017 at 14:10
I understand that the RSA keys $pk$ and $sk$ are chosen such that one is the multiplicative inverse of the other, in $\bmod \phi(n)$
Mostly correct. Actually, they're always multiplicative inverses modulo $\lambda(n) = \phi(n) / \gcd(n)$; selecting them as inverses modulo $\phi(n)$ does work (they will encrypt and decrypt properly), but using the smaller $\lambda(n)$ also works, and yields smaller key values.
$pk$ and $sk$ must also be multiplicative inverse to each other in $\bmod n$, right?
That is wrong; $pk \cdot sk \bmod n$ need not be any specific value; there's no reason it needs to be 1. One trivial example is $n = 85, e = 5, d = 13$; in this case, $e$ is not relatively prime to $n$, hence it doesn't have a multiplicative inverse.
After all, when you look at ${m^{pk}}^{sk} \bmod n$, you're not multiplying by $pk$ or $sk$; you're raising $m$ to the power of $pk, sk$; those are different operations.
• Hmm it looks like I am still misunderstanding something crucial. I mean..${m^{pk}}^{sk}$ is the same as $m^{sk*pk}$, and that must be $m^1$, so $pk * sk$ must be 1, right?This is why I thought pk must be the inverse of sk. Commented Apr 6, 2017 at 14:25
• @user66875: yes, however $m ^ {pk \cdot sk} = m ^ {pk \cdot sk \bmod \lambda(n)}$, and so the idea that they are inverses modulo $\lambda(n)$ already captures that notion... Commented Apr 6, 2017 at 14:27
Theorem. Let $a$ be an element of a finite group $G$ and $b,c$ be integers. Then $a^b = a^c$ in $G$ if and only if $b$ and $c$ are congruent modulo the order of $a$ in $G$ (that is, the smallest positive integer $k$ such that $a^k = 1$).
In the special case where $G$ is the multiplicative groups of integers modulo $n$, the order of $a$ is a divisor of $\varphi(n)$ (the order of the group), thus in order to have $(a^\mathrm{pk})^\mathrm{sk} = a^{\mathrm{pk}\cdot\mathrm{sk}} = a$ it is sufficient to have $\mathrm{pk}\cdot\mathrm{sk} \equiv 1 \pmod{\varphi(n)}$. This does not imply anything about how congruent they are modulo $n$.
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World Observer
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Transcript of World Observer
World Observer
Solution
1. 1 Meter = 100cm
15mm = 1.5 cm
100/4 = 25
1.5 x 2 = 3
25 - 3 = 22
22 x 22 = 484cm
So the area of the painting is 484cm.
2. We know the length of the frame is 25cm.
the width is 1.5cm. Find the area of each corner.
1.5 x 1.5 = 2.25. Multiply the corner area by 4 to find the area of all 4 corners.
2.25 x 4 = 9. Now we just need the area of the long middle pieces so we could add the areas.
25 - 3 = 22cm. 22 x 1.5 = 33
33 x 4 = 132.
132 + 9 = 141. The area of the frame is 141cm.
Formulas You'll Need to Know
Problem
John went to his local art museum. He found a
square
painting with a
square
frame and was told the perimeter of the frame is 1 Meter long,
and the width of the frame is 15mm.
1.Help John find the area of the painting.
If you finish #1 early:
2.Help John find the area of the frame
World Observer
Math is used in a multitude of ways in our world. Here is a problem on how you can use math in life.
A perimeter is the length of all sides added together. Ex.) Perimeter of a 5cm square is 20 (5+5+5+5)
Metric System:
10mm = 1cm
100cm = 1 M
Area of a square or rectangle = L x W
Full transcript
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# What is 300 grams of flour in cups?
## What is 300 grams of flour in cups?
Volume of 300 Grams of All Purpose Flour
300 Grams of All Purpose Flour =
2.07 U.S. Cups
1.72 Imperial Cups
1.96 Metric Cups
489.49 Milliliters
## How much is 300 grams sugar in cups?
Volume of 300 Grams of Sugar
300 Grams of Sugar =
1.50 U.S. Cups
1.25 Imperial Cups
1.42 Metric Cups
354.88 Milliliters
How much grams are in a cup?
More knowledge from the unit converter How many grams in 1 cups? The answer is We think you are changing between gram [water] and cup [US]. You can view extra main points on every dimension unit: grams or cups The SI derived unit for volume is the cubic meter.
How do I measure 300 grams of flour?
If we speak about all-purpose flour, then 300 grams are equivalent to 2-3/8 cups.
### How much is 100 grams of flour in cups?
Volume of 100 Grams of All Purpose Flour
100 Grams of All Purpose Flour =
0.69 U.S. Cups
0.57 Imperial Cups
0.65 Metric Cups
163.16 Milliliters
### How much is 300 grams of flour in ounces?
Common Gram to Ounce Conversion Table
Grams Ounces
150
200 7
250 9
300 11
What’s 250g of flour in oz.?
8
Is 250 grams equal to 8 ounces?
An ounce is a unit of weight equal to one/Sixteen th of a pound or about 28.35 grams. We conclude that 250 grams is an identical to eight.oz: 250 grams = 8.oz. The ounce (abbreviation: oz) is a unit of mass with several definitions, probably the most popularly used being equal to roughly 28 grams.
125 grams
## How many grams are in a 8 oz. cup?
Honey, Molasses & Syrup
Cups Grams Ounces
1/Four cup 85 g Three oz.
1/Three cup 113 g 4 oz
1/2 cup a hundred and seventy g 6 oz.
2/Three cup 227 g 8 ounces
How many cups is 8 oz of flour?
Flour Weight to Volume Conversion Table
Ounces Cups (A.P. Flour) Cups (Cake Flour)
Eight oz 1 3/Four c 2 1/Four c
9 oz. 2 1/16 c 2 1/2 c
10 oz. 2 1/4 c 2 3/Four c
Eleven oz 2 1/2 c Three 1/8 c
How many cups is 14 oz. sweetened condensed milk?
Most are 14 oz now. I take advantage of this for sticky liquids reminiscent of condensed milk, jam, honey, and many others. alter a cup . I’ve Three sizes, but use the two cup probably the most….
How Do You Measure Sweetened Condensed Milk?
Author Message
306 grams
### How many cups is 7 ounces of sweetened condensed milk?
Condensed milk Conversion Chart Near 6.8 oz.
ounces to US cups of Condensed milk
7 oz. = 0.649 ( 5/8 ) US cup
7.1 oz. = 0.658 ( 5/8 ) US cup
7.2 oz. = 0.667 ( 5/8 ) US cup
7.3 oz = 0.677 ( 5/8 ) US cup
What is 200g of condensed milk in cups?
150 grams of condensed milk equals to 0.49 ( ~ 1/2 ) US cup. (*) or more precisely 0.US cup. All figures are approximate….Gram to US cup Conversion Chart Near 150 grams.
Grams to US cups Conversion Chart
200 grams 0.654 US cup
210 grams 0.686 US cup
220 grams 0.719 US cup
230 grams 0.752 US cup
Is it OK to make use of expired sweetened condensed milk?
Does sweetened condensed milk pass dangerous? Sweetened condensed milk will pass bad in the end, but it will ultimate a good year beyond any printed date. Eagle Brand puts a “best possible via” date on their cans for 2 years beyond production, but the product is good past this date.
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A082552 Number of sets of distinct primes, the greatest of which is prime(n), whose arithmetic mean is an integer. 1
1, 1, 2, 5, 6, 12, 21, 31, 58, 111, 184, 356, 665, 1223, 2260, 4227, 7930, 15095, 28334, 53822, 102317, 195012, 373001, 714405, 1370698, 2633383, 5067643, 9765457, 18846711, 36413982, 70431270, 136391723, 264384100, 512959093, 996173830 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS The sum of the first 23 primes gives 874 = 23*38, see A045345. - Alois P. Heinz, Aug 02 2009 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..100 EXAMPLE a(4) = 5: prime(4) = 7 and the five sets are (5+7)/2 = 6, 7/1 = 7, (3+7)/2 = 5, (2+3+7)/3 = 4, (3+5+7)/3 = 5. MAPLE b:= proc(t, i, m, h) option remember; if h=0 then `if` (t=0, 1, 0) elif i<1 or h>i then 0 else b (t, i-1, m, h) +b((t+ithprime(i)) mod m, i-1, m, h-1) fi end: a:= n-> add(b(ithprime(n) mod m, n-1, m, m-1), m=1..n): seq (a(n), n=1..40); # Alois P. Heinz, Aug 02 2009 MATHEMATICA f[n_] := Block[{c = 0, k = n, lst = Prime@ Range@n, np = Prime@n, slst}, While[k < 2^n, slst = Subsets[lst, All, {k}]; If[Last@slst == np && Mod[Plus @@ slst, Length@slst] == 0, c++ ]; k++ ]; c]; Do[ Print[{n, f@n} // Timing], {n, 24}] (* Robert G. Wilson v *) CROSSREFS Cf. A051293, A072701. Sequence in context: A108365 A064765 A257805 * A243798 A057683 A277012 Adjacent sequences: A082549 A082550 A082551 * A082553 A082554 A082555 KEYWORD nonn AUTHOR Naohiro Nomoto, May 03 2003 EXTENSIONS a(22)-a(24) from Robert G. Wilson v, Jan 19 2007 Corrected a(23) and extended by Alois P. Heinz, Aug 02 2009 STATUS approved
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0
# What is the center of this hyperbola -4x2 plus 9y2 plus 16x-18y plus 29 equals 0?
Updated: 9/19/2023
Wiki User
13y ago
-4x2 + 9y2 + 16x - 18y +29 = 0 (I hope this is what you meant)
9y2 - 18y - (4x2 -16x) +29 = 0 (group together the x's and y's, and make the x2 term positive.
9(y2 - 2y) - 4(x2 - 4x) +29 = 0 (remove common factors)
9(y - 1)2 - 1 - 4(x - 2)2 - 4 + 29 = 0 (complete the squares)
9(y - 1)2 - 4(x - 2)2 = -25 (move constants to right-hand side)
[(y - 1)2] - 4[(x - 2)2]/9 = -25/9 (divide by coefficient in front of x or y bracket, in this case 9)
[(y - 1)2]/4 - [(x - 2)2]/9 = -25/36 (divide by coefficient in front of other bracket, in this case 4)
As you can see, we now have our equation in standard form. Our center points satisfy [(x - 2 = 0) , (y - 1 = 0)], thus our center is (2,1).
Wiki User
13y ago
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Related questions
### What 2xy-3y plus 1 equals 0?
It's an hyperbola equation.
hyperbola
### Is y2-5y plus 2x-x2-120 equals 0 a circle or something else?
y2 - 5y + 2x - x2 - 120 is not a circle. It is a hyperbola rotated through 90 degrees.
56
### Find the center and radius of x2 plus y2 equals 49?
Center is (0, 0) . . . the origin.Radius is 7.
### Is the center of the circle given by the equation x plus 2 2 plus y - 52 equals 25?
At the center, (x, y) = (-2, 5)
### Is the center of the circle given by the equation x plus 2 2 plus y - 5 2 equals 25?
At the center, (x, y) = (-2, 5)
### What is the center of the circle given by the equation x plus 2 2 plus y - 5 2 equals 25?
At the center, (x, y) = (-2, 5)
### Is y2-5y plus 2x-x2-120 equals 0 a circle or another conic?
This not a circle, both the squared terms must have the same coefficients if it might be a circle. Different signs indicate a hyperbola.
(-4,-6)
(-7,5)
(-4,3)
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Ex.2. ABC is a triangle. The bisect...
Question
Ex.2. ABC is a triangle. The bisectors of the internal angle ZB and external angle ZC intersect at D. If ZBDC= 50°, then ZA is (1) 100° (2) 90° (3) 120° (4) 60°
11th - 12th Class
Maths
Solution
107
4.0 (1 ratings)
So (2) ( A B=A C=A D ) ( Rightarrow angle mathrm{ABC}=angle mathrm{ACB}=30^{circ} ) ( Rightarrow angle B A C=180^{circ}-60^{circ}=120^{circ} ) ( mathrm{Now}, angle mathrm{DAC}=180^{circ}-120^{circ}=60^{circ} ) ( Rightarrow angle A D C+angle A C D=120^{circ} ) ( therefore angle A C D=frac{120^{circ}}{2}=60^{circ} ) ( therefore angle mathrm{BCD}=angle mathrm{ACB}+angle mathrm{ACD} ) ( =30^{circ}+60^{circ}=90^{circ} )
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# Tape Equilibrium
I picked the first test (Tape Equilibrium) from Codility here.
Question:
A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non−empty parts:
A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of:
|(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.For example, consider array A such that:
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
We can split this tape in four places:
P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7
Write a function: int solution(int A[], int N); that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
This is how I implemented it. I got 50% with complexity N*N. How could I make it cleaner?
// you can also use imports, for example:
import java.math.*;
import java.util.*;
import java.lang.*;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 7
int sizeOfArray = A.length;
int smallest = Integer.MAX_VALUE;
int result = 0;
for(int i=1;i<sizeOfArray;i++){
int difference = Math.abs(sumOfArray(subArray(0,i,A))-
sumOfArray(subArray(i,sizeOfArray,A)));
//System.out.println("difference"+difference);
result = Math.min(smallest,difference);
smallest = result;
}
return result;
}
public int sumOfArray(int[] arr) {
int sum=0;
for(int i:arr) {
sum += i;
}
return sum;
}
public int[] subArray(int begin, int end, int[] array) {
return Arrays.copyOfRange(array, begin, end);
}
}
Naming
The Java naming convention for arguments is camelCase. You should rename this argument A here:
public int solution(int[] A)
It would be best to make it look like the following code, or use a better name that would fit your needs:
public int solution(int[] arrayA)
I don't have anything against a variable named arr for an int [], but you're using arr and array in different methods. I would recommend to stick to one name, or try to find less generic name if they mean different things.
Formatting
Your formatting is very good in general, and you're consistent. Some times, you could use a little bit of white-space.
for(int i=1;i<sizeOfArray;i++)
You could add some spaces to clearly define the three part of the for-loop:
for(int i=1; i<sizeOfArray; i++)
I will not evaluate your algorithm, since this is not my forte.
• I would have gone for array Mar 14 '14 at 17:51
You should be able to do it in $O(n)$ time and $O(1)$ space. Keep a running total of the left sum and the right sum. As you test each $p$, deduct a number from one side and credit it to the other.
public static int minDiff(int[] a) {
int leftSum = 0, rightSum = 0;
for (int ai : a) {
leftSum += ai;
}
int minDiff = Integer.MAX_VALUE;
for (int p = a.length - 1; p >= 0; p--) {
rightSum += a[p];
leftSum -= a[p];
int diff = Math.abs(leftSum - rightSum);
if (diff == 0) {
return 0;
} else if (diff < minDiff) {
minDiff = diff;
}
}
return minDiff;
}
• Basically, the same solution as @konijn's. Mar 14 '14 at 18:31
• This solution will not work when the input contains negative values... The sum of the two sides will not be increasing Mar 14 '14 at 18:31
• @rolfl Could you provide a counterexample? Mar 14 '14 at 18:32
• Never mind, I misread your description vs your implementation. Your description says 'keep a running total of the left sum', but that is not what your code does, your code calculates the complete total of all values, and then subtracts things. Mar 14 '14 at 18:37
• Except that my last Java code was in 1.4, so I wrote some JavaScript ;) Mar 14 '14 at 18:39
From a once over,
you seem to call sumofArray a number of times, you should only have to call it once.
At the very beginning, you call it once for the entire array, results should be 13
Then for position 0 (value 3), you substract 3 and get 3 and 10 ( 13 - 10)
Then for position 1 (value 1). you add 1 to 3 and subtract 1 from 10 giving 4 and 9
Then for position 2 (value 2), you add 2 to 4 and substract 2 from 9 giving 6 and 7
This way you access all elements twice, if I count correctly.
As I mentioned in a comment, I would rename arr -> array
I am no Java expert, but JavaScript is close enough that you should be able to follow:
var A = [3,1,2,4,3];
function sumArray( array )
{
var sum = 0, index = array.length;
while(index--)
sum += array[index];
return sum;
}
function tapeEquilibrium( array )
{
var left = sumArray( array ),
right = 0,
smallest = left,
index = array.length,
difference;
while( index-- )
{
right += array[index];
left -= array[index];
difference = Math.abs( right-left );
if( difference < smallest )
smallest = difference;
}
return smallest;
}
console.log( tapeEquilibrium( A ) );
The added advantage is that very little extra memory is required when very large arrays need to be examined.
The trick to this problem is in the algorithm. A solution of $O(n)$ complexity is available if you process the data in a 'clever' way. Consider the following algorithm:
• create a new array of the same size, call it sum
• populate the sum array with the sum of all values to the left in the original data array
• Once the sum array is fully populated, you will know what the total sum is for the array.
• this allows you to determine what the balance-point-sum is, it will be half of the total.
• you may be tempted to just binary search the sum array for the place that is half the total, but this will fail if there are negative values in the input array.
• the only solution is to scan the sums looking for half the value, with some short-circuit if there is an exact half found.
Putting it together as code, it looks like:
public static final int tapeEquilibrium(int[] data) {
if (data.length < 3) {
// rules indicate 0 < P < N which implies at least 3-size array
throw new IllegalStateException("Need minimum 3-size array input");
}
int[] sums = new int[data.length];
for (int i = 1; i < sums.length; i++) {
sums[i] = sums[i - 1] + data[i - 1];
}
int total = sums[sums.length - 1] + data[data.length - 1];
int min = Integer.MAX_VALUE;
for (int i = 0; i < sums.length; i++) {
int diff = Math.abs((total - sums[i]) - sums[i]);
if (diff == 0) {
return 0;
}
if (diff < min) {
min = diff;
}
}
return min;
}
Just a minor note to add about the original code which was not mentioned earlier:
The code uses subArray only for summarizing:
int difference = Math.abs(sumOfArray(subArray(0,i,A))-
sumOfArray(subArray(i,sizeOfArray,A)));
It would be faster with the original array:
public int sumOfArray(int[] array, int begin, int end) {
int sum = 0;
for (int i = begin; i < end; i++) {
sum += array[i];
}
return sum;
}
+1: An extra tab in the second line above the would make it clear that it's one statement:
int difference = Math.abs(sumOfArray(subArray(0,i,A))-
sumOfArray(subArray(i,sizeOfArray,A)));
This makes the code easier to read (and maintain).
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Prime Perimeter (Posted on 2005-03-11)
You and two other people have numbers written on your foreheads. You are all told that the three numbers are primes and that they form the sides of a triangle with a prime perimeter. You see 5 and 7 on the other two heads and both of the other people agree that they cannot deduce the number on their own foreheads.
See The Solution Submitted by Erik O. Rating: 2.8235 (17 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): full solution as understood | Comment 21 of 89 |
(In reply to re: full solution as understood by Sachin)
"Scenario 1. A=5, B =5 and C =7
In this case C will see 2 fives which means the only possible values for C are 3 or 7. If C's value is 3, B could have guessed his numbers as he is seeing 3 and 5, and only possible third number is 5. But B couldn't, which leaves C's number as 7. Since even C could not guess his number as 7. So scenario 1 is not possible."
But if B were seeing 3 and 5 he could assume his own was either 3 or 5, not just 5.
"Scenario 2. A = 7, B =5 and C =7
In this case B will see 2 sevens which means the only possible values for B are 3 or 5. If B's value is 3, C could have guessed his number as he is seeing 3 and 7, and only possible third number is 7. But C couldn't, which leaves B's number as 5. Since even B could not guess his number as 5. So scenario 2 is also not possible."
If C were seeing 3 and 7, then 3 would be another possibility for C to consider possible besides the 7.
So is still think it would take at least two rounds of thinking, that included the third person (me, you, depending on who's reading--Yes I mistakenly said narrator before--I meant the person doing this thinking and coming to a conclusion).
Posted by Charlie on 2005-03-12 03:01:36
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Soil Fertility Spotlight: Know how many pounds of actual nitrogen you’re applying with any source.
Tom J. Bechman, Editor, Indiana Prairie Farmer
January 17, 2023
REAL RATE: If you’re applying 30 gallons of 28% N in this pass, how many pounds of N are you applying? Based on the table below, you would be applying 90 pounds per acre.Tom J. Bechman
Two farming partners reached a revelation. They applied about 20 pounds of nitrogen per acre less in 2022 than they thought. Why? Because the partner determining rates made a basic error in converting gallons of liquid nitrogen to pounds of nitrogen per acre.
Tight margins don’t allow room for error. This new series will focus on basic key questions about fertilizer that deserve solid, straightforward answers. Jim Camberato, Purdue Extension soil fertility specialist, provides answers in this inaugural article. To submit questions, email [email protected] or mail them to 599 N., 100 W., Franklin, IN 46131.
How can you calculate pounds of nitrogen supplied from gallons of liquid fertilizer? The percentage of the nutrient and the density of the liquid in pounds per gallon are the only two pieces of information needed to calculate how many pounds of nutrient are in a gallon of liquid fertilizer. Refer to the table to see percentages of nitrogen and density in 28% N and 32% N.
For example: Consider 28% urea ammonium nitrate. It is 28% N and has a density of 10.6 pounds per gallon. To find the amount of actual N per gallon of 28%, simply multiply 28 by 10.6, then divide by 100 to get pounds per gallon of 28%. The math is: 28 x 10.6 / 100 = 2.97. For practical purposes, round it off to 3 pounds of N per gallon, as noted in the table.
So, if I apply 20 gallons of 28% liquid N, how many pounds am I applying? It’s straightforward math. If you apply 20 gallons times 3 pounds per gallon, you applied 60 pounds per acre. Technically, it’s 59.4 pounds of actual N, but 60 is likely close enough for your purposes.
I apply 32% liquid N. What does the math look like for me? The density is 11.06 pounds per gallon. So, the math is: 32 x 11.06 / 100 = 3.54, or about 3.5 pounds per gallon. If you applied the same 20 gallons with 32% N, you would apply about 70 pounds per acre.
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# Normal, self-adjoint, or neither linear operator
• Apr 11th 2008, 08:35 AM
Normal, self-adjoint, or neither linear operator
In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$
Is T normal, self-adjoint, or neither?
In $M_{2x2} ( \mathbb {R} )$, define T by $T(A) = A^t$
Same problem here.
• Apr 11th 2008, 10:30 AM
Opalg
Quote:
In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$
Is T normal, self-adjoint, or neither?
In $M_{2x2} ( \mathbb {R} )$, define T by $T(A) = A^t$
Same problem here.
I don't know how to find T* from here.
The adjoint operator is given by $\langle T^*f,g\rangle = \langle f,Tg\rangle$. If Tg = g', and the inner product is given by integration over the unit interval, then $\langle f,Tg\rangle = \int _{0}^{1} f(t)g'(t)\,dt$. You want to write this in the form $\int _{0}^{1} ???g(t)\,dt$, where "???" will be T*(f).
• Apr 11th 2008, 01:23 PM
Can I say $T^*(f) = \frac {f(t)g(t)}{g'(t)}$?
And for the second one, I have:
$ = $
$==tr(B^t,A^t)$
and $==tr(???^t,A)$
Then I'm bit lost, how I can find something that would fit ???
• Apr 12th 2008, 09:30 AM
Opalg
For the second one, you're almost there. If the inner product is given by $\langle A,B\rangle = \text{tr}(B^{\textsc T}A)$, and $T(A) = A^{\textsc T}$, then $\langle T(A),B\rangle = \text{tr}(B^{\textsc T}A^{\textsc T})$. But this is the same as $\langle A,T(B)\rangle$ (since $\text{tr}(M) = \text{tr}(M^{\textsc T})$). So T is selfadjoint.
The first one is a good deal more complicated than I first thought, and I wonder if there is something wrong with the question. My original idea was to use integration by parts to express $\textstyle\int fg'$ in terms of $\textstyle\int f'g$. But when you do the integration by parts, with the limits 0 and 1, you actually get $\langle f',g\rangle = f(1)g(1)-f(0)g(0) - \langle f,g'\rangle$. The awkward terms f(1)g(1)-f(0)g(0) prevent you from getting a neat expression for the adjoint of the differentiation operator. That seems to have the effect of making this question unreasonably difficult.
• Apr 15th 2008, 05:24 AM
I asked my professor, and he said we should use this theorem:
T is normal if and only if there exist an orthonormal basis for V consisting of eigenvectors of T.
So, to find eigenvectors of T, I have $T(f)=f'= \lambda f$ So I need to find some polynomials f, are there any?
the standard ordered basis of V is $\{ 1,x,x^2 \}$
So the matrix $[T]_{ \beta } = \begin{pmatrix} 0 && 0 && 0 \\ 0 && 1 && 2 \\ 0 && 0 && 0 \end{pmatrix}$
So the eigenvalue of T is 0,1.
Well, then any polynomial to the first power would satisfy T, right?
• Apr 15th 2008, 11:04 AM
Opalg
Quote:
In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$
Is T normal, self-adjoint, or neither?
Just to be clear about this, I assume that $P_{2}( \mathbb {R} )$ means polynomials of degree at most 2 over the real numbers.
Quote:
I asked my professor, and he said we should use this theorem:
T is normal if and only if there exist an orthonormal basis for V consisting of eigenvectors of T.
Interesting suggestion. I hadn't thought of doing it that way. There's a snag, though. That theorem works if the scalars are the complex numbers, but you have to be careful how to use it if the scalars are the reals. For example, the matrix $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is normal, but has no real eigenvalues or eigenvectors.
To use the theorem here, you must first show that the complex space $P_{2}( \mathbb {C} )$ does not have a base consisting of eigenvectors of T, and then deduce that since T is not normal as an operator on that space it must also fail to be normal when we restrict to real scalars.
Quote:
So, to find eigenvectors of T, I have $T(f)=f'= \lambda f$ So I need to find some polynomials f, are there any?
the standard ordered basis of V is $\{ 1,x,x^2 \}$
So the matrix $[T]_{ \beta } = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ Not quite; the 1 should be at the bottom of the middle column.
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http://narviacademy.in/Topics/Competitive%20Exams/Quantitative%20Aptitude/Quantitative%20Aptitude%20Notes/permutation-and-combination-quantitative-aptitude-notes.php
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# Permutation and Combination
Factorial: The factorial of an integer is the product of all less than or equal positive integers. It is denoted by $$n!$$, where n is any positive integer. The factorial of a negative integer is not possible.$$n! = n \ (n - 1) \ (n - 2)......3 \times 2 \times 1$$
Example: $$4! = 4 \times 3 \times 2 \times 1 = 24$$
Example: $$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$= 720$$
Note: Remember, $$0! = 1$$, $$1! = 1$$.
Permutation: The arrangement of a number of objects in an ordered manner is called permutation. Here 'order' is important. It is represented by $$nP_r$$.$$nP_r = \frac{n!}{(n - r)!}$$
Where, n = Total number of objects.
P = Permutation.
r = Number of selected objects at a time from the total number of things.
For example let there are 2 Chocolates of different brands for two children P and Q, then Chocolates can be distributed in $$2!$$ ways PQ and QP.$$2! = 2 \times 1 = 2 \ ways$$
Example(1): If there are one Apple, one Banana, and one Orange for three boys P, Q, and R, then three fruits can be distributed in $$3!$$ ways.$$3! = 3 \times 2 \times 1 = 6 \ ways$$ 6 ways are PQR, PRQ, QPR, QRP, RPQ, RQP.
Example(2): If there are 6 Biscuits of different brands for 4 persons, then how many ways Biscuits can be distributed among them?
Solution: 6 Biscuits can be distributed among 4 persons by $$6P_4$$ ways.$$nP_r = \frac{n!}{(n - r)!}$$ $$6P_4 = \frac{6!}{(6 - 4)!}$$ $$= \frac{6!}{2!}$$ $$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$$ $$= \frac{720}{2} = 360 \ ways$$
Example(3): If there are 4 chairs for 3 girls, then how many ways the girls can sit on the chairs?
Solution: The girls can sit by $$4P_3$$ ways.$$nP_r = \frac{n!}{(n - r)!}$$ $$4P_3 = \frac{4!}{(4 - 3)!}$$ $$= \frac{4!}{1!}$$ $$= \frac{4 \times 3 \times 2 \times 1}{1}$$ $$= \frac{24}{1} = 24 \ ways$$
Combination: The selection of a number of objects in which order does not matter is called combination. Here 'order' is not important. It is represented by $$nC_r$$.$$nC_r = \frac{n!}{(n - r)! \times r!}$$
Where, n = Total number of objects.
C = Combination.
r = Number of selected objects at a time from the total number of things.
Example: If three students are selected out of 4 students A, B, C, and D, to perform on stage, then how many ways the students can be selected?
Solution: The students can be selected by $$4P_3$$ ways.$$nC_r = \frac{n!}{(n - r)! \times r!}$$ $$4C_3 = \frac{4!}{(4 - 3)! \times 3!}$$ $$= \frac{4!}{3!}$$ $$= \frac{4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}$$ $$= \frac{24}{6} = 4 \ ways$$ 4 ways are ABC, BCD, CDA, and DAB. Here order does not matter. One thing you have to remember that Permutation is an arrangement whereas Combination is a selection, like in this case, Once three students ABC are selected then they can not be selected again as BCA or ACB, because they all are the same persons. But in the permutation, they can be arranged in three ways ABC, BCA, and ACB.
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What are p-adic numbers? They are a different set of numbers first introduced by Kurt Hensel in 1897. The motivation at that time was an attempt to bring the ideas and methods of power series methods into number theory. They have been used in proving Fermat’s Last Theorem and have other applications in number theory. See http://mathworld.wolfram.com/p-adicNumber.html for more information.
A little terminology needs to be introduced. The p in p-adic represents any prime number. For each prime, there is a new and different set of p-adic numbers. Q2 identifies the 2-adics, Q5 represents the 5-adics, Q17 represents the 17-adics. To keep the same notation, Q will represent the real numbers.
Another term to consider is “close.” The basic idea is that given a number n, it is close to 0 if it is highly divisible by a prime p. Consider the numbers 25 and 625. Relatively speaking, 625 has more factors of 5 than does 25, or in other words, 625 has a higher divisibility by 5. Therefore 625 is closer to 0 than 25. This idea will be made a little bit clearer in future postings.
## 8 thoughts on “Quick Intro to p-adics”
1. Samuel Coskey
Great, I had always wondered what was the original motivation behind the p-adic numbers!
One funny consequence of this new definition of “closeness” is the following: The numbers 25, 50, and 75 all have the same number of factors of 5 in them, so even though they are different they are all equally close to 0! This contrasts with the real numbers where two numbers that are equally close to 0 have to be the same or else negatives of one another.
Looking forward to more!
2. Kenny
Does it generally hold that bigger numbers are “closer” to 0? Or like Sam said, do there exists numbers that share “closeness”?
1. Ken Coiteux Post author
Perhaps I may have overgeneralized a little bit. In the 5-adics, 25 is closer to 0 than 26, but that is because 26 has no factors of 5 in it. However, 25, 50, 75, and 100 all are equally close to 0 for they all contain the same powers of 5 in their factors. This will become a little clearer in the post on absolute values (http://math.boisestate.edu/m287/?p=536).
Looking at the P-adic series seems a bit difficult without the explanation of what the -adic is in the series. Does the -adic define a type of series, or function, that generates these numbers? Or is it in reference to the closeness that was discussed?
1. Samuel Coskey
From the OED:
To be close to 0, it seems to be necessary that the number be large. But this is not sufficient, for example, with $p=5$, the number 626 has no factors of 5 so it is not close to 0 at all (even though 625 is)!
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# Physics 120Lab 6: a Study of Air Resistance
Union CollegeWinter 2015
Physics 120Lab 6: A study of Air Resistance
Name:______
In earlier labs, we modeled the motion of a projectile ignoring its interaction with the air. We know, though, that the force of air resistance can be significant, as a sheet of paper, we all know, falls more slowly than a metal ball bearing. In this lab, you will experimentally determine the strength of the force of air resistance Fair.
One interesting (and useful) consequence of air resistance is the concept of “terminal velocity.” At the start of a body’s fall it speeds up. At some point, it stops accelerating and falls at a constant speed – this is the body’s the terminal velocity.
1. What is the net force acting on a body that is falling at terminal velocity?
2. Take a flat-bottom coffee filter, stand on the table top, hold the coffee filter as high up as you can, right-side up, as it would be in a coffee maker. Letgo of the coffee filter, letting it drop to the floor, and watch. Does it reach terminal velocity? How quickly?
3. Now drop a pack of 8 coffee filters. Does the pack reach terminal velocity?
4. Drop one coffee filter and pack of 8 at the same time. Do they reach the same terminal velocity? Do they reach the ground at the same time?
5. Considering the answer to #1, calculate the magnitude of the force of air resistance in both cases. (There is a measurement that you’ll need to make to complete the calculation.)
6. Are the magnitudes of the air resistance forces the same?
7. Why might the strengths of air resistance forces in the two cases be different? What parameter(s) differs in the two cases that could have an effect on the force of air resistance? Obviously the masses of the falling bodies are not the same, but this is not a reasonable answer, since there is no reasonable explanation for how a body’s mass (without changing its size) affects the body’s interaction with air molecules. (Note that the physical size and shape of the falling objects are the same in both cases.) Hint: what is the force of air resistance on a coffee filter that is hanging by a string and not moving? When your group is confident with an answer tell your instructor. If your idea is good, your instructor will give you another handout with specific instructions.
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Test: Statistics - 2 - UPSC MCQ
# Test: Statistics - 2 - UPSC MCQ
Test Description
## 20 Questions MCQ Test CSAT Preparation - Test: Statistics - 2
Test: Statistics - 2 for UPSC 2024 is part of CSAT Preparation preparation. The Test: Statistics - 2 questions and answers have been prepared according to the UPSC exam syllabus.The Test: Statistics - 2 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Statistics - 2 below.
Solutions of Test: Statistics - 2 questions in English are available as part of our CSAT Preparation for UPSC & Test: Statistics - 2 solutions in Hindi for CSAT Preparation course. Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free. Attempt Test: Statistics - 2 | 20 questions in 20 minutes | Mock test for UPSC preparation | Free important questions MCQ to study CSAT Preparation for UPSC Exam | Download free PDF with solutions
Test: Statistics - 2 - Question 1
### If mode of a grouped data is 10 and mean is 4, then median will be
Detailed Solution for Test: Statistics - 2 - Question 1
Concept use:
The relationship between mean, median, and mode in a "perfectly" symmetrical distribution is given by the empirical relationship:
Mode = 3(Median) - 2(Mean)
Calculations:
Median = (Mode + 2 × Mean) / 3
Median = (10 + 2 × 4) / 3 = 18/3 = 6
Test: Statistics - 2 - Question 2
### In Uni-model distribution, if mode is less than mean, then the distribution will be_________.
Detailed Solution for Test: Statistics - 2 - Question 2
A unimodal distribution refers to a distribution that has a single peak or mode. It means that there is one value in the dataset that occurs more frequently than any other value.
When the mode is less than the mean, it indicates that the majority of the data points are located towards the higher values, while the tail extends towards the lower values. This pattern is commonly observed in negatively skewed distributions, also known as left-skewed distributions.
In a negatively skewed distribution, the tail extends towards the left side, while the bulk of the data is concentrated towards the right side. The mode represents the most frequently occurring value, and when it is lower than the mean, it suggests that the distribution is pulled towards the higher values.
To visualize this, imagine a dataset representing the test scores of a group of students. If the distribution is negatively skewed and the mode is less than the mean, it implies that there are a few students who scored exceptionally high, which extends the tail towards the left. However, the majority of the students have lower scores, resulting in the mode being lower than the mean.
Therefore, based on the given information, in a unimodal distribution where the mode is less than the mean, the distribution will most likely be negatively skewed. The tail of the distribution is longer on the left side, indicating a concentration of values towards the right side.
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Test: Statistics - 2 - Question 3
### In a symmetrical distribution, mean is ____________ mode.
Detailed Solution for Test: Statistics - 2 - Question 3
In a symmetrical distribution, the data points are evenly distributed around a central value, resulting in a mirror image when the distribution is folded along its center. This means that the left and right sides of the distribution are symmetrically balanced.
Given this symmetry, the mean, median, and mode will all have the same value in a perfectly symmetrical distribution.
The mean represents the average value of the dataset, calculated by summing all the values and dividing by the total number of values. The mode represents the most frequently occurring value in the dataset.
Since a symmetrical distribution has equal frequencies on both sides of the mode, the mode will be the value that occurs most often and therefore represents the highest peak in the distribution.
In a symmetrical distribution, the balance of the data on both sides of the mode implies that the mean will be the same as the mode.
Therefore, in a symmetrical distribution, the mean is equal to the mode.
Test: Statistics - 2 - Question 4
In symmetrical distribution, mean, median, and mode are__________
Detailed Solution for Test: Statistics - 2 - Question 4
In a perfectly symmetrical distribution, the data points are evenly distributed around a central value, resulting in a mirror image when the distribution is folded along its center. This means that the left and right sides of the distribution are symmetrically balanced.
Given this symmetry, the mean, median, and mode will all have the same value in a perfectly symmetrical distribution.
The mean represents the average value of the dataset, calculated by summing all the values and dividing by the total number of values. The median represents the middle value when the dataset is arranged in ascending or descending order. The mode represents the value or values that occur most frequently in the dataset.
In a perfectly symmetrical distribution, the balance of the data on both sides of the mode implies that the mean and median will be located at the center of the distribution, which is also where the mode will be located.
Therefore, in a symmetrical distribution, the mean, median, and mode are equal. They all have the same value, reflecting the central tendency and balance of the data points in the distribution.
Test: Statistics - 2 - Question 5
The values of mean, median and mode can be________.
Detailed Solution for Test: Statistics - 2 - Question 5
The mean, median, and mode are three measures of central tendency that provide insights into the location or center of a dataset. While they can be equal in certain situations, it is not a universal rule that they will always be equal.
In some distributions, the mean, median, and mode may have the same value. This occurs in perfectly symmetrical distributions, such as the normal distribution, where the data points are evenly distributed around a central value.
However, in many distributions, the mean, median, and mode can have different values. This is especially true in distributions that are skewed or have multiple modes.
For example, in a positively skewed distribution, the mean will be greater than the median, and both of these may differ from the mode. Similarly, in a negatively skewed distribution, the mean will be less than the median, and the mode may be different as well.
It's important to consider the shape and characteristics of the specific distribution when determining the relationship between the mean, median, and mode. While they can be equal in some cases, it is not a guaranteed or universal outcome.
Therefore, the values of mean, median, and mode can be some times equal, but it is not always the case.
Test: Statistics - 2 - Question 6
If mean, median, and mode are all equal then distribution will be________
Detailed Solution for Test: Statistics - 2 - Question 6
In a symmetrical distribution, the data points are evenly distributed around a central value, resulting in a mirror image when the distribution is folded along its center. This means that the left and right sides of the distribution are symmetrically balanced.
When the mean, median, and mode are equal, it indicates that the distribution is perfectly balanced and there is no skewness or bias towards either side. Each side of the distribution has an equal number of data points, resulting in a symmetrical shape.
A symmetrical distribution is also known as a normal distribution or Gaussian distribution. It follows a characteristic bell-shaped curve, where the mean, median, and mode are all located at the center of the distribution.
In summary, if the mean, median, and mode are all equal in a distribution, it indicates that the distribution is symmetrical. The balance of the data points on both sides of the distribution implies a lack of skewness or bias towards either side.
Test: Statistics - 2 - Question 7
The middle value of an ordered array of numbers is the________.
Detailed Solution for Test: Statistics - 2 - Question 7
The median is a measure of central tendency that represents the middle value in a dataset when the values are arranged in ascending or descending order. It divides the dataset into two equal halves, with an equal number of values above and below it.
To find the median in an ordered array of numbers, you simply identify the value located at the center position. If the total number of values in the array is odd, there will be one middle value that is the median. If the total number of values is even, the median is typically calculated as the average of the two middle values.
For example, consider the ordered array [2, 4, 7, 9, 12]. The middle value is 7, so 7 would be the median of this array.
The median is particularly useful when dealing with skewed distributions or datasets with outliers. Unlike the mean, which is influenced by extreme values, the median provides a robust measure of central tendency that is less affected by outliers.
In summary, the middle value of an ordered array of numbers is the median. It represents the central value that divides the dataset into two equal halves when the values are arranged in ascending or descending order.
Test: Statistics - 2 - Question 8
Which of the following describe the middle part of a group of numbers?________.
Detailed Solution for Test: Statistics - 2 - Question 8
The measure of central tendency provides insights into the typical or representative value of a dataset. It aims to identify a single value that represents the central or middle part of the data.
Common measures of central tendency include the mean, median, and mode. The mean is calculated by summing all the values and dividing by the total number of values. The median represents the middle value when the dataset is arranged in ascending or descending order. The mode represents the most frequently occurring value in the dataset.
These measures help us understand where the "center" or "middle" of the dataset lies. They provide a summary or representative value that characterizes the central tendency of the group of numbers.
In contrast, measures of variability (such as the range, variance, and standard deviation) quantify the spread or dispersion of the data points. Measures of association are used to assess the relationship or correlation between two or more variables. Measures of shape describe the overall pattern or distribution of the data.
Therefore, the measure of central tendency is the one that describes the middle part of a group of numbers. It provides insights into the representative value that characterizes the central tendency of the dataset.
Test: Statistics - 2 - Question 9
Find the median of the following data: 160, 180, 200, 280, 300, 320, 400_______.
Detailed Solution for Test: Statistics - 2 - Question 9
To find the median of the given data: 160, 180, 200, 280, 300, 320, 400, we first need to arrange the data in ascending order:
160, 180, 200, 280, 300, 320, 400
Since the total number of values in the dataset is odd (7 values), the median will be the middle value.
The middle value in this case is the fourth value, which is 280.
Therefore, the median of the given data is 280.
Test: Statistics - 2 - Question 10
________ is the measure of average which can have more than one value.
Detailed Solution for Test: Statistics - 2 - Question 10
The mode is the value or values in a dataset that occur most frequently. In some cases, there may be multiple values with the same highest frequency, resulting in multiple modes. When this occurs, the dataset is described as having multiple modes or being multimodal.
For example, consider a dataset of exam scores: 75, 80, 85, 90, 90, 95, 95, 95. In this dataset, the value 95 occurs three times, which is the highest frequency. Therefore, the mode(s) of this dataset is 95. This dataset is said to have a mode of 95.
However, in some cases, a dataset may not have any repeated values, or all values may have the same frequency. In such cases, the dataset is considered to have no mode.
On the other hand, the mean, median, and harmonic mean are measures of central tendency that typically yield a single value. The mean is the average calculated by summing all the values and dividing by the total number of values. The median is the middle value when the dataset is arranged in ascending or descending order. The harmonic mean is a type of average used for rates or ratios.
Therefore, the measure of average that can have more than one value is the mode.
Test: Statistics - 2 - Question 11
Which of the following cannot be less than zero (negative)?
Detailed Solution for Test: Statistics - 2 - Question 11
The geometric mean is a measure of central tendency that is commonly used for a set of positive numbers. It is calculated by taking the nth root of the product of n positive values.
Since the geometric mean involves taking the root of positive values, it cannot be negative. This is because taking the root of a negative number or zero is not defined in standard mathematical operations.
On the other hand, the median, arithmetic mean, and harmonic mean can be negative under certain circumstances. For example, if a dataset contains negative values, the median and arithmetic mean can be negative if the negative values outweigh the positive values.
Therefore, among the options given, the measure that cannot be less than zero (negative) is the geometric mean. It is specifically designed for positive values and does not yield negative results.
Test: Statistics - 2 - Question 12
To find the average speed of a journey which is the appropriate measure of central tendency____________
Detailed Solution for Test: Statistics - 2 - Question 12
The harmonic mean is specifically designed for rates or ratios, making it suitable for calculating average speeds. It is calculated by taking the reciprocal of each value, finding their arithmetic mean, and then taking the reciprocal of that result.
When finding the average speed of a journey, it is common to have different segments or intervals with varying speeds. The harmonic mean is useful in this scenario because it gives more weight to the slower speeds.
The harmonic mean ensures that the calculated average reflects the overall time taken for the journey, considering the different speeds and distances traveled in each segment. By taking the reciprocal of the speeds, finding their arithmetic mean, and then taking the reciprocal again, the harmonic mean effectively balances the impact of different speeds on the average.
On the other hand, the mean, geometric mean, and weighted mean are not as appropriate for finding the average speed of a journey. The mean does not account for the different speeds and distances traveled. The geometric mean is more suitable for multiplicative relationships rather than additive ones like average speeds. The weighted mean involves assigning different weights to each value, which may not be necessary unless there are specific considerations for certain segments of the journey.
Therefore, to find the average speed of a journey, the appropriate measure of central tendency is the harmonic mean. It accounts for the varying speeds and ensures the average reflects the overall time taken for the journey.
Test: Statistics - 2 - Question 13
If any of the value in the data set is zero then it is not possible (i.e. impossible) to compute_________.
Detailed Solution for Test: Statistics - 2 - Question 13
The harmonic mean is a measure of central tendency that is calculated by taking the reciprocal of each value, finding their arithmetic mean, and then taking the reciprocal of that result.
When computing the harmonic mean, taking the reciprocal of a value means dividing 1 by that value. However, division by zero is undefined in mathematics. Therefore, if any value in the data set is zero, it becomes impossible to compute the harmonic mean because it would involve division by zero.
On the other hand, the mode, median, and mean can still be computed even if there is a zero value in the data set.
• The mode represents the most frequently occurring value, which can still be determined even if there is a zero value present.
• The median is the middle value when the data set is arranged in ascending or descending order. If there is a zero value, it will still have a position in the order, and the median can be determined accordingly.
• The mean can be computed by summing all the values and dividing by the total number of values. The presence of a zero value does not prevent the mean from being calculated.
Therefore, if any value in the data set is zero, it is not possible to compute the harmonic mean, but it is still possible to compute the mode, median, and mean.
Test: Statistics - 2 - Question 14
If any of the value in data set is negative then it is impossible to compute___________
Detailed Solution for Test: Statistics - 2 - Question 14
The geometric mean is a measure of central tendency that is calculated by taking the nth root of the product of n positive values. It is commonly used when dealing with multiplicative relationships or ratios.
While the geometric mean is typically used with positive values, it can still be computed with negative values present in the data set. However, it is important to note that the presence of negative values can affect the interpretation and usefulness of the geometric mean.
Negative values in the data set will result in complex or imaginary numbers when calculating the nth root of the product. This means that the geometric mean may not always have a meaningful interpretation when negative values are involved.
On the other hand, the arithmetic mean, harmonic mean, and mode can still be computed even if there are negative values in the data set.
• The arithmetic mean, or simply the mean, is calculated by summing all the values and dividing by the total number of values. The presence of negative values does not prevent the mean from being calculated.
• The harmonic mean is calculated by taking the reciprocal of each value, finding their arithmetic mean, and then taking the reciprocal of that result. The presence of negative values does not prevent the harmonic mean from being computed.
• The mode represents the most frequently occurring value, which can still be determined even if there are negative values present.
Therefore, if any value in the data set is negative, it is not impossible to compute the arithmetic mean, harmonic mean, and mode. However, it is important to consider the implications and limitations when negative values are involved. The geometric mean may still be computed, but the interpretation may not be meaningful in such cases.
Test: Statistics - 2 - Question 15
Data must be arranged either in ascending or descending order if some want to compute________.
Detailed Solution for Test: Statistics - 2 - Question 15
The median is a measure of central tendency that represents the middle value of a dataset. To find the median, the data points need to be arranged in order from least to greatest (ascending order) or from greatest to least (descending order).
By arranging the data in this manner, it becomes easier to identify the middle value. If the dataset contains an odd number of values, the middle value is the one that falls exactly in the center. If the dataset contains an even number of values, the median is typically calculated as the average of the two middle values.
For example, let's consider the following dataset: 5, 10, 3, 8, 2, 9. To find the median, we need to arrange the data in ascending or descending order:
Ascending order: 2, 3, 5, 8, 9, 10 Descending order: 10, 9, 8, 5, 3, 2
By arranging the data in either ascending or descending order, we can easily identify the middle value or values to calculate the median.
In summary, if someone wants to compute the median, it is necessary to arrange the data either in ascending or descending order. This allows for the determination of the middle value or values, which represent the median of the dataset.
Test: Statistics - 2 - Question 16
Which of the following Measure of averages is affected by extreme (very small or very large) values in the data set?
Detailed Solution for Test: Statistics - 2 - Question 16
The arithmetic mean, also known as the mean, is calculated by summing all the values in the data set and dividing by the total number of values. It represents the balance point or center of the data.
Extreme values in the data set can have a significant impact on the arithmetic mean because they contribute to the overall sum. If there are extreme values that are very small or very large, they can pull the mean towards those extreme values.
For example, consider the following data set: 1, 2, 3, 4, 1000. The arithmetic mean of this data set is (1 + 2 + 3 + 4 + 1000) / 5 = 202. If we remove the extreme value of 1000, the mean becomes (1 + 2 + 3 + 4) / 4 = 2.5. The presence of the extreme value significantly affects the arithmetic mean.
On the other hand, the geometric mean, median, and harmonic mean are less influenced by extreme values.
• The geometric mean is calculated by taking the nth root of the product of n values. Since extreme values contribute to the product rather than the sum, their effect is mitigated.
• The median represents the middle value when the data set is arranged in ascending or descending order. Extreme values do not impact the position of the middle value, making the median less affected by them.
• The harmonic mean is calculated by taking the reciprocal of each value, finding their arithmetic mean, and then taking the reciprocal of that result. Extreme values have a smaller influence on the harmonic mean due to the reciprocal operations involved.
In summary, the measure of average that is affected by extreme values in the data set is the arithmetic mean. Extreme values can significantly alter the mean due to their contribution to the overall sum.
Test: Statistics - 2 - Question 17
Which of the following Measure of Averages is not based on all the values given in the data set___________
Detailed Solution for Test: Statistics - 2 - Question 17
The mode is the value or values that occur most frequently in the data set. It represents the most common observation(s) or the peak of the distribution.
Unlike the arithmetic mean, geometric mean, and median, the mode does not take into account all the values in the data set. Instead, it focuses solely on identifying the value(s) with the highest frequency.
For example, consider the following data set: 2, 4, 4, 6, 6, 6, 8, 8, 8. In this case, the mode is 6 because it occurs three times, which is more frequently than any other value. The mode is determined by counting the occurrences of each value, rather than considering the entire range of values.
On the other hand:
• The arithmetic mean is calculated by summing all the values in the data set and dividing by the total number of values. It incorporates all the values in the calculation.
• The geometric mean is calculated by taking the nth root of the product of n values. It also considers all the values in the data set.
• The median represents the middle value when the data set is arranged in ascending or descending order. It includes all the values and identifies the middle observation(s).
Therefore, among the options given, the measure of average that is not based on all the values given in the data set is the mode. It focuses on identifying the most frequently occurring value(s) rather than considering all the values in the data set.
Test: Statistics - 2 - Question 18
The most repeated (popular) value in a data set is called_______.
Detailed Solution for Test: Statistics - 2 - Question 18
The mode is a measure of central tendency that represents the value or values in a data set that occur most frequently. It is the observation(s) with the highest frequency.
In other words, the mode represents the most popular or commonly occurring value in the data set. It is the value that appears more often than any other value.
For example, consider the following data set: 3, 5, 5, 7, 7, 7, 9, 9, 9. In this case, the mode is 7 and 9 because they both occur three times, which is the highest frequency. Both 7 and 9 are the most repeated values in the data set.
The mode is particularly useful when you want to identify the value(s) that have the highest occurrence or when you are interested in the most typical observation in the data set.
On the other hand:
• The median represents the middle value when the data set is arranged in ascending or descending order.
• The mean, also known as the arithmetic mean, is calculated by summing all the values and dividing by the total number of values.
• The geometric mean is calculated by taking the nth root of the product of n positive values.
Therefore, among the options given, the most repeated (popular) value in a data set is called the mode. It represents the value(s) that occur with the highest frequency in the data set.
Test: Statistics - 2 - Question 19
The Geometric Mean of -2, 4, 03, 6, 0 will be__________.
Detailed Solution for Test: Statistics - 2 - Question 19
The geometric mean is calculated by taking the nth root of the product of n positive values. However, it is important to note that the geometric mean is only defined for positive values. It cannot be calculated when negative values or zero are present in the dataset.
In the given values -2, 4, 03, 6, 0, we have a negative value (-2) and a zero (0). Since the geometric mean cannot be computed with negative values or zero, we cannot find the geometric mean for this dataset.
Therefore, the geometric mean of -2, 4, 03, 6, 0 is cannot be computed.
Test: Statistics - 2 - Question 20
The calculation of mean and variance is based on________.
Detailed Solution for Test: Statistics - 2 - Question 20
Both the mean and variance are statistical measures that provide insights into different aspects of a dataset.
The mean, also known as the arithmetic mean or average, is calculated by summing all the values in the dataset and dividing by the total number of values. It represents the central tendency or average value of the dataset. To obtain an accurate mean, all values in the dataset are considered and included in the calculation.
The variance is a measure of the dispersion or spread of the dataset. It quantifies the average squared deviation from the mean. To calculate the variance, each value in the dataset is subtracted from the mean, squared, and then summed. Again, all values in the dataset are taken into account in the variance calculation.
Both the mean and variance require consideration of all values in the dataset to provide meaningful and accurate results. Excluding any values would lead to an incomplete representation of the data and could potentially introduce biases or inaccuracies in the calculations.
Therefore, the calculation of mean and variance is based on all values in the dataset. It is important to include all values to obtain reliable and comprehensive measures of central tendency and dispersion.
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# calculating pH - is my answer correct?
exercice:
1/Calculate the ionization coefficient and pH (with two methods) of a 0.1M ammonia solution of pkb = 4.7
2/ to one liter of the previous solution, an equal volume of water is added. Calculate the new ionization coefficient and deduce the pH of this new solution
1/ For the first part my answer is :
$\ce{NH3 + H2O = NH4+ + OH^-}\tag{1}$
$Ka = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3+]}}\tag{2}$
The coefficient of ionization $\alpha$
$\alpha = \sqrt{Ka/C°}\tag{3}$
$Pkb = \log{kb}\tag{4}$
$Kb= 10-pkb\tag{5}$
$\alpha =0,014\tag{6}$
Two methods:
First one:
$pH= \frac{1}{2} (pka + 14 + \log{C°})\tag{7}$
$pka + pkb=14\tag{8}$
$pH = 11,15\tag{9}$
second method:
$14 + \log (\alpha C°) = 11,15\tag{10}$
for the second part, i am lost
Thank you
• I tried to edit this to make sense of it, but I left the OP's mistakes and sloppy terminology. Also note that OP uses a comma for a decimal point which I left in. – MaxW Jul 22 '18 at 15:23
• Trying to clear up a terminology problem. Is this problem from a book written in English or did you translate it? I'm assuming that ionization coefficient was mean to mean the fraction of the ammonia that ionizes. The common term for that in English is ionization degree. – MaxW Jul 23 '18 at 22:12
The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.
The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then,
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )}$$
Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} }$$
This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$ and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation. But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b} }{ \mathrm{[OH^-]} }$$
We can now get
$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}}$$
from which
$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$
and so you can obtain the pH.
When you dilute, the concentration of base $c_\mathrm{b}$ is halved.
The first part of the question ask for the calculation of the pH two different ways and the ionization coefficient for a 0.1 molar ammonia solution. Also given that $pK_\beta = 4.7$.
First calculation of pH using $K_\beta$ of ammonia
$K_\beta = 10^{-pK_\beta} = 2.00\times10^{-5}\tag{1}$
Note: I'm using two extra significant figures for intermediate calculations. The answer should only have one since only one is given $pK_\beta$.
$\ce{NH3 + H2O <=> NH4+ + OH^-}\tag{2}$
$K_\beta = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3]}}\tag{3}$
Little of the ammonia will ionize to form the ammonium cation, but enough will that the ionization of water itself doesn't need to be considered. So $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So
$K_\beta = \dfrac{\ce{[OH^-]^2}}{\ce{[NH3]}}\tag{4}$
$\ce{[OH-]} =\sqrt{\ce{K_\beta[NH3]}} = \sqrt{(2.0\times10^{-5})(0.1)} = 1.414\times10^{-3}\tag{5}$
!!Check!!
We assumed that little of the ammonia would be ionized. There was 0.1 moles originally and $1.414\times10^{-3}$ did ionize. So there is 0.098586 moles of ammonia left. So the assumption is reasonable, especially since the given $pK_\beta$ only had one significant figure.
$\ce{[H^+]} = \dfrac{K_w}{\ce{[OH^-]}}= \dfrac{1.0\times10^{-14}}{1.414\times10^{-3}} = 7.072\times10^{-12}\tag{6}$
$pH = - \log{\ce{[H^+]}} = - \log{7.072\times10^{-12}} = 11.1504 \ce{->[rounding(1)] = 11.2}\tag{7}$
Second calculation of pH using $K_\beta$ of ammonia
For this calculation let $x$ be the molarity of $\ce{NH4^+}$. Again assume that autoionization of water is negligible. Hence $\ce{[OH^-]} = x$ also, and $\ce{[NH3] = 0.1 - x}$. Substituting into equation (3) yields:
$2.0\times10^{-5} = \dfrac{x^2}{0.1 - x}\tag{8}$
rearranging we get the quadratic equation
$x^2 + (2.0\times10^{-5})x- 2.0\times10^{-6} \tag{9}$
for which the real solution is $x = 1.424\times10^{-3}$. Calculating the pH using method given above yields $pH = 11.2$ again.
Calculation of pH using $K_\alpha$ of ammonium
$pK_\alpha = 14.0 - pK_\beta= 14.0 - 4.7 = 9.3\tag{10}$
$K_\alpha = 10^{-pK_\alpha} = 10^{-9.3} = 5.012\times10^{-10}\tag{11}$
$\ce{NH4^+ <=> NH3 + H^+}\tag{12}$
$K_\alpha = \dfrac{\ce{[NH3][H^+]}}{\ce{[NH4^+]}}\tag{13}$
but $\ce{[H+]} = \dfrac{K_w}{\ce{[OH^-]}}$ so
$K_\alpha = \dfrac{\ce{[NH3]K_w}}{\ce{[NH4^+][OH^-]}}\tag{14}$
Again $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So rearranging and solving for $\ce{[OH^-]}$ gives
$\ce{[OH^-]} = \sqrt{\dfrac{\ce{[NH3]K_w}}{K_a}} = \sqrt{\dfrac{(0.1)(1.0\times10^{-14})}{5.012\times10^{-10}}} = 1.413\times10^{-3}\tag{15}$
Solving for pH as above will yield 11.2.
Solving for the "ionization coefficient"
I don't remember encountering the term ionization coefficient ($\alpha$) before so I'm assuming it means the fraction of the ammonia which is ionized. Let $C_N$ be the nominal concentration of ammonia. Then the mass balance of ammonia species yields:
$\ce{C_N = [NH3] + [NH4^+]}\tag{16}$
so the desired calculation is:
$\alpha = \dfrac{[NH4^+]}{\ce{C_N}}\tag{17}$
Solving equation (2) for $\ce{[NH3]}$ and substituting into equation (16) yields
$\ce{C_N} = \dfrac{\ce{[NH4^+][OH^-]}}{K_\beta} + \ce{[NH4^+]}\tag{18}$
$\ce{K_\beta C_N = [NH4^+][OH^-] + [NH4]K_\beta}\tag{19}$
$\dfrac{\ce{[NH4^+]}}{C_N} = \dfrac{K_\beta}{\ce{[OH^-]} + K_\beta} = \dfrac{2.000\times10^{-5}}{1.414\times10^{−3} + 2.000\times10^{-5}} = 1.4\%\tag{20}$
For the second part the 0.1M ammonia solution has been diluted to 0.05 molar. Just redo calculations above.
• Maxw: According to the charge balance equation:$$\ce{[H+] +[NH4+]= [OH-]}$$ Why we assuming $\ce{[NH4+]= [OH-]}$ – Adnan AL-Amleh May 30 at 17:55
• @AdnanAL-Amleh - $\ce{[NH4^+] \gg [H^+]}$ – MaxW May 30 at 18:00
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How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Cut it Out
##### Stage: 2 Challenge Level:
Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?
### Roll over the Dice
##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
### Snake Coils
##### Stage: 2 Challenge Level:
This challenge asks you to imagine a snake coiling on itself.
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Steps to the Podium
##### Stage: 2 and 3 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Spirals, Spirals
##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### Sticky Triangles
##### Stage: 2 Challenge Level:
Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### Up and Down Staircases
##### Stage: 2 Challenge Level:
One block is needed to make an up-and-down staircase, with one step up and one step down. How many blocks would be needed to build an up-and-down staircase with 5 steps up and 5 steps down?
### Lost Books
##### Stage: 2 Challenge Level:
While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book?
### Growing Garlic
##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
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A113533 Ascending descending base exponent transform of the infinite Fibonacci word (A003842). 2
%I
%S 1,3,6,5,7,12,10,15,14,14,23,16,20,27,21,30,27,25,40,28,37,38,32,49,
%T 36,40,53,39,54,49,43,68,45,55,66,50,71,60,56,83,57,74,75,61,92,67,73,
%U 94,68,93,84,72,113,75,94,101,79,116,89,91,122,86,115,108,90
%N Ascending descending base exponent transform of the infinite Fibonacci word (A003842).
%C The infinite Fibonacci word b(n) is the fixed point of the morphism 1->12, 2->1, starting from b(1) = 2. This transform a(n) of that sequence b(n) satisfies n <= a(n) <= 4*n, but that is not a tight bound.
%H G. C. Greubel, <a href="/A113533/b113533.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = Sum_{k=1..n} A003842(k)^(A003842(n-k+1)). - _G. C. Greubel_, May 18 2017
%e a(1) = A003842(1)^A003842(1) = 1^1 = 1.
%e a(2) = A003842(1)^A003842(2) + A003842(2)^A003842(1) = 1^2 + 2^1 = 3.
%e a(3) = 1^1 + 2^2 + 1^1 = 6.
%e a(4) = 1^1 + 2^1 + 1^2 + 1^1 = 5.
%e a(5) = 1^2 + 2^1 + 1^1 + 1^2 + 2^1 = 7.
%e a(6) = 1^1 + 2^2 + 1^1 + 1^1 + 2^2 + 1^1 = 12.
%e a(7) = 1^2 + 2^1 + 1^2 + 1^1 + 2^1 + 1^2 + 2^1 = 10.
%e a(8) = 1^1 + 2^2 + 1^1 + 1^2 + 2^1 + 1^1 + 2^2 + 1^1 = 15.
%e a(9) = 1^1 + 2^1 + 1^2 + 1^1 + 2^2 + 1^1 + 2^1 + 1^2 + 1^1 = 14.
%e a(10) = 1^2 + 2^1 + 1^1 + 1^2 + 2^1 + 1^2 + 2^1 + 1^1 + 1^2 + 2^1 = 14.
%t A003842[n_] := n + 1 - Floor[((1 + Sqrt[5])/2)*Floor[2*(n + 1)/(1 + Sqrt[5])]]; Table[Sum[A003842[k]^(A003842[n - k + 1]), {k, 1, n}], {n, 1, 50}] (* _G. C. Greubel_, May 18 2017 *)
%Y Cf. A003842, A005408, A087316, A113122, A113153, A113154, A113208, A113231, A113257, A113258, A113271, A113320, A113336, A113498.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Jan 13 2006
%E Corrected and extended by _Giovanni Resta_, Jun 13 2016
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407 views
What will be the solution of the following recurrence?
$$T(n)=3T\sqrt{n}+\log(n)$$
edited
yeah $O(logn^{1.58})$ to be precise $O(logn^{1.584963})$
Yeah it is ,thanks
You did it by master theorem
right :), check out my answer below
$T(n) = 3T \sqrt{n} + logn$
Now, assuming,
$q = logn \\ Or, n = 2^q$
$∴ T(n) = T(2^q) \\ = 3T \sqrt{2^q} + q \\ = 3T. (2^\left(q/2 \right) )+ q$
Now, taking
$X(q ) = T(2^q) \\ ∴ X(q) = 3.X \left (\dfrac{q}{2} \right ) + q$
Now, applying Master Method,
According to Master Method,
Recurrence relation should be in this form,
$$aT \left ( \dfrac{n}{b} \right ) + f(n) , \text{where, } a \geq 1 \hspace{0.1cm} \& \hspace{0.1cm} b>1$$
& we have the equation as : $X(q) = 3.X.(\dfrac{q}{2})+q$
Here, $a = 3, b = 2$
According to master theorem
$\text{If } f(n) = \Theta(n^c), \text{ where }, c < \log_b a, \text{ then, } T(n ) = \Theta (n^ {\log_b a})$
& if $a > b^c$ & $c<\log_b a$, then it will be satisfy $Case-1)$ of Master method
∴ $X(q) = \Theta(q^{\log_2 3 }) \\ \qquad = \Theta(q^{1.58})$
Now, putting back the value of $q$ which is $\log n$
∴ $T(n) = \Theta(\log n ^{1.58})$
'c' is missing in first equation?
is it ok now ?
check this
Thanks Mam for the effort
i was happy to help :)
1 vote
1
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https://brainmass.com/math/integrals/triple-integral-specific-solid-57005
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Explore BrainMass
# Triple Integral on Specific Solid
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Compute a triple integral over a specific solid that would be very difficult in rectangular coordinates, but easy in parabolic cylindrical coordinates, u, v, z, where
x=(1/2)(u^2-v^2) y=uv z=z
You must come up with the solid. Remember that the Jacobian determinant (u^2+v^2) must be used when transforming an integral to this coordinate system"
The function of the solid should be given in Cartesian coordinates, and please show the work done to convert to parabolic cylindrical coordinates and to compute the actual integral. It should also include a graph of the solid.
https://brainmass.com/math/integrals/triple-integral-specific-solid-57005
#### Solution Preview
Mathematics, Calculus
Year 4
Triple Integral
Compute a triple integral over a specific solid that would be very difficult in rectangular coordinates, but easy in parabolic cylindrical coordinates, u, v, z, where
x=(1/2)(u^2-v^2) y=uv z=z
You must come up with the solid. ...
#### Solution Summary
This shows how to construct a triple integral for a solid for the given conditions. A triple integral over a specific solids are determined.
\$2.49
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# 10.2: Hap
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## Thomas Hardy
[1]
If but some vengeful god would call to me
From up the sky, and laugh: “Thou suffering thing,
Know that thy sorrow is my ecstasy,
That thy love’s loss is my hate’s profiting!”
Then would I bear it, clench myself, and die,
Steeled by the sense of ire[2] unmerited;
Half-eased in that a Powerfuller than I
Had willed and meted[3] me the tears I shed.
But not so. How arrives it joy lies slain,
And why unblooms the best hope ever sown?
—Crass Casualty[4] obstructs the sun and rain,
And dicing Time for gladness casts a moan. . . .
Blisses about my pilgrimage as pain.
— 1898
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# How do you solve the inequality 1+4k<-7?
It is true for $k < - 2$
#### Explanation:
We have that
$1 + 4 k < - 7 \implies 4 k < - 8 \implies k < - 2$
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The Quantum Search Algorithm
2018-07-01
The algorithm for finding solutions to the search problem using a quantum computer finds a solution with a much smaller number of steps than a classical computer.
In the case of a quantum computer, when there is only one solution to the problem and $$N$$ elements to search (for example, we want to find a password from $$N$$ possible passwords), the number of operations required is $$O(\sqrt{N})$$, while for classical computers it is $$O(N)$$.
For example, if we have a password consisting of $$6$$ characters, the number of which is equal to $$50$$, then on a classic computer we would need to check $$50^{6}=15625000000$$ possible passwords, whereas on a quantum computer only approximately $$50^{3}=125000$$ iterations are needed to find correct password.
Definitions
Suppose we have a space of $$N$$ elements that we will search in order to find exactly $$M$$ solutions for a given problem, we assume that $$M \in [1 ,N/2]$$. Instead of looking for elements we will look for the indexes of these elements in the range from $$0$$ do $$N−1$$. In addition, we will simplify the situation assuming that $$N =2^{n}$$, that is, index is stored in $$n$$ bits. The solution of the problem can be described by the function $$f(x)$$, which takes the index $$x$$ as an argument and returns the value $$1$$ if the index matches the solution or $$0$$, if the given index is not a solution to the problem.
Let’s start with defining the operation $$O$$ as follows:
$$$$\left \vert x\right \rangle \left \vert q\right \rangle \overset{O}{ \longrightarrow }\left \vert x\right \rangle \left \vert q \oplus f(x)\right \rangle$$$$
where $$\oplus$$ denotes addition modulo $$2$$. That is, a single qubit $$\left \vert q\right \rangle$$ which is equal to $$\left \vert 0\right \rangle$$ or$$\left \vert 1\right \rangle$$ will be changed only if $$f(x) =1$$, which means that $$x$$ is the solution.
Below we will use qubit:
$$$$\left \vert q\right \rangle =\frac{1}{\sqrt{2}}(\left \vert 0\right \rangle -\left \vert 1\right \rangle )$$$$
which we get from qubit $$\left \vert 0\right \rangle$$ after operation $$HX$$, where:
\begin{align}X =\left (\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right ) \\ \end{align}
and
\begin{align} H =\frac{1}{\sqrt{2}}\left (\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right )\end{align}
where $$H$$ is called the Hadamard operation.
Therefore, $$O$$ operation can also be defined as follows:
$$$$\left \vert x\right \rangle \left \vert q\right \rangle \overset{O}{ \longrightarrow }( -1)^{f(x)}\left \vert x\right \rangle \left \vert q\right \rangle$$$$
Algorithm
The goal of the algorithm is to find a solution at the smallest possible amount of the operation of $$O$$.
We start from the computer in state $$\left \vert 0\right \rangle ^{ \otimes n}\left \vert 0\right \rangle$$ ($$\left \vert 0\right \rangle ^{ \otimes n}$$ corresponds to the index $$x=0$$), applying the Hadamard transform $$H^{ \otimes n}$$ to state $$\left \vert 0\right \rangle ^{ \otimes n}$$ and $$HX$$ to last qubit. We get equal superposition state:
$$$$\left \vert \psi \right \rangle =\frac{1}{N^{1/2}}\sum \limits _{x =0}^{N -1}\left \vert x\right \rangle \left \vert q\right \rangle$$$$
The quantum algorithm consists of the repeated application of the $$G$$ operator, whose single application can be described as follows:
1. Apply $$O$$
2. Apply the Hadamard transform $$H^{ \otimes n}$$
3. Phase change to $$-1$$ in each state $$\left \vert x\right \rangle \left \vert q\right \rangle$$, except $$\left \vert 0\right \rangle \left \vert q\right \rangle$$:
$$\left \vert x\right \rangle \left \vert q\right \rangle \longrightarrow -( -1)^{\delta _{x0}}\left \vert x\right \rangle \left \vert q\right \rangle$$
4. Apply the Hadamard transform $$H^{ \otimes n}$$
By combining steps 2, 3 and 4 we get:
$$$$H^{ \otimes n}(2\vert 0\rangle \left \vert q\right \rangle \left \langle q\right \vert \left \langle 0\right \vert -I)H^{ \otimes n} =2\left \vert \psi \right \rangle \left \langle \psi \right \vert -I$$$$
Therefore, the $$G$$ operator can be expressed by means of a formula $$G =(2\left \vert \psi \right \rangle \left \langle \psi \right \vert -I)O$$.
Let’s define two orthonormal states:
\begin{align}\left \vert \alpha \right \rangle =\frac{1}{\sqrt{N -M}}\sum _{x}\left \vert x\right \rangle \left \vert q\right \rangle \\\left \vert \beta \right \rangle =\frac{1}{\sqrt{M}}\sum _{y}\left \vert y\right \rangle \left \vert q\right \rangle \end{align}
where sum over $$y$$ is sum over all indexes which are solutions to the search problem, and sum over $$x$$ is sum over all indexes which are not solutions to the search problem. It is easy to show that the following equality occurs:
$$\left \vert \psi \right \rangle =\sqrt{\frac{N -M}{N}}\left \vert \alpha \right \rangle +\sqrt{\frac{M}{N}}\left \vert \beta \right \rangle$$
By calculating the matrix elements $$G$$ in the basis states $$\left \vert \alpha \right \rangle$$ and $$\left \vert \beta \right \rangle$$ we obtain, that $$G$$ is the operator of rotation in the two-dimensional space spanned by these states:
$$G =\left (\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right )$$
where $$\cos \theta =(N -2M)/N$$. State $$\left \vert \psi \right \rangle$$ can be expressed by $$\theta$$ as:
$$\left \vert \psi \right \rangle =\cos \frac{\theta }{2}\left \vert \alpha \right \rangle +\sin \frac{\theta }{2}\left \vert \beta \right \rangle$$
Applying of $$G$$ $$k$$ times we obtain:
$$G^{k}\left \vert \psi \right \rangle =\cos \left (\frac{2k +1}{2}\theta \right )\left \vert \alpha \right \rangle +\sin \left (\frac{2k +1}{2}\theta \right )\left \vert \beta \right \rangle$$
Therefore, by applying the operator $$G$$ many times, we turn the state $$\left \vert \psi \right \rangle$$ close to the state $$\left \vert \beta \right \rangle$$, so observation yields the correct answer with high probability.
Because the initial state has the form $$\left \vert \psi \right \rangle =\sqrt{\frac{N -M}{N}}\left \vert \alpha \right \rangle +\sqrt{\frac{M}{N}}\left \vert \beta \right \rangle$$, so in the space of two basis states $$\left \vert \alpha \right \rangle$$ and $$\left \vert \beta \right \rangle$$ rotation of the initial state by an angle $$\arccos \sqrt{M/N}$$ transform it into the state $$\left \vert \beta \right \rangle$$. Let $$[z]$$ be the nearest integer for the real number z. Thus, the number of rotations made by $$G$$:
$$$$R =\left [\genfrac{}{}{}{}{\arccos \sqrt{M/N}}{\theta }\right ]$$$$
transform the state $$\left \vert \psi \right \rangle$$ close to the solution $$\left \vert \beta \right \rangle$$. Because $$R\leq [\pi /2\theta ]$$, so an upper bound on $$R$$ is taken from a lower bound on $$\theta$$. Because $$\theta /2 \geq \sin \theta /2 =\sqrt{\frac{M}{N}}$$, hence we get the upper bound of the required amount of iteration:
$$R \leq \left [\frac{\pi }{4}\sqrt{\frac{N}{M}} \, \right ]$$
Finally we get, that $$R =O(\sqrt{N/M})$$. Therefore only for one solution, $$M=1$$, $$R =O(\sqrt{N})$$, in the classic case we would get $$R=O(N)$$.
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#### What is 38 percent of 669?
How much is 38 percent of 669? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 38% of 669 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 38% of 669 = 254.22
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating thirty-eight of six hundred and sixty-nine How to calculate 38% of 669? Simply divide the percent by 100 and multiply by the number. For example, 38 /100 x 669 = 254.22 or 0.38 x 669 = 254.22
#### How much is 38 percent of the following numbers?
38% of 669.01 = 25422.38 38% of 669.02 = 25422.76 38% of 669.03 = 25423.14 38% of 669.04 = 25423.52 38% of 669.05 = 25423.9 38% of 669.06 = 25424.28 38% of 669.07 = 25424.66 38% of 669.08 = 25425.04 38% of 669.09 = 25425.42 38% of 669.1 = 25425.8 38% of 669.11 = 25426.18 38% of 669.12 = 25426.56 38% of 669.13 = 25426.94 38% of 669.14 = 25427.32 38% of 669.15 = 25427.7 38% of 669.16 = 25428.08 38% of 669.17 = 25428.46 38% of 669.18 = 25428.84 38% of 669.19 = 25429.22 38% of 669.2 = 25429.6 38% of 669.21 = 25429.98 38% of 669.22 = 25430.36 38% of 669.23 = 25430.74 38% of 669.24 = 25431.12 38% of 669.25 = 25431.5
38% of 669.26 = 25431.88 38% of 669.27 = 25432.26 38% of 669.28 = 25432.64 38% of 669.29 = 25433.02 38% of 669.3 = 25433.4 38% of 669.31 = 25433.78 38% of 669.32 = 25434.16 38% of 669.33 = 25434.54 38% of 669.34 = 25434.92 38% of 669.35 = 25435.3 38% of 669.36 = 25435.68 38% of 669.37 = 25436.06 38% of 669.38 = 25436.44 38% of 669.39 = 25436.82 38% of 669.4 = 25437.2 38% of 669.41 = 25437.58 38% of 669.42 = 25437.96 38% of 669.43 = 25438.34 38% of 669.44 = 25438.72 38% of 669.45 = 25439.1 38% of 669.46 = 25439.48 38% of 669.47 = 25439.86 38% of 669.48 = 25440.24 38% of 669.49 = 25440.62 38% of 669.5 = 25441
38% of 669.51 = 25441.38 38% of 669.52 = 25441.76 38% of 669.53 = 25442.14 38% of 669.54 = 25442.52 38% of 669.55 = 25442.9 38% of 669.56 = 25443.28 38% of 669.57 = 25443.66 38% of 669.58 = 25444.04 38% of 669.59 = 25444.42 38% of 669.6 = 25444.8 38% of 669.61 = 25445.18 38% of 669.62 = 25445.56 38% of 669.63 = 25445.94 38% of 669.64 = 25446.32 38% of 669.65 = 25446.7 38% of 669.66 = 25447.08 38% of 669.67 = 25447.46 38% of 669.68 = 25447.84 38% of 669.69 = 25448.22 38% of 669.7 = 25448.6 38% of 669.71 = 25448.98 38% of 669.72 = 25449.36 38% of 669.73 = 25449.74 38% of 669.74 = 25450.12 38% of 669.75 = 25450.5
38% of 669.76 = 25450.88 38% of 669.77 = 25451.26 38% of 669.78 = 25451.64 38% of 669.79 = 25452.02 38% of 669.8 = 25452.4 38% of 669.81 = 25452.78 38% of 669.82 = 25453.16 38% of 669.83 = 25453.54 38% of 669.84 = 25453.92 38% of 669.85 = 25454.3 38% of 669.86 = 25454.68 38% of 669.87 = 25455.06 38% of 669.88 = 25455.44 38% of 669.89 = 25455.82 38% of 669.9 = 25456.2 38% of 669.91 = 25456.58 38% of 669.92 = 25456.96 38% of 669.93 = 25457.34 38% of 669.94 = 25457.72 38% of 669.95 = 25458.1 38% of 669.96 = 25458.48 38% of 669.97 = 25458.86 38% of 669.98 = 25459.24 38% of 669.99 = 25459.62 38% of 670 = 25460
1% of 669 = 6.69 2% of 669 = 13.38 3% of 669 = 20.07 4% of 669 = 26.76 5% of 669 = 33.45 6% of 669 = 40.14 7% of 669 = 46.83 8% of 669 = 53.52 9% of 669 = 60.21 10% of 669 = 66.9 11% of 669 = 73.59 12% of 669 = 80.28 13% of 669 = 86.97 14% of 669 = 93.66 15% of 669 = 100.35 16% of 669 = 107.04 17% of 669 = 113.73 18% of 669 = 120.42 19% of 669 = 127.11 20% of 669 = 133.8 21% of 669 = 140.49 22% of 669 = 147.18 23% of 669 = 153.87 24% of 669 = 160.56 25% of 669 = 167.25
26% of 669 = 173.94 27% of 669 = 180.63 28% of 669 = 187.32 29% of 669 = 194.01 30% of 669 = 200.7 31% of 669 = 207.39 32% of 669 = 214.08 33% of 669 = 220.77 34% of 669 = 227.46 35% of 669 = 234.15 36% of 669 = 240.84 37% of 669 = 247.53 38% of 669 = 254.22 39% of 669 = 260.91 40% of 669 = 267.6 41% of 669 = 274.29 42% of 669 = 280.98 43% of 669 = 287.67 44% of 669 = 294.36 45% of 669 = 301.05 46% of 669 = 307.74 47% of 669 = 314.43 48% of 669 = 321.12 49% of 669 = 327.81 50% of 669 = 334.5
51% of 669 = 341.19 52% of 669 = 347.88 53% of 669 = 354.57 54% of 669 = 361.26 55% of 669 = 367.95 56% of 669 = 374.64 57% of 669 = 381.33 58% of 669 = 388.02 59% of 669 = 394.71 60% of 669 = 401.4 61% of 669 = 408.09 62% of 669 = 414.78 63% of 669 = 421.47 64% of 669 = 428.16 65% of 669 = 434.85 66% of 669 = 441.54 67% of 669 = 448.23 68% of 669 = 454.92 69% of 669 = 461.61 70% of 669 = 468.3 71% of 669 = 474.99 72% of 669 = 481.68 73% of 669 = 488.37 74% of 669 = 495.06 75% of 669 = 501.75
76% of 669 = 508.44 77% of 669 = 515.13 78% of 669 = 521.82 79% of 669 = 528.51 80% of 669 = 535.2 81% of 669 = 541.89 82% of 669 = 548.58 83% of 669 = 555.27 84% of 669 = 561.96 85% of 669 = 568.65 86% of 669 = 575.34 87% of 669 = 582.03 88% of 669 = 588.72 89% of 669 = 595.41 90% of 669 = 602.1 91% of 669 = 608.79 92% of 669 = 615.48 93% of 669 = 622.17 94% of 669 = 628.86 95% of 669 = 635.55 96% of 669 = 642.24 97% of 669 = 648.93 98% of 669 = 655.62 99% of 669 = 662.31 100% of 669 = 669
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# Momentum question watch
1. a one-tonne spacecraft which is travelling at a speed of 9500 m/s is hit by a piece of debris of mass 0.5kg travelling at right angles to it at the same speed. As a result of the impact, the debris melts into the wall of the spacecraft leaving a 12cm hole.
Estimate the average force acting during the impact
Here is what I attempted to do:
F = m(v-u) / t
Momentum of spacecraft before: 1000*9500
Momentum of debris before: 0.5*9500
Velocity of spacecraft after: 9495 m/s (because 1000.5/(1000*9500))
Velocity of debris after: 13431.5 m/s
Momentum spaceship after: 1000*9495
Momentum debris after: 0.5*12431.5
Change in momentum of spaceship: mv-mu = 5000
Change in momentum of debris: -1965.75
Total change in momentum = 3034.25
Time: 0.12/9500
This gives me a force of 240 MN but the answer at the back of the book is 376 MN. Where have I gone wrong?
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# Solving Angle Problems
The Zoom desktop client and Zoom Mobile App are both available for free download.= 32'1 minute(minute hand) = 6 degrees32 minutes(minute hand) = 32*6 = 192 degrees For the 32 minutes, the hour hand moves by an additional 16 degrees = 32*0.5So Hour hand at 32 minutes = 90 16=106 degrees Degrees between minute hand and hour hand at ' = 192-106=86 degrees1) A clock makes two right angles between 2 hours2) The clock does not makes 48 right angles in 24 hours( 1 day)P. Note: angle A is opposite side a, B is opposite b, and C is opposite c.
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AAA triangles are impossible to solve further since there are is nothing to show us size ... In this case we find the third angle by using Angles of a Triangle, then use The Law of Sines to find each of the other two sides. This means we are given two sides and the included angle.
For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. This means we are given two sides and one angle that is not the included angle.
Description CTY's Problem Solving courses sharpen investigative skills, broaden mathematical understanding of concepts, and enhance reasoning skills.
Designed around performance objectives that reflect national and state mathematical standards and drawing on video lectures provided by Thinkwell, these courses demonstrate how mathematical issues arise out of real-life situations.
Upon successful completion of this course, a student will be able to understand definitions and theorems as well as solve problems related to: Sample Video Technical Requirements This course requires a properly maintained computer with high-speed internet access and an up-to-date web browser (such as Chrome or Firefox).
The student must be able to communicate with the instructor via email.
There may be cases when our downloadable resources contain hyperlinks to other websites.
These hyperlinks lead to websites published or operated by third parties.
Concepts are assessed through challenging quizzes and chapter tests.
In Problem Solving in Geometry, students apply geometric concepts to solve sets of word problems at varying levels of difficulty.
## Comments Solving Angle Problems
• ###### Geometry I Build a foundation for geometry with angles.
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In this course, you'll solve delightful geometry puzzles and build a solid foundation of skills for problem-solving with angles, triangles, and polygons. You'll also.…
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May 12, 2009. Checklist of questions was developed by the CIA as a problem solving tool to encourage agents to look at a challenge from different angles.…
• ###### Problem-Solving Worksheets Problem Sheet 2 – Shape and.
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Problem-Solving Worksheets. Problem Sheet 2 – Shape and Angle Problems. 1. The solid shown rests on a flat surface. It is made from 1cm cubes placed, but.…
• ###### How to Solve GMAT Clock Angle Problems -
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This tutorial gives you a clear insight on cracking clock based angle and units problems.…
• ###### Vertical Angles Proof and Problem-Solving In this lesson.
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In this lesson, students learn about the relationship between vertical angles by making inferences and proving the Vertical Angle Theorem. They then use this.…
• ###### Problem Solving in Geometry Johns Hopkins Center for.
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In Problem Solving in Geometry, students apply geometric concepts to solve sets. Points, Lines, Planes, and Angles; Angle Relationships in Polygons; Triangle.…
• ###### Solving Problems Geometry Of Straight Lines Siyavula
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In Grade 8 you identified relationships between angles on straight lines. In this chapter, you will revise all of the angle relationships and write clear descriptions.…
• ###### MATH G7 Angle Problems and Solving Equations - UnboundEd
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Students describe an angle relationship and set up and solve an equation that models it.…
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# How many maximum moles of AgCl are obtained?
Contents
## How many moles of AgCl will be obtained when agno3?
2 moles because 2 Cl- ions are present as counter ion in the complex.
## How many moles of AgCl will be precipitated when an excess of agno3 is added to a molar solution of Cr h2o 5cl cl2?
Assuming that 1 litre of solution is taken.. It would give 2 moles of Cl . Hence, adding excess of AgNO3 will ppt out 2 moles of AgCl..
## Which gives only 25% mole of AgCl when reacts with AgNO3?
3 NH 3 or [ Pt ( NH 3 ) 3 Cl 3 ] Cl gives only 25 % of AgCl as we can see than here 4 Cl are present , but only 1 mol of AgCl is obtained i . e . 1 4 × 100 = 25 % . This is because here the 3 Cl atoms are inside the coordination sphere while 1 Cl is outside , therefore 1 Cl is only available for AgNO 3 .
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## What is the number of moles of AgCl precipitated when one mole of CoCl3 4NH3 is treated with excess of silver nitrate?
5nh3 was treated with excess of silver nitrate solution, 2 mol of agcl was precipitated.
## How many moles of AgCl will be obtained when AgNO3 solution is completely react with solution?
of moles of AgCl obtained when excess AgNO_(3) is added to one mole of [Cr(NH_(3))_(4)Cl_(2)]Cl. [Cr(NH3)4Cl2]Cl ionises gives one Cl- ion. So it gives 1 mole AgCl ppt.
## What fraction of chlorine will be precipitated as AgCl?
– So, only these 2 chloride ions can be precipitated out by AgNO3 as AgCl. – So, the fraction of chloride precipitated is 2/3.
## How many grams of AgCl are produced after the reaction?
Therefore, the mass of AgCl formed by the reaction is 0.84 g.
## How many moles are in Ag’in AgCl?
Divide the given mass m by the molecular weight W of AgCl, which is given as 143 g/mole. That gives the number of moles of AgCl. Since there is one atom of Ag in each molecule of AgCl, the number of moles of Ag equals the number of moles of AgCl.
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posted by .
Please explain the steps because I have 5 to do for homework-Thank you-I'm really stuck on this
I thought the answer would be x^2/4 + y^2/40 = 1 but that can't be because the choices are:
x^2/1 + y^2/40 = 1 or x^2/1 + y^2/36
I'm really confused
• Algebra II Please check- -
I forgot to put that I think b= 2
c=6, so therefore
a^2-2^2 = 6^2
so a= 40, that' how I got to my equation
• Algebra II Please check- -
• Algebra II-it can't be correct- Please check- -
It can't be correct my equation because that isn't a choice I was given-that's what I don't understand-thanks for trying to help though
• Algebra II Please check- -
Jodi,my answer was exactly identical to
yours.
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Please help-What is the equation of the ellipse with foci (0,6), (0,-6) and co-vertices (2,0),(-2,0) My answer and logic is below but it can't be correct I thought the answer would be x^2/4 + y^2/40 = 1 but that can't be because the …
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# Difference Between Tree and Graph Data structure
In this tutorial you will learn about the Difference Between Tree and Graph Data structure and its application with practical example.
## What Is Tree Data Structure?
Trees are the most common and a kind of hierarchical data structure which stores data or information in sequential manner, and this all occurs because of a group of nodes which is basically a hierarchical bonding, it is a kind of non-linear data structure which works on the principle of parent-child relationship and each data item is noted as node and the top most node is called a parent node, and these nodes have their child nodes also. In data structure tree always grow downside, due to this characteristics searching and transverse can easily be done and insertion deletion and searching can be done real quick. Trees may be of following type, (1) General Tree, (2) Binary Tree, (3) Binary Search Tree, (4) AVL Tree, (5) B Tree.
## What Is Graph Data Structure?
It is that kind of data structure, where the groups of nodes are present which is called as vertex and these vertices are so connected to each other in formation of bridge pattern and these patterns are called as edges. These edges works as a communication link between the two nodes. Graphs are a kind of non-primitive and non-linear data structure where the bunch of nodes vertices and edges are present and these may vary in directional as well as unidirectional way. Graphs are of following types (1) Directed graphs, (2) Undirected graphs, (3) Weighted graphs, (4) Non-weighted graphs.
## Tree Vs Graph Data structure
TREES GRAPHS Here elements are arranged at multiple levels due to is non-linear structure. It is also a kind of non-linear and non-primitive data structure. Collection of nodes and edges are present. Collection of vertices and edges are present. Edges and nodes are denoted as N,E respectively and written as T={N,E}. Edges and vertices are denoted as N,E respectively and written as T={V,E}. The topmost node is representing as parent node and rest are their child nodes. There is no requirement of nodes. Loop or cycle formation is not needed. Loop or cycle formation is needed. It works on hierarchical model where nodes are arranged at multiple level. It works on network model and is used dynamically for SNS websites. n-1 number of edges are present if there are nodes present in it as in directed edges. Dependencies of edges are present on the structure of graphs as in both directed & non-directed edges. Operations like Insertion, deletion, searching can easily be done. It is used to find the shortest path for the network model.
In this tutorial we have learn about the Difference Between Tree and Graph Data structure and its application with practical example. I hope you will like this tutorial.
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https://www.aqua-calc.com/one-to-all/entropy/preset/hectojoule-per-kelvin/1
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# Convert hectojoules per kelvin [hJ/K] to other units of entropy
## hectojoules/kelvin [hJ/K] entropy conversions
1 hJ/K = 1 × 10+20 attojoules per kelvin hJ/K to aJ/K 1 hJ/K = 10 000 centijoules per kelvin hJ/K to cJ/K 1 hJ/K = 1 000 decijoules per kelvin hJ/K to dJ/K 1 hJ/K = 10 dekajoules per kelvin hJ/K to daJ/K 1 hJ/K = 1 × 10-16 Exajoule per kelvin hJ/K to EJ/K 1 hJ/K = 1 × 10+17 femtojoules per kelvin hJ/K to fJ/K 1 hJ/K = 1 × 10-7 Gigajoule per kelvin hJ/K to GJ/K 1 hJ/K = 0.1 kilojoule per kelvin hJ/K to kJ/K 1 hJ/K = 0.0001 Megajoule per kelvin hJ/K to MJ/K 1 hJ/K = 100 000 000 microjoules per kelvin hJ/K to µJ/K 1 hJ/K = 100 000 millijoules per kelvin hJ/K to mJ/K 1 hJ/K = 100 000 000 000 nanojoules per kelvin hJ/K to nJ/K 1 hJ/K = 100 joules per kelvin hJ/K to J/K 1 hJ/K = 1 × 10-13 Petajoule per kelvin hJ/K to PJ/K 1 hJ/K = 1 × 10+14 picojoules per kelvin hJ/K to pJ/K 1 hJ/K = 1 × 10-10 Terajoule per kelvin hJ/K to TJ/K 1 hJ/K = 1 × 10+26 yoctojoules per kelvin hJ/K to yJ/K 1 hJ/K = 1 × 10-22 Yottajoule per kelvin hJ/K to YJ/K 1 hJ/K = 1 × 10+23 zeptojoules per kelvin hJ/K to zJ/K 1 hJ/K = 1 × 10-19 Zettajoule per kelvin hJ/K to ZJ/K
#### Foods, Nutrients and Calories
MOMEN (REGULAR), UPC: 016778000012 contain(s) 82 calories per 100 grams (≈3.53 ounces) [ price ]
11 foods that contain Inositol. List of these foods starting with the highest contents of Inositol and the lowest contents of Inositol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Tahitian Moon weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Vaterite [CaCO3] weighs 2 711 kg/m³ (169.2422 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Coconut oil with temperature in the range of 30°C (86°F) to 110°C (230°F)
#### Weights and Measurements
kilogram per meter (kg/m) is a metric measurement unit of linear or linear mass density.
Momentum is a vector quantity of motion of a body that is given by the product of its mass and velocity.
t/yd³ to troy/US c conversion table, t/yd³ to troy/US c unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a hollow cylinder and its surface area
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https://convertoctopus.com/5533-cubic-inches-to-fluid-ounces
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## Conversion formula
The conversion factor from cubic inches to fluid ounces is 0.55411255411377, which means that 1 cubic inch is equal to 0.55411255411377 fluid ounces:
1 in3 = 0.55411255411377 fl oz
To convert 5533 cubic inches into fluid ounces we have to multiply 5533 by the conversion factor in order to get the volume amount from cubic inches to fluid ounces. We can also form a simple proportion to calculate the result:
1 in3 → 0.55411255411377 fl oz
5533 in3 → V(fl oz)
Solve the above proportion to obtain the volume V in fluid ounces:
V(fl oz) = 5533 in3 × 0.55411255411377 fl oz
V(fl oz) = 3065.9047619115 fl oz
The final result is:
5533 in3 → 3065.9047619115 fl oz
We conclude that 5533 cubic inches is equivalent to 3065.9047619115 fluid ounces:
5533 cubic inches = 3065.9047619115 fluid ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 0.000326167992047 × 5533 cubic inches.
Another way is saying that 5533 cubic inches is equal to 1 ÷ 0.000326167992047 fluid ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five thousand five hundred thirty-three cubic inches is approximately three thousand sixty-five point nine zero five fluid ounces:
5533 in3 ≅ 3065.905 fl oz
An alternative is also that one fluid ounce is approximately zero times five thousand five hundred thirty-three cubic inches.
## Conversion table
### cubic inches to fluid ounces chart
For quick reference purposes, below is the conversion table you can use to convert from cubic inches to fluid ounces
cubic inches (in3) fluid ounces (fl oz)
5534 cubic inches 3066.459 fluid ounces
5535 cubic inches 3067.013 fluid ounces
5536 cubic inches 3067.567 fluid ounces
5537 cubic inches 3068.121 fluid ounces
5538 cubic inches 3068.675 fluid ounces
5539 cubic inches 3069.229 fluid ounces
5540 cubic inches 3069.784 fluid ounces
5541 cubic inches 3070.338 fluid ounces
5542 cubic inches 3070.892 fluid ounces
5543 cubic inches 3071.446 fluid ounces
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https://jp.mathworks.com/matlabcentral/cody/problems/44372-polarisation/solutions/1313015
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Cody
# Problem 44372. Polarisation
Solution 1313015
Submitted on 24 Oct 2017 by Matthew Eicholtz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 5; y_correct = 0.5; assert(abs(polarised(x)-y_correct) < 1e-10)
2 Pass
x = 180; y_correct = 0.5; assert(abs(polarised(x)-y_correct) < 1e-10)
3 Pass
x = 365; y_correct = 0.5; assert(abs(polarised(x)-y_correct) < 1e-10)
4 Pass
x = [91, 1]; y_correct = 0; assert(abs(polarised(x)-y_correct) < 1e-10)
5 Pass
a = randi([-360, 360]); b = 90*(1+2*randi([-3, 3])); x = [a, a+b]; y_correct = 0; assert(abs(polarised(x)-y_correct) < 1e-10)
6 Pass
a = randi([-360, 360]); b = 90*(1+2*randi([-3, 3])); x = [a, a+b]; y_correct = 0; assert(abs(polarised(x)-y_correct) < 1e-10)
7 Pass
x = [0, 22.5]; y_correct = 0.85355339059/2; assert(abs(polarised(x)-y_correct) < 1e-10)
8 Pass
x = [0, -45]; y_correct = 0.25; assert(abs(polarised(x)-y_correct) < 1e-10)
9 Pass
x = [5, 140]; y_correct = 0.25; assert(abs(polarised(x)-y_correct) < 1e-10)
10 Pass
x = 5 + (1:5)*22.5; y_correct = 0.53079004294/2; assert(abs(polarised(x)-y_correct) < 1e-10)
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https://www.teachthis.com.au/products/dividing-decimals-by-powers-of-10-2
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Curriculum Focus
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# Dividing Decimals by Powers of 10
Description
The Dividing Decimals by Powers of 10 Board Game offers an interactive and dynamic approach for students to enhance their proficiency in dividing decimals by powers of 10.
Suitable for
• Relief Teachers
Lesson Structure
• Individual Activity
• Rotations / Group Work
• Class Activity
Curriculum Codes
AC9M6N06 9
Multiply and divide decimals by multiples of powers of 10 without a calculator, applying knowledge of place value and proficiency with multiplication facts; using estimation and rounding to check the reasonableness of answers
ACMNA130 8.4
Multiply and divide decimals by powers of 10
ACMNA129 8.4
Multiply decimals by whole numbers and perform divisions by non-zero whole numbers where the results are terminating decimals, with and without digital technologies
VCMNA216
Multiply and divide decimals by powers of 10
VCMNA215
Multiply decimals by whole numbers and perform divisions by non-zero whole numbers where the results are terminating decimals, with and without digital technologies
MA3-1WM old
Describes and represents mathematical situations in a variety of ways using mathematical terminology and some conventions
MA3-2WM old
Selects and applies appropriate problem- solving strategies, including the use of digital technologies, in undertaking investigations
MA3-3WM old
Gives a valid reason for supporting one possible solution over another
MA3-7NA old
Compares, orders and calculates with fractions, decimals and percentages
MA3-RN-01 new
Applies an understanding of place value and the role of zero to represent the properties of numbers
MA3-RN-02 new
Compares and orders decimals up to 3 decimal places
MA3-RN-03 new
Determines percentages of quantities, and finds equivalent fractions and decimals for benchmark percentage values
MA3-MR-01 new
Selects and applies appropriate strategies to solve multiplication and division problems
MA3-MR-02 new
Constructs and completes number sentences involving multiplicative relations, applying the order of operations to calculations
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https://www.quizover.com/online/course/4-12-lab-probability-topics-probability-topics-by-openstax
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# 4.12 Lab: probability topics
Page 1 / 1
This module presents students with a lab exercise allowing them to apply their understanding of Probability. In an experiment using M&Ms candies, students will calculate and compare the theoretical and empirical probabilities of drawing particular color candies at random, with and without replacement.
Class time:
Names:
## Student learning outcomes:
• The student will use theoretical and empirical methods to estimate probabilities.
• The student will appraise the differences between the two estimates.
• The student will demonstrate an understanding of long-term relative frequencies.
## Do the experiment:
Count out 40 mixed-color M&M’s® which is approximately 1 small bag’s worth (distance learning classes using the virtual lab would want to count out 25 M&M’s®). Record the number of each color in the "Population" table. Use the information from this table to complete the theoretical probability questions. Next, put the M&M’s in a cup. The experiment is to pick 2 M&M’s, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of the empirical table. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of the "Empirical Results" table. After you record the pick, put both M&M’s back. Do this a total of 24 times, also. Use the data from the "Empirical Results" table to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions.
Population
Color Quantity
Yellow (Y)
Green (G)
Blue (BL)
Brown (B)
Orange (O)
Red (R)
Theoretical probabilities
With Replacement Without Replacement
$P\left(\text{2 reds}\right)$
$P\left({R}_{1}{B}_{2}\text{OR}{B}_{1}{R}_{2}\right)$
$P\left({R}_{1}\text{AND}{G}_{2}\right)$
$P\left({G}_{2}\text{|}{R}_{1}\right)$
$P\left(\text{no yellows}\right)$
$P\left(\text{doubles}\right)$
$P\left(\text{no doubles}\right)$
Empirical results
With Replacement Without Replacement
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ ) ( __ , __ ) ( __ , __ )
Empirical probabilities
With Replacement Without Replacement
$P\left(\text{2 reds}\right)$
$P\left({G}_{2}\text{|}{R}_{1}\right)$
$P\left(\text{no yellows}\right)$
$P\left(\text{doubles}\right)$
$P\left(\text{no doubles}\right)$
## Discussion questions
1. Why are the “With Replacement” and “Without Replacement” probabilities different?
2. Convert $\text{P(no yellows)}$ to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to 4 decimal places.
• Theoretical “With Replacement”: $\text{P(no yellows)}=$
• Empirical “With Replacement”: $\text{P(no yellows)}=$
• Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why?
3. If you increased the number of times you picked 2 M&M’s to 240 times, why would empirical probability values change?
4. Would this change (see (3) above) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know?
5. Explain the differences in what $P\left({G}_{1}\text{AND}\phantom{\rule{2pt}{0ex}}{R}_{2}\right)$ and $P\left({R}_{1}\text{|}{G}_{2}\right)$ represent. Hint: Think about the sample space for each probability.
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# What's the Difference between Mean, Median, and Mode?
Okay math students, here’s a head-scratcher for you. What is the difference between the mean, the median and the mode? StraighterLine has created a terrific video that explains it all.
## Mean, Median and Mode
All three are referred to as measures of central tendency. That’s another way of saying that they indicate central tendencies that are present within a set of numbers. (In order to determine mean, median or mode, you need a set of numbers, like 12, 8, 7, 15, and 7. Those numbers could represent the lengths of fish you just caught, the number of parking spaces on five different streets – just about anything.)
## Okay, here we go . . .
• The Mean – To arrive at the mean of a set of numbers, you add the numbers up and then divide the sum by the number of numbers you were dealing with. (This is just like calculating what is called the “average.”) If your number set is 12, 8, 7, 15, and 7, for example, you would first add all those numbers together (equaling 49) and divide that sum by the number of numbers in your set (5). The result is 9.8. That’s your mean!
• The Median – This is the “middle” number in the set of numbers that you are dealing with. Think of it this way. Half of the numbers in the set are bigger than the median; half of the numbers are smaller. So to arrive at the median, simply put all the numbers in ascending order, with the smallest first. The number that is smack dab in the middle is the median. Aha you say, what if you have an even number of numbers in your set, not an odd number, so two numbers are in the middle? Well, the solution to that problem is easy; if that is the case, the median is midway between the two numbers that are in the middle of the set. So if your two middle two numbers are 7 and 9, your median is 8 (halfway in between).
• The Mode – Okay, this is pretty simple. The mode is the number in a set that appears most frequently. If you are dealing with those numbers we cited at the outset - 12, 8, 7, 15, and 7 – the mode is 7, because 7 appears twice, more than any other number. What if you have a set of numbers in which every number appears just once? Simple! You then have a set of numbers that doesn’t have a mode.
## Putting It All Together
That’s really all you need to know. But here’s an extra credit question for you. Can you use the words mean, median and mode in one sentence? Like these . . .
• “The officer made a mean face at me, which made me drive over the median, which was not my usual mode of driving.”
• “The median number of women in France wears the latest modes, but they look mean anyway.”
See what you can do, and post your sentences here. Happy mathing!
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Proof
• We first show that every language $L$ accepted by a deterministic finite automaton $DFA$ is regular.
• Let $DFA=(C,\Sigma,\delta,F,s_0)$ be a deterministic finite automaton.
• We construct a grammar $G=(V,T,R,S)$ corresponding to the DFA as follows: * Set $T=\Sigma$, i.e. the terminal symbols of the grammar are the alphabet $\Sigma.$
* Set $A_0\in V$ to correspond to the starting state $s_0.$ * Interprete all other states $s\in C$ of the DFA as some other variables $A_i$ of $G$ i.e. for every $s_i\in C$ there is a variable $A_i\in V$ and vice versa (and $A_0$ stands for the starting state $s_0.$) * Every state transition $s_{i+1}=\delta(s_i,\sigma)$ of the DFA can be interpreted as a production rule $A_i\to \sigma A_{i+1}\in R.$ * If $s_{i+1}\in F$ is a final state, we add $A_{i+1}\to\epsilon$ to the rules, in order to stop the construction of the word in $G$ exactly at the same time when the DFA stops at a final state. * By this construction, the grammar $G$ generates the language of the given DFA, i.e. $L(G)=L(\operatorname{DFA}).$ * Moreover, it is a grammar of a regular language, since the conclusion $Q$ of every production $P\to Q\in R$ is either the empty string $\epsilon$ or it is a concatenation of some non-empty string of terminals ending with a variable $\sigma A_{i+1}$ (in other words, $G$ is right-linear).
• Now, we show that for every regular language there is a DFA accepting it.
• Let $G=(V,T,R,S)$ be a grammar of a regular language,
• It suffices to find a non-deterministic finite automaton $NFA=(C,\Sigma,\Delta,F,s_0)$ which accepts the same language. It will then follow from Rabin-Scott theorem, than there is an appropriate DFA. * Set $\Sigma=T$, i.e. the alphabet of the NFA are exactly the terminal symbols of the grammar $G$. * Create a separate state $s_i\in C$ of the NFA or every variable $A_i\in V$ of the grammar $G.$ * Since $G$ is regular, every production rule is left-linear or right-linear. * Without loss of generality1, let $G$ be right-linear, i.e. let every of its production rules be either of the form $A_i\to\epsilon$ or $A_i\to \sigma A_{i+1}.$ * In the case $A_i\to \sigma A_{i+1},$ we mark the transition from the state $s_i$ to the state $s_{i+1}$ of the NFA with $\sigma.$ * In the case $A_i\to \epsilon,$ the state $s_i$ is a final state of the $NFA,$ i.e. we set $F=\{s_k\mid (A_k\to\epsilon)\in R\}.$ * Note, that $NFA$ is in general non-deterministic since the grammar $G$ can have more than one production rule for a variable $A_i$ and an input character $\sigma.$
Thank you to the contributors under CC BY-SA 4.0!
Github:
References
Bibliography
1. Erk, Katrin; Priese, Lutz: "Theoretische Informatik", Springer Verlag, 2000, 2nd Edition
2. Hoffmann, Dirk: "Theoretische Informatik, 3. Auflage", Hanser, 2015
Footnotes
1. A similar construction can be found for left-linear grammars. It is left for the reader as an exercise to recognize that the argument works also for left-linear grammars.
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# In $\Delta ABC$, if $r=1,R=3,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}s=5$, then the value of ${a}^{2}+{b}^{2}+{c}^{2}$ is _____
Video Solution
Text Solution
Verified by Experts
## $r=\frac{\Delta }{s}$ $s=5\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}a+b+c=10$ $\Delta =\frac{abc}{4R}\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}abc=60$ Now ${\Delta }^{2}=s\left(s-a\right)\left(s-b\right)\left(s-c\right)$ or $5=\left(5-a\right)\left(5-b\right)\left(5-c\right)$ $=125-25\left(a+b+c\right)+5\left(ab+bc+ca\right)-abc$ $\therefore Ab+bc+ca=38$ or ${a}^{2}+{b}^{2}+{c}^{2}={\left(a+b+c\right)}^{2}-2\left(38\right)=24$
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If in a ΔABC, right angled at B, s - a = 3, s - c = 2, then the values of a and c respectively are
A2,3
B3,4
C4,3
D6,8
• Question 2 - Select One
## In ΔABC, if 2R+r=r2 then ∠B=
Aπ3
Bπ4
Cπ6
Dπ2
• Question 3 - Select One
## In ΔABC,if 2s=a+b+c, then the value of s(s−a)bc−(s−b)(s−c)bc=
AsinA
BcosA
CtanA
DcosA
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
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### Instructions for Choices between Mixed Gambles
In this study, you are asked to choose between two gambles that each have three equally likely outcomes. You can imagine that there are two containers, one for each gamble that have exactly 99 tickets in them. There are 33 tickets for each outcome, meaning that each of the three outcomes is equally likely. Because there are three equally likely outcomes in each gamble, the probability to receive the first, second, or third outcome is 1/3, 1/3, and 1/3.
For example, consider the first choice below:
W1.
First Gamble Second Gamble LOSE \$100 LOSE \$100 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
In this case the First Gamble (on the left) has equal chances of losing \$100, winning \$10, or winning \$98. The second gamble (on the right) has equal chances of losing \$100, winning \$36, or winning \$40. Some people will prefer the first gamble and some will prefer the second gamble. If you choose the first gamble, click the button on the left underneath it (on the left); to choose the second gamble, click the second button (on the right). BREAK EVEN means that you neither win or lose. WHen you see BREAK EVEN twice in a gamble, that means there are two chances to BREAK EVEN in that gamble.
#### Warmup Trials Choices between Mixed Gambles
W2.
First Gamble Second Gamble WIN \$100 WIN \$50 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
W3.
First Gamble Second Gamble LOSE \$80 LOSE \$100 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
W4.
First Gamble Second Gamble WIN \$100 WIN \$55 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
W5.
First Gamble Second Gamble WIN \$100 WIN \$10 WIN \$10WIN \$98 WIN \$3BREAK EVEN First Gamble Second Gamble
1.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$44WIN \$44 LOSE \$4WIN \$4 First Gamble Second Gamble
2.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$44WIN \$44 BREAK EVENBREAK EVEN First Gamble Second Gamble
3.
First Gamble Second Gamble LOSE \$100 LOSE \$100 WIN \$10WIN \$98 WIN \$44WIN \$48 First Gamble Second Gamble
4.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
5.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$44WIN \$44 LOSE \$4WIN \$4 First Gamble Second Gamble
6.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
7.
First Gamble Second Gamble LOSE \$4 LOSE \$4 WIN \$10WIN \$98 WIN \$48WIN \$52 First Gamble Second Gamble
8.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$98LOSE \$10 LOSE \$40LOSE \$36 First Gamble Second Gamble
9.
First Gamble Second Gamble LOSE \$100 LOSE \$100 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
10.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$98LOSE \$10 LOSE \$48LOSE \$44 First Gamble Second Gamble
11.
First Gamble Second Gamble WIN \$100 WIN \$100 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
12.
First Gamble Second Gamble LOSE \$4 LOSE \$4 WIN \$10WIN \$98 WIN \$40WIN \$44 First Gamble Second Gamble
13.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
14.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$44WIN \$44 LOSE \$12LOSE \$4 First Gamble Second Gamble
15.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$98LOSE \$10 LOSE \$52LOSE \$48 First Gamble Second Gamble
16.
First Gamble Second Gamble WIN \$100 WIN \$100 WIN \$10WIN \$98 WIN \$44WIN \$48 First Gamble Second Gamble
17.
First Gamble Second Gamble LOSE \$4 LOSE \$4 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
18.
First Gamble Second Gamble BREAK EVEN BREAK EVEN WIN \$10WIN \$98 WIN \$44WIN \$48 First Gamble Second Gamble
19.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$98LOSE \$10 LOSE \$52LOSE \$48 First Gamble Second Gamble
20.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$44WIN \$44 BREAK EVENBREAK EVEN First Gamble Second Gamble
21.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
22.
First Gamble Second Gamble WIN \$4 WIN \$4 WIN \$10WIN \$98 WIN \$48WIN \$52 First Gamble Second Gamble
23.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$44WIN \$44 BREAK EVENBREAK EVEN First Gamble Second Gamble
24.
First Gamble Second Gamble LOSE \$4 LOSE \$4 WIN \$10WIN \$98 WIN \$44WIN \$48 First Gamble Second Gamble
25.
First Gamble Second Gamble BREAK EVEN BREAK EVEN WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
26.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$98LOSE \$10 LOSE \$44LOSE \$40 First Gamble Second Gamble
27.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$98LOSE \$10 LOSE \$44LOSE \$40 First Gamble Second Gamble
28.
First Gamble Second Gamble BREAK EVEN BREAK EVEN WIN \$10WIN \$98 WIN \$40WIN \$44 First Gamble Second Gamble
29.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$44WIN \$44 BREAK EVENBREAK EVEN First Gamble Second Gamble
30.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$98LOSE \$10 LOSE \$40LOSE \$36 First Gamble Second Gamble
31.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$44WIN \$44 BREAK EVENBREAK EVEN First Gamble Second Gamble
32.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$98LOSE \$10 LOSE \$44LOSE \$40 First Gamble Second Gamble
33.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$98LOSE \$10 LOSE \$48LOSE \$44 First Gamble Second Gamble
34.
First Gamble Second Gamble WIN \$4 WIN \$4 WIN \$10WIN \$98 WIN \$44WIN \$48 First Gamble Second Gamble
35.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$44WIN \$44 LOSE \$4WIN \$4 First Gamble Second Gamble
36.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$44WIN \$44 LOSE \$12LOSE \$4 First Gamble Second Gamble
37.
First Gamble Second Gamble WIN \$4 WIN \$4 WIN \$10WIN \$98 WIN \$40WIN \$44 First Gamble Second Gamble
38.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$98LOSE \$10 LOSE \$44LOSE \$40 First Gamble Second Gamble
39.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$98LOSE \$10 LOSE \$40LOSE \$36 First Gamble Second Gamble
40.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$44WIN \$44 LOSE \$12LOSE \$4 First Gamble Second Gamble
41.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$98LOSE \$10 LOSE \$44LOSE \$40 First Gamble Second Gamble
42.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$98LOSE \$10 LOSE \$48LOSE \$44 First Gamble Second Gamble
43.
First Gamble Second Gamble LOSE \$100 LOSE \$100 WIN \$10WIN \$98 WIN \$40WIN \$44 First Gamble Second Gamble
44.
First Gamble Second Gamble WIN \$100 WIN \$100 WIN \$10WIN \$98 WIN \$40WIN \$44 First Gamble Second Gamble
45.
First Gamble Second Gamble BREAK EVEN BREAK EVEN WIN \$10WIN \$98 WIN \$48WIN \$52 First Gamble Second Gamble
46.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$98LOSE \$10 LOSE \$52LOSE \$48 First Gamble Second Gamble
47.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$98LOSE \$10 LOSE \$52LOSE \$48 First Gamble Second Gamble
48.
First Gamble Second Gamble WIN \$100 WIN \$100 LOSE \$98LOSE \$10 LOSE \$40LOSE \$36 First Gamble Second Gamble
49.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$44WIN \$44 WIN \$4WIN \$12 First Gamble Second Gamble
50.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$44WIN \$44 LOSE \$4WIN \$4 First Gamble Second Gamble
51.
First Gamble Second Gamble LOSE \$100 LOSE \$100 WIN \$10WIN \$98 WIN \$48WIN \$52 First Gamble Second Gamble
52.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$98LOSE \$10 LOSE \$52LOSE \$48 First Gamble Second Gamble
53.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$98LOSE \$10 LOSE \$40LOSE \$36 First Gamble Second Gamble
54.
First Gamble Second Gamble LOSE \$4 LOSE \$4 LOSE \$44WIN \$44 LOSE \$12LOSE \$4 First Gamble Second Gamble
55.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$98LOSE \$10 LOSE \$48LOSE \$44 First Gamble Second Gamble
56.
First Gamble Second Gamble WIN \$4 WIN \$4 WIN \$10WIN \$98 WIN \$36WIN \$40 First Gamble Second Gamble
57.
First Gamble Second Gamble WIN \$100 WIN \$100 WIN \$10WIN \$98 WIN \$48WIN \$52 First Gamble Second Gamble
58.
First Gamble Second Gamble BREAK EVEN BREAK EVEN LOSE \$44WIN \$44 LOSE \$12LOSE \$4 First Gamble Second Gamble
59.
First Gamble Second Gamble LOSE \$100 LOSE \$100 LOSE \$44WIN \$44 LOSE \$4WIN \$4 First Gamble Second Gamble
60.
First Gamble Second Gamble WIN \$4 WIN \$4 LOSE \$98LOSE \$10 LOSE \$48LOSE \$44 First Gamble Second Gamble
61. Email Address: years.
62. Gender: Are you Male or Female?
Female
Male
63. Age:
64. Nationality (country of birth):
65. CWID:
Please check your answers. When you are done, push the button below.
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# Finding point of collision (moving circles + time)
For two circles moving linearly, it's easy enough to calculate the time of the collision: http://twobitcoder.blogspot.com/2010/04/circle-collision-detection.html
This assumes that the circles have fixed starting points, and fixed movement paths, and calculates the time of the collision.
Is it possible to do it the other way around:
Circle 1: Starting point X1,Y1 velocity VX1,VY1 (fixed starting point, fixed linear movement path), radius R1 Circle 2: Starting point X2,Y2 velocity scalar (1m/sec, etc) (fixed starting point, fixed speed, unknown direction), radius R2
Is it possible to determine the collision position of the two circles for a minimum travel time?
I.E. Circle 1 starts at 0,0 and moves at speed 1,0 (1 unit to the right per time) Circle 2 starts at 5,5 and can move 1 unit per time What would the collision position be (or the VX2,VY2 circle 2 would need to move in) in order for the 2 circles to collide at the lowest time T. Radius of both circles is 1
In this example, a solution would be somewhere around Circle 1 being at point 3,0 at time 3. The question feels fairly complex as you have unknown variables: collision point, collision time, VX2, VY2. Although VX2 and VY2 would be constrained by |VX1|+|VX2| = 1.
The reason for the question is to tell circle 2 where it should move to in order to 'catch' circle 1.
The brute force solution would be to check the position of circle 1 at every time interval, and calculate if circle 2 would collide with circle 1 if told to move to that point - but you could miss collision points of the circles were moving fast, or get a sub-optimal point, etc.
-
There are two keys to solving this simply:
• Firstly the points reachable from `x2` in time `t` form a circle centered on `x2`.
• Secondly the first moment that the circles can touch they must be touching tangentially.
These combine to tell us that the points `x2(0)`, `x2(T)`, the contact point and `x1(T)` are all colinear.
If we draw a diagram showing this we get a single quadratic equation in t:
``````|| x2(0) - x1(0) - v1 t ||^2 = (r1+r2+t)^2
``````
Which can easily be solved for t.
To get the direction for `v2` we just need to use the unit vector in the direction `x1(T)-x2(0)`.
-
Michael, would you please indicate which variables are 'a', 'b', & 'c' of the quadratic? I assume the 'x' is t which we're trying to solve for. Thanks! – Steve H Sep 19 '11 at 16:13
I've put a clearer description in PDF form here sites.google.com/site/therobotsbrain/files/… hopefully that will make things clearer (writing math on stack overflow is not easy..) – Michael Anderson Sep 20 '11 at 2:18
thanks. much appreciated – Steve H Sep 20 '11 at 12:36
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mersenneforum.org Sequence
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2005-09-13, 07:00 #1 Citrix Jun 2003 157610 Posts Sequence Generate a sequence where the N+1 term is the product of all previous terms +1 and the t(N+1) is divisible byp(N+1) ie the Nth prime. SO t(1) divisible by 1 t(2)=t(1)+1 and divisble by 2 t(3)=t(2)*t(1)+1 and divisble by 3 t(4)=t(3)*t(2)*t(1)+1 and divisble by 5. and so on... Good luck this is a hard problem, lets see who can generate the longest sequence. Can you find a algorithm to generate this sequence to some prime p? Citrix
2005-09-13, 07:24 #2 Citrix Jun 2003 157610 Posts There are multiple solution till each p but the smallest t(1) will be the correct one. Citrix
2005-09-14, 13:07 #3 Jushi Sep 2005 UGent 6010 Posts Let T = t(1). Then: t(2) = T + 1 t(3) = t(2)*t(1) + 1 = (T + 1)*T + 1 = T^2 + T + 1 t(4) = t(3)*t(2)*t(1) + 1 = (T^2 + T + 1)*(T + 1)*T + 1 = T^4 + 2T^3 + 2T^2 + T + 1 By looking modulo 5, the condition that t(4) is divisible by 5 is impossible. The best sequence is thus: t(1) = 1 t(2) = 2 t(3) = 3
2005-09-14, 23:00 #4 Citrix Jun 2003 110001010002 Posts cool solution, but the formula you generated for t4 is prime for 2,4,16,256. All 2^2^n. Pretty cool. Are there any other primes of the form 2^2^n? Citrix
2005-09-14, 23:23 #5 Citrix Jun 2003 23×197 Posts Also prime for 2^(2^6)
2005-09-14, 23:33 #6 Citrix Jun 2003 30508 Posts no primes upto n=18
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# How to write a justification for Line 2, Line 3, and Line 4? Explain why you chose each justification. Line 1 5x – 2 = 3x + 7 Line 2 2x – 2 = 7 Line 3 2x = 9 Line 4 x = 4.5
Oct 9, 2015
See Explanation (below) for Justification of each derived line
#### Explanation:
Line 1:color(white)("XXX") 5x – 2 = 3x + 7
$\textcolor{w h i t e}{\text{XXXXXXXX}}$Subtract $3 x$ from both sides of Line 1
Line 2:color(white)("XXX")2x – 2 = 7
$\textcolor{w h i t e}{\text{XXXXXXXX}}$Add $2$ to both sides of Line 2
Line 3:$\textcolor{w h i t e}{\text{XXX}} 2 x = 9$
$\textcolor{w h i t e}{\text{XXXXXXXX}}$Divide both sides of Line 3 by $2$
Line 4:$\textcolor{w h i t e}{\text{XXX}} x = 4.5$
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Every Question Helps You Learn
# Practice - Algebra Expressions and Equations - 04
Hey there, genius kids! Ready to test your maths skills and prove your knowledge in algebraic expressions and equations? Don't worry, this is not a 'take-away' but a 'plus' addition to your learning side. Get your pencils, papers and thinking caps ready. It's time to solve some algebraic puzzles.
1.
What is the expression for the statement: 'Three times a number n, increased by four'?
n + 3 + 4
n + 12
3n - 4
3n + 4
'Three times a number n' is 3n, and 'increased by four' translates into the operation +4. So, the expression is 3n + 4.
2.
Solve for y: 10y + 5 = 25.
2
3
6
8
Subtract 5 from both sides. It gives you 10y = 20. Dividing both sides by 10, we get y = 2.
3.
Solve for y: y ÷ 3 + 4 = 7
6
7
8
9
First subtract 4 from both sides, getting y ÷ 3 = 7 - 4. Then, multiply both sides by 3 to get y = 9.
4.
Simplify the expression: 2(5x - 3) + 4
10x + 2
10x - 6
10x - 2
10x - 6 + 4
Expand using the distributive property to get 10x - 6 + 4, then combine like terms to get 10x - 2.
5.
If a = 6, b = 5 and c = 3. What is the value of the expression 2a + 3b - c?
10
17
27
24
Substitute a = 6, b = 5, c = 3 in 2a + 3b - c. It becomes (2 x 6) + (3 x 5) - 3, = 12 + 15 - 3 = 24.
6.
What is the solution to the equation 3x + 2 = 2x + 5?
1
2
3
4
Re-arranging the equation gives x = 5 - 2, that is, 3. So, the solution to the equation is 3.
7.
What will be the result when y = 3 is substituted into the equation 2y2 - 4 =?
14
16
18
20
Substitute 'y' with '3' in the equation. Therefore, 2 x (32) - 4, which is 2 x 9 - 4, which results in 18 - 4, and it's 14.
8.
Find the value of 3x - 4 when x = 2.
1
2
6
10
Substitute 'x' with '2' in 3x - 4. Therefore, (3 x 2) - 4 = 6 - 4, which is 2.
9.
Solve 12a + 2 = 5.
4
6
8
10
Subtract 2 from both sides, getting 12a = 3. Multiply both sides by 2 to solve for a, so a = 6
10.
Expand the expression: 4(2x + 3)
8x + 12
6x + 7
8x + 9
10x + 12
By the distributive law, 4(2x + 3) becomes (4 x 2x) + (4 x 3), which is 8x + 12.
Author: Graeme Haw
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微积分笔记（42）——多变量函数的微分学（5）
Contents
多变量函数的微分学
带约束条件的极值问题
实例
$$S = xy + \frac{2(x + y)V}{xy}$$
条件极值问题
$$\begin{cases} \max f(\mathbf{x}, \mathbf{y}) \\ \pmb{\Phi}(\mathbf{x}, \mathbf{y}) = \mathbf{0} \end{cases} \text{或} \begin{cases} \min f(\mathbf{x}, \mathbf{y}) \\ \pmb{\Phi}(\mathbf{x}, \mathbf{y}) = \mathbf{0} \end{cases} \tag{M}$$
问题分析（$m = 1$）
$$\frac{\partial \Phi}{\partial y}(\mathbf{x}_0, y_0) \not = 0$$
$$\frac{\partial F}{\partial x_i}(\mathbf{x}_0) = \frac{\partial f}{\partial x_i} (\mathbf{x}_0, y_0) + \frac{\partial f}{\partial y}(\mathbf{x}_0, y_0) \frac{\partial y}{\partial x_i} (\mathbf{x}_0) = 0,\ \ i = 1, \cdots, n$$
$$\frac{\partial y}{\partial x_i} (\mathbf{x}_0) = -\frac{\partial \Phi}{\partial x_i} (\mathbf{x}_0, y_0) \Big / \frac{\partial \Phi}{\partial y} (\mathbf{x}_0, y_0)$$
$$\frac{\partial f}{\partial x_i} (\mathbf{x}_0, y_0) – \frac{\partial f}{\partial y}(\mathbf{x}_0, y_0) \frac{\partial \Phi}{\partial x_i} (\mathbf{x}_0, y_0) \Big / \frac{\partial \Phi}{\partial y} (\mathbf{x}_0, y_0) = 0$$
$$\lambda = – \frac{\partial f}{\partial y}(\mathbf{x}_0, y_0) \Big / \frac{\partial \Phi}{\partial y} (\mathbf{x}_0, y_0) = 0$$
$$\frac{\partial f}{\partial x_i} (\mathbf{x}_0, y_0) + \lambda \frac{\partial \Phi}{\partial x_i} (\mathbf{x}_0, y_0) = 0$$
$$J f(\mathbf{x}_0, y_0) + \lambda J \Phi(\mathbf{x}_0, y_0) = 0$$
（必要条件）
定理（条件极值必要定理）
$$J f(\mathbf{x}_0, \mathbf{y}_0) + \Lambda J \pmb{\Phi}(\mathbf{x}_0, \mathbf{y}_0) = \mathbf{0}$$
：令 $\pmb{\Phi} = \begin{pmatrix}\varphi_1 & \cdots & \varphi_m\end{pmatrix}^T, \Lambda = \begin{pmatrix}\lambda_1 & \cdots & \lambda_m\end{pmatrix}$，上式化为：
$$J(f + \lambda_1 \varphi_1 + \cdots + \lambda_m \varphi_m) = \mathbf{0}$$
（在 $P$ 点）
Lagrange 乘数法
Lagrange 数乘法（引入辅助函数）
$$L(\mathbf{z}, \Lambda) = f(\mathbf{z}) + \Lambda \pmb{\Phi}(\mathbf{z}), (\mathbf{z}, \Lambda) \in D \times \mathbb{R}^m$$
$$J_\mathbf{z} L(\mathbf{z}_0, \Lambda) = J_\mathbf{z} f(\mathbf{z}_0) + \Lambda \Phi(\mathbf{z}_0) = \mathbf{0}$$
$$J_\Lambda L(\mathbf{z}_0, \Lambda) = \pmb{\Phi}(\mathbf{z}_0) = \mathbf{0}$$
$$J L(\mathbf{z}_0, \Lambda) = \mathbf{0}$$
——满足 L-方程的临界点方程（$n + 2m$ 个方程，$n + 2m$ 个未知数）
条件极值的充分条件
$$H_\mathbf{z} L(\mathbf{z}) = H_\mathbf{z} f(\mathbf{z}) + \Lambda H_\mathbf{z} \pmb{\Phi}(\mathbf{z})$$
定理（条件极值的充分条件）
$$L(\mathbf{z}, \Lambda) = f(\mathbf{z}) + \Lambda \pmb{\Phi}(\mathbf{z})$$
$$A := H_\mathbf{z} L(P, \Lambda) = H_\mathbf{z} L(\mathbf{z}) = H_\mathbf{z} f(P) + \Lambda H_\mathbf{z} \pmb{\Phi}(P)$$
1. 若 $A$ 正定，则 $f(P)$ 为严格条件极小值；
2. 若 $A$ 负定，则 $f(P)$ 为严格条件极大值。
：与无条件极值问题不同，在 $A$ 不定时 $f(P)$ 也有可能取得条件极值。
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Q1) EXAM PAPER 2018. In a bag, there are 6 marbles. Five are red and one is green. Marbles are randomly selected from the bag and not replaced.
a) What is the probability that the green marble will appear on the first draw? [8]
b) What is the probability that the green marble will appear on the fourth draw? [7]
Assume now that there are 10 marbles in the bag: 5 red, 3 greens and 2 blues. Three marbles are randomly selected.
c) Find the probability that the three marbles are of the same colour if marbles are not replaced after each draw. [6]
d) Find the probability that the three marbles are of the same colour if they are replaced in the bag after each draw. [6]
Q2) The University of Southampton IT service has found that on average 6% of the computers on campus need to be repaired each month. Assume that there are 100 computers on campus.
a) What is the probability that exactly 3 computers will break down this month? [7]
b) Use the Poisson approximation to the binomial distribution to estimate the probability that at least 2 computers will stop working this month [6]
c) Assume now that the number of computer failures during a month has a Poisson distribution with mean 1. A computer has just failed. Find the probability that at least 15 days will elapse before there is a further computer failure (you can assume that there are 30 days in a month). [6]
d) Suppose that 15 days has already passed since the last failure. What is the probability that at least 15 days will elapse before there is another computer failure. [6]
Q3) X is a random variable which follows a Binomial distribution. The mean of the distribution is 20 and the standard deviation is 4. The number of trials is 100.
a) Calculate the probability that X is equal to 15. [6]
b) Calculate the probability that X is equal to 100. [6]
c) What is the probability that X is less than or equal to 62? [6]
d) A researcher chooses a value of X called ‘A’ such that the 20% of the distribution lies below A. What is the value of A? [6]
Q4) A probability density function f(x) is given as follows:
f(x)={█(hx-2h for 2≤x≤3@4h-hx for 3≤x≤4@0 otherwise)┤
a) Find the value of h. [8]
b) What is the mean of the distribution? [4]
c) What is the value of F(2.5)? [4]
d) What is the value of F(1)? [4]
e) What is the value of F(4.5)? [4]
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# geometry
Author Message
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Intern
Joined: 20 Feb 2012
Posts: 42
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### Show Tags
12 Mar 2012, 14:16
how to find the length of arcs of circles at the vertices. for example in the figure if radius of circles is given. how to get length of major arcs?
Kudos [?]: 489 [0], given: 6
Current Student
Joined: 12 Sep 2011
Posts: 900
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Concentration: Finance, Finance
GMAT 1: 710 Q48 V40
### Show Tags
12 Mar 2012, 17:00
Could you clarify the question slightly?
If what I understand is correct, you are trying to find the length of the arc of the circles that are located inside the triangle? If this is the case, you would need to know the angles of the vertices of the triangle. For example, lets say your picture is an equalateral triangle (all angles are 60), then to find the length of the arc you would have to be atleast given the radius of the circle and know that the vertices of the triangle lie in the center of the circles. Once you have the radius you would use the formula: 2(pi)r to get the circumference, and that you would multiply the circumference by 60/360 (b/c this is the portion of the circumference of the circle within the 60 degrees of the triangle).
For example, if the radius of the circle was 1, then the circumference of the circle would be 2(pi). You'd multiply the 2(pi) by 60/360 (aka 1/6) to get the length of the arc of 1/3(pi) inside the triangle. If you are looking for the arc length on the outside of the trinable then you would have instead multiplied the 2(pi) by 300/360 to get 1 and 2/3(pi)
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Kudos [?]: 957 [0], given: 114
Re: geometry [#permalink] 12 Mar 2012, 17:00
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# Limit questions
• Feb 15th 2013, 08:33 AM
asilvester635
Limit questions
For (f) I know that the limit as x approaches 0 is 2 for f(x), and the limit as x approaches 0 for g(x) is 0
So is this how you put them together????
2(2) + 3(0)
4 + 0 = 4
• Feb 15th 2013, 09:10 AM
HallsofIvy
Re: Limit questions
Yes, there should be a theorem in your book that says that if $\lim_{x\to a} f(x)= F$ and $\lim_{x\to a}g(x)= G$, then $\lim_{x\to a}(f+ g)(x)= F+ G$.
There should also be theorems that say that, under the same hypotheses,
$\lim_{x\to a}(f- g)(x)= F- G$.
$\lim_{x\to a}(fg)(x)= FG$.
and, provided g(x) is not 0 in some neighborhood of a,
$\lim_{x\to a}(\frac{f}{g})(x)= \frac{F}{G}$.
There is another limit theorem that is not emphasised nearly enough:
If f(x)= g(x) for all x except a, the $\lim_{x\to a}f(x)= \lim_{x\to a}g(x)$
So that if, for example $f(x)= x^2$ for all x and $g(x)= x^2$ for all x except x= 1 and g(1)= 1000, then $\lim_{x\to 1}g(x)= \lim_{x\to 1}f(x)= 1$.
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Search
# Pre-K Math Shapes Worksheets and Activities
Updated: Jan 27, 2022
Are you looking for worksheets and activities to review basic shapes for your pre-k children?
Then, this downloadable worksheets set is perfect for you! It includes 15 basic shapes worksheets on circle, square, triangle, rectangle and oval with amazing illustrations. You will also find free sample pages that you can download at the end of this post.
You can use these worksheets as revision to reach the math objectives:
• Recognise and name basic shapes (circle, square, triangle, rectangle and oval)
• Match identical shapes
• Finding shapes in simple pictures
Your children can learn to recognise the basic shapes through various activities like tracing, coloring, counting, etc.
• 5 individual shape worksheets on circle, square, triangle, rectangle and oval. These shapes worksheets help children to recognise and identify each shape in its geometrical form as well as in objects or things found around them. You can see two sample pages on circle and square below.
You can use the worksheets as reference to some of the objects that you can show your children when introducing the shapes. As an example for the shape circle, you can point out the face of a clock is circular, a donut is also round and the hole in the donut is a circle too, serve circle cookies during snack time or bake your own cookies, etc.
• 5 shapes coloring or tracing worksheets on circle, square, triangle, rectangle and oval that aim to help children visually discriminate each shape and practice counting one - to - one corresponding at the same time.
• 3 variations of shapes matching activities
• 2 finding shapes worksheets that are suitable for your children who are very confident in naming all the shapes and are able to identify different shapes in simple pictures
This basic shapes worksheets set is going to save you a lot time in your planning and it covers all the shapes that pre-k level should know by the end of the year. In addition, the worksheets and activities are of varying level of difficulties giving you choices for suitable activity for your child.
### click the pictures below to download some of the free sample pages from the set.
Please take not that these worksheets are meant for revision and I do not recommend using them as your main activity. Please make sure that you have done enough hands - on activities on shapes before doing the worksheets with your child or students.
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# If n is a positive integer, what is the remainder when 2n is divided
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
We need to determine the remainder when 2n is divided by 8.
Statement One Alone:
n, when divided by 6, leaves remainder 5.
Thus, we see n can be a number such as 5, 11, 16, 21, 26, 31, 35, etc.
When 2(5) = 10 is divided by 8, the remainder is 2.
When 2(11) = 22 is divided by 8, the remainder is 6.
Statement one alone is not sufficient to answer the question.
Statement Two Alone:
3n, when divided by 6, leaves remainder 3.
Thus, we see that 3n can be a number such as 3, 9, 15, 21, 28, etc.
When 3n is 3, n is 1; when 3n is 9, n is 3; when 3n is 15, n is 5; etc.
In other words n will always be an odd number: 1, 3, 5, 7, ...
When 2(1) = 2 is divided by 8, the remainder is 2.
When 2(3) = 6 is divided by 8, the remainder is 6.
Statement two alone is not sufficient to answer the question.
Statements One and Two Together:
Using our two statements, we see the first value for n that satisfies both statements is 5. We also see that in statement two, n can be any odd number. So, another number that would match is n = 11.
When 2(5) = 10 is divided by 8, the remainder is 2.
When 2(11) = 22 is divided by 8, the remainder is 6.
We see that the statements together are still not sufficient to answer the question.
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Asked: If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
n = 6k + 5
2n = 12k +10
The remainder when 2n is divided by 8 = {6, 2}
NOT SUFFICIENT
(2) 3n, when divided by 6, leaves remainder 3.
3n = 6k + 3
n = 2k + 1
2n = 4k + 2
The remainder when 2n is divided by 8 = {6,2}
NOT SUFFICIENT
(1) + (2)
(1) n, when divided by 6, leaves remainder 5.
n = 6k + 5
2n = 12k +10
The remainder when 2n is divided by 8 = {6, 2}
(2) 3n, when divided by 6, leaves remainder 3.
3n = 6k + 3
n = 2k + 1
2n = 4k + 2
The remainder when 2n is divided by 8 = {6,2}
Combining, we get
The remainder when 2n is divided by 8 = {6,2}
NOT SUFFICIENT
IMO E
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Hmmm this question was a little tricky.
So basically the two statements are same
1) n = 6a + 5
2) 3n = 6b + 3
we can get statement 2 from 1.
3*n = 3*6a +3*5
3n = 18a + 15
3n = 18a + 12 + 3
3n = 6(3a+2) + 3
3n = 6b + 3 where b=3a +2.
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Remainders by 6 aren't really related to remainders by 8, so there's no reason to think the statements should be sufficient here. Using Statement 1, n can be 5 and 11, and those values also work with Statement 2. So using both Statements, n can be 5 and 11, and 2n can thus have a remainder of 2, when n=5, or 6, when n=11, when we divide by 8, and the answer is E.
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Question: $$Remainder(\frac{2n}{8}) =$$ ?
Statement 1: n, when divided by 6, leaves remainder 5.
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
Statement 2: 3n, when divided by 6, leaves remainder 3
i.e. 3n = 3, 9, 15, 21, 27, ... etc
i.e. n = 1, 3, 5, 7, 9, 11, 13, 15... etc
i.e. 2n = 2, 6, 10, 14, 18... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
COmbining teh statements
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
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If n is a positive integer, what is the remainder when 2n is divided [#permalink]
GMATinsight wrote:
Bunuel wrote:
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5.
(2) 3n, when divided by 6, leaves remainder 3.
Question: $$Remainder(\frac{2n}{8}) =$$ ?
Statement 1: n, when divided by 6, leaves remainder 5.
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
Statement 2: 3n, when divided by 6, leaves remainder 3
i.e. 3n = 3, 9, 15, 21, 27, ... etc
i.e. n = 1, 3, 5, 7, 9, 11, 13, 15... etc
i.e. 2n = 2, 6, 10, 14, 18... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc. This is incorrect
NOT SUFFICIENT
COmbining teh statements
i.e. n = 5, 11, 17, 23, 29... etc
i.e. 2n = 10, 22, 35, 46, 58... etc
i.e. $$Remainder(\frac{2n}{8}) =2, 6, 3...$$ etc.
NOT SUFFICIENT
There is a typo that affected your calculation as highlighted
when n =17, then 2n=34..............hence $$Remainder(\frac{2n}{8}) =2...$$
There is no 3 at all in any reminder. It is all 2 &6
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
If n is a positive integer, what is the remainder when 2n is divided by 8?
(1) n, when divided by 6, leaves remainder 5 --> insuff: n = 6p+5, if p =0, n = 5, 2n =10, so 2n divided by 8, reminder = 2, but if p =1, n = 11, 2n =22, so 2n divided by 8, reminder = 6
(2) 3n, when divided by 6, leaves remainder 3--> insuff: 3n = 6q+3 => n = 2q+1 (=2r+5), if q =0, n = 1, 2n =2, so 2n divided by 8, reminder = 2, but if q =1, n = 3, 2n =6, so 2n divided by 8, reminder = 6
Combining (1) & (2) we get,
n=6p+5, similar as (1), so not sufficient
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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Re: If n is a positive integer, what is the remainder when 2n is divided [#permalink]
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# Newton 2 on a spreadsheet (2013)
last edited by 10 years, 10 months ago
Class content > Newton's Laws > Newton's Laws as Foothold Principles > Newton's 2nd > Newton 2 as a stepping rule
Prerequisites
Once we view Newton's second law as a pair of stepping rules for position and velocity,
it's really easy to set it up on a spreadsheet. This allow you to use N2 to predict the future motion of an object if you know the initial position and velocity and what forces it will encounter as it moves. Here's how to do it.
1. Set up a column for the times, the positions, the velocities, and the forces.
2. Set up cells that will contain the mass, m, the time interval, dt, and any parameters that the forces depend on.*
3. Put starting values for the time, position, and velocity in the first row below the variable names.
4. In the next row put equations to update the variables.
5. Copy these down to as many rows as you want to calculate.
Here's a specific example. Suppose the force is "-kx". That is, it is proportional to the position and pushes back by an amount proportional to x. This is like a spring. Set up cells that contain the parameters dt (time interval), m (mass), and k (spring constant). If you have
• time in column A
• position in column B
• velocity in column C
• force in column D
• acceleration in column E
then here is what your second row would look like. The cell label is shown in parentheses first. This is not to be included; it's just to show you where this equation goes. We assume your first data is in row 2 since labels are in row 1. Note that if you name the cells with parameters in them, you can use the names -- dt, k, m -- in your equations.* If you don't, you have to use the cell label but with dollar signs to keep it from changing the cell ID when you copy down. (Like this: if "dt" is in cell H3, you have to write it as \$H\$3.)
(A3) = A2 + dt (B3) = B2 + C2*dt (C3) = C2 + E2*dt (D3) = -k*A2 (E3) = D3/m
This is how these are read conceptually:
New time = old time + dt New position = old position + current velocity * dt New velocity = old velocity + old acceleration * dt Force = function of old position New acceleration = force/m
For this to work, the time step has to be small enough that the force doesn't change a lot over the step. We are basically assuming that the force is a constant over the time step and if the time step is too big that may not work. To check if your time step is good enough, take a smaller step -- say half the size of your original step -- and do twice as many steps. If your original time step was small enough this should give you the same answer at the same final time.
* In Excel, it's easy to actually give the cells containing parameters names so you can refer to them by name in your equations. Decide which cell you want to put the information in. (I always put the name of the parameter with an = in the cell to the left so I remember what's what.) Put your number in that cell. For example, in the function line in the clip from a spreadsheet shown below, the function box shows that the value in the selected cell is 0.1 and the name box (at the top left) shows that the cell is cell labeled "H3".
If you go into the name box at the upper left where it says H3, select it and now type in "dt", it will give that name to the value in the cell. You can then use that name in your formulas. Once you've done that, it looks like this.
Joe Redish 9/17/11
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# Time dilation (special relativity) problem
palex3
## Homework Statement
This is a pretty classical problem. A plane goes from city A to city B. The distance between the two cities (measured on the ground) is 500 km. The plane is traveling at 0.2c. How long does the trip take for the pilot and what is the distance between the two cities for the pilot?
## Homework Equations
I know that
$$L = L_0/\gamma$$
and
$$\Delta T = \gamma \Delta T_0$$
## The Attempt at a Solution
The proper distance is 500 km, so I can work out that the pilot sees a distance of approximately 489.9 km. My problem is with the time. I know that I can divide 489.9 by 0.2c to get the time for the pilot: 0.00817 seconds. This is the correct answer. But why is it wrong for me to divide 500 km by 0.2 c (=0.00833s) and consider that the proper time, after which I transform it by multiplying by gamma (the result is 0.0085018 s)? I noticed that if I consider the time on the ground to be the "moving" time and the time for the pilot to be the proper time, using the formula gives me the correct time. But it doesn't make sense to me, since I started considering the ground as the rest frame. Why the switch? My guess would be that it has something to do with needing two clocks on the ground and one in the air, but I still don't really understand the asymetry.
Homework Helper
One common idea for proper time is that it is the time interval between two events in the reference frame at which they both occur at the same position coordinate.
Do you see how this means we can't set 0.00833 to be the proper time? (What are the x coordinates for the two events (plane is at A) and (plane is at B) in the ground reference frame? What are the x coordinates for the two events in the planes reference frame?)
Mentor
My guess would be that it has something to do with needing two clocks on the ground and one in the air, but I still don't really understand the asymetry.
That's it exactly. The two events in this problem are: (1) Plane passes city A, and (2) Plane passes city B. On the plane, the time between those events is measured on a single clock. To the earth, that plane clock is moving and the "time dilation" formula can be applied. ("Moving clocks run slow.")
On the earth, the times of those events are measured on two different clocks. According to the plane observer, those Earth clocks are not synchronized (clock B is ahead of clock A), so the plane observer disagrees with the Earth observers as to how much time elapsed on Earth clocks during his trip.
But Distance = speed * time always works, as long as you stick to a single frame.
palex3
That clears it up, thanks a lot to both of you!
My problem was that I didn't see the difference between proper time (and lenght). Or rather, I didn't see what was special about it. In my head, time was time, no matter where you measure it (as long as it's in the same reference frame). Part of the cause for that missunderstanding is the naming convention in my physics book, part was me not being completely in the ''relativity'' mindset and part was my teacher using youtube videos to try to explain the issue.
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Thought for the day: If you look through your teacher's materials (or even just look around you in the world!), you may find your own ideas for Creative Math Prompts.
Concepts: interior angles in polygons; finding angles by reasoning (instead of measuring); symmetry; properties of isosceles triangles; analyzing and extending patterns; describing patterns algebraically
Beginning
I notice a pentagon in the middle of a five-pointed star.
I notice that the angles at the tips of the stars are marked.
I notice that the star is symmetrical (order-5 rotational symmetry).
I notice that the star is made of five congruent isosceles triangles attached to a pentagon.
I notice five larger isosceles triangles that include the pentagon.
I notice that I could extend the picture outward or inward forever.
I wonder if I can calculate the angles at the tips of the star.
Exploring
I notice that once I know the angles in the pentagon, I can find the other angles without measuring.
I wonder what the angles at the star-tips would be if I built the star from other regular polygons.
I wonder if there is a pattern to the star-tip angles as the number of sides increases.
I wonder if I can create a formula for the star-tip angle in terms of the number of sides of the polygon.
I notice that polygons with seven or more sides have multiple layers of stars.
Creating
As they notice and wonder, students may create their own drawings of stars from polygons with more and more sides. Their drawings may be inspired by things that they have wondered about. For example, they may
Add to the drawing in the prompt in order to analyze it and make new discoveries.
Create stars from hexagons, heptagons, octagons, etc.
Extend their drawings inward and outward to create more polygons and more stars.
Try to create stars from irregular polygons.
For each drawing they create, they may calculate angles (or other measurements), look for patterns, develop equations, etc.
Reflecting and Extending
I notice that the star-tip angle for the 5-pointed star is 36°.
I notice that there are many ways to calculate this angle.
I notice that the star-tip angles increase more and more slowly as the number of sides increases.
I notice that the star-tip angles seem to get closer and closer to 180°, but never get there.
I notice that different strategies for finding star-tip angles lead to different formulas.
I wonder if I can use algebra to prove that all of the expressions in these formulas are equivalent.
I wonder if can find patterns and formulas for the angles in the multiple layers of stars (for polygons with 7 or more sides).
I wonder how the different side lengths in the picture compare (what their ratios are).
I wonder how the areas of the pentagon compare to the areas of the triangles.
I wonder how the sides lengths and areas would compare if I extended the picture outwards or inwards by making more pentagons and stars.
Notes
This image can inspire endless observations and questions. In these notes, I focus on angles. For a more structured activity built around these ideas, see Exploration 3: Starstruck! in my book Advanced Common Core Math Explorations: Measurement and Polygons.
Students who have learned about sums of interior angles in polygons will know (or be able to figure out) that each interior angle in the pentagon measures 108°. From there, they may use many different strategies (involving vertical angles, supplementary angles, interior angles in triangles, symmetry, etc.) to determine that the angles at the star-tips are 36°. Some may notice that the sum of the star-tip angles is 36 • 5 = 180°—the same as the sum of the interior angles of a triangle, and they may wonder why this happens.
From this point, you may explore stars built by extending the sides of other regular polygons. For example, stars built from hexagons have star-tip angles of 60°, and stars built from regular octagons have star-tip angles of 90°. As the number of sides increases, the star-tip angles increase ever more slowly, gradually approaching—but never quite reaching—180°. Depending on the strategies you use and the observations you make, you can find many different formulas that calculate the angles of the star-tips from the number of sides of the regular polygon. Most often, students find a very complex formula:
Sometimes, they discover simpler equivalent formulas such as
The possibilities for further questions and discoveries are nearly endless! For example, when you extend the sides of polygons that have more sides, you begin to get multiple layers of stars to explore within a single picture.
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# A juice manufacturer organized taste-testing sessions
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A juice manufacturer organized taste-testing sessions featuring four brands of orange juice, A, B C and D. All customers who participated told the organizer which variety they thought was the most similar to freshly-squeezed orange juice. 61% preferred brand A and exactly half as many preferred brand B. Only 65 chose brand C. Which of the following could be the number of customers who preferred brand D?
(A) 1 (B) 8 (C) 14 (D) 20 (E) 27
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30 Sep 2006, 19:08
% who prefer either C or D = 100 -(61 +61/2) = 8.5%
# who prefer C = 65
# who prefer D = x
so 65+x = 8.5/100 (total)
Alos RHS has to be an integer as # of people cannot be fractions.
so smallest interger value for RHS = 85
=> x = 20
anyone else?
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# rectangle
• Jan 25th 2010, 11:11 AM
sri340
rectangle
A rectangular floor measures 3.5 meters by 4.5 meters. If 1 inch = 2.54 cm, what
is the area of the floor, in square inches? Express your answer as a decimal to the
nearest tenth
• Jan 25th 2010, 12:52 PM
masters
Quote:
Originally Posted by sri340
A rectangular floor measures 3.5 meters by 4.5 meters. If 1 inch = 2.54 cm, what
is the area of the floor, in square inches? Express your answer as a decimal to the
nearest tenth
Hi sri340,
Let's find the area of the floor in square meters first.
$\displaystyle A=(3.5)(4.5)=15.75 m^2$
Now, for a little conversion.
$\displaystyle \frac{15.75 m^2}{1}\cdot \frac{(100 cm)^2}{1 m^2} \cdot \frac{(1in)^2}{(2.54 cm)^2}=$ your answer in square inches.
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## Wednesday, May 23, 2007
### Area of Circles Homework
Here is the work from today.
## Tuesday, May 22, 2007
### Circumference
Here is the homework for today. Please have it done by tomorrow.
## Thursday, May 17, 2007
### More Surface area Practice
Here are some more examples of surface area for you to practice.
Do not forget
1. total shapes
2. units of measurement
Good Luck Happy Weekend.....
## Tuesday, May 15, 2007
### Geometry Work
We are really flying through the Geometry unit. Here is some extra work you can do to become better mathematicians.
Thank you BBC Skillwise for the links
Area of a Rectangle
1. Key vocabulary
2. How to Calculate Area
3. Online Quiz 1
4. Online Quiz 2
5. Online Quiz 3 (may 15th Homework)
Worksheets that will make you think.
Area of a Triangle
1. Triangle Online Quiz(May 15th Homework)
Area of a Circle
1. Definition of How to calculate the area of a circle
2. Quick Online quiz
3. Pdf worksheet on Circle area
Area of Mixed Shapes
1. Introduction
2. Quick online quiz
3. PDF worksheet on Compound Shapes
All of this moves us towards finding total surface area. Please leave comments behind in the chat box or as comments.
## Thursday, May 10, 2007
### Pythagoras Word Problems
Try answering these word problems. You can answer them here at the blog or at your class blog.
1. Tony has got his kite stuck at the top of a very tall tree. He knows the string on his kite is 20 m long. When he pulls the string tight and holds the very end on the ground it touches 17 m from the bottom of the tree. If the ground is flat, how tall is the tree?
2. Susan is making a jump to ride her bike over. She uses a 1m plank of wood to make the jump, and raises the end 50 cm off the ground with a second piece of wood. How far along the ground from the end of the plank that touches the ground is the second piece of wood?
3. Robert is using a 5m ladder to climb in his upstairs bedroom window. He finds that if he puts the base of the ladder 3.3m from the wall the top leans on the windowsill. How high from the ground is the windowsill?
4. The guy wires holding up an 11m radio transmitter join 2m from the top of the aerial, and are anchored to the ground 7m from the base of the aerial. How long are the guy wires?
5. Tim is standing across a river from Mary. They discover that if they stand exactly opposite each other they can each hold one end of a 5m rope. How long would the rope have to be to reach if Tim moved 8m downstream?
### Some Pythagoras work
Here are some questions you can do to practice your work.
Draw a picture to solve this question
Robert is using a 5m ladder to climb in his upstairs bedroom window. He finds that if he puts the base of the ladder 3.3m from the wall the top leans on the windowsill. How high from the ground is the windowsill?
## Monday, May 07, 2007
### Video about Bullying
How does this video about bullying make you feel?
## Wednesday, May 02, 2007
### Your last Unproject
Here are the instructions for your final unproject. Pay attention to the due dates and ask questions on Tuesday.
The home wiki of this project is spunproject07.pbwiki.com
# Your Final Unproject
Important Dates
May 7th Unproject is Launched
May 26th Part 1 is due. (This will provide you with great material to study from for your exam)
June 14th Part 2 is due.(You will have access to the Math lab during the week of June 11th to 14th)
This unproject is part of the Term 4. It will be worth 10% of that terms mark.
### Instructions
Your job is to be an expert in 2 topics that we have learned this year. You will need to check your math portfolio to check the topics that we have covered.
1. Data Managemant and Graphs
2. Probability
3. Square Roots
4. Percent
5. Ratio
6. Proportional Reasoning
7. Fractions (only if you did not participate in the first Unproject)
8. Algebra
9. Geometry
You will be responsible to create a background page that explains both of your topics. It must be dynamic and hook the audience. You may create the background information at
• your class blogsite (just create the necessary link)
• here at the Final Unproject
• wikispaces (Just create the necessary link)
Your background page must include
• all needed math information
• pictures, graphics and other visual aides
• One or more word problems that are solved correctly.
Your Final Unproject is your last math assignment of Grade 8. It is a celebration of what you have learned. Just like the first unproject you choose what it will be. It can be anything but you must check with me first.
Previous unprojects were
### The timing of this project is to help you study for your final exam. If you choose to you can use this project as a way to study for exams.
You may work with a partner but you are both responsible for the final project. Once you have chosen a partner there is no changeing. There will be no groups of 3. Sorry.
Scoring Rubric
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# Translation:On the Apparent Mass of the Ions
On the Apparent Mass of the Ions (1900)
by Hendrik Lorentz, translated from German by Wikisource
In German: Über die scheinbare Masse der Ionen, Physikalische Zeitschrift. 2, 1900/1, pp. 78-80
H. A. Lorentz (Leiden)
On the apparent mass of the ions.
It is known that by observations of cathode rays we were able to derive the ratio ${\displaystyle {\tfrac {e}{m}}}$, i.e. the ratio between the charge of an ion ${\displaystyle e}$ and its mass ${\displaystyle m}$. The question arises, what is meant by that mass. In any case we must attribute an apparent mass to the ion, as it generates a certain energy in the ether by virtue of its motion. This apparent mass will be denoted by ${\displaystyle m_{0}}$. It is possible that the ion also possesses a real mass in the ordinary sense of the word; in this case, ${\displaystyle m_{0}. If this is not the case, then ${\displaystyle m_{0}=m}$.
So we have the inequality
${\displaystyle {\frac {e}{m_{0}}}>{\frac {e}{m}}{,}}$
when there still is a real mass besides the apparent mass; otherwise
${\displaystyle {\frac {e}{m_{0}}}={\frac {e}{m}}.}$
So we want to write
${\displaystyle {\frac {e}{m_{0}}}\geqq {\frac {e}{m}}{,}}$
where ${\displaystyle {\tfrac {e}{m}}=10^{7}}$ is.
Now
${\displaystyle m_{0}={\frac {8}{3}}\pi R\sigma e{,}}$
if we conceive the ion as a sphere, ${\displaystyle R}$ is the radius of this sphere, and ${\displaystyle \sigma }$ means the surface density of the charge.
This formula allows for an interesting conclusion on the radius of the ions. If, namely, we substitute for ${\displaystyle m_{0}}$ the now specified value into the inequality, we obtain an inequality for the radius. We have
${\displaystyle 4\pi R^{2}\cdot \sigma =e{,}}$
thus
${\displaystyle m_{0}={\frac {8}{3}}\pi R\sigma e={\frac {8}{3}}\pi Re\cdot {\frac {e}{4\pi R^{2}}}={\frac {2e^{2}}{3R}}}$
and thus
${\displaystyle {\frac {e}{m_{0}}}={\frac {3R}{2e}}{,}}$
and
${\displaystyle {\frac {3R}{2e}}\geq 10^{7}}$
and
${\displaystyle R>10^{7}\cdot {\frac {2}{3}}e.}$
The magnitude ${\displaystyle e}$ is unfortunately not known. If we take the charge of an ion in a cathode ray to be as great as in an electrolytic hydrogen, and presuppose the size of a hydrogen molecule, we obtain for ${\displaystyle R}$ a magnitude of order ${\displaystyle 10^{-12}}$ cm, that is certainly not an arbitrarily small magnitude, but a lower limit.
The question of whether or not a real mass exists besides the apparent mass of an ion, is extremely important; because by that we touch the question of the relation of ponderable matter with ether and electricity. I am far away to announce a decision, but I would like to cite but a few questions whose resolution can potentially bring us further in that question.
The first question is whether an ion rotates in a magnetic field. Actually, we should expect that. Since if an ion is present, and if a magnetic field is caused, then a rotation arises, as it can easily be derived from the formation of induced currents. Of course this is also the case when the ion flies into an already existing magnetic field. The velocity of rotation will depend on the magnitude of the mass; if only apparent mass is present, and even a corresponding moment of inertia, then the rotation velocity has a certain value. If, however, a real moment of inertia is added, the rotation is slowing down. Unfortunately I can not find any phenomenon, from which we could conclude anything about this rotation.
A second means by which we maybe could decide the question of the relationship between the apparent and real mass is the following:
The value for the apparent mass was given above only in first approximation. If the velocity is such that it is comparable to the velocity of light, then additional magnitudes will be added. For a straight path of the ion we can calculate the intensity of the field and the size of the energy and deduce from that the mass factor. In general, the trajectory will be curvilinear through the influence of the magnetic field, e.g. circular; then the calculation of the mass factor will become more complicated, but it can be carried out. If we denote by ${\displaystyle m_{0}}$ the expression above and ${\displaystyle q}$ is defined as the ratio of the ion velocity to that of light, it follows in second approximation for the apparent mass of the ion in linear motion:
${\displaystyle m_{0}\left(1+{\frac {6}{5}}q^{2}\right){,}}$
while in a circular motion the term with ${\displaystyle q^{2}}$ yields a different coefficient.
These terms of the second order could now perhaps become observable, because the velocity of cathode rays increases up to a third of that of light, hence ${\displaystyle q={\tfrac {1}{3}}}$ and ${\displaystyle q^{2}={\tfrac {1}{9}}}$. To come to a decision, we could think of experiments as they were done by Lenard, to examine the influence of electric forces on the velocity of cathode rays. He has shown that the magnetic deflectability of the cathode rays, which is of course the smaller, the greater the speed, will change when the rays can pass through the space between two charged capacitor plates in the direction of the electric force lines.
We could measure the magnetic deflection in the case of an uncharged capacitor, then in the case of charge in one direction and then for the other direction. Thus we would obtain three different values of deflectability, between which a simple relation should exist, if the terms of second order could be neglected. If we measure each time the magnetic field-force required for a particular deflection, then the squares of these three field forces should form an arithmetic row. A deviation from this relationship would indicate that the terms with ${\displaystyle q^{2}}$ shall not be neglected, and that therefore in any case the apparent mass is noticeable. Detailed specifications could decide concerning the ratio between the real and the apparent mass, and concerning the question whether a real mass exists. It turns out that by Lenard's experiments we were near to decide about the existence of terms of the second order.
(Self-lecture of the lecturer.)
Discussion. (Reviewed by the participants.)
W. Wien. I was recently concerned with similar issues, and would like to stress that Lenard has observed cathode rays at low velocities, triggered under the influence of ultraviolet light. There, he found a small value for the ratio of mass to charge, namely the decrease lies in the sense which is required by the theory.
I have tried to transcend over Lorentz's position, by posing me the question, whether it would suffice when we only consider the apparent mass and omit the inertial mass, and replace it with the electromagnetically defined apparent mass to present the mechanical and electromagnetic phenomena in an uniform way. Because the magnetic and mechanical phenomena are only connected by the energy principle so far. I've tried to pose the question as to whether we could try by Maxwell's theory, to involve mechanics as well. The possibility of an electromagnetic explanation of mechanics was given, after Lorentz has developed a conception of the law of gravity, according to which it would be very similar to electrostatic forces. We would have to think of matter as only composed of very small positive and negative charges, which are within a certain distance from each other. By this condition, the ponderable mass is not constant but depends on the velocity, and namely we obtain terms, depending on even powers of the ratio of velocity to the velocity of light. The numerical factor by which the second term is multiplied, depends on the curvature of the trajectory, but also on the shape of the electric charge. Depending on which different way we choose the form of electrified molecules, we come to other numerical factors. Concerning the ordinary motions on earth, it vanishes because the velocity is very small. Concerning planetary motions we probably can achieve something; because we reach velocities at which we have to consider the terms of second order. On the assumption of a specific type of charge, leading to the simplest electromagnetic field, these terms become relevant in a way, so that the accelerations of two bodies by gravitation are the same up to a slightly different numerical factor, as if the bodies attract each other with constant mass according to Weber's laws. The electromagnetically defined mass comes into play, as if not Newton's, but Weber's law would apply.
Lorentz. In essence, we agree; but Wien already wants to go further than I do. Anyway, it seemed of interest to me to look for means, by which we can come to a decision on the issue discussed. One more thing I would like to add: I made the assumption that the sphere, which forms an ion, is rigid. But perhaps one might think that the sphere would be transformed into an ellipsoid when in motion. This has some similarity with the diversity, that was pointed out by Wien.
Voigt. I would like to pose the question to the lecturer, concerning the reflection of cathode rays; should a rotating ion not be reflected differently, as a non-rotating one?
Lorentz. Certainly, if one imagines that the reflection happens on a surface. But if you look at the reflection, which is more likely to me, as caused by forces that occur at some distance from the surface of the ion, then those surely act on the center, and then the influence of rotation vanishes.
Warburg. What does the theory say about the velocity of the ions during reflection? Does it remain the same?
Lorentz. As far as I know, yes. I have not elaborated on this.
Warburg. Merritt has found that the velocity of reflection has not changed. But the experiments of Cady on the energy of cathode rays are in contradiction to this, so I've thought that the experiments of Merritt may not be completely correct, and maybe we could obtain a velocity change. I wanted to ask if the theory says something in this respect.
Lorentz. I can not say this right now.
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## Brain teaser: dice
Consider two perfectly weighted six sided dice.
What are the chances of rolling a 12?
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1 in 32
3. 1 in 21 and 1 in 7?
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Originally Posted by oldironmudder
1 in 32
you are correct sir!
5. Originally Posted by BTAutoMag
you are correct sir!
I know mine are wrong, but checking the sites I found, neither is this.
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the odds of me being wrong twice in one day is highly unlikly... but I am tired and math is rusty
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In order to roll a 12 you have to roll a 6 on each die
The odds of rolling a 6 on the first one is 1/6, the same on the second. This comes out to a 1/36 chance of rolling a 6 on both of them together.
With a seven it does not matter what is rolled on the first one. 1-6 will still allow you to get to seven with the roll of the other one. That means you are only concerned about one of them and the chance is 1/6 that it will hit the right number to add up to 7
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OMFG i wasnt wrong... I'm just blind
that was what I came up with I just saw the 32 and thought 36
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Actually a 7 can be by 6/1, 5/2, 4/3. So how does this change the odds? There are three combos, so does it increase or not affect the probability?
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Originally Posted by 1stTarget
Actually a 7 can be by 6/1, 5/2, 4/3. So how does this change the odds? There are three combos, so does it increase or not affect the probability?
A seven can be had regardless of what is rolled on the first dice.
The first one can be a 1, 2, 3, 4, 5, 6. This gives you 6/6 chances of rolling one of the two numbers you need. Then you just need the right number on the second one to get a total of seven which you have a 1/6 chance of rolling.
Out of the 36 possible combinations 6 of them result in seven - ie 6/36 or 1/6
6/6 on the first one multipled by 1/6 on the second one gives you 6/36
Rolling for a six the first one needs to be either a 1, 2, 3, 4, or 5. This gives you 5/6 possibilities. The second one then just has to match correctly meaning you have a 1/6. This gives you a 5/36 odds of rolling a 5.
I used this to demonstrate to my daughter odds the other night. It is between the simplistic coin flips (ie whats the chances of flipping two heads in a row) and between figuring odds in cards. Tonight we worked on figuring odds in cards and adjusting odds on the fly as more cards were in play and known.
Next week we aer going to discuss statistical sampling (ie if you flip a coin 50 times and get 49 heads there is a good chance the coin is not a naturally weighted coin). However in order to do that you need to understand probability first.
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War Maths – Projectile Motion
Despite maths having a reputation for being a somewhat bookish subject, it is also an integral part of the seamier side of human nature and has been used by armies to give their side an advantage in wars throughout the ages. Military officers all need to have a firm grasp of kinematics and projectile motion – so let’s look at some War Maths.
Cannons have been around since the 1200s – and these superseded other siege weapon projectiles such as catapults which fired large rocks and burning tar into walled cities. Mankind has been finding ever more ingenious ways of firing projectiles for the best part of two thousand years.
The motion of a cannon ball can be modeled as long as we know the initial velocity and angle of elevation. If the initial velocity is Vi and the angle of elevation is θ, then we can split this into vector components in the x and y direction:
Vxi = Vicosθ (Vxi is the horizontal component of the initial velocity Vi)
Vyi = Visinθ (Vyi is the vertical component of the initial velocity Vi)
Next we know that gravity will affect the motion of the cannonball in the y direction only – and that gravity can be incorporated using g (around 9.8 m/s2 ) which gives gravitational acceleration. Therefore we can create 2 equations giving the changing velocity in both the x direction (Vx) and y direction (Vy):
Vx = Vicosθ
Vy = Visinθ – gt
To now find the distance traveled we use our knowledge from kinematics – ie. that when we integrate velocity we get distance. Therefore we integrate both equations with respect to time:
Sx = x = (Vicosθ)t
Sy = y = (Visinθ)t – 0.5gt2
We now have all the information needed to calculate cannon ball projectile questions. For example if a cannon aims at an angle of 60 degrees with an initial velocity of 100 m/s, how far will the cannon ball travel?
Step (1) We find out when the cannon ball reaches maximum height:
Vy = Visinθ – gt = 0
100sin60 – 9.8(t) = 0
t ≈ 8.83 seconds
Step (2) We now use the fact that a parabola is symmetric around the maximum – so that after 2(8.83) ≈ 17.7 seconds it will hit the ground. Therefore substitute 17.7 seconds into the equation for Sx = (Vicosθ)t.
Sx = (Vicosθ)t
Sx = (100cos60).17.7
Sx ≈ 885 metres
So the range of the cannon ball is just under 1km. You can use this JAVA app to model the motion of cannon balls under different initial conditions and also factor in air resistance.
There are lots of other uses of projectile motion – the game Angry Birds is based on the same projectile principles as shooting a cannon, as is stunt racing – such as Evel Knieval’s legendary motorbike jumps:
If you enjoyed this post you might also like:
Bridge Building Lesson Plan. A lesson to introduce a practical example of maths and engineering.
Langton’s Ant – Order out of Chaos How computer simulations can be used to model life.
Essential resources for IB students:
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
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# Tatiana wants to give friendship bracelets to her 32 classmates. She already has 5 bracelets, and she can buy more bracelets in packages of 4. Write an inequality to determine the number of packages, p, Tatiana could buy to have enough bracelets.
## To determine the number of packages Tatiana could buy to have enough bracelets, we need to first calculate the total number of bracelets needed.
Tatiana wants to give friendship bracelets to her 32 classmates, and she already has 5 bracelets. So the total number of bracelets needed will be 32 - 5 = 27 bracelets.
Each package contains 4 bracelets. So, to get the minimum number of packages required, we need to find the ceiling value of the division between the total number of bracelets needed and the number of bracelets in each package.
Let p represent the number of packages. We can write the inequality as:
4p ≥ 27
This inequality states that the product of the number of packages (p) and the number of bracelets in each package (4) must be greater than or equal to the total number of bracelets needed (27).
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# Bitwise math
(Redirected from Bit rotation)
Bitwise math is the foundation of all binary math and most mathematic operations performed in assembly. Many of the bitwise mathematics primitive operators can be found in a variety of compiled languages and interpreted languages as well.
Language specific information can be found for Perl, Python, and Ruby
## Introduction to Binary
Binary math and hexadecimal math have slight differences to decimal math but the same principles apply. For example, in the decimal number 1234, the ‘4’ is in the “ones” placeholder, the ‘3’ is in the “tens” placeholder, the ‘2’ in the “hundreds” placeholder and the ‘1’ in the “thousands” placeholder.
Example 1: Decimal number 1234
Example 1: Decimal number 1234
Thousands (1x10^3) Hundreds (1x10^2) Tens (1x10^1) Ones (1x10^0)
1 2 3 4
Binary operates a little bit differently, instead of having 1’s, 10’s, 100’s, etc, it has 1’s, 2’s, 4’s, 8’s, etc (That is to say, decimal math operates on a base of 10, while binary math operates on a base of 2). So analysing this for a moment, the binary number 1010, has a ‘1’ in the “eights” placeholder and a ‘1’ in the “twos” placeholder. Add these together and 10 is obtained in decimal numbers.
Example 2: Binary number 1010 = 8+2 = Decimal number 10
Eights (1x2^3) Fours (1x2^2) Twos (1x2^1) Ones (1x2^0)
1 0 1 0
Another example is 1111:
Example 3: Binary number 1111 = 8+4+2+1 = Decimal number 15
Eights (1x2^3) Fours (1x2^2) Twos (1x2^1) Ones (1x2^0)
1 1 1 1
Eights (1x2^3) Fours (1x2^2) Twos (1x2^1) Ones (1x2^0) 1 1 1 1 8 4 2 1
Through the use of the binary table, the binary values can each be multiplied (1’s) by the above multipliers (1x2^3 1x2^2 1x2^1 1x2^0) or “8, 4, 2, 1”. This will then give us, 8+4+2+1=15.
Much like the decimal system, binary numbers can be added together. For this example the binary numbers 0110 and 0010 are going to be used and added.
Eights Fours Twos Ones Total 0 1 1 0 6 0 4 2 0 6
Eights Fours Twos Ones Total 0 0 1 0 2 0 0 2 0 2
Thus by using decimal addition, 4+2+2=8, the value of the addition of two binary numbers can be determined as 8.
The past exercises have featured working with 4 bits at once (4 values ranging from 0-1, e.g. 0001). This is known as a nybble in hexadecimal. A byte is made of two nybbles (8 bits make a byte).
In hexadecimal, there is a 1’s placeholder and a 16’s placeholder. Hexadecimal is 0 through 9 and A through F. A nybble can hold 16 unique values but the highest value is 15 because one of the values is 0. A nybble is a single hex digit. So, A = 10, B = 11, so on and so forth, F = 15.
In hex, AF is obtained as a byte.
AF = 175 in decimal because A is in the 16’s placeholder
• A = 10, 10*16=160,
Plus F which is in the 1’s placeholder,
• F = 15, 15*1=15
Therefore 160(A)+15(F) = 175.
## NOT, AND, OR and XOR
Note: all examples in this section will be using hexadecimal and binary.
### NOT
NOT is a bitwise operator that takes only ONE operand. It inverts or reverses the binary value.
Example:
``` A = 1010 in binary or 10 in decimal. NOT A results in the inversion of 1010 which is 0101. Therefore NOT A = 5.
```
### AND
AND returns “TRUE” per true bit in both of the required two operands. True is 1 and false is 0. It operates bit by bit, just like NOT.
#### AND rules
AND compares each bit and if both bits per placeholder are true, then it returns a true for that placeholder, all else gets turned into 0.
• 1 and 1 = 1
• 1 and 0 = 0
• 0 and 0 = 0
#### AND properties
• anything AND'd by itself results in itself
• anything AND'd with 0xF results in itself
• anything AND'd with 0x0 results in 0x0
#### AND example
Example: 0x6 AND 0x5 = 0x4
and
6
5
0110
0101
The second and third bits are true.
The second and fourth bits are true.
=
4 0100 The second bit is the only one true in both instances (5 and 6).
#### AND logic table
0 1 2 3 4 5 6 7 8 9 A B C D E F AND 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 2 2 0 0 2 2 0 0 2 2 0 0 2 2 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 0 0 0 4 4 4 4 0 0 0 0 4 4 4 4 0 1 0 1 4 5 4 5 0 1 0 1 4 5 4 5 0 0 2 2 4 4 6 6 0 0 2 2 4 4 6 6 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 8 8 8 8 8 8 8 8 0 1 0 1 0 1 0 1 8 9 8 9 8 9 8 9 0 0 2 2 0 0 2 2 8 8 A A 8 8 A A 0 1 2 3 0 1 2 3 8 9 A B 8 9 A B 0 0 0 0 4 4 4 4 8 8 8 8 C C C C 0 1 0 1 4 5 4 5 8 9 8 9 C D C D 0 0 2 2 4 4 6 6 8 8 A A C C E E 0 1 2 3 4 5 6 7 8 9 A B C D E F
### OR
OR will return true bits for the respective placeholder if any of the bits are true.
#### OR rules
OR determines if any bits are true from the given binary operands
• 1 or 1 = 1
• 1 or 0 = 1
• 0 or 0 = 0
#### OR properties
• Anything OR'd with itself results in itself
• Anything OR'd with 0xF results in 0xF
• Anything OR'd with 0x0 results in itself
#### OR example
Example: 5 OR C = D
or
5
C
0101
1100
The second and fourth bits are true.
The first and second bits are true.
=
D 1101 The first, second and fourth bits are true in at least one instance.
#### OR logic table
0 1 2 3 4 5 6 7 8 9 A B C D E F OR 0 1 2 3 4 5 6 7 8 9 A B C D E F 1 1 3 3 5 5 7 7 9 9 B B D D F F 2 3 2 3 6 7 6 7 A B A B E F E F 3 3 3 3 7 7 7 7 B B B B F F F F 4 5 6 7 4 5 6 7 C D E F C D E F 5 5 7 7 5 5 7 7 D D F F D D F F 6 7 6 7 6 7 6 7 E F E F E F E F 7 7 7 7 7 7 7 7 F F F F F F F F 8 9 A B C D E F 8 9 A B C D E F 9 9 B B D D F F 9 9 B B D D F F A B A B E F E F A B A B E F E F B B B B F F F F B B B B F F F F C D E F C D E F C D E F C D E F D D F F D D F F D D F F D D F F E F E F E F E F E F E F E F E F F F F F F F F F F F F F F F F F
### XOR
XOR is a bitwise comparison operator that returns a true bit if the compared bits in question are different. If they are the same, it returns a false bit.
#### XOR rules
XOR determines which bits differ in the two binary numbers used as operands.
• 1 xor 1 = 0
• 1 xor 0 = 1
• 0 xor 0 = 0
#### XOR properties
• Anything xor'd with itself results in 0
• Anything xor'd with 0xF is the same as a "not"
• Anything xor'd with zero results in itself
#### XOR example
Example: A xor F = 5
xor
A
F
1010
1111
The first and third bits are true.
The first, second, third, and fourth bits are true.
=
5 0101 The second and fourth bits are true in ONLY one instance as opposed to two.
• The 8’s and 2's placeholders are the same so they return 0.
• The 4’s and 1’s placeholders are different, therefore return true.
#### XOR logic table
0 1 2 3 4 5 6 7 8 9 A B C D E F XOR 0 1 2 3 4 5 6 7 8 9 A B C D E F 1 0 3 2 5 4 7 6 9 8 B A D C F E 2 3 0 1 6 7 4 5 A B 8 9 E F C D 3 2 1 0 7 6 5 4 B A 9 8 F E D C 4 5 6 7 0 1 2 3 C D E F 8 9 A B 5 4 7 6 1 0 3 2 D C F E 9 8 B A 6 7 4 5 2 3 0 1 E F C D A B 8 9 7 6 5 4 3 2 1 0 F E D C B A 9 8 8 9 A B C D E F 0 1 2 3 4 5 6 7 9 8 B A D C F E 1 0 3 2 5 4 7 6 A B 8 9 E F C D 2 3 0 1 6 7 4 5 B A 9 8 F E D C 3 2 1 0 7 6 5 4 C D E F 8 9 A B 4 5 6 7 0 1 2 3 D C F E 9 8 B A 5 4 7 6 1 0 3 2 E F C D A B 8 9 6 7 4 5 2 3 0 1 F E D C B A 9 8 7 6 5 4 3 2 1 0
## Shift and rotate
Each of these operations can occur either to the right or to the left. Examples will use nibbles as operands (4 bits or 0.5 bytes).
### Logical Shifts
For logical shifts, the bytes are assumed to be in big endian.
Using A or 1010 as an example:
Example: A shifted left by 1 = 4
Left Shift
A
1
1010
0001
shifted to the left once, the 1 at the beginning gets shifted off, thus the binary value becomes 0100, or 4 in hexadecimal.
= 4 0100 The end value 0 is always used to replace the residual numbers.
Example: A shifted left by 2 = 8
Left Shift
A
2
1010
0001
The first `10' is shifted off and the remaining two placeholders get zeroed out.
= 8 1000 The end value 0 is always used to replace the residual numbers.
Example: A shifted right by 1 = 5
Right Shift
A
1
1010
0101
a single shift to the right is division by two for even numbers.
= 5 0101 The end value 0 is always used to replace the residual numbers.
If 'A' is shifted to the right twice it goes from 1010 to 0010
Right Shift
A
2
1010
0001
shifted to the right twice, the one at the beginning gets shifted to the 2nd digit and the one in the 2nd digit place gets shifted off, thus the binary value becomes 0010, or 2 in hexadecimal.
= 2 0010 The end value 0 is always used to replace the residual numbers.
At this point, understanding should be emerging into real hacking and cryptography. If understanding is not attained regarding this knowledge, further extension topics will prove difficult. Thus, it is considered a baseline knowledge.
### Exercises:
1) Solve for B left shift one
Left Shift One
B
1
"(B in Binary)"
0001
= "??" "???"
2) Solve for F right shift three
Right Shift three
F
3
"(F in Binary)"
0011
= "??" "???"
1) 0110 or 6.
Left Shift One
B
1
1011
0001
= "6" "0110"
2) 0001 or 1.
Right Shift three
F
3
1110
0011
= 0001 1
## Circular Shift or Bit Rotation
Circular shifts, or bit rotation occurs in processors which involves rotate /with carry/. Circular shifts are the same as rotate /without/ carry. So, continuing the usage of a nibble and A (1010) as the example...
If 6 is rotated to the left by 1 it becomes 0101 or 5
Left Rotate
6
1
1010
0001
shifted to the left once, the 1 at the beginning gets shifted to the first digit, thus the binary value becomes 0101, or 5 in hexadecimal.
= 5 0101 The highest value "loops" back around to the lowest, like a clock/modular arithmetic.
Instead of replacing a value with zero, like a normal shift, the value is taken and shifted off of one side and applied to the other side. So for 1100 (C)
Example: C rotated left by 1, becomes 9 (1001)'
Left Rotate
C
1
1100
0001
shifted to the left once, the 1 at the beginning gets shifted to the first digit, thus the binary value becomes 1001, or 9 in hexadecimal.
= 9 1001 The highest value "loops" back around to the lowest, like a clock/modular arithmetic.
Example: C rotated right by 1, becomes 6 (0110)'
Right Rotate
C
1
1100
0001
shifted to the right once, the 1 at the beginning gets shifted to the third digit, thus the binary value becomes 0110, or 6 in hexadecimal.
= 6 0110 Something a little more intuitive
Based upon these examples, a circular shift is not the same thing as a logical shift. Proceeding further, two's complement and something small about binary will be explained.
Remember in the previous lesson, a nibble (four bits) can hold 16 values (the uppermost being 15 because one of these values is 0). Four bits maximum value also = 2^4 - 1. 4 is taken from the number of bits and the -1 from the zero placeholder. If a full byte was going to be shifted, it would be 2^8 - 1. The maximum value of which is 255.
## Negative Numbers
There are several ways to represent a negative number:
• Sign and Magnitude
• Two's Complement
### Sign and Magnitude
The first bit is considered the sign bit, which if set, translates the number to it's negative counterpart. While this is very straightforward, it requires a a new operation to be supported by the hardware (subtraction), and generates two representations for the 0 value (10000000 and 00000000). As a result, the sign bit is usually used on the largest endian number (called the two's compliment implementation), so that there can be a negative value one value higher or lower than the highest positive value, for example:
• 1000 = -8, while 0111 = 7
### Two's Complement
A two's complement is basically a NOT operation performed on the nibble. Any binary number is converted between positive and negative by computing its two's complement. The leftmost bit is called the sign bit. There are two ways. (taking nibbles as an example) -1 could be 1001. This is not useful, as now there are two ways to represent 0, 1000 for -0 and 0000 for ordinary 0. Also the addition and subtraction has to be changed for this. This is all not nice. This is why people invented two's complement, two's complement is basically using 1111 for -1 and 1000 for -8.
The first operation performed is a NOT operation on the nibble, -1 has to be done. To determine the two complements the positive representation of the number is required. If it's negative, complement it and add one. Thus, to determine the two complements, positive representation of the number is required. If the original value is negative, determine the compliment and do +1. If its positive, the easy way to do it, is to tack zeroes on in the right number of bits. If negative tack 1's instead.
Just using one byte. It has 256 possible values Doing something like:
-42 -42 + 256 -42 + 255 + 1 255 - 42 + 1 213 + 1 214
The binary representation of both numbers /without/ a signing bit is the same.
Computers use this stuff to represent signing amongst other things. When doing arithmetic stuff gets messy. This gets even more messy when doing arithmetic shifts and rotates with carry on a CPU, since that's what modern day CPU's use.
Taking the example of -42. First, its positive representation is taken. So 42 in binary is: 00101010 (32+8+2). All of those placeholders have 1's and the rest have 0's.
Next step, as mentioned is to perform a NOT operation. So NOT(00101010) WILL BE 11010101. Replace 1 by 0 and vice-versa. Final step is to add 1 to the number obtained in above step. Thus:
11010101 00000001 + =
What can also be done, is go at the bits one by one, starting at the right, and copy all the zeroes, until a 1 is obtained. Copy that 1 too, then flip all the remaining bits. So 42 is obtained again, which is 00101010. Then move from right-to-left, and get 11010110.
## Overflows
Now, overflow occurs in a "signed" integer when it increments too far. Using a nibble for an example,
-8 4 2 1
The highest positive value is 7, 0111, the lowest negative value is -8, 1000. On a processor, the increment instruction(inc) changes this. If -8 is incremented, -7 is obtained, since it is incremented. Basically when moving to 7 and incremented, the next step in binary is 1000 and it lands on -8.
In an arithmetic shift, this is typically what processors do since signed numbers are worked with most of the times. Now of course, there are ways to use unsigned numbers, but this will be left for later. Most languages use an arithmetic shift, which is also called a 'signed shift'. This is because the number is "signed", that is, it has a + or - attatched to its value. Thus, when reading the hex, the way this works isn't going to change so terribly.
Example: 6 incremented left by 1, becomes -8 (1100)'
Left Shift
6
1
0110
0001
shifted to the left once, the largest digit gets a 1. In this case, this gives the whole value a -8, not an eight.
= "-8" 1100 The highest value "loops" back around to the lowest, like a clock/modular arithmetic.
The difference here is if that first bit is -8 and not an 8. A positive number has just been turned into a negative number. 6 left shift 1 = -4
Binary number 1100 = -8 + 4 = decimal number -4
Negative Eights -(1x2^3) Fours (1x2^2) Twos (1x2^1) Ones (1x2^0)
-8 4 2 1
1 1 0 0
It's now -4.
If shifted to the right, a single shift will /always/ change the sign. If shifted to the left on any /unsigned/ binary number (shift by n), it is the same as multiplying it by 2^n. If shifted to the right on a two's complement signed number it is the same as dividing it by 2^n and always rounds down.
Example:
-8 right shift 1 or 1000 right shift 1 becomes 4. In other words, any signed number right shifted becomes positive immediately. Regardless of whether it was positive or not, it will always be positive once a right shift has been applied.
As stated before, x right shift n on a signed number is the same as x/(2^n). The problem that this could cause is that it always rounds down. So, if it always rounds down from 2^n (i.e. a fraction is obtained) it will round down until the fraction is gone.
The flow on problem from this is that if it rounds down and n is signed, the processor can run into a divide by zero error.
## Rotate With Carry
Moving on now to rotate with carry. Rotate with carry occurs on a computer when a number is too large to fit into a single register. For example: On 32 bit systems the largest value an unsigned variable can hold is 2^32 - 1. An 8 bit system is being used just for sanity:
2^8 1 - =
That would be the largest value that can be held. If wanting to hold 256, it would need two pieces of processor memory to do this so rotate with carry. There's a bit on the CPU called CF or carry flag. It is just a little bit, 1 or 0. So on an 8 bit system a 16 bit number would want to be rotated.
16 bit number 1011 1000 1010 0011
Now, the registers only contain 8 bits, so:
• register 1 = 1011 1000
• register 2 = 1010 0011
To connect these during the shift operation rather the rotate operation, there is a small bit called the carry flag (CF) on a CPU, that will store the value as the bits are shifted between the two registers. It will take the first "1" from register 2 into the CF bit then move all of register2 1 bit to the left, and so on and so forth.
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# Results forrational numbers review
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A set of 10 Review Stations all about Rational Numbers! Practice with negative and positive rational numbers in the form of fractions, decimals and percents, word problems, converting forms and more! I love to use these after we have learned the content and as a wrap up activity. Students should fe
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Test Review with multiple choice that answers a puzzle at the end for rational numbers and other previous skills. Includes; multiplying, dividing decimals, unit rates, comparing, ordering and converting rational numbers, and unit rates
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This is a card game used to review operations with rational numbers. There are 42 different cards with adding, subtracting, multiplying, and dividing integers, decimals, and fractions. There are also word problem cards. The idea for the game is to play it like the card game War. Each student pla
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This lesson includes a power point with a warm up/do now problem, Examples of Multiple Choice, Multiple Response, Constructed Response, and Extended Constructed Response guided practice problems with test taking strategies and answer key included. There is a guided note taking handout that goes alon
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AIM: To review for tomorrow’s Unit Assessment by solving problems containing integers and completing a partner-and-self-check activity. Directions for the game are included in the materials. This product also includes a spiraled review homework to prepare students for an end of unit assessment con
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7th Grade Standard: Unit 1 - Operations with Rational Numbers Task cards that serve as a good review for this unit Includes problems with the following: -Absolute value -Properties of math -Additive inverses -Adding & subtracting integers -Multiplying & dividing integers -Order of operatio
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21 questions set up as a scavenger hunt for students to practice rational numbers, converting between fractions and decimals, as well as a quick review on coordinate plane. Answer Key available
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Q:
# Shen’s mother gave him $10 to go buy groceries. He picked up a gallon of milk for$3.49, three pounds of oranges that cost $1.14 per pound, and a box of cereal for$3.46. Shen used front-end estimation to determine whether he had enough money for the groceries. How much did he estimate the total cost would be?
Accepted Solution
A:
Shen’s mother gave him $10 to go buy groceries. He picked up a gallon of milk for$3.49, three pounds of oranges that cost $1.14 per pound, and a box of cereal for$3.46. Shen used front-end estimation to determine whether he had enough money for the groceries. How much did he estimate the total cost would be?He estimated that he had about $7 worth of groceries.He estimated that he had about$9 worth of groceries.He estimated that he had about $10.30 worth of groceries.He estimated that he had about$10.40 worth of groceries.What is the value of –112.84 – 54.14 – (–29.18)?Total cost would be $10.37 which is$(10.37 - 10) = \$0.37 higher than the amount given to him by his mother.
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You are viewing an older version of this Concept. Go to the latest version.
Chemical Reaction Rate
Calculating how fast a reaction occurs
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Chemical Reaction Rate
Two for the price of one
Drag racing is a sport where two cars start from a dead stop and drive as fast as they can down a quarter-miles trip. At the end of the strip are timers that determine both elapsed time (how long did it take for them to cover the quarter-mile) and top speed (how fast were they going as they went through the timer chute). Both pieces of data are important. One car may accelerate faster and get ahead that way, while the other care may be slower off the line, but can get up to a higher top speed at the end of the run.
Chemical Reaction Rate
Chemical reactions vary widely in the speeds with which they occur. Some reactions occur very quickly. If a lighted match is brought in contact with lighter fluid or another flammable liquid, it erupts into flame instantly and burns fast. Other reactions occur very slowly. A container of milk in the refrigerator will be good to drink for weeks before it begins to turn sour. Millions of years were required for dead plants under Earth’s surface to accumulate and eventually turn into fossil fuels such as coal and oil.
Chemists need to be concerned with the rates at which chemical reactions occur. Rate is another word for speed. If a sprinter takes 11.0 s to run a 100 m dash, his rate or speed is given by the distance traveled divided by the time.
$\text{speed}=\frac{\text{distance}}{\text{time}}=\frac{100 \text{ m}}{11.0 \text{ s}}=9.09 \text{ m/s}$
The sprinter’s average running rate for the race is 9.09 m/s. We say that it is his average rate because he did not run at that speed for the entire race. At the very beginning of the race, while coming from a standstill, his rate must be slower until he is able to get up to his top speed. His top speed must then be greater than 9.09 m/s so that taken over the entire race, the average ends up at 9.09 m/s.
Runner.
Chemical reactions can’t be measured in units of meters per second, as that would not make any sense. A reaction rate is the change in concentration of a reactant or product with time. Suppose that a simple reaction were to take place in which a 1.00 M aqueous solution of substance $A$ was converted to substance $B$ .
$A(aq) \ \rightarrow \ B(aq)$
Suppose that after 20.0 seconds, the concentration of $A$ had dropped from 1.00 M to 0.72 M as $A$ was slowly being converted to $B$ . We can express the rate of this reaction as the change in concentration of $A$ divided by the time.
$\text{rate}=-\frac{\Delta [A]}{\Delta t}=-\frac{[A]_{\text{final}}-[A]_{\text{initial}}}{\Delta t}$
A bracket around a symbol or formula means the concentration in molarity of that substance. The change in concentration of $A$ is its final concentration minus its initial concentration. Because the concentration of $A$ is decreasing over time, the negative sign is used. Thus, the rate for the reaction is positive and the units are molarity per second or M/s.
$\text{rate}=-\frac{0.72 \text{ M}-1.00 \text{ M}}{20.0 \text{ s}}=-\frac{-0.28 \text{ M}}{20.0 \text{ s}}=0.014 \text{ M/s}$
The molarity of $A$ decreases by an average rate of 0.014 M every second. In summary, the rate of a chemical reaction is measured by the change in concentration over time for a reactant or product. The unit of measurement for a reaction rate is molarity per second (M/s).
Summary
• The reaction rate indicates how fast the reaction proceeds.
Practice
1. Why is the rate of disappearance a negative value?
2. What is the average rate of reaction?
3. What is the instantaneous rate of reaction?
Review
1. What is another word for rate?
2. What does [ ] stand for?
3. What are the units of reaction rate?
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# Unit tangent, normal, binormal etc.
• Dec 4th 2009, 02:10 AM
SirOJ
Unit tangent, normal, binormal etc.
Find the vectors
T, N and B as well as the curvature and the torsion of;
r(t) = (t^3)/3i + (t^2)/2j for t>0
I parametrized the curve and got
r(s) = ((3s+1)^2/3 - 1)^3/2)/2i + ((3s+1)^2/3 - 1)/2i
and I then proceeded to find T(s) for which I calculated;
T(s) = (3s+1)^5/3((3s+1)^2/3 - 1)^5/2)i + (3s+1)^5/3j
Is this the right answer? or Am i going the right way about it this at all...? because using this answer for the unit tangent vector its going to take me ages to calculate N, B, curvature & torsion..
Any help would be greatly appreciated
• Dec 4th 2009, 04:20 AM
Calculus26
No need to reparameterize
T = r'(t)/||r'(t)||
N = T '(t)/||T'(t)||
B = T x N
The curvature can be computed by k = ||T'(t)||/||r'(t)||
• Dec 4th 2009, 04:30 AM
SirOJ
Quote:
Originally Posted by Calculus26
No need to reparameterize
T = r'(t)/||r'(t)||
N = T '(t)/||T'(t)||
B = T x N
The curvature can be computed by k = ||T'(t)||/||r'(t)||
Cheers for the help..
In our class notes we were given the formula's for T,N,B,k etc. using reparametrizations.. so i'm not sure how happy lecturere would be using a different method.. Although i guess a correct answer is a correct answer no matter how you get it...?
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# Search by Topic
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Broad Topics > Transformations and constructions > Compound transformations
### Transforming the Letters
##### Age 7 to 11 Challenge Level:
What happens to these capital letters when they are rotated through one half turn, or flipped sideways and from top to bottom?
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Simplifying Transformations
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How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Trees and Friezes
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This problem is based on the idea of building patterns using transformations.
### Sorting Symmetries
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Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Maurits Cornelius Escher
##### Age 7 to 14
Have you ever noticed how mathematical ideas are often used in patterns that we see all around us? This article describes the life of Escher who was a passionate believer that maths and art can be. . . .
### Going Places with Mathematicians
##### Age 7 to 14
This article looks at the importance in mathematics of representing places and spaces mathematics. Many famous mathematicians have spent time working on problems that involve moving and mapping. . . .
### Mathematical Patchwork
##### Age 7 to 14
Jenny Murray describes the mathematical processes behind making patchwork in this article for students.
### Friezes Using Logo
##### Age 11 to 14 Challenge Level:
Experimenting with variables and friezes.
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Making Rectangles, Making Squares
##### Age 11 to 14 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
### Transformations Tables
##### Age 7 to 11 Challenge Level:
These grids are filled according to some rules - can you complete them?
### Grouping Transformations
##### Age 11 to 18
An introduction to groups using transformations, following on from the October 2006 Stage 3 problems.
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Screwed-up
##### Age 11 to 14 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
### Square Tangram
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This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### Cut and Make
##### Age 7 to 11 Challenge Level:
Cut a square of paper into three pieces as shown. Now,can you use the 3 pieces to make a large triangle, a parallelogram and the square again?
### Who Is the Fairest of Them All ?
##### Age 11 to 14 Challenge Level:
Explore the effect of combining enlargements.
### Twice as Big?
##### Age 7 to 11 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
##### Age 7 to 11 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Midpoint Triangle
##### Age 7 to 11 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Triangular Tantaliser
##### Age 11 to 14 Challenge Level:
Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles.
### Counting Triangles
##### Age 11 to 14 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Square to L
##### Age 7 to 11 Challenge Level:
Find a way to cut a 4 by 4 square into only two pieces, then rejoin the two pieces to make an L shape 6 units high.
### Cutting Corners
##### Age 7 to 11 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Flight of the Flibbins
##### Age 11 to 14 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### Bow Tie
##### Age 11 to 14 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
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Math 3013 Quiz 3 Solutions
# Math 3013 Quiz 3 Solutions - Math 3013 Linear Algebra Fall...
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Math 3013 Linear Algebra, Fall 2012 Quiz 3 Oct. 10, 2012 1. Determine if the following matrix is invertible or not. If it is invertible, find its inverse. A = 1 - 1 - 2 2 - 3 - 5 - 1 3 5 Solution ± A I ² = 1 - 1 - 2 1 0 0 2 - 3 - 5 0 1 0 - 1 3 5 0 0 1 r 2+ r 3 -→ 1 - 1 - 2 1 0 0 1 0 0 0 1 1 - 1 3 5 0 0 1 r 1 r 2 1 0 0 0 1 1 1 - 1 - 2 1 0 0 - 1 3 5 0 0 1 r 2 - r 1 ,r 3+ r 1 1 0 0 0 1 1 0 - 1 - 2 1 - 1 - 1 0 3 5 0 1 2 ( - 1) × r 2 1 0 0 0 1 1 0 1 2 - 1 1 1 0 3 5 0 1 2 r 3 - 3 r 2 1 0 0 0 1 1 0 1 2 - 1 1 1 0 0 - 1 3 - 2 - 1 ( - 1) × r 3 1 0 0 0 1 1 0 1 2 - 1 1 1 0 0 1 - 3 2 1 r 2 - 2 r 3 1 0 0 0 1 1 0 1 0 5 - 3 - 1 0 0 1 - 3 2 1 2. Compute the LU factorization of A = 2 - 1 0 0 6 - 4 5 - 3 8 - 4 1 0 4 - 1 0 7 Solution A = LU can be written into 2 - 1 0 0 6 - 4 5 - 3 8 - 4 1 0 4 - 1 0 7 = 1 0 0 0 * 1 0 0 * * 1 0 * * * 1 * * * * 0 * * * 0 0 * * 0 0 0 * 1. The first row of U . By using matrix multiplication and the first row of A , we have 2 - 1 0 0 6 - 4 5 - 3 8 - 4 1 0 4 - 1 0 7 = 1 0 0 0 *
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# Contact Force Calculator
## About Contact Force Calculator (Formula)
A Contact Force Calculator is a tool used to calculate the force exerted between two objects that are in contact with each other. Contact force is a fundamental concept in physics that describes the interaction between objects through direct physical contact, such as pushing, pulling, or pressing against each other.
The formula for calculating contact force depends on the specific scenario and the forces involved, such as friction, normal force, tension, compression, etc.
For example, one common scenario is calculating the normal contact force between a solid object and a surface. In this case, the formula for the normal contact force is:
Normal Contact Force (F) = Mass (m) * Gravitational Acceleration (g)
Where:
• Normal Contact Force (F) is the force exerted perpendicular to the surface due to the object’s weight.
• Mass (m) is the mass of the object, measured in kilograms (kg).
• Gravitational Acceleration (g) is the acceleration due to gravity, which is approximately 9.81 m/s² on Earth.
To use the Contact Force Calculator formula, follow these steps:
1. Determine the mass of the object (m) in kilograms.
2. Plug the value of mass (m) into the formula: Normal Contact Force (F) = Mass (m) * Gravitational Acceleration (g).
3. Calculate the normal contact force. The result represents the force exerted by the object on the surface due to its weight.
Different scenarios might involve calculating other types of contact forces, such as frictional force, tension force, or compressive force, which would require specific formulas based on the forces at play.
Contact force calculations are fundamental to understanding how objects interact in various situations, from everyday scenarios to more complex engineering and physics problems.
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There is a long time debate in China about the following question: Which city is the financial center in China? If you raised this question 10 years ago, the answer is definitely Hong Kong, but now Shanghai is catching up really fast, and Shenzhen, a city near Hong Kong, is grabbing more attention. Shanghai used to be the financial center of China for a long time. When I travelled to Shanghai I heard quite a lot people talking about the old Shanghai, they said the city used to be called the Eastern Paris in the early 20th century. That city became history when the Communist took power and shut down all stock activities. The old stock dealers were worried about their asset being captured by the government, they chose to leave Shanghai to Taiwan or to Hong Kong, which made this small city full of financial talent. Due to the immigrants coming to Hong Kong, the city developed really fast after the second world war, and made it become a am
• Buy Wholesale Promotional Armani Suits from China, How to Calculate Sea Freight?
International ocean shipping is a practice used to trade goods all over the world. Items are packed into shipping containers and loaded onto a vessel before traveling across the sea to an arrival port. Large machinery and project material that is too large to fit into ocean containers are known as break-bulk and travels uncovered, aboard vessels. Calculating sea freight is the process that determines a shipment's mass, which is used to rate the cost of either an LCL (less than container-load) or a break-bulk shipment. Instructions LCL shipments 1. For LCL shipments, Measure the length, width, and height of your cargo (in inches). Calculate the volume of your cargo using the formula: Length x Width x Height = Volume (in cubic inches). 2. Create the denominator by multiplying 12 x 12 x 12. There are 12 inches in a foot: (12 x 12 x 12) = 1,728 cubic inches. 3. Divide the volume by 1,728 to get your cubic footage volume: Volume / 1,728 = cubic feet. 4. Multiply by 0.0283168466 to
• What to Look Out For in Wholesale Drop Shipping Companies
If you are planning to start an online business as product retailer, it helps to know that you can work with many other companies to maximize your business profit. For example, you can work with wholesale drop shipping companies to provide efficient product supply and delivery service to customer. It is a form of retailing where the goods are kept by wholesaler instead of retailer. Customer's order and shipment details are received by retailer and passed to drop shipping companies. The job of preparing and shipping of good is done by wholesaler. Both sides profit in different ways. Retailer's inventory management risk is passed on to wholesaler and wholesaler gets to rely on retailer to promote products on their behalf. There are lists of legitimate wholesale companies that you can find online but there are also many scams involving bogus companies or individuals. They claim to be able to help you deliver goods to customers and ask for payment
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# Factorize a^3 + b^3 + c^3 - 3abc
Let f(a)= a3 +b3 +c3 −3abc be a function in a.
Now, putting a = -(b+c), we get f( -(b+c)) =
b 3 +c 3 – (b+c)3 + 3bc(b+c) = (b+c)3 – (b+c)3 = 0
Hence, by factor theorem, (a+b+c) is a factor.
a³ + b³ + c³ -3abc =a³ + (b³ + c³) -3abc
= a³ + (b +c)³ -3bc(b+c) -3abc
[ given, a+ b + c =0 => b+c = -a put it ]
= a³ + (b+c)³ -3bc(-a) -3abc
= a³ + (b + c)³ +3abc -3abc
[ use, formula , x³ + y³ = (x + y)(x² + y² -xy )
= {a + (b + c)}{a² + (b+c)² -a(b + c)}
=(a + b + c)(a² + b² + c² + 2bc – ab – ac)
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# Next Greater Element | LeetCode | GeeksforGeeks
## Next Greater Element
Given an array arr[ ] of size N having distinct elements, the task is to find the next greater element for each element of the array in order of their appearance in the array.
Next greater element of an element in the array is the nearest element on the right which is greater than the current element.
If there does not exist next greater of current element, then next greater element for current element is -1. For example, next greater of the last element is always -1.
Example 1:
```Input:
N = 4, arr[ ] = [1 3 2 4]
Output:
3 4 4 -1
Explanation:
In the array, the next larger element
to 1 is 3 , 3 is 4 , 2 is 4 and for 4 ?
since it doesn't exist, it is -1.
```
Example 2:
```Input:
N = 5, arr[ ] [6 8 0 1 3]
Output:
8 -1 1 3 -1
Explanation:
In the array, the next larger element to
6 is 8, for 8 there is no larger elements
hence it is -1, for 0 it is 1 , for 1 it
is 3 and then for 3 there is no larger
element on right and hence -1.```
This is a function problem. You only need to complete the function nextLargerElement() that takes list of integers arr[ ] and N as input parameters and returns list of integers of length N denoting the next greater elements for all the corresponding elements in the input array.
Expected Time Complexity : O(N)
Expected Auxiliary Space : O(N)
Constraints:
1 ≤ N ≤ 106
1 ≤ Ai ≤ 1018
### Solution:
The brute force solution can be that we will run a for loop and then at every index, we will find its next greater element by another for a loop.
We can optimize the code by using stack. We will check whether the current element is greater that the element at the top of the stack, if yes then we will pop the element and we will store the current arr[i] in the answer array.
See the code below:
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mat h
posted by on .
ms. buxx invested a total of \$2000 in two savings accounts. The first account pays 3% interest per year and the second account pays 5% interest per yera. If the interest from both accounts totals \$84 per yera, how much is invested in each account?
• mat h - ,
x = amt at 3%
y = amt at 5%
x + y = 2000
.03x + .05y = 84
solve simultaneously for x, y
give it a try, post if you need for help
x = 800, y = 1200 if you are right
• math - ,
the answer is the first account has 800 and the second account has 1200
• mat h - ,
I need the steps to this equation
• mat h - ,
x+y=2000
.03x+.05y=84
x+y=2000
-x -x
y=-x+2000 y=-(800)+2000
.03x+.05(-x+2000)=84 y=1200
.03x+ -.05x +100=84
-.02x+100=84
-100 -100
-.02x=-16
-----------
-.02x
x= 800
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http://everything2.com/title/absolute+value
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The absolute value of a real number x, often written |x| in mathematics or abs(x) in various programming languages is x if x>=0, and -x if x < 0.
The magnitude of a real number, disregarding its positive or negative sign: the absolute value of -4 or +4, written |-4| or |+4|, is 4.
The absolute value of a complex number z=(r,i) is |z| or abs(z), and can be found using good old trigonometry: |z| = sqrt(r2 + i2)
The absolute value of an n-dimentional vector v = (x1, x2, ... , xn) can similarly be found thus: |v| = sqrt(x12 + x22 + ... + xn2)
In simple terms, the absolute value of a number is its "distance" from the number .
With real numbers, this is quite simple, but when you have the "2-dimensionality" of a complex number (represented as a + bi, where ii=-1), you must use the pythagorean theorem to find the hypotenuse, which is the distance from 0.
|a + bi| = sqrt(aa + bb)
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https://courses.lumenlearning.com/wm-macroeconomics/chapter/calculating-gdp/
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## Calculating GDP
### Learning Objectives
• Describe how GDP it is measured as a component of total expenditure (demand)
• Explain how gross domestic product can be broken down and measured as different types of product
## The Expenditure Approach: GDP Measured by Components of Demand
If we know that GDP is the measurement of everything that is produced, we should also ask the question, who buys all of this production? This demand can be divided into four main parts:
1. consumer expenditure (consumption)
2. investment expenditure
3. government expenditure on goods and services
4. net export expenditure
Table 1 shows how these four components of demand added up to the GDP in 2016.
Table 1. Components of U.S. GDP in 2016: From the Expenditure Side
Components of GDP (in trillions of dollars) Percentage of Total
Consumption $12.8 68.7% Investment$3.0 16.3%
Government $3.3 17.6% Net Exports -$.50 -2.7
Exports $2.2 12.0% Imports –$2.7 –14.7%
### Government Spending
• The only part of government spending counted in GDP is government purchases of goods or services produced in the economy. Examples include the government buying a new fighter jet for the Air Force (federal government spending), building a new highway (state government spending), or a new school (local government spending).
• Government expenditure in the United States is about 20% of GDP, and includes spending by all three levels of government: federal, state, and local
• What’s NOT included: A significant portion of government budgets are transfer payments, like unemployment benefits, veteran’s benefits, and Social Security payments to retirees. These payments are excluded from GDP because the government does not receive a new good or service in return or exchange. Instead they are transfers of income from taxpayers to others. What taxpayers spend their income on is captured in consumer expenditure.
### Net Exports, or Trade Balance
• When thinking about the demand for domestically produced goods in a global economy, it is important to count spending on exports—domestically produced goods that are purchased by foreigners. By the same token, we must also subtract spending on imports—goods produced in other countries that are purchased by residents of this country. The net export component of GDP is equal to the value of exports (X) minus the value of imports (M), (X – M).
• The gap between exports and imports is also called the trade balance. If a country’s exports are larger than its imports, then a country is said to have a trade surplus. In the United States, exports typically exceeded imports in the 1960s and 1970s, as shown in Figure 2(b).
Since the early 1980s, imports have typically exceeded exports, and so the United States has experienced a trade deficit in most years. Despite growth in the trade deficit in the late 1990s and in the mid-2000s, the deficit typical remains less than 5% of GDP. Figure 2(b) also shows that imports and exports have both risen substantially in recent decades, even after the declines during the Great Recession between 2008 and 2009. As noted before, if exports and imports are equal, foreign trade has no effect on total GDP. However, even if exports and imports are balanced overall, foreign trade might still have powerful effects on particular industries and workers by causing nations to shift workers and physical capital investment toward one industry rather than another.
## GDP Measured using Components of Demand
Based on these four components of demand, GDP can be measured as:
GDP = Consumption + Investment + Government Spending + Net Exports
GDP = C + I + G + (X – M)
### Try It
Understanding how to measure GDP is important for analyzing connections in the macro economy and for thinking about macroeconomic policy tools.
### How do Statisticians Measure GDP?
Government economists at the Bureau of Economic Analysis (BEA), within the U.S. Department of Commerce, construct estimates of GDP from a variety of sources.
• Consumption. Once every five years, in the second and seventh year of each decade, the U.S. Census Bureau carries out a detailed census of businesses throughout the United States. In between, it carries out a monthly survey of retail sales. These figures are adjusted with foreign trade data to account for exports that are produced in the United States and sold abroad and for imports that are produced abroad and sold here. Once every ten years, the Census Bureau also conducts a comprehensive survey of housing and residential finance. Together, these sources provide the main basis for figuring out what is produced for consumers.
• Investment. For investment, the Census Bureau carries out a monthly survey of construction and an annual survey of expenditures on physical capital equipment.
• Government Spending. For what is purchased by the federal government, the statisticians rely on the U.S. Department of the Treasury. An annual Census of Governments gathers information on state and local governments. Because a lot of government spending at all levels involves hiring people to provide services, a large portion of government spending is also tracked through payroll records collected by state governments and by the Social Security Administration.
• Net Exports: With regard to foreign trade, the Census Bureau compiles a monthly record of all import and export documents. Additional surveys cover transportation and travel, and adjustment is made for financial services that are produced in the United States for foreign customers.
Many other sources contribute to the estimates of GDP. Information on energy comes from the U.S. Department of Transportation and Department of Energy. Information on healthcare is collected by the Agency for Health Care Research and Quality. Surveys of landlords provide information about rental income. The Department of Agriculture collects statistics on farming.
All of these bits and pieces of information arrive in different forms, at different time intervals. The BEA melds them together to produce estimates of GDP on a quarterly basis (every three months). These numbers are then “annualized” by multiplying by four. That is, quarterly GDP estimates what annual GDP would be if the trend over the three months continued for twelve months. As more information comes in, these estimates are updated and revised. The “advance” estimate of GDP for a certain quarter is released one month after a quarter. The “preliminary” estimate comes out one month after that. The “final” estimate is published one month later, but it is not actually final. In July, roughly updated estimates for the previous calendar year are released. Then, once every five years, after the results of the latest detailed five-year business census have been processed, the BEA revises all of the past estimates of GDP according to the newest methods and data, going all the way back to 1929. Read FAQs on the BEA site. You can even email your own questions!
## GDP Measured by Type of Product
The Expenditure Approach divides GDP based on who is doing the spending: Consumption (households), Investment (businesses and households), Government Spending (governments) and Net Exports (the rest of the world).
GDP can also be measured by examining what is produced, instead of what is demanded. Everything that is purchased must be produced first. Table 2 breaks down GDP a different way, based on the type of output produced: durable goodsnondurable goods, services, structures, and the change in inventories.
Consumer Expenditure (from the expenditure approach you read about above) consists of spending on durable goods, nondurable goods, and services. The same thing is true of Government and Net Export Expenditures. Investment Expenditures is a combination of durable goods (like business equipment) and structures e.g. factories, office buildings, retail stores and residential construction).
Table 2. GDP by Type of Product in 2016
Components of GDP by Type of Product (in trillions of dollars) Percentage of Total
Goods
Durable goods $3.0 16.3% Nondurable goods$2.5 13.1%
Services $11.6 62.4% Structures$1.5 8.0%
Changes in inventories $0.2 1.1% Total GDP$18.6 100%
Source: http://bea.gov Table 1.2.5 GDP by Major Type of Product
Figure 3 provides a visual representation of the five categories used to measure GDP by type of product.
Figure 3. GDP by type of product, 2016.
Note that whether you decompose GDP into expenditure components or by type of product the total is exactly the same. Figure 4 shows the components of GDP by Type of Product, expressed as a percentage of GDP, since 1960. Services are the largest single component of GDP, representing over half. Nondurable goods used to be larger than durable goods, but in recent years, nondurable goods have been dropping closer to durable goods, which is about 15% of GDP. Structures hover around 10% of GDP, though they’ve been declining in recent years. The change in inventories, the final component of aggregate supply, is not shown here; it is typically less than 1% of GDP.
Figure 4. Types of Product.
Let’s take a closer look at these components of GDP:
### Goods and Services
In thinking about what is produced in the economy, many non-economists immediately focus on solid, long-lasting goods, like cars and computers. Goods that last three or more years are called durable goods. Goods that last less than three years are called nondurable goods. By far the largest part of GDP, however, is services. Moreover, services have been a growing share of GDP over time. A detailed breakdown of the leading service industries would include healthcare, education, and legal and financial services. It has been decades since most of the U.S. economy involved making solid objects. Instead, the most common jobs in a modern economy involve a worker looking at pieces of paper or a computer screen; meeting with co-workers, customers, or suppliers; or making phone calls.
Even within the overall category of goods, Table 2 shows that long-lasting durable goods like cars and refrigerators are about the same share of the economy as short-lived nondurable goods like food and clothing.
### Structures
The category of structures includes everything from homes, to office buildings, shopping malls, and factories. The new structures that were built, or produced, during a time period are counted in this measure of GDP, which is another way of looking at investment, as it was discussed above in focusing on demand to measure GDP.
### Change in Inventories
Inventories is a small category that refers to the goods that have been produced by one business but have not yet been sold to consumers, and are still sitting in warehouses and on shelves. The amount of inventories sitting on shelves tends to decline if business is better than expected, or to rise if business is worse than expected. When a business produces output but fails to sell it, the increase in inventory is treated as an investment expenditure.
## Another Way to Measure GDP: The National Income Approach
GDP is a measure of what is produced in a nation. The primary way GDP is estimated is with the Expenditure Approach we discussed above, but there is another way.
Everything a firm produces, when sold, becomes revenues to the firm. Businesses use revenues to pay their bills: Wages and salaries for labor, interest and dividends for capital, rent for land, profit to the entrepreneur, etc. So adding up all the income produced in a year provides a second way of measuring GDP. This is why the terms GDP and national income are sometimes used interchangeably. The total value of a nation’s output is equal to the total value of a nation’s income.
### Try It
These questions allow you to get as much practice as you need, as you can click the link at the top of the first question (“Try another version of these questions”) to get a new set of questions. Practice until you feel comfortable doing the questions.
### Glossary
durable good:
a good that last three years or more, such as a car or refrigerator
gross domestic product (GDP):
the value of the output of all final goods and services produced within a country in a year
inventory:
good that has been produced, but not yet been sold
national income:
includes all income earned: wages, profits, rent, and profit income
nondurable good:
a good that lasts less than three years, such as food and clothing
service:
product which is intangible (in contrast to goods) such as entertainment, healthcare, or education
structure:
building used as residence, factory, office building, retail store, or for other purposes
gap between exports and imports
exists when a nation’s imports exceed its exports and is calculated as imports –exports
exists when a nation’s exports exceed its imports and is calculated as exports – imports
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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# Write the Value of Lim X → C F ( X ) − F ( C ) X − C - Mathematics
Write the value of $\lim_{x \to c} \frac{f(x) - f(c)}{x - c}$
#### Solution
$\text{ Using the definition of derivative, we have }:$
$\lim_{x \to c} \frac{f\left( x \right) - f\left( x \right)}{x - c} = f'\left( c \right)$
Concept: The Concept of Derivative - Algebra of Derivative of Functions
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 1 | Page 46
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Author Archive | yuliy
parabola construction
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Consider the optimal control problem
$\dot{x}=u; x(0)=x_o; \int_0^T u^2/2dt+M(x(T))\to\min.$
The task is to find the cost function $$S(x_o)$$ (the optimal achievable value).
It can be solved as follows. Assume for …
ece515, homework 5
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Due by midnight of december 4.
Use MATLAB or octave if needed.
1. Consider the optimal control problem
${x}”’=u; \int_0^\infty u^2+x^2 dt\to\min.$
• Write this system as a LQR;
• Solve the ARE;
november 21
1. Find sin transform of the function on $$[0,1]$$
given by
1. $$1$$;
2. $$\cos(\pi x)$$;
3. $$x$$.
2. Solve the heat equation
$u_t=\frac{1}{2} u_{xx},$
on $$[0,1]$$ with boundary conditions $$u(0,t)=u(1,t)=0$$ and initial values
1. u(x,0)=$$1$$;
2. u(x,0)=$$\cos(\pi x)$$;
3. u(x,0)=$$x$$.
3. Solve the heat equation
$math285, week of november 17 Read the textbook, chapters 9.5 and 9.7. Homework (due by Monday, 12.1): 1. Find the cosine transform of $$\sin(x)$$ on $$[0,\pi]$$. 2. Using separation of variables, solve the heat equation on the interval $$[0,\pi]$$: \[u_{t}=u_{xx}; \quad u_x(0,t)=u_x(\pi,t)=0; u(x,0)=\sin(x) november 14 1. Find the solutions of the wave equation at time $$t=1/6,1/3,1/2$$ \[ u_{tt}=u_{xx}, u(0,t)=u(1,t)=0,$
on $$[0,1]$$
with initial data $$u_t(x,0)=0$$ and $$u(x,0)$$ is as shown:
2. Same question, but the initial data are
$$u(x,0)=0$$ and $$u_t(x,0)$$ is as shown:
3. Sketch the
math 285, week of november 10
Read the textbook, chapters 9.4 and 9.6.
Videos to watch.
Homework (due by Monday, 11.17):
1. Find sine transform of the function (\T(x)\) on $$[0,1]$$ equal to $$x$$ for $$0\leq x\leq 1/2$$ and to $$1-x$$ for $$1/2\leq x \leq1$$.
2. Using
november 6
• Find Fourier series for $$2\pi$$-periodic function defined on $$-\pi,\pi$$ by
1. $\cos(2x)\sin^2(x);$
2. $\cos^3(x);$
3. $f(x)=\left\{\begin{array}{rcl} 1& \mathrm{if}& -\pi\leq x<0;\\ -1 & \mathrm{if}& 0\leq x \leq \pi.\\ \end{array}\right.$
• Find the Fourier series for the function which is
• notes for lectures 10-14
here
Many thanks to James!…
here
ece515, homework 4
$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}} \def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$
Due by midnight of november 8.
1. Consider the system $$\dot{x}=Ax+bu$$,
$A=\left( \begin{array}{ccc} 1&1&0\\ 1&1&1\\ 0&1&1\\ \end{array}\right); b=\left( \begin{array}{c} b_1\\ b_2\\ b_3\\ \end{array}\right).$
• Derive conditions on $$b$$ making the system controllable.
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Algebra 2 1-5 Complete Lesson: Solving Inequalities
starstarstarstarstarstarstarstarstarstar
5 (1 rating)
by Matthew Richardson
| 28 Questions
Note from the author:
A complete formative lesson with embedded slideshow, mini lecture screencasts, checks for understanding, practice items, mixed review, and reflection. I create these assignments to supplement each lesson of Pearson's Common Core Edition Algebra 1, Algebra 2, and Geometry courses. See also mathquest.net and twitter.com/mathquestEDU.
The outlined content above was added from outside of Formative.
1
1
10 pts
Solve It! You want to download some new songs on your MP3 player. Each song will use about 4.3 MB of space. The amount of space available on your MP3 player is shown in the image. At most, how many songs can you download? Hint: 1 GB = 1000 MB
Enter only a whole number without a comma.
2
2
10 pts
Problem 1 Got It?
A
B
C
D
3
10 pts
Problem 2 Got It? What is the solution of the inequality?
x < 8
x ≤ -8
x > 8
x ≥ -8
4
10 pts
Problem 2 Got It? Graph the solution from the previous item on the canvas.
Include relevant details, label points of interest and establish scale.
5
5
10 pts
Problem 3 Got It?
A
B
C
D
6
6
10 pts
Problem 4 Got It?
A
B
C
7
10 pts
Problem 5 Got It? What is the solution of the inequality? Graph the solution on the canvas.
Include relevant details, label points of interest and establish scale.
8
10 pts
Problem 5 Got It? Reasoning: Is the compound inequality in Problem 5 always, sometimes, or never true?
Always
Sometimes
Never
9
10 pts
Problem 6 Got It? What is the solution of the inequality? Graph the solution on the canvas.
Include relevant details, label points of interest and establish scale.
10
10 pts
Problem 6 Got It? What is the solution of the inequality? Graph the solution on the canvas.
Include relevant details, label points of interest and establish scale.
11
10 pts
Write an inequality that represents the sentence. Rachel's hair is at least as long as Julia's.
R ≥ J
R < J
R > J
R ≤ J
12
10 pts
Write an inequality that represents the sentence. The wind speeds of tropical storms are at least 40 mi/h, but less than 74 mi/h.
w ≥ 40 and w ≤ 74
w ≤ 40 and w < 74
w ≥ 40 and w < 74
w > 40 and w < 74
13
10 pts
Solve the inequality. Graph the solution. Include relevant graph detail.
14
10 pts
Solve the inequality. Graph the solution. Include relevant graph detail.
15
10 pts
Solve the inequality. Graph the solution. Include relevant graph detail.
16
10 pts
Reasoning: Make up an example to help explain why you must reverse the inequality symbol when you multiply or divide by a negative number.
17
10 pts
Compare and Contrast: Describe how the properties of inequality are similar to the properties of equality and how they differ.
18
10 pts
Provide an Example: Write an inequality for which the solution is the set of all real numbers.
19
10 pts
Error Analysis: Your classmate says that you cannot write a compound inequality that has no solution. Do you agree? If so, explain why. If not, give a counterexample.
20
10 pts
Review Lesson 1-3: Simplify the expression.
12a
12
7a + 5
-2a + 14
21
10 pts
Review Lesson 1-3: Simplify the expression.
-2x + 14y
8x + 14y
2x + 2y - 2
-2x + 2y
22
10 pts
Review Lesson 1-3: Simplify the expression.
23
10 pts
Review Lesson 1-4: Match each equation with its solution. Check your answers.
• no solution
• -20
• 4
24
10 pts
Vocabulary Review: Classify each item in the left column as an equation or as an inequality.
• 15 > 32x
• -27 ≠ 3x
• 45 < 46
• 17x = 34
• Equation
• Inequality
25
10 pts
Use Your Vocabulary: A student uses the word compound in three different sentences. The sentences are listed in the right column. Tag sentences that correctly use the word with a √ symbol. Tag sentences that do not with a ⊗ symbol.
• In chemistry class we learned about various chemical compounds.
• Compound inequalities contain more than one inequality symbol.
• To simplify an expression, compound like terms.
26
10 pts
Use Your Vocabulary: Match the appropriate inequality symbol with each phrase on the right.
• <
• >
• x is at most 25
• x is not equal to 25
• x is greater than 25
• x is at least 25
27
100 pts
Notes: Take a clear picture or screenshot of your Cornell notes for this lesson. Upload it to the canvas. Zoom and pan as needed.
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https://quantumcomputing.stackexchange.com/questions/27336/multiple-hadamard-gates-transformation-on-n-qbits
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# Multiple Hadamard gates Transformation on N qbits
I am newbie to quantum computing and having a bit confusion regarding the action of Hadamard gate on multiple qbits which are already in superposed state (I well understand how it works for qbits which are in |00..00> state) i.e. in the middle of the circuit.
For example in Grover's algorithim with three qbits |ψ2>=1√8(|000⟩+|001⟩+|010⟩+|011⟩+|100⟩−|101⟩−|110⟩+|111⟩) this state when given to Hadamard gates (on all three qibts) generates state |ψ3a⟩=12(|000⟩+|011⟩+|100⟩−|111⟩). Also when given |ψ3a⟩=12(-|000⟩+|011⟩+|100⟩−|111⟩), it generates |ψ3a⟩=12(-|101⟩-|110⟩). Now when I apply the truth table I might get the intended answers but is their any logical way or a generalized formula to infer the output without diving into applying Hadamard to each single ket and xoring them to get the results, as its not fun for larger number of qbits.
Thanks.
• Hi and welcome to Quantum Computing SE. Please use MathJax for typesetting mathematical expressions. Jul 15, 2022 at 5:25
• sure and thanks Jul 15, 2022 at 9:50
There is a formula for the Hadamard transform on $$N$$ qubits. $$H^{\otimes N}=\frac{1}{\sqrt{2^N}}\sum_{x,y\in\{0,1\}^N}(-1)^{x\cdot y}|y\rangle\langle x|$$ In this case, $$x\cdot y=\sum_{i=1}^Nx_iy_i$$ is just like thinking of the binary strings as vectors and taking their inner product.
For the 3-qubit cases that you're describing, this just expands to an $$8\times 8$$ matrix. If you take it and apply it to any input state, it'll tell you the output state.
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# Proving That a Quadrilateral is a Parallelogram
## Presentation on theme: "Proving That a Quadrilateral is a Parallelogram"— Presentation transcript:
Proving That a Quadrilateral is a Parallelogram
Lesson 6-3 Lesson Quiz Find the values of the variables for which GHIJ must be a parallelogram. 1. 2. x = 6, y = 0.75 a = 34, b = 26 Determine whether the quadrilateral must be a parallelogram. Explain. 3. 4. 5. Yes; the diagonals bisect each other. Yes; one pair of opposite sides is both congruent and parallel. No; both pairs of opposite sides are not necessarily congruent. 6-4
Special Parallelograms
Lesson 6-4 Notes A segment bisects an angle if and only if it divides the angle into two congruent angles. 6-4
Special Parallelograms
Lesson 6-4 Notes Proof: ABCD is a rhombus, so its sides are all congruent. Therefore, ABC ADC by the SSS Postulate. 12 and 34 by CPCTC. Therefore, bisects BAD and BCD by the definition of bisect. You can show similarly that bisects ABC and ADC. 6-4
Special Parallelograms
Lesson 6-4 Notes In the rhombus above, points B and D are equidistant from A and C. By the Converse of the Perpendicular Bisector Theorem, they are on the perpendicular bisector of segment AC. 6-4
Special Parallelograms
Lesson 6-4 Notes 6-4
Special Parallelograms
Lesson 6-4 Notes Statements Reasons 1. Rectangle ABCD 1. Given 2. ABCD is a 2. Definition of rectangle 3. → opp. sides 4. DAB & CBA are right s 4. Definition of rectangle 5. DABCBA 5. All right s are 6. Reflexive POC 7. DABCBA 7. SAS 8. CPCTC 6-4
Special Parallelograms
Lesson 6-4 Notes 6-4
Special Parallelograms
Lesson 6-4 Additional Examples Finding Angle Measures Find the measures of the numbered angles in the rhombus. Theorem 6–9 states that each diagonal of a rhombus bisects two angles of the rhombus, so m 1 = 78. Theorem 6-10 states that the diagonals of a rhombus are perpendicular, so m 2 = 90. Because the four angles formed by the diagonals all must have measure 90, 3 and ABD must be complementary. Because m ABD = 78, m 3 = 90 – 78 = 12. Finally, because BC = DC, the Isosceles Triangle Theorem allows you to conclude So m 4 = 78. Quick Check 6-4
Special Parallelograms
Lesson 6-4 Additional Examples Finding Diagonal Length One diagonal of a rectangle has length 8x + 2. The other diagonal has length 5x Find the length of each diagonal. By Theorem 6-11, the diagonals of a rectangle are congruent. 5x + 11 = 8x + 2 Diagonals of a rectangle are congruent. 11 = 3x + 2 Subtract 5x from each side. 9 = 3x Subtract 2 from each side. 3 = x Divide each side by 3. 5x + 11 = 5(3) + 11 = 26 8x + 2 = 8(3) + 2 = 26 Substitute. The length of each diagonal is 26. Quick Check 6-4
Special Parallelograms
Lesson 6-4 Additional Examples Identifying Special Parallelograms The diagonals of ABCD are perpendicular. AB = 16 cm and BC = 8 cm. Can ABCD be a rhombus or rectangle? Explain. Use indirect reasoning to show why ABCD cannot be a rhombus or rectangle. Suppose that ABCD is a parallelogram. Then, because its diagonals are perpendicular, ABCD must be a rhombus by Theorem 6-12. But AB = 16 cm and BC = 8 cm. This contradicts the requirement that the sides of a rhombus are congruent. So ABCD cannot be a rhombus, or even a parallelogram. Quick Check 6-4
Special Parallelograms
Lesson 6-4 Additional Examples Real-World Connection Explain how you could use the properties of diagonals to stake the vertices of a play area shaped like a rhombus. By Theorem 6-7, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. By Theorem 6-13, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. One way to stake a play area shaped like a rhombus would be to cut two pieces of rope of any lengths and join them at their midpoints. Then, position the pieces of rope at right angles to each other, and stake their endpoints. Quick Check 6-4
Special Parallelograms
Lesson 6-4 Lesson Quiz 1. The diagonals of a rectangle have lengths 4 + 2x and 6x – 20. Find x and the length of each diagonal. 6; each diagonal has length 16. Find the measures of the numbered angles in each rhombus. m 1 = 90, m 2 = 20, m 3 = 20, m 4 = 70 3. 2. m 1 = 62, m 2 = 62, m 3 = 56 Determine whether the quadrilateral can be a parallelogram. If not, write impossible. Explain. 4. Each diagonal is 15 cm long, and one angle of the quadrilateral has measure 45. 5. The diagonals are congruent, perpendicular, and bisect each other. Impossible; if diagonals of a parallelogram are congruent, the quadrilateral is a rectangle, but a rectangle has four right angles. Yes; if diagonals of a parallelogram are congruent, the quadrilateral is a rectangle, and if diagonals of a parallelogram are perpendicular, the quadrilateral is a rhombus, and a rectangle that is a rhombus is a square. 6-4
Special Parallelograms
Lesson 6-4 Check Skills You’ll Need (For help, go to Lesson 6-2.) PACE is a parallelogram and m PAC = 109. Complete each of the following. 1. EC = ? 2. EP = ? 3. m CEP = ? 4. PR = ? 5. RE = ? 6. CP = ? 7. m EPA = ? 8. m ECA = ? 9. Draw a rhombus that is not a square. Draw a rectangle that is not a square. Explain why each is not a square. Check Skills You’ll Need 6-4
Special Parallelograms
Lesson 6-4 Check Skills You’ll Need Solutions 1. EC = AP = EP = AC = m CEP = m PAC = 109 4. PR = CR = RE = AR = CP = 2RC = 2(4.75) = 9.5 7. Since PACE is a parallelogram, AC || PE . By the same-side Interior Angles Theorem, PAC and EPA are supplementary. So, m EPA = 180 – m PAC = 180 – 109 = 71. 8. From Exercise 7, m EPA = 71. Since PACE is a parallelogram, opposite angles are congruent. So, m ECA = m EPA = 71. 9. Answers may vary. Samples given. The rhombus is not a square because it has no right s. The rectangle is not a square because all 4 sides aren’t . 6-4
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http://blogs.sas.com/content/graphicallyspeaking/2014/07/08/spirals/
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# Spirals
Spirals are cool. And useful. We use them every day without thinking about it. Every time the road turns from a straight line to a curve, we go through a transition spiral. Spirals allow us to change curvature in a steady increasing or decreasing fashion. Without a spiral, this transition would be abrupt.
For our purpose, spirals can also be useful for visualizing data that is cyclical in nature. If you are visualizing daily high temperature over a 2 year period, you could plot it on a straight X axis as shown on the right.
The cyclical nature of the data is evident in the graph. But, this same cyclical behavior may be easier to understand on a spiral with one cycle per year. So, that is the plan for today.
A visit to Wikipedia page on spirals reveals there are many kinds of spirals including logarithmic, hyperbolic and many more. Let us start with the simple Archimedean Spiral. This has the simple formula: R = A + B * theta.
Experimenting with this equation yields these different spirals for various values of A and number of cycles.
A=0, Cycles=1
The x and y points are computed using the spiral equation, for theta values up to N * 360, where N is the number of cycles. The curve is plotted using the series statement of the SGPLOT procedure.
proc sgplot data=spiral nowall noborder pad=0;
series x=x y=y / smoothconnect;
xaxis min=-&max max=&max grid display=none;
yaxis min=-&max max=&max grid display=none;
run;
A=0.5, Cycles=2
The first spiral starts from the center (A=0) and turns through 360 degree cycle. The second spiral starts at an offset of 0.5 the radius and turns through 720 degrees.
Once the spiral is drawn, now we need to map the data on to the spiral so that the time axis is along the spiral and the response values are drawn normal to the spiral, towards the center.
The graph on the right shows the basic idea. At each point along the time axis, we compute the theta and then find the (x, y) point on the spiral. The direction vector (cx, cy) for the response (the arrows) is towards the center of the spiral (0, 0). In the example on the right, all arrow heights are half the spiral spacing. So, we can compute the (x2, y2) location of the arrow heads from (x, y) and (cx, cy) as shown in the program.
I have used some SAS 9.4 features to draw smooth curves and remove background wall and border. A SAS 9.3 version program is also included.
With real time series data, we would normalize the response over the entire range, and plot the data one side of the spiral by using the abs() value and a color to signify sign. Then, scale the vector by response and plot. A simulated example is shown on the right.
We will cover the mapping real world time series data (as shown in the first graph on top) on to the spirals in next article on Spirals.
SAS 9.4 Program: Spiral_Macro_94
SAS 9.3 Program: Spiral_Macro_93
tags: Series, SGPLOT, SmoothConnect, Vector
1. Steve Bloom
Posted July 31, 2014 at 11:51 am | Permalink
In your future spiral post, could you do a heat map? Here is an example of using a donut chart, but It would be better as a spiral so December 2013, would match up with January 2014.
https://www.dropbox.com/s/i7f7clipc4umj3m/Calendar%20as%20Donut.png
• Steve Bloom
Posted July 31, 2014 at 2:54 pm | Permalink
Here is my intial attemp of using your spiral macro to make a gmap map file. Now I need to figure how and where to label months and years.
https://www.dropbox.com/s/4azvar5ynjas02t/spiral.sas
• Sanjay Matange
Posted July 31, 2014 at 3:01 pm | Permalink
Very cool. This is very close to what I did to create the Spiral Heat Map from the weather data set. I will write it up for the next article. I used the new Polygon Plot, but you can easily use GMAP too.
1. By Splines - Graphically Speaking on September 21, 2014 at 10:21 pm
[…] you can go a step beyond, and use these systems to create completely non-standard graphs such as the Spiral Plot, the Polar Graph, the Euler Diagram and […]
Welcome to Graphically Speaking, a blog by Sanjay Matange focused on the usage of ODS Graphics for data visualization in SAS. The blog will cover topics related to the Statistical Graphics procedures, the Graph Template Language and the ODS Graphics Designer. Sanjay and Dan are the co-authors of Statistical Graphics Procedures by Example: Effective Graphs Using SAS
Sanjay is the author of: Getting Started with the Graph Template Language in SAS
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https://www.geeksforgeeks.org/minimum-steps-in-which-n-can-be-obtained-using-addition-or-subtraction-at-every-step/?ref=rp
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# Minimum steps in which N can be obtained using addition or subtraction at every step
• Difficulty Level : Medium
• Last Updated : 08 Jun, 2022
Given N, print the sequence of a minimum number of steps in which N can be obtained starting from 0 using addition or subtraction of the step number.
Note: At each step, we can add or subtract a number equal to the step number from the current position. For example, at step 1 we can add 1 or -1. Similarly, at step 2 we add 2 or -2 and so on.
The below diagram shows all possible positions that can be reached from 0 in 3 steps by performing the specified operations.
Examples :
```Input: n = -4
Output: Minimum number of Steps: 3
Step sequence: 1 -2 -3
Explanation:
Step 1: At step 1 we add 1 to move from 0 to 1.
Step 2: At step 2 we add (-2) to move from 1 to -1.
Step 3: At step 3 we add (-3) to move from -1 to -4.
Input: n = 11
Output: Minimum number of steps = 4
Step sequence: 1 -2 3 4 5 ```
Approach: The approach to solve the above problem is to mark the step numbers where we have to subtract or add where if N is positive or negative respectively. If N is positive, add numbers at every step, until the sum exceeds N. Once the sum exceeds N, check if sum-N is even or not. If sum-N is even, then at step number (sum-N)/2, subtraction is to be done. If sum-N is an odd number, then check if the last step at which sum exceeded N was even or odd. If it was odd, perform one more step else perform two steps. If sum = N at any step, then addition or subtraction at every step will give the answer.
Let N = 11, then 1+2+3+4+5=15 exceeds 11. Subtract 15-11 to get 4, which is equivalent to performing subtraction at step 2. Hence the sequence of steps is 1 -2 3 4 5
Let N=12, then 1+2+3+4+5=15 exceeds 11. Subtract 15-12 to get 3, which cannot be performed at any step. So add two more steps, one is the 6th step and 7th step. The target is to make sum-N even, so perform addition at 6th step and subtraction at 7th step, which combines to subtract 1 from the sum. Now sum-N is even, 14-12=2 which is equivalent to performing subtraction at step 1. Hence the sequence of steps are -1 2 3 4 5 6 -7
Let N=20, then 1+2+3+4+5+6 exceeds 20. Subtract 21-20 to get 1, so add 7 to 21 to get 28. Performing addition at next step will do as (sum-n) is odd. sum-N gives 8 which is equivalent to performing subtraction at step 4. Hence the sequence of steps is 1 2 3 -4 5 6 7.
Below is the illustration of the above approach:
## C++
`// C++ program to print the sequence``// of minimum steps in which N can be``// obtained from 0 using addition or``// subtraction of the step number.``#include ``using` `namespace` `std;` `// Function to return the vector``// which stores the step sequence``vector<``int``> findSteps(``int` `n)``{`` ``// Steps sequence`` ``vector<``int``> ans;` ` ``// Current sum`` ``int` `sum = 0;` ` ``// Sign of the number`` ``int` `sign = (n >= 0 ? 1 : -1);`` ``n = ``abs``(n);` ` ``int` `i;`` ``// Basic steps required to get sum >= required value.`` ``for` `(i = 1; sum < n; i++) {`` ``ans.push_back(sign * i);`` ``sum += i;`` ``}`` ``cout << i << endl;` ` ``// Reached ahead of N`` ``if` `(sum > sign * n) {` ` ``// If the last step was an odd number`` ``if` `(i % 2) {`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % 2) {`` ``ans.push_back(sign * i);`` ``sum += i++;`` ``}`` ``// subtract the equivalent sum-n`` ``ans[(sum / 2) - 1] *= -1;`` ``}`` ``else` `{`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % 2) {` ` ``// since addition of next step and subtraction`` ``// at the next step will give sum = sum-1`` ``sum--;`` ``ans.push_back(sign * i);`` ``ans.push_back(sign * -1 * (i + 1));`` ``}`` ``// subtract the equivalent sum-n`` ``ans[(sum / 2) - 1] *= -1;`` ``}`` ``}`` ``// returns the vector`` ``return` `ans;``}` `// Function to print the steps``void` `printSteps(``int` `n)``{`` ``vector<``int``> v = findSteps(n);` ` ``// prints the number of steps which is the size of vector`` ``cout << ``"Minimum number of Steps: "` `<< v.size() << ``"\n"``;` ` ``cout << ``"Step sequence:"``;` ` ``// prints the steps stored`` ``// in the vector`` ``for` `(``int` `i = 0; i < v.size(); i++)`` ``cout << v[i] << ``" "``;``}` `// Driver Code``int` `main()``{`` ``int` `n = 20;`` ``printSteps(n);`` ``return` `0;``}`
## Java
`// Java program to print the``// sequence of minimum steps``// in which N can be obtained``// from 0 using addition or``// subtraction of the step``// number.``import` `java.util.*;` `class` `GFG``{` `// Function to return the``// Arraylist which stores``// the step sequence``static` `ArrayList findSteps(``int` `n)``{`` ``// Steps sequence`` ``ArrayList ans = ``new` `ArrayList();` ` ``// Current sum`` ``int` `sum = ``0``;` ` ``// Sign of the number`` ``int` `sign = (n >= ``0` `? ``1` `: -``1``);`` ``n = Math.abs(n);` ` ``int` `i;`` ``// Basic steps required to`` ``// get sum >= required value.`` ``for` `(i = ``1``; sum < n; i++)`` ``{`` ``ans.add(sign * i);`` ``sum += i;`` ``}`` ``System.out.println( i );` ` ``// Reached ahead of N`` ``if` `(sum > sign * n)`` ``{` ` ``// If the last step`` ``// was an odd number`` ``if` `(i % ``2` `!= ``0``)`` ``{`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % ``2` `!= ``0``)`` ``{`` ``ans.add(sign * i);`` ``sum += i++;`` ``}`` ` ` ``// subtract the`` ``// equivalent sum-n`` ``ans.set((sum / ``2``) - ``1``,`` ``ans.get((sum / ``2``) - ``1``) * -``1``);`` ``}`` ``else`` ``{`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % ``2` `!= ``0``)`` ``{` ` ``// since addition of next`` ``// step and subtraction at`` ``// the next step will`` ``// give sum = sum-1`` ``sum--;`` ``ans.add(sign * i);`` ``ans.add(sign * -``1` `* (i + ``1``));`` ``}`` ` ` ``// subtract the`` ``// equivalent sum-n`` ``ans.set((sum / ``2``) - ``1``,`` ``ans.get((sum / ``2``) - ``1``) * -``1``);`` ``}`` ``}`` ` ` ``// returns the Arraylist`` ``return` `ans;``}` `// Function to print the steps``static` `void` `printSteps(``int` `n)``{`` ``ArrayList v = findSteps(n);` ` ``// prints the number of steps`` ``// which is the size of Arraylist`` ``System.out.println(``"Minimum number "` `+`` ``"of Steps: "` `+`` ``v.size());` ` ``System.out.print(``"Step sequence:"``);` ` ``// prints the steps stored`` ``// in the Arraylist`` ``for` `(``int` `i = ``0``; i < v.size(); i++)`` ``System.out.print(v.get(i) + ``" "``);``}` `// Driver Code``public` `static` `void` `main(String args[])``{`` ``int` `n = ``20``;`` ``printSteps(n);``}``}``// This code is contributed``// by Arnab Kundu`
## C#
`// C# program to print the``// sequence of minimum steps``// in which N can be obtained``// from 0 using addition or``// subtraction of the step``// number.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to return the``// Arraylist which stores``// the step sequence``static` `List<``int``> findSteps(``int` `n)``{`` ``// Steps sequence`` ``List<``int``> ans = ``new` `List<``int``>();` ` ``// Current sum`` ``int` `sum = 0;` ` ``// Sign of the number`` ``int` `sign = (n >= 0 ? 1 : -1);`` ``n = Math.Abs(n);` ` ``int` `i;`` ` ` ``// Basic steps required to`` ``// get sum >= required value.`` ``for` `(i = 1; sum < n; i++)`` ``{`` ``ans.Add(sign * i);`` ``sum += i;`` ``}`` ``Console.WriteLine( i );` ` ``// Reached ahead of N`` ``if` `(sum > sign * n)`` ``{` ` ``// If the last step`` ``// was an odd number`` ``if` `(i % 2 != 0)`` ``{`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % 2 != 0)`` ``{`` ``ans.Add(sign * i);`` ``sum += i++;`` ``}`` ` ` ``// subtract the`` ``// equivalent sum-n`` ``ans[(sum / 2) - 1]=`` ``ans[(sum / 2) - 1] * -1;`` ``}`` ``else`` ``{`` ``sum -= n;` ` ``// sum-n is odd`` ``if` `(sum % 2 != 0)`` ``{` ` ``// since addition of next`` ``// step and subtraction at`` ``// the next step will`` ``// give sum = sum-1`` ``sum--;`` ``ans.Add(sign * i);`` ``ans.Add(sign * -1 * (i + 1));`` ``}`` ` ` ``// subtract the`` ``// equivalent sum-n`` ``ans[(sum / 2) - 1]=`` ``ans[(sum / 2) - 1] * -1;`` ``}`` ``}`` ` ` ``// returns the Arraylist`` ``return` `ans;``}` `// Function to print the steps``static` `void` `printSteps(``int` `n)``{`` ``List<``int``> v = findSteps(n);` ` ``// prints the number of steps`` ``// which is the size of Arraylist`` ``Console.WriteLine(``"Minimum number "` `+`` ``"of Steps: "` `+`` ``v.Count);` ` ``Console.Write(``"Step sequence:"``);` ` ``// prints the steps stored`` ``// in the Arraylist`` ``for` `(``int` `i = 0; i < v.Count; i++)`` ``Console.Write(v[i] + ``" "``);``}` `// Driver Code``public` `static` `void` `Main(String []args)``{`` ``int` `n = 20;`` ``printSteps(n);``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
## Python3
`# Python3 program to print the sequence``# of minimum steps in which N can be``# obtained from 0 using addition or``# subtraction of the step number.` `# Function to return the``# which stores the step sequence``def` `findSteps( n):`` ``# Steps sequence`` ``ans``=``[]` ` ``# Current sum`` ``sum` `=` `0` ` ``# Sign of the number`` ``sign ``=` `1` `if` `n >``=` `0` `else` `-``1`` ``n ``=` `abs``(n)`` ``i``=``1`` ``# Basic steps required to get sum >= required value.`` ``while` `sum`` sign ``*` `n) :` ` ``# If the last step was an odd number`` ``if` `(i ``%` `2``) :`` ``sum` `-``=` `n` ` ``# sum-n is odd`` ``if` `(``sum` `%` `2``) :`` ``ans.append(sign ``*` `i)`` ``sum` `+``=` `i`` ``i``+``=``1`` ` ` ``# subtract the equivalent sum-n`` ``ans[``int``((``sum` `/` `2``) ``-` `1``)] ``*``=` `-``1`` ` ` ``else` `:`` ``sum` `-``=` `n` ` ``# sum-n is odd`` ``if` `(``sum` `%` `2``) :` ` ``# since addition of next step and subtraction`` ``# at the next step will give sum = sum-1`` ``sum``-``=``1`` ``ans.append(sign ``*` `i)`` ``ans.append(sign ``*` `-``1` `*` `(i ``+` `1``))`` ` ` ``# subtract the equivalent sum-n`` ``ans[``int``((``sum` `/` `2``) ``-` `1``)] ``*``=` `-``1`` ` ` ` ` ``# returns the`` ``return` `ans` `# Function to pr the steps``def` `prSteps(n):`` ``v ``=` `findSteps(n)` ` ``# print the number of steps which is the size of`` ``print``(``"Minimum number of Steps:"``,``len``(v))` ` ``print``(``"Step sequence:"``,end``=``"")` ` ``# print the steps stored`` ``# in the`` ``for` `i ``in` `range``(``len``(v)):`` ``print``(v[i],end``=``" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``n ``=` `20`` ``prSteps(n)`
Output :
```7
Minimum number of Steps: 7
Step sequence:1 2 3 -4 5 6 7```
Time Complexity : O(sqrt(N))
Auxiliary Space : O(sqrt(N))
Note: sum = i*(i+1)/2 is equivalent or greater than N, which gives i as sqrt(N).
My Personal Notes arrow_drop_up
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# Number of possible sum?
Given a list of integers $a_1,a_2,\dotsc,a_n$, where $0\le a_i\le 100$ for every $i$, where $n\le 100$, find all the distinct possible sums that can be obtained by taking any number of elements from the list and adding them.
Example: for $1,2,3$, the answer is $0,1,2,3,4,5,6$.
My approach was a brute force in which called a recurcive method for an index $i$ and inside that since there are two possibilities (to add a number or to leave it) I once added and again called the recursion and called the recursion again without adding it. This approach is time consuming; can anyone please suggest a faster method because total possible sum is only $10001$ ($0+100\cdot 100$). Thanks.
• If all numbers are different, I would say you have: i)The "empty sum", ii)The "1-sums" $a_1,a_2,..,a_{100}$ , then iii)The "2-sums" $a_i+a_j$ , etc. – user99680 Dec 1 '13 at 6:20
• @user99680 But that is inefficient since you would then need to do $2^{100}$ such sums, whereas the total candidates are only $100^2$. – user17762 Dec 1 '13 at 6:23
• @user 17762: but aren't we also considering adding triples, quadruples, etc. of numbers? – user99680 Dec 1 '13 at 6:24
• @user99680 Yes, we are. But if $1\leq a_1, a_2,\ldots,a_m \leq n$, then the number of possible sums without $a_i$'s repeating is bounded by $mn$, whereas if we look at number of $r$ sums, we would have $\dbinom{m}r$ of them and letting $r$ from $0$ to $m$ would give us a total of $2^m$ sums to look at. – user17762 Dec 1 '13 at 6:26
• @user17762: Ah, I see; you're right. – user99680 Dec 1 '13 at 6:27
## 2 Answers
I would start with an array of length $10001$, filled with zeros to indicate you have not yet found a way to make that total. Set $array[0]$ to $1$ to indicate you can sum to zero with the empty subset Then for each element, loop through the array. Add the element to the indices that are already $1$ and set those indices to $1$. For example, if $a_i=3$ and $array[6]=1$, set $array[9]$ to $1$. This requires $10000n$ loops.
• +1. A minor obvious comment. For a given element $a_i$, first set $array[a_i]=1$ and then proceed. – user17762 Dec 1 '13 at 6:36
• @Ross Millikan nice :) thanks – Ashesh Vidyut Dec 1 '13 at 9:59
Here is my take on this:
hash[0]=true //sum=0 can be obtained by any empty subset
Now,let SUM=sum of all numbers in array
//Iterate through the entire array
for(i=0 to n-1)
//When i-th element is included in sum,minimum (sum)=a[i],maximum(sum)=SUM
for(j=sum;j>=a[i];j--)
//Now,if sum=j-a[i],is a possible sum value then j would also be a possible value,just add a[i]
if(hash[j-a[i]]==true)
hash[j]=true
//Count number of all possible sum using hash
for(i=0;i<=SUM;i++) //remember,we just need to go upto SUM
{
if(hash[i]==true)
count++;
}
print count
• @Templar:thanks,it looks better, i didn't knew how to make it look so,my first answer. – charany1 Apr 25 '14 at 16:13
• no problem, just select a text which you want to output as code or something similar and click { } icon called preformatted text or press Ctrl + k – Templar Apr 25 '14 at 16:43
• @Templar:ok,got it. – charany1 Apr 25 '14 at 16:47
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+0
# need help
0
37
1
An element with a mass of 700 grams decays by 6.1% per minute. To the nearesttenth of a minute, how long will it be until there are 220 grams of the element remaining?
Sep 16, 2020
#1
0
220 = 700 x (0.939)^m
220/700 = (0.939)^m
m = Log(220/700) / Log(0.939)
m =18.4 minutes
Sep 16, 2020
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## A dragster takes off from rest and crosses the finish line 320 m away. If the dragster is able to accelerate at 10LaTeX: \frac{m}{s^2}m s 2
Question
A dragster takes off from rest and crosses the finish line 320 m away. If the dragster is able to accelerate at 10LaTeX: \frac{m}{s^2}m s 2 then how long does it take to finish?
in progress 0
3 years 2021-07-19T16:34:57+00:00 2 Answers 6 views 0
## Answers ( )
1. Answer:
t = 8 s
Explanation:
In order to find the time taken by the dragster we will use equations of motion. Here, we will use second equation of motion:
s = Vi t + (1/2)at²
where,
s = distance covered = 320 m
Vi = Initial Velocity = 0 m/s (Since, dragster starts from rest)
t = time taken = ?
a = acceleration of dragster = 10 m/s²
Therefore,
320 m = (0 m/s)t + (1/2)(10 m/s²)t²
t² = (320 m)(2)/(10 m/s²)
t = √(64 s²)
t = 8 s
2. Answer:
t = 8 s
Explanation:
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# PSSA Survivor – Math Outwit Lesson. comparing grades placing events in date order comparing the cost of two or more objects.
## Presentation on theme: "PSSA Survivor – Math Outwit Lesson. comparing grades placing events in date order comparing the cost of two or more objects."— Presentation transcript:
PSSA Survivor – Math Outwit Lesson
comparing grades placing events in date order comparing the cost of two or more objects
If the numbers are all integers (numbers without decimals), we just place them in order. However, most likely the numbers we will have to place in order will be either decimals, fractions, or a combination of both. Therefore, we will look into all three scenarios.
Step 1: Line up the decimal points of each number Step 2: Make sure there are the same number of decimal places for each number Step 3: Compare each digit from left to right until a difference is found
Place 3.498, 3.49, 3.4892, and 3.4981 in order from least to greatest Step 13.498 3.49 3.4892 3.4981 Step 23.4980 3.4900 3.4892 3.4981 Step 33.4892, 3.49, 3.4980, 3.4981
Step 1: Change everything to a decimal (use division of the top number by the bottom number) Step 2: Follow steps from the all decimals directions
Order from least to greatest Step 1: Step 2:.25.66…..50.16…. Answer:
Step 1: Change the fractions to decimals Step 2: Follow the rules to decimals
Order from least to greatest Step 1: Step 2: 5.66 7.00 5.33… Answer:
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# How to find a “better description” (e.g. recurrence relation) for this sequence?
My solution to a problem in Project Euler required to solve this subproblem: find values of $k\in\mathrm{N}$ such that $3k^2+4$ is a perfect square.
As I was writing a computer program, I just tried all $k$ and checking if $3k^2+4$ is a perfect square. I solved the problem, but this is not efficient and it doesn't really answer the question.
It turns out that this sequence is http://oeis.org/A052530, there is an easy recurrence relation ($k_n = 4k_{n-1} - k_{n-2}$), and some closed-form formulas for $k_n$ (e.g. $k_n = \left((2+\sqrt{3})^n-(2-\sqrt{3})^n\right)/\sqrt{3}$).
Now I know some answers, but I still don't see how to derive them from the definition. Also, I wasn't able to prove that the recurrence relation works (given that $k_{n-2}$ and $k_{n-1}$ are to consecutive terms of the sequence, prove that $4k_{n-1} - k_{n-2}$ is a term in the sequence, and that it is next term).
So my question is: given the definition of the sequence ($k\in\mathrm{N}$ such that $3k^2+4=n^2$), how can I find a recurrence relation for this sequence?
I will be very happy if can use the same procedure for other similar sequences.
-
The keyphrase is "Pell's equation". – Gerry Myerson Feb 25 '13 at 0:33
@dbarbosa You must define $k_0$ and $k_1$. My answer to this question describes the method to find a closed formula in a particular example. The method used is the one described here under Linear. If you happen to guess the closed formula for a given recurrence relation you can use induction to prove it. – Git Gud Feb 25 '13 at 0:55
@GitGud You can use $k_0 = 0$ and $k_1 = 2$ to match with oeis definition. I am happy with the recurrence or the closed formula (for the context of the original problem that I was solving, the recurrence is actually better). Guessing + induction can be helpful, however in this case I couldn't prove that the recurrence relation was really the same sequence with induction (and I didn't guess the answer before solving the problem). I will read your other answer. – dbarbosa Feb 25 '13 at 1:46
@GitGud the answer that you pointed describes how to find the closed formula for a recurrence relation. Here I am actually trying to find the recurrence relation (or the closed formula) from this definition: $k$ such that $3k^2 + 4$ is a perfect square. – dbarbosa Feb 25 '13 at 1:52
@dbarbosa oh, ok. Sorry. I thought I misunderstood your question. – Git Gud Feb 25 '13 at 7:01
Suppose $$3k^2+4=m^2$$ so that $$m^2-3k^2=4$$ or $$(m+\sqrt3k)(m-\sqrt3k)=4$$
and we are also able to find a solution to $$p^2-3q^2=1$$$$(p+\sqrt3q)(p-\sqrt3q)=1$$
Then $$(p+\sqrt3q)(p-\sqrt3q)(m+\sqrt3k)(m-\sqrt3k)=(p+\sqrt3q)(m+\sqrt3k)(p-\sqrt3q)(m-\sqrt3k)=4$$which becomes$$\left((pm+3kq)+(pk+qm)\sqrt3\right)\left((pm+3kq)-(pk+qm)\sqrt3\right)=4$$so that $$(pm+3kq)^2-3(pk+qm)^2=4$$
We note that $p=2, q=1$ works (and is the minimal solution), so that given a solution $(m,n)$, we have another solution $(2m+3k, 2k+m)$.
-
$3k^2+4=n^2$ ; as $n\to \infty$, $3k^2\cong n^2$ ; so $\sqrt 3\cong n/k, \sqrt 3-(n/k)\cong 0$ ; taking $\sqrt 3=x ;kx-n\cong (+/-)0, (kx-n)^m\to 0$, as $m\to \infty$
$2-x=2-1.732051=0.267949$ $(2-x)^2=(2^2+x^2-2\times 2\times x)=(7-4x)=0.071797$
$(7^2+4^2\times 3-2\times 7\times 4x)=(97-56x)=0.005155$
$(a_1-b_1\times x)\cong 0$;then $a_1^2+3\times b_1^2-2\times a_1\times b_1\times x = (a_1-b_1\times x)^2 \cong 0$or$\to a_2-b_2\times x\cong 0$ , where $a_2=a_1^2+3\times b_1^2 ;b_2=2\times a_1\times b_1$, in this case $7^2-4^2\times 3 =1 ;97^2-3\times 56^2 =9409-9408=1$
This process can be repeated.
-
This is pretty much unreadable (and no justification is given for it working). – Gerry Myerson Mar 1 '13 at 12:10
To derive the solution to your recurrence, you assume there is a solution of the form $r^n$. Plugging that into the recurrence gives the characteristic polynomial $r^2=4r-1$ with solutions $r=2\pm \sqrt 3$. Because your equation is linear and homogenous, any linear combination of the solutions is again a solution and the general solution is $k_n=a(2+\sqrt 3)^n+b(2-\sqrt 3)^n$ You evaluate $a,b$ from your initial conditions $k_0=0, k_1=2$
You can observe another recurrence: if $(k_n,m_n)=\left(k_n,\sqrt{3k_n^2+4}\right)$ is a pair in your series, $(k_{n+1},m_{n+1})=(2k_n+m_n,2m_n+3k_n)$ is the next.
Once you have this recurrence, you can observe that $m_{n+1}^2-3k_{n+1}^2=(2m_n+3k_n)^2-3(m_n+2k_n)^2=4m_n^2+12m_nk_n+9k_n^2-3m_n^2-12m_nk_n-12k_n^2=m_n^2-3k_n^2$ which shows $3k_{n+1}^2+4$ is a square because $3k_n^2+4$ was.
This doesn't show how to find your recurrence, nor that this gets all the $(k,m)$ pairs.
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CC-MAIN-2016-26
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http://forum.enjoysudoku.com/a-help-is-needed-t31641.html
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## A help is needed
Post the puzzle or solving technique that's causing you trouble and someone will help
### A help is needed
I'm very new to these kind of methods, which i read about them only recently: hidden/naked pairs/triples/quads..
and i kinda jumped to deep water and trying to solve 5 sudokus combined.
tell me if im right, please. if im wrong, i'd like an explanation
in pic 1 BELOW: the LEFT block - is there a hidden quad of 2-5-7-9 in cells: 2, 3, 8, 9? which means i can rule out 2-5-7-9 in any other cell?
the digit 3 is being discovered in cell 5, therfore 6 is being discovered in cell 7. true or false?
in pic 1 BELOW: the RIGHT block - is there a hidden quad of 2-4-5-9 in cells: 1, 5, 7, 8? which means i can rule out 2-4-5-9 in any other cell?
the digit 1 is being discovered in cell 3, therfore the digit 6 is being discovered in cell 6 (no other 6 candidate) and 8 is being discovered on cell 9. true or false?
http://www.siz.co.il/my.php?i=mjvmmjilggze.jpg
111.JPG (278.35 KiB) Viewed 444 times
in pic 2 BELOW: the LEFT block - is there a hidden quad of 2-5-8-9 in cells: 3, 4, 5, 9? which means i can rule out 2-5-8-9 in any other cell?
i only get the result of two digits (1, 4) in cells 7 and 8. true or false?
http://www.siz.co.il/my.php?i=a1jyjz0yeyey.jpg
222.JPG (294.18 KiB) Viewed 444 times
in pic 3 BELOW: the SECOND row - is there a hidden quad of 2-5-8-9 again in row 2, collumns 4, 5, 6, 8? which means i can rule out 5, 9 in collumn 9?
the digit 6 is being discovered in collumn 9. true or false?
http://www.siz.co.il/my.php?i=mnct5nkty1in.jpg
333.JPG (119.92 KiB) Viewed 444 times
Last edited by syxQn on Mon Feb 24, 2014 2:52 am, edited 1 time in total.
syxQn
Posts: 4
Joined: 23 February 2014
### Re: A help is needed
I don't see any pictures, so can't see the puzzles you are commenting on. However, there does appear to be something wrong with your logic. You appear to be mixing up properties of hidden and naked quads.
A hidden quad, four candidates that only appear in four cells of a house, allows you to eliminate all other candidates in the four cells that make up a quad.
A naked quad, four candidates sharing four cells in a house with no other candidates, allows you to eliminate the candidates in the quad from all other cells in the house that are not part of the quad.
JasonLion
2017 Supporter
Posts: 641
Joined: 25 October 2007
Location: Silver Spring, MD, USA
### Re: A help is needed
Now that I can see the pictures, there are not any hidden quads in any of those pictures. Read the definition I gave above. In all of your examples the specified candidates also appear in other cells, which violates the definition.
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