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[ "1,183 Pages\n\nA figurate number is a number represented as a regular and discrete geometric pattern (e.g. dots) such as a polygonal number or a polyhedral number.\n\n## Terminology\n\nThe triangular numbers can be generalized in various ways. As used here, and in a number other places, the term figurate number is meant to be broadly inclusive.\n\nIn historical works about Greek mathematics the preferred term is figured number. For example A history of Greek Mathematics by T. Heath and Greek Mathematical Philosophy by E.A. Maziarz.\n\nIn a use going back to Jakob Bernoulli's Ars Conjectandi, the term figurate number is used for triangular numbers made up of successive integers, tetrahedral numbers made up of successive triangular numbers, etc. In other words binomial coefficients. In this usage the square numbers 4,9,16,25 would not be considered figurate numbers when viewed as arranged in a square. This is the sense in which L.E. Dickson uses the term in his History of the Theory of Numbers.\n\nA number of other sources use the term figurate number as synonymous for the polygonal numbers, either just the usual kind or both those and the centered polygonal numbers.\n\n## Triangular Numbers\n\nThe first few triangular numbers can be built from rows of 1, 2, 3, 4, 5, and 6 items:\n\nThese are seen to be the binomial coefficients", null, ". This is the case r=2 of the fact that the rth diagonal of Pascals Triangle for", null, "consists of the figurate numbers for the r-dimensional analogs of triangles (simplices).\n\nPolytopic numbers for r = 2, 3, and 4 are:\n\n•", null, "(triangular numbers).\n•", null, "(tetrahedral numbers).\n•", null, "(pentatopic numbers).\n\nOur present terms square number and cubic number derive from their geometric representation as a square or cube. The difference of two positive triangular numbers is a trapezoidal number.\n\n## Gnomon\n\nFigurate numbers were a concern of Pythagorean geometry, since Pythagoras is credited with initiating them, and the notion that these numbers are generated from a gnomon or basic unit. The gnomon is the piece added to a figurate number to transform it to the next bigger one.\n\nFor example, the gnomon of the square number is the odd number, of the general form 2n + 1, n = 0, 1, 2, 3, ... . The square of size 8 composed of gnomons looks like this:\n\n8   8   8   8   8   8   8   8\n8   7   7   7   7   7   7   7\n8   7   6   6   6   6   6   6\n8   7   6   5   5   5   5   5\n8   7   6   5   4   4   4   4\n8   7   6   5   4   3   3   3\n8   7   6   5   4   3   2   2\n8   7   6   5   4   3   2   1\n\nTo transform from the n-square (the square of size n) to the (n + 1)-square, one adjoins 2n + 1 elements: one to the end of each row (n elements), one to the end of each column (n elements), and a single one to the corner. For example, when transforming the 7-square to the 8-square, we add 15 elements; these adjunctions are the 8s in the above figure.\n\nNote that this gnomonic technique also provides a proof that the sum of the first n odd numbers is n2; the figure illustrates 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 82.\n\n## Demonstration of mathematical properties\n\nSchool children construct figurate numbers from pebbles, bottle caps, etc. As a bonus, children can use figurate numbers to discover the commutative law and associative law for addition and multiplication — laws usually dictated to them — by building rows and tables of dots.\n\nFor example, the additive commutativity of 2 + 3 = 3 + 2 = 5 becomes:\n\nAnd the multiplicative commutativity of 2 × 3 = 3 × 2 = 6 becomes:\n\nBesides the subtractive method, the additive method can also approximate square roots of positive integers and solve quadratic equations.\n\nThe concepts of figurate numbers and gnomon implicitly anticipate the modern concept of recursion.\n\nCommunity content is available under CC-BY-SA unless otherwise noted." ]

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[ "# Two explanations of how electric motor works: are they correct?\n\nLet's consider the simplest model for a DC electric motor:", null, "For simplicity, we assume that over a short period of time, the current through the motor is constant. All the discussion below refers to this time period.\n\nIn high school, apparently, two explanations of the mechanism of this motor are acceptable:\n\n1. \"The moving electrons in the wire experience a force in the magnetic field created by the permanent magnets.\" This is clearly talking about the Lorentz force.\n2. \"The coil in the armature is essentially an electromagnet. So the two magnets (permanent and electric) attract and repel each other, creating a torque.\"\n\nAfter learning some degree level electromagnetism, I am pretty certain that explanation 1 is valid. However, explanation 2 is quite strange, in that there is no simple formula in electromagnetics that describes \"the force between two magnets\". We have the electrical forces between charges and Lorentz force on charges due to the magnetic field, but no analogy for forces that only involve magnets and magnetic fields only; after all, all magnetic fields are supposed to be generated by electric currents.\n\nLet's have a closer look: the strength of the magnetic field generated by a wire should be approximately inversely proportional to the distance from the wire. The magnetic field on the wire is, therefore, undefined; thus we cannot really analyze how the magnetic field generated by the coil affects the motion of the coil, given that the current is constant. So, the second explanation above isn't really supported by any rigorous formulae and mathematics.\n\nCan I say the second explanation is unacceptable or wrong based on that?\n\n## 1 Answer\n\nNo, the second explanation is a qualitative explanation but you can't say it's wrong. I suppose if you needed to predict the torque quantitatively you could say the second explanation is unacceptable but if your goal is to explain how an electric motor works, particularly at a high school level, it's perfectly fine.\n\nIf someone, say a high school student, has hands-on experience with permanent magnets and knows how it feels when they attract and repel and different distances, and especially if they have had a chance to play around with electromagnets a bit, then I would expect the second explanation would be preferable." ]

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[ "# netket.sampler.MetropolisHamiltonian¶\n\nclass netket.sampler.MetropolisHamiltonian(hilbert, hamiltonian, *args, **kwargs)\n\nSampling based on the off-diagonal elements of a Hamiltonian (or a generic Operator). In this case, the transition matrix is taken to be:\n\n$T( \\mathbf{s} \\rightarrow \\mathbf{s}^\\prime) = \\frac{1}{\\mathcal{N}(\\mathbf{s})}\\theta(|H_{\\mathbf{s},\\mathbf{s}^\\prime}|),$\n\nwhere $$\\theta(x)$$ is the Heaviside step function, and $$\\mathcal{N}(\\mathbf{s})$$ is a state-dependent normalization. The effect of this transition probability is then to connect (with uniform probability) a given state $$\\mathbf{s}$$ to all those states $$\\mathbf{s}^\\prime$$ for which the Hamiltonian has finite matrix elements. Notice that this sampler preserves by construction all the symmetries of the Hamiltonian. This is in generally not true for the local samplers instead.\n\nThis sampler only works on the CPU. To use the Hamiltonian smapler with GPUs, you should use netket.sampler.MetropolisHamiltonianNumpy\n\nParameters\n• machine – A machine $$\\Psi(s)$$ used for the sampling. The probability distribution being sampled from is $$F(\\Psi(s))$$, where the function $$F(X)$$, is arbitrary, by default $$F(X)=|X|^2$$.\n\n• hamiltonian – The operator used to perform off-diagonal transition.\n\n• n_chains – The number of Markov Chain to be run in parallel on a single process.\n\n• sweep_size – The number of exchanges that compose a single sweep. If None, sweep_size is equal to the number of degrees of freedom (n_visible).\n\nExamples\n\nSampling from a RBM machine in a 1D lattice of spin 1/2\n\n>>> import netket as nk\n>>>\n>>> g=nk.graph.Hypercube(length=10,n_dim=2,pbc=True)\n>>> hi=nk.hilbert.Spin(s=0.5, N=g.n_nodes)\n>>>\n>>> # Transverse-field Ising Hamiltonian\n>>> ha = nk.operator.Ising(hilbert=hi, h=1.0, graph=g)\n>>>\n>>> # Construct a MetropolisHamiltonian Sampler\n>>> sa = nk.sampler.MetropolisHamiltonian(hi, hamiltonian=ha)\n>>> print(sa)\nMetropolisSampler(rule = HamiltonianRule(Ising(J=1.0, h=1.0; dim=100)), n_chains = 16, machine_power = 2, n_sweeps = 100, dtype = <class 'numpy.float64'>)\n\nReturn type\n\nMetropolisSampler" ]

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[ "# If the volume of a sphere is increased by 95 5/16% without changing the shape, what will be the percentage increase in the surface area?\n\nApr 26, 2018\n\ncolor(blue)(56.25%)\n\n#### Explanation:\n\nFirst we observe that, for similar figures. If a similar figure is increased by a scale factor $\\frac{a}{b}$, then all linear measurements will increase by $\\frac{a}{b}$. Quadratic measurements will increase by a factor ${\\left(\\frac{a}{b}\\right)}^{2}$ and cubic measurements will increase by a factor ${\\left(\\frac{a}{b}\\right)}^{3}$.\n\nWe require an increase of 1525/16%=61/64:\n\n$\\frac{61}{64}$\n\nRemember this will give us 1525/16%, we need to add 100% to this.,the 100% is the amount we have before adding 1525/16% to it. So:\n\n$1 + \\frac{61}{64} = \\frac{125}{64}$\n\nNow this is volume, so:\n\n${\\left(\\frac{a}{b}\\right)}^{3} = \\frac{125}{64}$\n\n$\\frac{a}{b} = \\sqrt{\\frac{125}{64}}$\n\nWe require the scale factor for area.So:\n\n${\\left(\\frac{a}{b}\\right)}^{2} = {\\left(\\sqrt{\\frac{125}{64}}\\right)}^{2}$\n\nIf we now subtract 1 form this, we will get the increase in area percentage:\n\n${\\left(\\sqrt{\\frac{125}{64}}\\right)}^{2} - 1 = 0.562500000$\n\nThis is:\n\n56.25%" ]

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[ "# 7791 (number)\n\n7,791 (seven thousand seven hundred ninety-one) is an odd four-digits composite number following 7790 and preceding 7792. In scientific notation, it is written as 7.791 × 103. The sum of its digits is 24. It has a total of 4 prime factors and 12 positive divisors. There are 4,368 positive integers (up to 7791) that are relatively prime to 7791.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 4\n• Sum of Digits 24\n• Digital Root 6\n\n## Name\n\nShort name 7 thousand 791 seven thousand seven hundred ninety-one\n\n## Notation\n\nScientific notation 7.791 × 103 7.791 × 103\n\n## Prime Factorization of 7791\n\nPrime Factorization 3 × 72 × 53\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1113 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 7,791 is 3 × 72 × 53. Since it has a total of 4 prime factors, 7,791 is a composite number.\n\n## Divisors of 7791\n\n1, 3, 7, 21, 49, 53, 147, 159, 371, 1113, 2597, 7791\n\n12 divisors\n\n Even divisors 0 12 6 6\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 12312 Sum of all the positive divisors of n s(n) 4521 Sum of the proper positive divisors of n A(n) 1026 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 88.2666 Returns the nth root of the product of n divisors H(n) 7.59357 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 7,791 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 7,791) is 12,312, the average is 1,026.\n\n## Other Arithmetic Functions (n = 7791)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 4368 Total number of positive integers not greater than n that are coprime to n λ(n) 1092 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 989 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 4,368 positive integers (less than 7,791) that are coprime with 7,791. And there are approximately 989 prime numbers less than or equal to 7,791.\n\n## Divisibility of 7791\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 0 7 6\n\nThe number 7,791 is divisible by 3 and 7.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (7791)\n\nBase System Value\n2 Binary 1111001101111\n3 Ternary 101200120\n4 Quaternary 1321233\n5 Quinary 222131\n6 Senary 100023\n8 Octal 17157\n10 Decimal 7791\n12 Duodecimal 4613\n20 Vigesimal j9b\n36 Base36 60f\n\n## Basic calculations (n = 7791)\n\n### Multiplication\n\nn×y\n n×2 15582 23373 31164 38955\n\n### Division\n\nn÷y\n n÷2 3895.5 2597 1947.75 1558.2\n\n### Exponentiation\n\nny\n n2 60699681 472911214671 3684451273501761 28705559871852219951\n\n### Nth Root\n\ny√n\n 2√n 88.2666 19.8243 9.39503 6.00231\n\n## 7791 as geometric shapes\n\n### Circle\n\n Diameter 15582 48952.3 1.90694e+08\n\n### Sphere\n\n Volume 1.98093e+12 7.62775e+08 48952.3\n\n### Square\n\nLength = n\n Perimeter 31164 6.06997e+07 11018.1\n\n### Cube\n\nLength = n\n Surface area 3.64198e+08 4.72911e+11 13494.4\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 23373 2.62837e+07 6747.2\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.05135e+08 5.57331e+10 6361.32" ]

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[ "• 一、知识目标\n• 二、知识点\n• 01、生成器\n• ①、生成器表达式\n• ②、yield函数\n• 02、迭代器\n• ①、iter函数\n• ②、next函数\n• ③、自定义迭代器\n\n• # 一、知识目标\n\n1、掌握可迭代对象、迭代器与生成器的概念\n2、理解可迭代对象的分类方法\n3、掌握迭代器与生成器的使用", null, "# 二、知识点\n\n## 01、生成器\n\n### ②、yield函数\n\n~~在python中使用了yield的函数被称为生成器（generator）", null, "~~将列表生成表达式中的一对中括号改为小括号即可得到生成器。", null, "~~如果生成元素的方法比较复杂，不适合用for循环方式实现，我们还可以借助yield关键字利用函数实现生成器的功能。例子：实现faclist函数，依次生成1的阶乘、2的阶乘、、、、n的阶乘", null, "使用yield实现斐波那契数列：\n\n``````import sys\n\ndef fibonacci(n): # 生成器函数-斐波那契\na, b, counter = 0, 1, 0\nwhile True:\nif (counter > n):\nreturn n\nyield a\na, b = b, a + b\ncounter += 1\n\nf = fibonacci(10) # f是一个迭代器，由生成器返回生成\nwhile True:\ntry:\n# \",\".join(next(f)) # 空暇在尝试，，此处有一个小bug\nprint(next(f), end=\",\")\nexcept StopIteration:\nsys.exit()\n``````", null, "## 02、迭代器\n\n### ①、iter函数\n\n迭代器（Iterator）是指可以通过next函数不断获取下一个值的对象", null, "例如：\n\n``````# 例子：字符串、列表或元组都可以用于创建迭代器\nlist = [1, 2, 3]\nit = iter(list) # 创建迭代器对象\nprint(next(it)) # 输出迭代器的下一个元素\nprint(next(it))\n# for i in it:\n# print(i,end=\"\")\n\n# 判断可迭代对象和迭代器\nfrom collections.abc import Iterator, Iterable\n\nls = [1, 2, 3, 4, 5] # 创建一个列表对象\ng = (x for x in range(1, 6)) # 创建一个生成器\nprint(\"ls是可迭代对象:\", isinstance(ls, Iterable))\nprint(\"g是可迭代对象:\", isinstance(ls, Iterable))\nprint(\"ls是迭代器:\", isinstance(ls, Iterator))\nprint(\"g是迭代器:\", isinstance(ls, Iterator))\n\n``````", null, "~~iter函数：对于可迭代对象，可以通过iter函数得到迭代器", null, "### ②、next函数\n\n~~对于迭代器，则可以使用next函数不断获取下一个元素，当所有元素都获取完毕后在调取next函数，就会引发StopIteration。", null, "~~处理StopIteration异常", null, "### ③、自定义迭代器\n\n使用__next__和__iter__\n\n``````from collections.abc import Iterator\n\nclass Faclist: # 定义Faclist类\ndef __init__(self): # 定义构造方法\nself.n = 1\nself.fac = 1\n\ndef __next__(self): # 定义next方法\nself.fac *= self.n\nself.n += 1\nreturn self.fac\n\ndef __iter__(self): # 定义iter方法\nreturn self\n\nif __name__ == '__main__':\nfacs = Faclist()\nprint('facs是迭代器：', isinstance(facs, Iterator))\nfor i in range(1, 6): # i在1至5范围依次取值\nprint('第%d个元素：' % i, next(facs))\n\n``````", null, "来源：Argonaut_" ]

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[ "# nLab model structure on sSet-operads\n\nContents\n\nmodel category\n\n## Model structures\n\nfor ∞-groupoids\n\n### for $(\\infty,1)$-sheaves / $\\infty$-stacks\n\n#### Higher algebra\n\nhigher algebra\n\nuniversal algebra\n\n# Contents\n\n## Idea\n\nThe operadic generalization of the model structure on sSet-categories: a presentation of (∞,1)Operad.\n\n## Definition\n\nAll operads considered here are multi-coloured symmetric operads (symmetric multicategories).\n\n###### Definition\n\nCall a morphism of simplicial operads $f : P \\to Q$\n\n###### Theorem\n\nThis defines on $sSet Operad$ the structure of a model category which is\n\nThis is (Cisinski-Moerdijk, theorem 1.14).\n\n###### Remark\n\nFor $C \\in$ Set, let $sSet Operad_C \\hookrightarrow sSet Operad$ be the full subcategory on operads with $C$ as their set of colours.\n\nThen $sSet Operad_C \\simeq (Operad_C)^{\\Delta^{op}}$ is the category of simplicial objects in $C$-coloured symmetric operads, and restricted to this the above model category structure is corresponding the model structure on simplicial algebras.\n\n###### Remark\n\nRestricted along the inclusion\n\n$j_! : sSet Cat \\hookrightarrow sSet Operad$\n\nthe above model structure restricts to the model structure on sSet-categories by Julie Bergner.\n\n## Properties\n\n###### Remark\n\nA morphism in $sSet Operad$ is an acyclic fibration precisely if it is componentwise an acyclic Kan fibration.\n\nThe entries of the following table display models, model categories, and Quillen equivalences between these that present the (∞,1)-category of (∞,1)-categories (second table), of (∞,1)-operads (third table) and of $\\mathcal{O}$-monoidal (∞,1)-categories (fourth table).\n\ngeneral pattern\nstrict enrichment(∞,1)-category/(∞,1)-operad\n$\\downarrow$$\\downarrow$\nenriched (∞,1)-category$\\hookrightarrow$internal (∞,1)-category\n(∞,1)Cat\nSimplicialCategories$-$homotopy coherent nerve$\\to$SimplicialSets/quasi-categoriesRelativeSimplicialSets\n$\\downarrow$simplicial nerve$\\downarrow$\nSegalCategories$\\hookrightarrow$CompleteSegalSpaces\n(∞,1)Operad\nSimplicialOperads$-$homotopy coherent dendroidal nerve$\\to$DendroidalSetsRelativeDendroidalSets\n$\\downarrow$dendroidal nerve$\\downarrow$\nSegalOperads$\\hookrightarrow$DendroidalCompleteSegalSpaces\n$\\mathcal{O}$Mon(∞,1)Cat\nDendroidalCartesianFibrations" ]

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[ "Word:\n\n# Sum\n\nn.1.The aggregate of two or more numbers, magnitudes, quantities, or particulars; the amount or whole of any number of individuals or particulars added together; as, the sum of 5 and 7 is 12.\nTake ye the sum of all the congregation.\n2.A quantity of money or currency; any amount, indefinitely; as, a sum of money; a small sum, or a large sum.\nWith a great sum obtained I this freedom.\n3.The principal points or thoughts when viewed together; the amount; the substance; compendium; as, this is the sum of all the evidence in the case; this is the sum and substance of his objections.\n4.Height; completion; utmost degree.\n5.(Arith.) A problem to be solved, or an example to be wrought out.\nA large sheet of paper . . . covered with long sums.\n Algebraic sum as distinguished from arithmetical sum, the aggregate of two or more numbers or quantities taken with regard to their signs, as + or -, according to the rules of addition in algebra; thus, the algebraic sum of -2, 8, and -1 is 5.\n In sum in short; in brief.- Rogers.\nv. t.1.To bring together into one whole; to collect into one amount; to cast up, as a column of figures; to ascertain the totality of; - usually with up.\n[imp. & p. p. Summed ; p. pr. & vb. n. Summing.]\n2.To bring or collect into a small compass; to comprise in a few words; to condense; - usually with up.\n3.(Falconry) To have (the feathers) full grown; to furnish with complete, or full-grown, plumage.\n Summing up a compendium or abridgment; a recapitulation; a résumé; a summary.\n Noun 1 sum - a quantity of money; \"he borrowed a large sum\"; \"the amount he had in cash was insufficient\"Synonyms: amount of money, sum of money, amount 2 sum - a quantity obtained by additionSynonyms: total, amount 3 sum - the final aggregate; \"the sum of all our troubles did not equal the misery they suffered\"Synonyms: sum total, summation 4 sum - the choicest or most essential or most vital part of some idea or experience; \"the gist of the prosecutor's argument\"; \"the heart and soul of the Republican Party\"; \"the nub of the story\" 5 sum - the whole amountSynonyms: total, aggregate, totality 6 sum - the basic unit of money in Uzbekistan 7 sum - a set containing all and only the members of two or more given sets; \"let C be the union of the sets A and B\"Synonyms: join, union Verb 1 sum - be a summary of; \"The abstract summarizes the main ideas in the paper\"Synonyms: summarise, summarize, sum up 2 sum - determine the sum of; \"Add all the people in this town to those of the neighboring town\"Synonyms: add together, summate, tot, tot up, tote up, total, add up, sum up, tally, add\n 1 (theory) sum - In domain theory, the sum A + B of two domains contains all elements of both domains, modified to indicate which part of the union they come from, plus a new bottom element. There are two constructor functions associated with the sum:inA : A -> A+B inB : B -> A+B inA(a) = (0,a) inB(b) = (1,b)and a disassembly operation:case d of isAThis can be generalised to arbitrary numbers of domains.See also smash sum, disjoint union. 2 (tool) sum - A Unix utility to calculate a 16-bit checksum of the data in a file. It also displays the size of the file, either in kilobytes or in 512-byte blocks. The checksum may differ on machines with 16-bit and 32-bit ints.Unix manual page: sum(1).\naccount, add, add up, addend, affective meaning, all, amount of money, amplitude, batch, be-all and end-all, bearing, body, box score, budget, bulk, bunch, cast, cast up, chunk, cipher up, clutch, coloring, compute, condense, connotation, consequence, core, count, count up, deal, denotation, detail, difference, digest, dose, drift, effect, entirety, entity, epitome, essence, extension, extent, figure, figure up, foot, foot up, force, gist, gob, grammatical meaning, grand total, gross, gross amount, group, heap, hunk, idea, impact, implication, import, integral, integrate, intension, inventory, itemize, large amount, lexical meaning, literal meaning, lot, lump sum, magnitude, main point, mass, matter, meaning, measure, measurement, meat, mess, number, numbers, nutshell, overtone, pack, parcel, part, pertinence, pith, plus, plus sign, point, portion, practical consequence, product, purport, quantity, quantum, range of meaning, ration, real meaning, recap, recapitulate, recapitulation, recite, reckon up, reckoning, recount, reference, referent, rehearse, relate, relation, relevance, resume, round sum, run-through, rundown, scope, score, score up, semantic cluster, semantic field, sense, significance, signification, significatum, signifie, small amount, span of meaning, spirit, strength, structural meaning, structure, substance, subtotal, sum and substance, sum total, sum up, summarize, summary, summate, summation, summing-up, symbolic meaning, synopsize, system, tale, tally, tally up, tenor, the amount, the bottom line, the story, the whole story, tot, tot up, total up, totality, totality of associations, tote, tote up, transferred meaning, unadorned meaning, undertone, value, whole amount, x number\nDefinitions Index: # A B C D E F G H I J K L M N O P Q R S T U V W X Y Z" ]

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[ "## WeBWorK Main Forum", null, "by Adam Z -\nNumber of replies: 3\n\nHello all:\n\nI am beginner at using this , this code takes long time to load , any idea of fixing this up:\n\nDOCUMENT();        # This should be the first executable line in the problem.\n\n\"PGstandard.pl\",     # Standard macros for PG language\n\n\"MathObjects.pl\",\n\n\"PGML.pl\",\n\n\"contextCurrency.pl\",\n\n\"PGcourse.pl\",\n\n);\n\nTEXT(beginproblem());\n\nContext(\"Currency\");\n\n$showPartialCorrectAnswers = 1; do{$a = Currency(random(15000,30000,5));\n\n$b = Currency(random(20000,40000,5));$c = Currency(random(50,300,5));\n\n$d = Currency(random(20,100,5));$y = random(1,1000,1);\n\n$m=$a+$c*$y;\n\n$n=$b+$d*$y;\n\n} until ($m==$n);\n\n@r = ('Mae','Suzan', 'Donna');\n\n$r_index = random(0,2,1);$r = $r[$r_index];\n\nBEGIN_PGML\n\n[$r] currently sells televisions for company A at a salary of [$a] plus a [$c] commission for each television she sells. Company B offers her a position with a salary of [$b] plus a [$d] commission for each television she sells. How many televisions would [$r] need to sell for the options to be equal?\n\n[$r] needs to sell [________]{$y} televisions.\n\nEND_PGML\n\nENDDOCUMENT();       # This should be the last executable line in the problem.", null, "by Danny Glin -\nOn average your loop is going to require a very large number if iterations before it finds five values that satisfy the condition.\n\nA much more efficient way to do this would be to randomly generate four of the five parameters, and then assign a value to the fifth parameter so that your desired condition is satisfied.\n\nIn your case you want $a+$c*$y to equal$b+$d*$y, so you could let $b=$a+($c-$d)*$y; You would just have to make sure that you always end up with a positive value of$b.\n\nOne option would be something like\n\n$a = Currency(random(15000,30000,5));$d = Currency(random(20,100,5));\n\n$c = Currency(random($d+5,300,5));\n\n$y = random(1,1000,1);$b = $a + ($c-$d)*$y;\n\nThis ensures that $c>$d, so the value of $a + ($c-$d)*$y will always be positive.", null, "In reply to Danny Glin\n\nThank you, Danny. Very helpful. I will implement this and see how it goes. One more question, in another question I assign  $x= Currency($a). But the output does not accept decimals how to sort out this issue and how to generate numbers with tenths only but written in hundredths e.g. 10.3 but need to be shown as 10.30 and so on.", null, "" ]

[ null, "https://webwork.maa.org/moodle/theme/image.php/classic/core/1584125254/u/f1", null, "https://webwork.maa.org/moodle/theme/image.php/classic/core/1584125254/u/f1", null, "https://webwork.maa.org/moodle/theme/image.php/classic/core/1584125254/u/f1", null, "https://webwork.maa.org/moodle/theme/image.php/classic/core/1584125254/u/f1", null ]

[ "February 11 2011\n\nScala\n\nIn an earlier post on Project Euler I mentioned that I was using the problems from the project as one way of picking up the language Scala.\nScala is quite interesting. It is a new language, but it utilizes the Java VM to run upon (Scala compiles to bytecode). This cool idea lets a programmer take advantage of the plethora of Java code that is already out there and call it quite easily from a Scala program.\nScala is object oriented, but more importantly (otherwise just use Java) Scala is a functional language. Every value is a object, and every function is a value. The syntax lets you do things in either a way that seems natural to someone who has been programming in Java (or C) or in a much more succinct functional style.\nFor example, the first problem in project Euler is to add all the natural numbers below one thousand that are multiples of 3 or 5.\nA simple way of doing this in Java would be:\n\nint accum = 0;\n\nfor (int i=1; i <= 1000; i++) { if ( i % 3 == 0 || i % 5 == 0) accum += i; } System.out.println(accum);\nThis is a pretty straightforward rendering of the problem.\n\nIn Scala we could do something like:\n\nprintln( (1 until 1000).filter(n => n%3 == 0 || n%5 == 0).foldLeft(0)(_+_) )\n\nThis looks kind of odd on first glance, but it is really doing something similar to the java example.\n(1 until 1000) defines the range of numbers from 1 to 1000. until is a method in this case, producing a Range. The filter method is then called, on that Range instance and selects all of the elements that satisfy the predicate (n%3 == 0 || n%5== 0) or in other words, all the elements that are divisible by 3 or 5. On this new Range, we then call foldLeft. This takes a starting value of 0 and applies the function \"+\" to successive elements (remember + is a value too). The result is then printed.\nIn this example, the Java code was pretty simple, so the advantage of using Scala may not be immediately apparent. It is kind of fun though. I'll share a more complex example later." ]

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[ "# Linear vs Non-Linear Relations\n\n## Differences between Linear and Quadratic Equations\n\n• Qudratic functions are expressed as $$y = x^2$$ while linear expressions are expressed as $$y = x$$\n• Quadratic functions have $$2$$ as the degree of their highest term whereas linear functions have $$1$$ as their highest degree\n• All quadratic functions increase and decrease regradless of the slope whereas linear equations either always increase (if the slope is positive) or decrease (if the slope is negative)\n• The slope of a quadratic function is constantly changing. The slope of a linear function is always constant\n• Linear functions have each input produce a unique output. Conversely, quadratic functions (with the exception of the vertex) have pairs of unique independent variables produce the same output (ie $$x$$ values of $$2$$ and $$-2$$ would produce the same result)\n\nScatter plots can also be used to identify the different relations between $$x$$ and $$y$$:\n\n• The relationship is considered linear when the points on a scatter plot follow a somewhat straight line pattern\n• The relationship is considered non-linear if the points on a scatter plot follow a pattern but not a straight line", null, "Linear", null, "Non-Linear\n\n## Identifying Linear and Non-Linear Equations\n\nAs stated above, a function can be determined by the degree of their highest term - also known as the Leading Term (ie a Linear Function's highest term is $$1$$ while a Quadratic Function's highest term is $$2$$). You will learn about other non-linear relations if you continue studying math!\n\nTry and fit the following functions into their correct description:", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

[ null, "https://app.myhomeworkrewards.com/images/g9MathPic/linear_Graph.png", null, "https://app.myhomeworkrewards.com/images/g9MathPic/nonLinear_Graph.png", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Linear1.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Linear2.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Linear3.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Linear4.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Quad1.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Quad2.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Quad3.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Quad4.PNG", null, "https://app.myhomeworkrewards.com/images/gr10MathPic/Quad5.PNG", null ]

[ "NO TIME TO DANCE\n\nRASPBERRY PI\n\nKõlar", null, "STONERIDGE LÜLITI\n\nSONIC PI\n\nOlemasolevate helifailide mängimine\n\n``````3.times do\nsample :ambi_sauna\nsleep 10\nsample :elec_bell\nsleep 6\nsample :bass_hit_c\nsleep 2\nend``````\n\nJuhuslikud toonid\n\n``````play rrand(50, 100)\n\n10.times do\nplay rrand(50, 100)\nsleep 0.5\nend``````\n\nTÄHEKE\n\nTARK LINN\n\nROBOTITE SUMO\n\nSumo 3kg\n\nEdison\n\nARDUINO\n\nMootorid\n\n``````//Arduino DC motor controlling\n\n#define motor_pin 3\n\nvoid setup() { //setup function\n// put your setup code here, to run once:\nSerial.begin(9600);\nSerial.println(\"Arduino Motor controller\");\n\npinMode(motor_pin, OUTPUT);\n}\n\nvoid loop() { //loop function\n// put your main code here, to run repeatedly:\nSerial.println(\"Motor ON\");\ndigitalWrite(motor_pin, HIGH); //turn ON motor\n\ndelay(1000); //wait 1second\n\nSerial.println(\"Motor 0FF\");\ndigitalWrite(motor_pin, LOW); //turn OFF motor\n\ndelay(1000); //wait 1second\n}``````\n\nROBOTEX EELVÕISTLUS 04.11.2022\n\nOn hea rõõm teatada, et meie meeskond The Beast võitis Lego sumo eelvõistluse kuni 13.aastaste vanusegrupis ja pääseb osalema Robotex International võistlusel.\n\nSuurimad õnnesoovid Karl Kasparile ja Hendrik-Johanile!\n\nSPHERO\n\nRaja läbimine ja topside püüdmine\n\nRASPBERRY PI\n\nLED lambi ühendamine\n\nLED lamp ühendada GND (3.pin) ja GPIO18 (6.pin)\nTeine LED lamp ühendada GPIO23 (8.pin)\nLambi lühem jalg läheb takisti poole", null, "KOBRAS viktoriin\n\nhttps://viktoriinid.ee\n\nPYTHON\n\nProgrammeerimine\n\nLahenda erinevaid ülesandeid kasutades Thonny programmi.\n\npj4t-modules.eu/et/\n\n## Nutikas Manta ringitund 28.05.18\n\nHOVERBOARD NAGU AUTO\n\nLENDAV SEGWAY\n\nKAHOOT", null, "RASPBERRY PI\n\nEsimene tutvus Raspberry PI’ga", null, "LIBLIKA ROBOT", null, "## TÜG vanemate ringitund 16.05.18\n\nARDUINO\n\nServo motor", null, "", null, "", null, "", null, "RASPBERRY SENSE HAT\n\nVärviline Digipurk", null, "projects.raspberrypi.org/en/projects/getting-started-with-the-sense-hat\n\nRGB värvid\n\n 255 0 0 Punane 0 255 0 Roheline 0 0 255 Sinine 255 255 0 Kollane 255 0 255 Lilla 0 255 255 Siniroheline\n\nwww.w3schools.com/colors/colors_picker.asp?colorhex=ff00ff\n\nRGB kalkulaator\n\nwww.w3schools.com/colors/colors_rgb.asp\n\nTekst\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nsense.show_message(“Hello world”)\n\nRGB värv\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nr = 255\ng = 255\nb = 255\nsense.clear((r, g, b))\n\nLiikuv tekst\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nblue = (0, 0, 255)\nyellow = (255, 255, 0)\nsense.show_message(“Astro Pi is awesome!”, text_colour=yellow, back_colour=blue, scroll_speed=0.05)\n\nErinevat värvi tähed\n\nfrom sense_hat import SenseHat\nfrom time import sleep\nsense = SenseHat()\nred = (255, 0, 0)\nblue = (0, 0, 255)\ngreen = (0, 255, 0)\nwhite = (255, 255, 255)\nyellow = (255, 255, 0)\nsense.show_letter(“L”, red)\nsleep(1)\nsense.show_letter(“a”, blue)\nsleep(1)\nsense.show_letter(“u”, green)\nsleep(1)\nsense.show_letter(“r”, white)\nsleep(1)\nsense.show_letter(“a”, yellow)\n\nSuvalised värvid\n\nfrom sense_hat import SenseHat\nfrom time import sleep\nfrom random import randint\nsense = SenseHat()\nred = (255, 0, 0)\nblue = (0, 0, 255)\ngreen = (0, 255, 0)\nwhite = (255, 255, 255)\nyellow = (255, 255, 0)\n# Generate a random colour\ndef pick_random_colour():\nrandom_red = randint(0, 255)\nrandom_green = randint(0, 255)\nrandom_blue = randint(0, 255)\nreturn (random_red, random_green, random_blue)\nsense.show_letter(“L”, pick_random_colour())\nsleep(1)\nsense.show_letter(“a”, pick_random_colour())\nsleep(1)\nsense.show_letter(“u”, pick_random_colour())\nsleep(1)\nsense.show_letter(“r”, pick_random_colour())\nsleep(1)\nsense.show_letter(“a”, pick_random_colour())\nsleep(1)\nsense.clear()\n\nPikslid\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nblue = (0, 0, 255)\nred = (255, 0, 0)\nsense.set_pixel(0, 2, blue)\nsense.set_pixel(7, 4, red)\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nsense.set_pixel(2, 2, (0, 0, 255))\nsense.set_pixel(4, 2, (0, 0, 255))\nsense.set_pixel(3, 4, (100, 0, 0))\nsense.set_pixel(1, 5, (255, 0, 0))\nsense.set_pixel(2, 6, (255, 0, 0))\nsense.set_pixel(3, 6, (255, 0, 0))\nsense.set_pixel(4, 6, (255, 0, 0))\nsense.set_pixel(5, 5, (255, 0, 0))\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\n# Define some colours\ng = (0, 255, 0) # Green\nb = (0, 0, 0) # Black\n# Set up where each colour will display\ncreeper_pixels = [\ng, g, g, g, g, g, g, g,\ng, g, g, g, g, g, g, g,\ng, b, b, g, g, b, b, g,\ng, b, b, g, g, b, b, g,\ng, g, g, b, b, g, g, g,\ng, g, b, b, b, b, g, g,\ng, g, b, b, b, b, g, g,\ng, g, b, g, g, b, g, g\n]\n# Display these colours on the LED matrix\nsense.set_pixels(creeper_pixels)\n\nÕhurõhk\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nsense.clear()\npressure = sense.get_pressure()\nprint(pressure)\n\nTemperatuur\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nsense.clear()\ntemp = sense.get_temperature()\nprint(temp)\n\nÕhuniiskus\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nsense.clear()\nhumidity = sense.get_humidity()\nprint(humidity)\n\nÕhurõhk, temperatuur ja õhuniiskus\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nwhile True:\n# Take readings from all three sensors\nt = sense.get_temperature()\np = sense.get_pressure()\nh = sense.get_humidity()\n# Round the values to one decimal place\nt = round(t, 1)\np = round(p, 1)\nh = round(h, 1)\n# Create the message\n# str() converts the value to a string so it can be concatenated\nmessage = “Temperature: ” + str(t) + ” Pressure: ” + str(p) + ” Humidity: ” + str(h)\n# Display the scrolling message\nsense.show_message(message, scroll_speed=0.05)\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\n# Define the colours red and green\nred = (255, 0, 0)\ngreen = (0, 255, 0)\nwhile True:\n# Take readings from all three sensors\nt = sense.get_temperature()\np = sense.get_pressure()\nh = sense.get_humidity()\n# Round the values to one decimal place\nt = round(t, 1)\np = round(p, 1)\nh = round(h, 1)\n# Create the message\n# str() converts the value to a string so it can be concatenated\nmessage = “Temperature: ” + str(t) + ” Pressure: ” + str(p) + ” Humidity: ” + str(h)\nif t > 18.3 and t < 26.7:\nbg = green\nelse:\nbg = red\n# Display the scrolling message\nsense.show_message(message, scroll_speed=0.05, back_colour=bg)\n\nLiikumine", null, "from sense_hat import SenseHat\nsense = SenseHat()\nsense.clear()\no = sense.get_orientation()\npitch = o[“pitch”]\nroll = o[“roll”]\nyaw = o[“yaw”]\nprint(“pitch {0} roll {1} yaw {2}”.format(pitch, roll, yaw))\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nwhile True:\nacceleration = sense.get_accelerometer_raw()\nx = acceleration[‘x’]\ny = acceleration[‘y’]\nz = acceleration[‘z’]\nx=round(x, 0)\ny=round(y, 0)\nz=round(z, 0)\nprint(“x={0}, y={1}, z={2}”.format(x, y, z))\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\n# Display the letter J\nsense.show_letter(“J”)\nwhile True:\nacceleration = sense.get_accelerometer_raw()\nx = acceleration[‘x’]\ny = acceleration[‘y’]\nz = acceleration[‘z’]\nx=round(x, 0)\ny=round(y, 0)\nz=round(z, 0)\nprint(“x={0}, y={1}, z={2}”.format(x, y, z))\n# Update the rotation of the display depending on which way up the Sense HAT is\nif x == -1:\nsense.set_rotation(180)\nelif y == 1:\nsense.set_rotation(90)\nelif y == -1:\nsense.set_rotation(270)\nelse:\nsense.set_rotation(0)\n\nJoystick – klaviatuuri noolte kasutamine\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\nwhile True:\nfor event in sense.stick.get_events():\nprint(event.direction, event.action)\n\nfrom sense_hat import SenseHat\nfrom time import sleep\nsense = SenseHat()\ne = (0, 0, 0)\nw = (255, 255, 255)\nsense.clear()\nwhile True:\nfor event in sense.stick.get_events():\n# Check if the joystick was pressed\nif event.action == “pressed”:\n# Check which direction\nif event.direction == “up”:\nsense.show_letter(“U”) # Up arrow\nelif event.direction == “down”:\nsense.show_letter(“D”) # Down arrow\nelif event.direction == “left”:\nsense.show_letter(“L”) # Left arrow\nelif event.direction == “right”:\nsense.show_letter(“R”) # Right arrow\nelif event.direction == “middle”:\nsense.show_letter(“M”) # Enter key\n# Wait a while and then clear the screen\nsleep(0.5)\nsense.clear()\n\nfrom sense_hat import SenseHat\nsense = SenseHat()\n# Define the functions\ndef red():\nsense.clear(255, 0, 0)\ndef blue():\nsense.clear(0, 0, 255)\ndef green():\nsense.clear(0, 255, 0)\ndef yellow():\nsense.clear(255, 255, 0)\n# Tell the program which function to associate with which direction\nsense.stick.direction_up = red\nsense.stick.direction_down = blue\nsense.stick.direction_left = green\nsense.stick.direction_right = yellow\nsense.stick.direction_middle = sense.clear # Press the enter key\nwhile True:\npass # This keeps the program running to receive joystick events\n\nProjekt – reaktsiooni kiiruse mäng\n\n# IMPORT the required libraries (sense_hat, time, random)\nfrom sense_hat import SenseHat\nfrom time import sleep\nfrom random import choice\n# CREATE a sense object\nsense = SenseHat()\n# Set up the colours (white, green, red, empty)\nw = (150, 150, 150)\ng = (0, 255, 0)\nr = (255, 0, 0)\ne = (0, 0, 0)\n# Create images for three different coloured arrows\narrow = [\ne,e,e,w,w,e,e,e,\ne,e,w,w,w,w,e,e,\ne,w,e,w,w,e,w,e,\nw,e,e,w,w,e,e,w,\ne,e,e,w,w,e,e,e,\ne,e,e,w,w,e,e,e,\ne,e,e,w,w,e,e,e,\ne,e,e,w,w,e,e,e\n]\narrow_red = [\ne,e,e,r,r,e,e,e,\ne,e,r,r,r,r,e,e,\ne,r,e,r,r,e,r,e,\nr,e,e,r,r,e,e,r,\ne,e,e,r,r,e,e,e,\ne,e,e,r,r,e,e,e,\ne,e,e,r,r,e,e,e,\ne,e,e,r,r,e,e,e\n]\narrow_green = [\ne,e,e,g,g,e,e,e,\ne,e,g,g,g,g,e,e,\ne,g,e,g,g,e,g,e,\ng,e,e,g,g,e,e,g,\ne,e,e,g,g,e,e,e,\ne,e,e,g,g,e,e,e,\ne,e,e,g,g,e,e,e,\ne,e,e,g,g,e,e,e\n]\n# Set a variable pause to 3 (the initial time between turns)\n# Set variables score and angle to 0\n# Create a variable called play set to True (this will be used to stop the game later)\npause = 3\nscore = 0\nangle = 0\nplay = True\nsense.show_message(“Keep the arrow pointing up”, scroll_speed=0.05, text_colour=[100,100,100])\n# WHILE play == True\nwhile play:\n# CHOOSE a new random angle\nlast_angle = angle\nwhile angle == last_angle:\nangle = choice([0, 90, 180, 270])\nsense.set_rotation(angle)\n# DISPLAY the white arrow\nsense.set_pixels(arrow)\n# SLEEP for current pause length\nsleep(pause)\nacceleration = sense.get_accelerometer_raw()\nx = acceleration[‘x’]\ny = acceleration[‘y’]\nz = acceleration[‘z’]\nx = round(x, 0)\ny = round(y, 0)\nprint(angle)\nprint(x)\nprint(y)\n# IF orientation matches the arrow…\nif x == -1 and angle == 180:\n# ADD a point and turn the arrow green\nsense.set_pixels(arrow_green)\nscore += 1\nelif x == 1 and angle == 0:\nsense.set_pixels(arrow_green)\nscore += 1\nelif y == -1 and angle == 90:\nsense.set_pixels(arrow_green)\nscore += 1\nelif y == 1 and angle == 270:\nsense.set_pixels(arrow_green)\nscore += 1\nelse:\n# SET play to `False` and DISPLAY the red arrow\nsense.set_pixels(arrow_red)\nplay = False\n# Shorten the pause duration slightly\npause = pause * 0.95\n# Pause before the next arrow\nsleep(0.5)\n# When loop is exited, display a message with the score\nmsg = “Your score was %s” % score\nsense.show_message(msg, scroll_speed=0.05, text_colour=[100, 100, 100])\n\nVeel ideid\n\n• Tell a joke on the LED screen.\n• If your Sense HAT is connected to the internet, you could use a Twitter API library to make it display incoming tweets.\n• Create your own images to display on the LED matrix.\n• Can you alternate between images to create an animation? Check out this Geek Gurl Diaries video for some inspiration.\n• Create an electronic dielike this one. Shaking the Pi triggers the roll of the die.\n• Create a simple graphical thermometer which outputs different colours or patterns depending on the temperature.\n• Write a program that displays an arrow (or other symbol) on screen; this symbol could be used to point to which way is down. This way, astronauts in low gravity always know where the Earth is.\n• Use the accelerometer to sense small movements — this could form part of a game, alarm system, or even an earthquake sensor.\n• Make use of the humidity sensor to detect breath and display a colour depending on the humidity.\n\n## TÜG nooremate ringitund 08.05.18\n\nInteraktiivne bowling\n\nTransformer robot\n\nSCRATCH\n\nSpeaker\n\nNäide, kuidas kasutada spraidi kõlarit ja sellega muusikat teha.", null, "_Melody =\n\n10-.125/9.5-.125/10-.125/9.5-.125/10-.125/7-.125/9-.125/8-.125/6-.375/1-.125/3-.125/6-.125/7-.375/3-.125/5.5-.125/7-.125/8-.375/3-.125/10-.125/9.5-.125/10-.125/9.5-.125/10-.125/7-.125/9-.125/8-.125/6-.375/1-.125/3-.125/6-.125/7-.375/3-.125/8-.125/7-.125/6-.375/7-.125/8-.125/9-.125/10-.375/5-.125/11-.125/10-.125/9-.375/4-.125/10-.125/9-.125/8-.375/3-.125/9-.125/8-.125/7-.375/3-.125/10-.125/9.5-.125/10-.125/9.5.125/10-.125/7-.125/9-.125/8-.125/6-.375/1-.125/3-.125/6-.125/7-.375/3-.125/8-.125/7-.125/6-.5/\n\nSound\n\nKõigepealt tuleb teha valmis spraidid erinevate kujunditega. Igal spraidil on erinev arv ruute, mis siis hiljem vahelduma hakkavad.", null, "Spraid kood.", null, "CUBROID", null, "www.cubroid.com\n\ncodingblocks.cubroid.com\n\nMaterjalid\n\nCubroid raamat\n\nCubroid juhend", null, "RASPBERRY HAT\n\nMinecrafti moodi", null, "", null, "", null, "" ]

[ null, "https://lh4.googleusercontent.com/xAM3hXxX6u_OfaBHRxeErEa6cZsqJgDZ2BKU_uqw2WL0qsx0vwyXDTGylC9uYhmjpQr8jDj5lumkzO24Tt5t377cCfjDPDCLnYe8HrBAIFyO6U2DE7bsFxHUmqKrf50XYHw2adAQY-fBhWhzPVEkrTjYpUEP3-OgY2ACQuy8GhB1e8GYw30bQ1LOrrpS4A=s2048", null, "https://lh3.googleusercontent.com/8U8V6wUprYpOFeZCvWz6X8zCjCv4f5IwHE7YFTSQnckKg2r8HuleDAKLZxnaPZ4pM1sebsSKRciaqf5vZY2moVayLf0LeqZPNisaq4z1jnJNz30lPrq3g8bcxhGMpgDMF6eKWDyFLU3kD5J0kAj0EuBQv4xlz3IfDktAn0kOzpyy5d6faQ-f1rTcIQ", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/kahoot.jpeg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0217.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_1953.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0123.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0125-1.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0126.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0122.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_8263.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/orientation.png", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/scratch-speaker-1.jpeg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/scratch-sound-4.jpeg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/scratch-sound-3.jpeg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_1843.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_4058.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0115.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_1300.jpg", null, "https://robootika.digipurk.ee/wp-content/uploads/2018/05/IMG_0116.jpg", null ]

[ "", null, "Burinrad Jan 13, 2021\n\n# The surface area of a cube is 216m^2 . Find its volume​\n\nThe time for answering the question is over\n\nTotal Surface Area of a Cube = 6a²\n\nVolume = a³\n\nThen,\n\na = 216 ÷ 6 = 36\n\na = √36 = 6\n\nThen,\n\na = 6\n\nVolume = a³ = 6³ = 216 cm³\n\nHope it helps !!!\n\n385", null, "Banks\nJan 13, 2021", null, "" ]

[ null, "https://expertinstudy.com/storage/users/profile/c/3695profile_ava.jpg", null, "https://expertinstudy.com/storage/users/profile/c/1898profile_ava.jpg", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAEALAAAAAABAAEAAAIBTAA7", null ]

[ "# Document (#3587)\n\nTitle\nKlassifikation\nImprint\nEssen : Stadtbibliothek\nYear\n1984 ff\nFootnote\nIn der FHBD-Bibliothek vorhanden: 1984: 1. Systematische Tafeln 2. alphabetisches Stich- und Schlagwortregister. - 1991: 3. aktualisierte Aufl. Band. 1.2.\nTheme\nKlassifikationssysteme\nObject\nSStBEssen\n\n## Similar documents (content)\n\n1. Dahlberg, I.: Klassifikation (1993) 6.27\n```6.269241 = sum of:\n6.269241 = weight(title_txt:klassifikation in 3284) [ClassicSimilarity], result of:\n6.269241 = score(doc=3284,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n0.15950893 = queryNorm\n6.2692413 = fieldWeight in 3284, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n1.0 = fieldNorm(doc=3284)\n```\n2. Lehmann, K.: Klassifikation (1984) 6.27\n```6.269241 = sum of:\n6.269241 = weight(title_txt:klassifikation in 3392) [ClassicSimilarity], result of:\n6.269241 = score(doc=3392,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n0.15950893 = queryNorm\n6.2692413 = fieldWeight in 3392, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n1.0 = fieldNorm(doc=3392)\n```\n3. Dorn, G.J.W.: Klassifikation (1980) 6.27\n```6.269241 = sum of:\n6.269241 = weight(title_txt:klassifikation in 4375) [ClassicSimilarity], result of:\n6.269241 = score(doc=4375,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n0.15950893 = queryNorm\n6.2692413 = fieldWeight in 4375, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n1.0 = fieldNorm(doc=4375)\n```\n4. Wolters, G.: Klassifikation (1980-) 6.27\n```6.269241 = sum of:\n6.269241 = weight(title_txt:klassifikation in 4403) [ClassicSimilarity], result of:\n6.269241 = score(doc=4403,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n0.15950893 = queryNorm\n6.2692413 = fieldWeight in 4403, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n1.0 = fieldNorm(doc=4403)\n```\n5. Manecke, H.-J.: Klassifikation (1997) 6.27\n```6.269241 = sum of:\n6.269241 = weight(title_txt:klassifikation in 285) [ClassicSimilarity], result of:\n6.269241 = score(doc=285,freq=1.0), product of:\n0.99999994 = queryWeight, product of:\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n0.15950893 = queryNorm\n6.2692413 = fieldWeight in 285, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2692413 = idf(docFreq=215, maxDocs=41962)\n1.0 = fieldNorm(doc=285)\n```" ]

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[ "# #01f4f3 Color Information\n\nIn a RGB color space, hex #01f4f3 is composed of 0.4% red, 95.7% green and 95.3% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 0.4% yellow and 4.3% black. It has a hue angle of 179.8 degrees, a saturation of 99.2% and a lightness of 48%. #01f4f3 color hex could be obtained by blending #02ffff with #00e9e7. Closest websafe color is: #00ffff.\n\n• R 0\n• G 96\n• B 95\nRGB color chart\n• C 100\n• M 0\n• Y 0\n• K 4\nCMYK color chart\n\n#01f4f3 color description : Vivid cyan.\n\n# #01f4f3 Color Conversion\n\nThe hexadecimal color #01f4f3 has RGB values of R:1, G:244, B:243 and CMYK values of C:1, M:0, Y:0, K:0.04. Its decimal value is 128243.\n\nHex triplet RGB Decimal 01f4f3 `#01f4f3` 1, 244, 243 `rgb(1,244,243)` 0.4, 95.7, 95.3 `rgb(0.4%,95.7%,95.3%)` 100, 0, 0, 4 179.8°, 99.2, 48 `hsl(179.8,99.2%,48%)` 179.8°, 99.6, 95.7 00ffff `#00ffff`\nCIE-LAB 87.569, -46.772, -13.19 48.536, 71.174, 95.969 0.225, 0.33, 71.174 87.569, 48.596, 195.749 87.569, -67.81, -13.786 84.365, -44.945, -8.39 00000001, 11110100, 11110011\n\n# Color Schemes with #01f4f3\n\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #f40102\n``#f40102` `rgb(244,1,2)``\nComplementary Color\n• #01f47a\n``#01f47a` `rgb(1,244,122)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #017cf4\n``#017cf4` `rgb(1,124,244)``\nAnalogous Color\n• #f47a01\n``#f47a01` `rgb(244,122,1)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #f4017c\n``#f4017c` `rgb(244,1,124)``\nSplit Complementary Color\n• #f4f301\n``#f4f301` `rgb(244,243,1)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #f301f4\n``#f301f4` `rgb(243,1,244)``\n• #02f401\n``#02f401` `rgb(2,244,1)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #f301f4\n``#f301f4` `rgb(243,1,244)``\n• #f40102\n``#f40102` `rgb(244,1,2)``\n• #01a8a7\n``#01a8a7` `rgb(1,168,167)``\n• #01c1c0\n``#01c1c0` `rgb(1,193,192)``\n• #01dbda\n``#01dbda` `rgb(1,219,218)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #10fefd\n``#10fefd` `rgb(16,254,253)``\n• #2afefd\n``#2afefd` `rgb(42,254,253)``\n• #43fefd\n``#43fefd` `rgb(67,254,253)``\nMonochromatic Color\n\n# Alternatives to #01f4f3\n\nBelow, you can see some colors close to #01f4f3. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01f4b6\n``#01f4b6` `rgb(1,244,182)``\n• #01f4cb\n``#01f4cb` `rgb(1,244,203)``\n• #01f4df\n``#01f4df` `rgb(1,244,223)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #01e1f4\n``#01e1f4` `rgb(1,225,244)``\n• #01cdf4\n``#01cdf4` `rgb(1,205,244)``\n• #01b8f4\n``#01b8f4` `rgb(1,184,244)``\nSimilar Colors\n\n# #01f4f3 Preview\n\nThis text has a font color of #01f4f3.\n\n``<span style=\"color:#01f4f3;\">Text here</span>``\n#01f4f3 background color\n\nThis paragraph has a background color of #01f4f3.\n\n``<p style=\"background-color:#01f4f3;\">Content here</p>``\n#01f4f3 border color\n\nThis element has a border color of #01f4f3.\n\n``<div style=\"border:1px solid #01f4f3;\">Content here</div>``\nCSS codes\n``.text {color:#01f4f3;}``\n``.background {background-color:#01f4f3;}``\n``.border {border:1px solid #01f4f3;}``\n\n# Shades and Tints of #01f4f3\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000a0a is the darkest color, while #f5ffff is the lightest one.\n\n• #000a0a\n``#000a0a` `rgb(0,10,10)``\n• #001d1d\n``#001d1d` `rgb(0,29,29)``\n• #003130\n``#003130` `rgb(0,49,48)``\n• #004444\n``#004444` `rgb(0,68,68)``\n• #005857\n``#005857` `rgb(0,88,87)``\n• #006b6b\n``#006b6b` `rgb(0,107,107)``\n• #017f7e\n``#017f7e` `rgb(1,127,126)``\n• #019292\n``#019292` `rgb(1,146,146)``\n• #01a6a5\n``#01a6a5` `rgb(1,166,165)``\n• #01b9b9\n``#01b9b9` `rgb(1,185,185)``\n• #01cdcc\n``#01cdcc` `rgb(1,205,204)``\n• #01e0e0\n``#01e0e0` `rgb(1,224,224)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\n• #0bfefd\n``#0bfefd` `rgb(11,254,253)``\n• #1efefd\n``#1efefd` `rgb(30,254,253)``\n• #32fefd\n``#32fefd` `rgb(50,254,253)``\n• #45fefd\n``#45fefd` `rgb(69,254,253)``\n• #59fefe\n``#59fefe` `rgb(89,254,254)``\n• #6cfefe\n``#6cfefe` `rgb(108,254,254)``\n• #80fefe\n``#80fefe` `rgb(128,254,254)``\n• #93fffe\n``#93fffe` `rgb(147,255,254)``\n• #a7fffe\n``#a7fffe` `rgb(167,255,254)``\n• #bafffe\n``#bafffe` `rgb(186,255,254)``\n• #ceffff\n``#ceffff` `rgb(206,255,255)``\n• #e2ffff\n``#e2ffff` `rgb(226,255,255)``\n• #f5ffff\n``#f5ffff` `rgb(245,255,255)``\nTint Color Variation\n\n# Tones of #01f4f3\n\nA tone is produced by adding gray to any pure hue. In this case, #728383 is the less saturated color, while #01f4f3 is the most saturated one.\n\n• #728383\n``#728383` `rgb(114,131,131)``\n• #698c8c\n``#698c8c` `rgb(105,140,140)``\n• #5f9696\n``#5f9696` `rgb(95,150,150)``\n• #569f9f\n``#569f9f` `rgb(86,159,159)``\n• #4ca9a8\n``#4ca9a8` `rgb(76,169,168)``\n• #43b2b2\n``#43b2b2` `rgb(67,178,178)``\n• #3abbbb\n``#3abbbb` `rgb(58,187,187)``\n• #30c5c4\n``#30c5c4` `rgb(48,197,196)``\n• #27cece\n``#27cece` `rgb(39,206,206)``\n• #1dd8d7\n``#1dd8d7` `rgb(29,216,215)``\n• #14e1e0\n``#14e1e0` `rgb(20,225,224)``\n• #0aebea\n``#0aebea` `rgb(10,235,234)``\n• #01f4f3\n``#01f4f3` `rgb(1,244,243)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01f4f3 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# #1755fe Color Information\n\nIn a RGB color space, hex #1755fe is composed of 9% red, 33.3% green and 99.6% blue. Whereas in a CMYK color space, it is composed of 90.9% cyan, 66.5% magenta, 0% yellow and 0.4% black. It has a hue angle of 223.9 degrees, a saturation of 99.1% and a lightness of 54.3%. #1755fe color hex could be obtained by blending #2eaaff with #0000fd. Closest websafe color is: #0066ff.\n\n• R 9\n• G 33\n• B 100\nRGB color chart\n• C 91\n• M 67\n• Y 0\n• K 0\nCMYK color chart\n\n#1755fe color description : Vivid blue.\n\n# #1755fe Color Conversion\n\nThe hexadecimal color #1755fe has RGB values of R:23, G:85, B:254 and CMYK values of C:0.91, M:0.67, Y:0, K:0. Its decimal value is 1529342.\n\nHex triplet RGB Decimal 1755fe `#1755fe` 23, 85, 254 `rgb(23,85,254)` 9, 33.3, 99.6 `rgb(9%,33.3%,99.6%)` 91, 67, 0, 0 223.9°, 99.1, 54.3 `hsl(223.9,99.1%,54.3%)` 223.9°, 90.9, 99.6 0066ff `#0066ff`\nCIE-LAB 43.993, 46.003, -87.874 21.488, 13.833, 95.298 0.165, 0.106, 13.833 43.993, 99.188, 297.633 43.993, -17.676, -129.557 37.193, 38.037, -125.881 00010111, 01010101, 11111110\n\n# Color Schemes with #1755fe\n\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #fec017\n``#fec017` `rgb(254,192,23)``\nComplementary Color\n• #17c9fe\n``#17c9fe` `rgb(23,201,254)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #4d17fe\n``#4d17fe` `rgb(77,23,254)``\nAnalogous Color\n• #c9fe17\n``#c9fe17` `rgb(201,254,23)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #fe4c17\n``#fe4c17` `rgb(254,76,23)``\nSplit Complementary Color\n• #55fe17\n``#55fe17` `rgb(85,254,23)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #fe1755\n``#fe1755` `rgb(254,23,85)``\n• #17fec0\n``#17fec0` `rgb(23,254,192)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #fe1755\n``#fe1755` `rgb(254,23,85)``\n• #fec017\n``#fec017` `rgb(254,192,23)``\n• #0136c8\n``#0136c8` `rgb(1,54,200)``\n• #013de1\n``#013de1` `rgb(1,61,225)``\n• #0144fa\n``#0144fa` `rgb(1,68,250)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #3068fe\n``#3068fe` `rgb(48,104,254)``\n• #4a7afe\n``#4a7afe` `rgb(74,122,254)``\n• #638dfe\n``#638dfe` `rgb(99,141,254)``\nMonochromatic Color\n\n# Alternatives to #1755fe\n\nBelow, you can see some colors close to #1755fe. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #178ffe\n``#178ffe` `rgb(23,143,254)``\n• #177cfe\n``#177cfe` `rgb(23,124,254)``\n• #1768fe\n``#1768fe` `rgb(23,104,254)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #1742fe\n``#1742fe` `rgb(23,66,254)``\n• #172ffe\n``#172ffe` `rgb(23,47,254)``\n• #171bfe\n``#171bfe` `rgb(23,27,254)``\nSimilar Colors\n\n# #1755fe Preview\n\nThis text has a font color of #1755fe.\n\n``<span style=\"color:#1755fe;\">Text here</span>``\n#1755fe background color\n\nThis paragraph has a background color of #1755fe.\n\n``<p style=\"background-color:#1755fe;\">Content here</p>``\n#1755fe border color\n\nThis element has a border color of #1755fe.\n\n``<div style=\"border:1px solid #1755fe;\">Content here</div>``\nCSS codes\n``.text {color:#1755fe;}``\n``.background {background-color:#1755fe;}``\n``.border {border:1px solid #1755fe;}``\n\n# Shades and Tints of #1755fe\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000102 is the darkest color, while #eef2ff is the lightest one.\n\n• #000102\n``#000102` `rgb(0,1,2)``\n• #000616\n``#000616` `rgb(0,6,22)``\n• #000b29\n``#000b29` `rgb(0,11,41)``\n• #00113d\n``#00113d` `rgb(0,17,61)``\n• #001650\n``#001650` `rgb(0,22,80)``\n• #001b64\n``#001b64` `rgb(0,27,100)``\n• #012078\n``#012078` `rgb(1,32,120)``\n• #01268b\n``#01268b` `rgb(1,38,139)``\n• #012b9f\n``#012b9f` `rgb(1,43,159)``\n• #0130b2\n``#0130b2` `rgb(1,48,178)``\n• #0136c6\n``#0136c6` `rgb(1,54,198)``\n• #013bd9\n``#013bd9` `rgb(1,59,217)``\n• #0140ed\n``#0140ed` `rgb(1,64,237)``\n• #0347fe\n``#0347fe` `rgb(3,71,254)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\n• #2b63fe\n``#2b63fe` `rgb(43,99,254)``\n• #3e72fe\n``#3e72fe` `rgb(62,114,254)``\n• #5280fe\n``#5280fe` `rgb(82,128,254)``\n• #658efe\n``#658efe` `rgb(101,142,254)``\n• #799dfe\n``#799dfe` `rgb(121,157,254)``\n• #8cabff\n``#8cabff` `rgb(140,171,255)``\n• #a0b9ff\n``#a0b9ff` `rgb(160,185,255)``\n• #b3c7ff\n``#b3c7ff` `rgb(179,199,255)``\n• #c7d6ff\n``#c7d6ff` `rgb(199,214,255)``\n• #dae4ff\n``#dae4ff` `rgb(218,228,255)``\n• #eef2ff\n``#eef2ff` `rgb(238,242,255)``\nTint Color Variation\n\n# Tones of #1755fe\n\nA tone is produced by adding gray to any pure hue. In this case, #838792 is the less saturated color, while #1755fe is the most saturated one.\n\n• #838792\n``#838792` `rgb(131,135,146)``\n• #7a839b\n``#7a839b` `rgb(122,131,155)``\n• #717fa4\n``#717fa4` `rgb(113,127,164)``\n``#687aad` `rgb(104,122,173)``\n• #5f76b6\n``#5f76b6` `rgb(95,118,182)``\n• #5672bf\n``#5672bf` `rgb(86,114,191)``\n• #4d6ec8\n``#4d6ec8` `rgb(77,110,200)``\n``#446ad1` `rgb(68,106,209)``\n• #3b66da\n``#3b66da` `rgb(59,102,218)``\n• #3261e3\n``#3261e3` `rgb(50,97,227)``\n• #295dec\n``#295dec` `rgb(41,93,236)``\n• #2059f5\n``#2059f5` `rgb(32,89,245)``\n• #1755fe\n``#1755fe` `rgb(23,85,254)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #1755fe is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Hazard and Hazard Ratio in Statistics\n\nHazard ratios are a specific type of relative risk that are calculated using a statistical technique known as survival analysis. Survival analysis keeps track of how many subjects do not experience a particular event during a specific time period. When the data is plotted over the entire time of the study, the results show a decreasing curve, which falls as fewer people are not affected by the event (time-to-event curve or Kaplan-Meier curve, figure 1.). If we have two identical groups that differ only in the treatment they are given, we might expect the two curves plotted for their survival to differ over time as the treatment has an effect. From the curves we could calculate the hazard, which is essentially the absolute risk over time. The hazard ratio is simply the value of the hazard calculated from the treatment curve, divided by the hazard calculated from the control curve. Based on the complexity, statistical software is required to make this calculation to estimate the hazard ratio.\n\n#####", null, "Figure 1. The time-to-event curve or Kaplin-Meier curve. As time progresses, percentage survival decreases in both groups. Plotting curves on the graphs allows statistical analysis to be performed to calculate the hazard (absolute risk over time) for each group. Dividing the hazard in the treatment group by the hazard in the control group produces the hazard ratio.\n\nIn simple terms we can therefore state that a hazard is the rate at which an event occurs (risk x time) and a hazard ratio is a the ratio of that rate from two differing groups. In other words, the hazard ratio is a relative risk, when there is an interest in the timing of that risk. For example, while a relative risk might not be able to show that a treatment has an effect because both groups suffered the same number of deaths, a hazard ratio might show a difference because the treatment delayed the rate of the deaths in that group. For this reason hazard ratios are used extensively in clinical trails for new pharmaceutical drugs where survival is a consideration. Except for this difference the hazard ratio is expressed much in the same way as the relative risk. A hazard ratio of 1 would indicate that there was no difference between treatments, whereas a hazard ratio of 2 would signify that the treatment group had twice the rate of an event, and a hazard ratio of 0.5 would signify that the treatment group had half the rate of an event." ]

[ null, "http://www.robertbarrington.net/wp-content/uploads/2012/09/Kaplan-Meier-Survival-Plot.jpg", null ]

[ "# Parameters for Fitzhugh-Nagumo model of Action Potential\n\nI'm a high school student who's currently in AP Biology, so my background in biology is quite limited. I'm interested in mathematically modelling action potentials.\n\nDoing a quick google search, I discovered the Fitzhugh-Nagumo model, a simplified case modelling the voltage of the cell during an action potential as a second order differential equation $C \\frac{d^2V}{dt^2} + \\epsilon(V^2-1) + \\frac{V}{L} = 0$. I want to solve this equation numerically, but I can't find the numerical values of $\\epsilon$, $L$, or $C$ online. What are some reasonable parameters for this model for a human neuron?\n\n• A high school student would wants to solve a differential equation numerically. Woah. I definitely did not have your level in math and programming when I was in high school. +1. Apr 28 '15 at 0:12\n• I didn't know about the Fitzhugh-Nagumo model. I quickly scrolled down this paper and this scholarpedia article. Maybe you can indicate where your differential equation come from and link to some info about the meaning of the parameters $\\epsilon$, $L$ and $C$. Apr 28 '15 at 0:13\n\nThis is actually not the Fitzhugh-Nagumo model of the a neuron, however it is highly related to it. I believe the the Equation you have there, is capable of oscilations for any positive values of $\\epsilon$, $C$, $L$. I generated this graph with the parameter values all set to 1. Note how it is very similar to a sine wave.", null, "I believe you where trying to write Van der Pol's equation here: $$C\\frac{d^2V}{dt^2} +\\epsilon (V^2-1) \\frac{dV}{dt}+ \\frac{V}{L}$$ Note the subtle difference of multiplying by the first derivative. Here is again a image with all parameters set to 1", null, "The Fitzhugh-Nagumo equation can be found on wikipedia here and on peer reviewed scholarpedia here. For reference it is a two state model described by the equation:\n\n$$\\frac{dV}{dt}=V-V^3/3-W+I$$ $$\\tau \\frac{dW}{dt}= V + a -bW$$\n\nA common value for the parameters are $a=.7$,$b=.8$,$\\tau=12.5$ $I$ can be anything really, I used $I=.5$ to generate the graph below. Also note that the model wont spike if I below a certain value. The Fitzhugh Equation is is slightly more general, but does simplify to the Van der pol under the assumption $a=b=0$. Here is the graph of it with the constants I described.", null, "" ]

[ null, "https://i.stack.imgur.com/Bf12R.png", null, "https://i.stack.imgur.com/ufBe6.png", null, "https://i.stack.imgur.com/q0aeu.png", null ]

[ "# Untitled", null, "unknown\nplain_text\n2 years ago\n7.9 kB\n9\nIndexable\nNever\n```package labs;\n\nimport static org.junit.Assert.assertArrayEquals;\n\nimport java.util.Scanner;\n\npublic class Lab01 {\n//get string for user input\npublic static String getString(String userQuestion) {\nScanner inKey = new Scanner(System.in);\nSystem.out.print(userQuestion);\nString input = inKey.nextLine();\n\nreturn input;\n\n}\n//Method to enter and put a space between each letter of a word to make it a diagonal\npublic static void diagonal(String word) {\nchar[] inputWord = word.toCharArray();\nString space = \" \";\nint spaceNum = 0;\nfor(int i =0; i<word.length(); i++) {\nfor (int o = 0; o<spaceNum; o++) {\nSystem.out.print(space);\n}\nSystem.out.println(inputWord[i]);\nspaceNum++;\n}\n\n}\n//Method to enter and put a space between each letter of a word to make it a diagonal\npublic static void diagonals(String userInput) {\n//if there is 2 words, resets it between the space\nString[] words = userInput.split(\" \");\nfor(int i = 0; i <words.length; i++) {\ndiagonal(words[i]);\n}\n}\n//makes a letter into a binary\npublic static String letterToBinary(char input) {\nint x = Integer.valueOf(input);\n\nString str = \"\";\n\nfor(int i =7; i>=0; i--) {\nif (( x / (int) Math.pow(2, i)) > 0) {\nstr += \"1\";\nx -= Math.pow(2, i);\n}\nelse {\nstr += \"0\";\n}\n}\nreturn str;\n}\n//makes a whole sentence into a binary\npublic static String sentenceToBinary(String userSentence) {\nchar[] binary = userSentence.toCharArray();\nString x = \"\";\nfor(int i =0; i<userSentence.length(); i++){\nx+=letterToBinary(binary[i]);\n\n}\n\nreturn x;\n}\n\n/*\n* ONLY PLAY WITH THIS FOR #2\n* ONLY PLAY WITH THIS FOR #2\n* ONLY PLAY WITH THIS FOR #2\n* ONLY PLAY WITH THIS FOR #2\n*/\npublic static void main(String[] args) {\n\n//For #2\n\n//\t\tScanner inKey = new Scanner(System.in);\n//\t\tSystem.out.print(\"Enter a single word: \");\n//\t\tString input = inKey.nextLine();\nString input = getString(\"Enter a single word: \");\ndiagonal(input);\n\n//System.out.print(\"Enter a sentence or Phrase: \");\\\nString input2 = getString(\"Enter a sentence or Phrase: \");\n\ndiagonals(input2);\n\nSystem.out.print(sentenceToBinary(getString(\"Enter a sentence: \")));\n\n}\n\n}\n\npackage labs;\n\nimport java.util.Scanner;\n\nprivate static int dealCard;\nprivate static int dealCard2;\nprivate static int totalDeal;\nprivate static int computerDeal;\nprivate static int totalComputer;\nprivate static int compterTotalHit;\n/*\n* Instance Variables (class parameters)\n* - Accessible AND changeable in ANY method\n* Very convenient, but dangerous.\n* - declared only!!\n* - You must initialize them before using\n*\n* Place them all here.\n*/\n\nprivate static int playerHandTotal;\nprivate static String playerHand;\n\n//welcome to the game\npublic static void welcomeMessage() {\nSystem.out.println(\"BLACK - JACK!\\n\"\n+ \"Come try your luck and\\n\"\n+ \"maybe take home a buck!\");\n\n}\n//start to play the game\npublic static void playGame() {\nplayerTurn();\ndrawCard();\ncomputerTurn();\nprintWinner();\nresetGame();\n\n}\n//Scanner to see what the use enters\npublic static String getString(String userQuestion) {\nScanner inKey = new Scanner(System.in);\nSystem.out.print(userQuestion);\nString input = inKey.nextLine().toLowerCase();\n\nreturn input;\n\n}\n\n//player turn\npublic static void playerTurn() {\n//random int for the card worth\nint player = (int) (Math.random() * 11 + 1);\nint computer = (int) (Math.random() * 11 + 1);\n//int for card deals\ndealCard = player;\nplayer = (int) (Math.random() * 11 + 1);\ndealCard2 = player;\nif (player >=22) {\ndealCard2 = 1;\n//total deal card\n}\ntotalDeal += (dealCard + dealCard2);\nSystem.out.println(\"Your Hand:\" + dealCard + \" \" + dealCard2 + \" = \"+ totalDeal);\ncomputer = (int) (Math.random() * 11 + 1);\nSystem.out.println(\"Comp Hand \" + computer);\ntotalComputer+=computer;\n\n}\n\npublic static void drawCard() {\n//System.out.print((getString(\"hit (h) or stay(s)\")));\nString sup = getString(\"hit (h) or stay(s)\");\n\nint player = (int) (Math.random() * 11 + 1);\n\n//see if user wants to hit card\nwhile(totalDeal<21) {\nsup = getString(\"hit (h) or stay(s)\");\nSystem.out.println(sup);\nif (sup.equals(\"h\")) {\ncompterTotalHit++;\nplayer = (int) (Math.random() * 11 + 1);\nint dealCard3 = player;\ntotalDeal+=dealCard3;\nSystem.out.println(\"You got a \" + dealCard3 + \" total deal = \" + totalDeal);\n}\n//if they don't want to, code breaks and moves on\nelse if(sup.equals(\"s\")){\nbreak;\n}\n//if they put another letter, tells them to put the correct letter\nelse {\nSystem.out.println(\"Wrong letter, try again\");\n}\n\n}\n\n}\n//computer turn\npublic static void computerTurn() {\nSystem.out.println(\"Computer's turn\");\n//random int generater to give card value\nint computer = (int) (Math.random() * 11 + 1);\n//computer card value, tells user what computer got\ncomputerDeal = computer;\nSystem.out.println(\"The computer got \" + computerDeal + \" and the computer total is \" + totalComputer);\ntotalComputer+=computerDeal;\n//while loop so computer will keep hitting till it gets higher then user, only if user is lower then 21\nwhile(totalComputer<totalDeal&&totalDeal<=21){\ncomputer = (int) (Math.random() * 11 + 1);\ncomputerDeal = computer;\n//total computer value\ntotalComputer+=computerDeal;\nSystem.out.println(\"The computer got \" + computerDeal + \" and the computer total is \" + totalComputer);\n}\n\n}\n//who won? code to see who won\npublic static void printWinner() {\nif(totalDeal>21) {\nSystem.out.println(\"You busted...computer wins...\");\n}\nelse if(totalDeal== totalComputer) {\nSystem.out.println(\"You tied...computer wins...\");\n}\nelse if(totalComputer>21) {\nSystem.out.println(\"Computer busted...you wins!!\");\n}\nelse if(totalDeal<totalComputer) {\nSystem.out.println(\"You lose... computer wins...\");\n}\n\nelse if(totalComputer<totalDeal) {\nSystem.out.println(\"You Win!!!\");\n}\n\n}\n//reset game void\npublic static void resetGame() {\n//ask user if they want to play\nString finish = getString(\"Good game, play again? Yes(y) or No(n)\");\n//if they say yes does the following\nif(finish.equals(\"y\")) {\n//resets the ints to reset the game\ndealCard=0;\ndealCard2=0;\ntotalDeal=0;\ncomputerDeal=0;\ntotalComputer=0;\ncompterTotalHit=0;\nplayerHandTotal=0;\n//resets game\nplayGame();\n}\n\n}\n\n/*\n* Keep the starter code\n* Feel free to change it to suit your needs.\n*/\npublic static void main(String[] args) {\n\n//initialize instance variables\nplayerHandTotal = 0;\n\nchar keepPlaying = 'y';\n\n//Welcome them to the game\n//Place welcome message method here!\nwelcomeMessage();\n\nwhile (keepPlaying == 'y') {\n\n//just puts a space after the welcome message\nSystem.out.println(\" \");\n\n//Start game here\nplayGame();\n\n//Ask if they want to play again?\n//Loop until they give a valid answer\nkeepPlaying = 'a'; //so the loop will start\nwhile (keepPlaying != 'y' && keepPlaying != 'n') {\n\n/*\n* You only need the first letter\n* So I...\n* \t- made it lower case\n* \t- then grabbed the first character as a char\n*/\nkeepPlaying = getString(\"\\nPlay again? (Y or N): \").toLowerCase().charAt(0);\n\nif (keepPlaying != 'y' && keepPlaying != 'n') {\n}\n\n}\n}\n\n//Say goodbye\nSystem.out.println(\"\\nIt's been fun!\");\nSystem.out.println(\"Come back soon!\");\n\n}\n\n}\n```" ]

[ null, "https://www.gravatar.com/avatar/9b5c56e0227d015b77ba7196ff4c34ad", null ]

[ "## Proportions and Rates\n\n### Learning Outcomes\n\n• Given the part and the whole, write a percent\n• Calculate both relative and absolute change of a quantity\n• Calculate tax on a purchase\n\nIf you wanted to power the city of Lincoln, Nebraska using wind power, how many wind turbines would you need to install? Questions like these can be answered using rates and proportions.\n\n## Rates\n\nA rate is the ratio (fraction) of two quantities.\n\nA unit rate is a rate with a denominator of one.\n\n### Example\n\nYour car can drive 300 miles on a tank of 15 gallons. Express this as a rate.\n\n## Proportion Equation\n\nA proportion equation is an equation showing the equivalence of two rates or ratios.\n\nFor an overview on rates and proportions, using the examples on this page, view the following video.\n\n### Example\n\nSolve the proportion $\\displaystyle\\frac{5}{3}=\\frac{x}{6}$ for the unknown value x.\n\n### Example\n\nA map scale indicates that ½ inch on the map corresponds with 3 real miles. How many miles apart are two cities that are $\\displaystyle{2}\\frac{1}{4}$ inches apart on the map?\n\n### Example\n\nYour car can drive 300 miles on a tank of 15 gallons. How far can it drive on 40 gallons?\n\nA worked example of this last question can be found in the following video.\n\nNotice that with the miles per gallon example, if we double the miles driven, we double the gas used. Likewise, with the map distance example, if the map distance doubles, the real-life distance doubles. This is a key feature of proportional relationships, and one we must confirm before assuming two things are related proportionally.\n\nYou have likely encountered distance, rate, and time problems in the past. This is likely because they are easy to visualize and most of us have experienced them first hand. In our next example, we will solve distance, rate and time problems that will require us to change the units that the distance or time is measured in.\n\n### Example\n\nA bicycle is traveling at 15 miles per hour. How many feet will it cover in 20 seconds?\n\nView the following video to see this problem worked through.\n\n### Try It\n\nA 1000 foot spool of bare 12-gauge copper wire weighs 19.8 pounds. How much will 18 inches of the wire weigh, in ounces?\n\n### Example\n\nSuppose you’re tiling the floor of a 10 ft by 10 ft room, and find that 100 tiles will be needed. How many tiles will be needed to tile the floor of a 20 ft by 20 ft room?\n\nOther quantities just don’t scale proportionally at all.\n\n### Example\n\nSuppose a small company spends $1000 on an advertising campaign, and gains 100 new customers from it. How many new customers should they expect if they spend$10,000?\n\nMatters of scale in this example and the previous one are explained in more detail here.\n\nSometimes when working with rates, proportions, and percents, the process can be made more challenging by the magnitude of the numbers involved. Sometimes, large numbers are just difficult to comprehend.\n\n### Examples\n\nThe 2010 U.S. military budget was $683.7 billion. To gain perspective on how much money this is, answer the following questions. 1. What would the salary of each of the 1.4 million Walmart employees in the US be if the military budget were distributed evenly amongst them? 2. If you distributed the military budget of 2010 evenly amongst the 300 million people who live in the US, how much money would you give to each person? 3. If you converted the US budget into$100 bills, how long would it take you to count it out – assume it takes one second to count one \\$100 bill.\n\n### Example\n\nCompare the electricity consumption per capita in China to the rate in Japan.\n\nWorking with large numbers is examined in more detail in this video." ]

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[ "5\n\n# Points) Consider the iuitial value problemV)6)(z _ 6)(1 _ 6)z6 = (2)f (8)f |For which of the following values of %o does there exist some t4 > 3 such that u(t4) ...\n\n## Question\n\n###### Points) Consider the iuitial value problemV)6)(z _ 6)(1 _ 6)z6 = (2)f (8)f |For which of the following values of %o does there exist some t4 > 3 such that u(t4) = 2yo? s0- Jo = 1 Yo = L5 Vu = 2.5 3.5(VII)points) For which of the following initial value problems docs the solution exist for all + > (? v(t) '(#)f 3(o) B. y () 2(16 = (0)f y(t)3 (0)6 V (6) v(')2e-y(t) 4(o)\n\npoints) Consider the iuitial value problem V) 6)(z _ 6)(1 _ 6)z6 = (2)f (8)f | For which of the following values of %o does there exist some t4 > 3 such that u(t4) = 2yo? s0- Jo = 1 Yo = L5 Vu = 2.5 3.5 (VII) points) For which of the following initial value problems docs the solution exist for all + > (? v(t) '(#)f 3(o) B. y () 2(16 = (0)f y(t)3 (0)6 V (6) v(')2e-y(t) 4(o)", null, "", null, "#### Similar Solved Questions\n\n##### Provideranappropriate nesponseMEEMEEKHRMRRHmR MnE HenmgmmErEE @hkeKBKBHARKB [email protected] KHKEHHHHBHHHHHHH\nProvideranappropriate nesponse MEEMEEKHRMRRHmR MnE HenmgmmErEE @hkeK B KBHAR KB B @BH KHKEHHHHBHHHHHHH...\n##### A solution contains 0.102 M potassium fluoride and 0.450 M hydrofluoric acid_The pH of this solution is\nA solution contains 0.102 M potassium fluoride and 0.450 M hydrofluoric acid_ The pH of this solution is...\n##### Score: 0 of 1 ptH11 of 25 (18 complete2.3.61 Rewrite the given equation Tx + 8y 56 = 0 slope-intercept form b. Give the slope and y-intercept Use the slope and y-intercept to graph the linear function. a. The slope-intercept form of the equation iS (Simplify your answer Use integers or fractions for any numbers in the equationa andmehcn Check AnswerTleOur ans '-\nScore: 0 of 1 pt H11 of 25 (18 complete 2.3.61 Rewrite the given equation Tx + 8y 56 = 0 slope-intercept form b. Give the slope and y-intercept Use the slope and y-intercept to graph the linear function. a. The slope-intercept form of the equation iS (Simplify your answer Use integers or fractions ...\n##### \"Cuestion 151; Resonant Frequencies When played in a certain manner; the lowest resonant frequency of a certain violin string 599 Hz. What is the frequency in Hz of the third hanonic?Answer: 1198hzYou did not give the corroct unit. The fundamental frequency is also called the first harmonic. The violin string is fixed at both ends_ The correct answer is: 1800 Hz\n\"Cuestion 151; Resonant Frequencies When played in a certain manner; the lowest resonant frequency of a certain violin string 599 Hz. What is the frequency in Hz of the third hanonic? Answer: 1198hz You did not give the corroct unit. The fundamental frequency is also called the first harmonic. ...\n##### An object mass moving witn speed the right on horionta frictionless surface when explodes into two Dleces Subsequenuy Iert. Tine speed the other piece the object is 3/2 Vopiecemass 2/5m movesSpeco1/3 Vo 7/5 Vo1/2 VoL(Zva\nAn object mass moving witn speed the right on horionta frictionless surface when explodes into two Dleces Subsequenuy Iert. Tine speed the other piece the object is 3/2 Vo piece mass 2/5m moves Speco 1/3 Vo 7/5 Vo 1/2 Vo L(Zva...\n##### Use the RelenncdICCLXYuuaneaenWrite the expression for the equilibrium constant Kp for the following reaction(Enclose pressures in parentheses and do NOT write the chemical fvrmula as subscript: For cxample, cntcr (PNHy)? ns (P NH;)}. If cither the numerator denominator is please enter 1,)PCIs(g)PCls(e)Clz(g)Submlt AnowutRotry Entlrg Groupmore group uttompta remalnlng\nUse the Relenncd ICCLX Yuua neaen Write the expression for the equilibrium constant Kp for the following reaction (Enclose pressures in parentheses and do NOT write the chemical fvrmula as subscript: For cxample, cntcr (PNHy)? ns (P NH;)}. If cither the numerator denominator is please enter 1,) PCIs...\n##### A fluid s velocity field is given by F(r,y, \" 2) = ci+2j+yk. Find the flow along the helix r(t) = (cost) i + (sint) j +tk, 0 < t < \"/2.\nA fluid s velocity field is given by F(r,y, \" 2) = ci+2j+yk. Find the flow along the helix r(t) = (cost) i + (sint) j +tk, 0 < t < \"/2....\n##### Emy on Jy Gui__. Ratio 0 Prac Inted1 Question 4o tollowng 01 ~Io 1 ar L 6 ee Select 8 tat Kacesle Olo ok53)\nemy on Jy Gui__. Ratio 0 Prac Inted 1 Question 4o tollowng 01 ~Io 1 ar L 6 ee Select 8 tat Kace sle Olo ok 5 3)...\n##### -5 0 -17 3 -5 5 11 -19 7 -13 5 -3Examine the 4x5 matrix A(2 points each)Find the rank of Ab) Find dim[NulA)]Find a basis for Col(A)d) Col(A) is a subspace of which (larger) vector space?e) Find a basis for Row(A)f) Row(A) is a subspace of which vector space?g) Find a basis for NulA)h) Nul(A) is a subspace of which vector space?\n-5 0 -17 3 -5 5 11 -19 7 -13 5 -3 Examine the 4x5 matrix A (2 points each) Find the rank of A b) Find dim[NulA)] Find a basis for Col(A) d) Col(A) is a subspace of which (larger) vector space? e) Find a basis for Row(A) f) Row(A) is a subspace of which vector space? g) Find a basis for NulA) h) Nul(...\n##### Check the divergence theorem for the vector fuunction V = cos 0 f + 72 cos @ 0 cos 0 sin @ @, using one octant of the sphere of radius R as your volume.\nCheck the divergence theorem for the vector fuunction V = cos 0 f + 72 cos @ 0 cos 0 sin @ @, using one octant of the sphere of radius R as your volume....\nUse the eighteen rules of inference to derive the conclusions of the following symbolized arguments. Do not use either conditional proof or indirect proof. 1. $(\\exists x) A x \\supset(x)[(A x \\vee E x) \\supset D x]$ 2. $(\\exists x) D x \\supset(x)[D x \\supset(C x \\vee \\sim B x)]$ 3. $A n \\cdot B n \\... 5 answers ##### Excrcisck:Let A be 3*3 matrix such that A = /\" 0 a Then,2) det(A) = a) - 2 b) 2 c) 3] d) 10 a) 0 b) Excrcisck: Let A be 3*3 matrix such that A = /\" 0 a Then, 2) det(A) = a) - 2 b) 2 c) 3] d) 1 0 a) 0 b)... 5 answers ##### Gtah?Wnstah _Suppose that currently r = 2 and X=0.6_ Assuming we can make only one intervention on r, what should we make the value of r if we want to maximize X in the long run?A. Leave where it is_Raise to about 30_C.Raise r to about 24D. Raise r to 15_ Gtah? Wnstah _ Suppose that currently r = 2 and X=0.6_ Assuming we can make only one intervention on r, what should we make the value of r if we want to maximize X in the long run? A. Leave where it is_ Raise to about 30_ C.Raise r to about 24 D. Raise r to 15_... 5 answers ##### QUESION4Supposed that the yearly rate of consumption of oil (in millions of barrels) is given byc' (t) = ke\"t ,where t is the time (in years), r is a constant and k is the consumption in the year when t = 0 corresponds to year 2020 In 2020, PETRONAS GROUP sold 1.2 million barrels of oil. Assume thatr = 0.04-HNT: letk = 42 since PETRONAS GRQUP sold 1.2 millionDISCLAIMER: The equation provided here is just & made-up equation for the purpose of teaching and learning for integrationFi QUESION4 Supposed that the yearly rate of consumption of oil (in millions of barrels) is given by c' (t) = ke\"t , where t is the time (in years), r is a constant and k is the consumption in the year when t = 0 corresponds to year 2020 In 2020, PETRONAS GROUP sold 1.2 million barrels of oi... 5 answers ##### Jiculzte Ine Strctcn 01 (e spring: Cnlculte the diffetcnce batween the recorded position for ccry added mass znd the refercace This Is the additiatal strctch of thc spring when mass 1S added Record the result _ posItion mcicts i Table Table Enter Ax inCalculate the force: The weight added at thc cnd ofthe spring sutctches the SpTing futher Calculate the due added masses Thi 1s the additional force exczted on the spng Recordehbe esult tatibc]] . folces rcsult mn the TablcB-i6July 17,2020Klechle Jiculzte Ine Strctcn 01 (e spring: Cnlculte the diffetcnce batween the recorded position for ccry added mass znd the refercace This Is the additiatal strctch of thc spring when mass 1S added Record the result _ posItion mcicts i Table Table Enter Ax in Calculate the force: The weight added at thc c... 5 answers ##### Tiffany invests$14,000.00 at 3.41% simple interest for25 months.How much interest is earned over the 25 month period?The interest earned over the 25 month period is ___________How much is in the account at the end of the 25 monthperiod?Tiffany will have _______ in the account at the end of the25 month period.\nTiffany invests \\$14,000.00 at 3.41% simple interest for 25 months. How much interest is earned over the 25 month period? The interest earned over the 25 month period is ___________ How much is in the account at the end of the 25 month period? Tiffany will have _______ in the account at the end of t..." ]

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[ "### Vedic Math - Square Root-2\n\nNote: Vedic Math Blog has been moved to http://vedicmath.vedantatree.com/. Please bookmark the new address for new and existing blogs.\n\nIn this article, we shall continue discussing the remaining part of discussion 'how to find the Square Root using Vedic Math'. In last article , we discussed the technique for 4-digit numbers, In this article, we shall discuss the another technique which is useful for bigger numbers. In this method, we shall use \"Duplex\" (mentioned in General Squaring).\n\nSo, First observation:\n• if number is 69563217 then n=8, Digits in the square root is 8/2=4, pairing is 69'56'32'17 and the first digit will be 8(82=64)\n• if number is 764613731 then n=9, Digits in the square root is (9+1)/2=5, pairing is 7'64'61'37'31 and the first digit will be 2 (22=4)\n\nRecall \"Duplex\"\n• for a single digit 'a', D = a2. e.g. D(4) = 16\n• for a 2-digit number of the form 'ab', D = 2( a x b ). e.g. D(23) = 2(2x3) = 12\n• for a 3-digit number like 'abc', D = 2( a x c ) + b2. e.g. D(231) = 2(2x1) + 32 = 13\n• for a 4-digit number 'abcd', D = 2( a x d ) + 2( b x c ) e.g. D(2314) = 2(2x4) + 2(3x1) = 22\n• for a 5-digit number 'abcde', D = 2( a x e ) + 2( b x d ) + c2 e.g. D(14235) = 2(1x5) + 2(4x3) + 22 = 38  and so on.\n\nAs we know how to calculate the duplex of a number, now we learn how to use it in calculating the square root of a number?\nWe will explain using an example.\n\nExample:  734449\n\nStep1: n=6, Digits in the square root is 6/2=3, pairing is 73'44'49. Rearrange the numbers of two-digit groups from right to left as follows:\n| 73 :  4  4  4  9\n.|    :\n-----------------\n.|    :\nAs you see, in above representation, we provide spaces in front of the numbers to perform straight division, if required.\n\nStep2: Now, find the perfect square less than the first group 73 i.e 64 and its square root is 8. Write down this 8 and the reminder 9 (73-64=9) as shown below:\n| 73 :   4   4   4   9\n16| 64 :9\n------------------\n| 8  :\n\nWe also calculate twice of number '8' (i.e. 8 x 2 = 16), and put that number to the left of the \"|\" on the second line as shown above. Here, number '16' is the divisor and which is always double of the quotient (here, quotient is 8).\n\nStep3: Next is the gross dividend, the number which we have written after the colon on the second line appended in front of the next digit of the square. Thus, our gross dividend is 94.\n\nSince there are no digits to the right of the \" \" on the answer line, we will not subtract anything here. If there are any digits on the answer line to the right of the \" \", then we calculate duplexes for that digit and subtract it from dividend. But here, without subtracting anything from the gross dividend, we divide 94 by the divisor 16 and put down the second Quotient digit 5 and the second reminder 14 in their proper place.\n\nStep4: Third gross dividend-unit is 144. From 144 subtract 25 [ Duplex value of the second quotient digit (number to the right of the \":\" on the answer line) D(5) = 25 ] ,get 119 as the actual dividend. Now, divide it by 16 and set the Quotient 7 and reminder 7 in their proper places.\nStep5: Fourth gross dividend-unit is 74. From 74 subtract Duplex D(57) [because D(57) = 2(5 X 7) = 70 ] obtain 4 , divide this 4 by 16 and put down Quotient as 0 and reminder 4 in their proper places\nWe put a decimal point after the third digit since we know that the square root of a 6-digit number has to have 3 digits before the decimal point (mentioned in Step1).\n\nStep6: Fifth gross dividend-unit is 49. From 49 subtract Duplex(570) = 49 and get 0.\nThis means the work has been completed, the given expression is a prefect square and 857 is its square root.\n\nNow, let us discuss some of the complicated cases.\n\nCase 1: Take an example, which is complicated.\nExample: 36481  (n=5, Digits in square root is (5+1)/2=3)\nStep1:\n| 3 :  6  4  8  1\n2 |    :2\n--------------------\n| 1 :\n\nStep2: Divisor 2, can fully divide 26 with quotient 13 and no remainder. But in the duplex method, we always restrict our quotients to be single digits. In other words, we add numbers to the answer row one digit at a time. Because of this, we put down 9 on the answer row as the quotient, and put down 8 as our next remainder (remember that 9*2 + 8 = 26).\n\n| 3 :  6   4   8   1\n2|    :2   8\n---------------------------\n| 1 : 9\n\nStep3: Next gross dividend is 84. From 84 subtract Duplex D(9),get 3. Divide this 3 by 2 and put down Quotient as 1 and reminder 1 in their proper places\n\n| 3 :  6   4    8    1\n2 |   :2    8   1\n---------------------------\n| 1 : 9  1\n\nStep4: Gross dividend is 18. From 18 subtract Duplex D(91), get 0. Divide this 0 by 2 and put down Quotient as 0 and reminder 0 in their proper places\n| 3 :  6    4    8    1\n2 |   : 2    8    1    0\n---------------------------\n| 1 :  9     1     0\n\nStep5: Gross dividend is 01. From 01 subtract Duplex D(910), get 0.\n| 3 : 6   4    8    1\n2 |   :2    8    1    0\n---------------------------\n| 1 : 9   1 .  0    0\nThis completes the procedure. The final answer is 191.\n\nCase 2: Now, we move to the next complication.\nExample: 16384   (n=5, Digits in square root is (5+1)/2=3)\nStep1:\n| 1 :  6   3   8   4\n2 |    :0   0\n-----------------\n| 1 :  3\nWe see that divisor 2, can fully divide 06 with quotient 3 and no remainder. This would then lead to a new gross dividend to 3, and a net dividend to -6 because the duplex of 3 is 9.\nThis type of complication occurs many times. To solve this problem, we reduce the second quotient to 2 and carry over a remainder of 2 to the next step.  As shown below:\n| 1 :  6   3   8   4\n2 |    :0   2\n-----------------\n| 1 :  2\n\nStep2: Next gross dividend is 23, and a net dividend is 19 (23 - the duplex of 2, which is 4).Divide this 19 by 2 and put down Quotient as 1 and reminder 1 in their proper places\n| 1 :  6   3   8   4\n2 |    :0   2   1\n-------------------\n| 1 :  2   9\nAgain, the same case arises. Divisor 2 divides 18 with quoitent 9 and reminder 0. And then the new gross dividend is 4 and net dividend is -32 (4 - D(29)= -32). So, we reduce the third quotient to 8 and carry over a reminder of 3 to the next step.\n| 1 :  6   3   8   4\n2 |    :0   2   3\n--------------------\n| 1 :  2   8\n\nStep3: Gross dividend is 38 and net dividend is 6 (38 - D(28) = 6). Divide this 6 by 2 and put down Quotient as 3 and reminder 0 in their proper places\n| 1 :  6   3   8   4\n2 |    :0   2   3   0\n------------------\n| 1 :  2   8 . 3\nBut again net dividend comes negative. so, we reduce the quotient to 2, and we get\n| 1 :  6   3   8   4\n2 |    :0   2   3   2\n------------------\n| 1 :  2   8 . 2\nHere gross dividend is 24 and net dividend which is again negative. Again we reduce the quotient to 1, but again the net dividend comes negative. So, now the new quotient is 0.\n| 1 :  6   3   8   4\n2 |    :0   2   3   6\n------------------\n| 1 :  2   8 . 0\n\nStep4: Gross dividend is 64. From 64 subtract Duplex D(280), gets 0.\n| 1 :  6   3   8   4\n2 |    :0   2   3   6\n---------------------\n| 1 :  2    8 . 0 0\n\nFollowing are few of the examples:\n\n(1)   552049   (n=6, Digits in square root is 6/2=3)\n|55 :  2    0    4    9\n14|     :6    6    2    0\n--------------------\n|  7 :  4   3 .   0   0  (A perfect Square)\n\n(2)   14047504  (n=8, Digits in square root is 8/2=4)\n|14 :  0    4     7     5    0    4\n6 |     :5     8   11   13    1\n---------------------------\n| 3 :  7   4    8   .   1 ....       ( Not a perfect square. As number of digits in square root is 4 and it didn't terminate after 4 digits )\n\n(3) 119716  (n=6, Digits in square root is 6/2=3)\n|11 :  9    7    1    6\n6 |     :2    5     5    3\n---------------------\n| 3 :  4    6 .  0    0   (A perfect Square)\n\nHopefully, this lesson will be helpful to handle the computation of square roots.\n\nIf you like the article, you may contribute by:\n\n• Posting the article link on Social Media using the Social Media Bookmark bar\n\nAnonymous said...\n\nplz solve this problem 25281....\n\nAnonymous said...\n\n14047504... is a perfect square root of 3748\n\nAnonymous said...\n\nbest complicated case is 114921\n\nJitendra Soni said...\n\nIt is solved easily\n\nFUN MATHEMATICS said...\n\nIn a six digit square number the hundred and unit place can be calculated easily. But how to say the tenth place or middle value instantly. Any clue?" ]

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[ "Home » Core Java » math » Generate random Integer within given range", null, "Byron is a master software engineer working in the IT and Telecom domains. He is an applications developer in a wide variety of applications/services. He is currently acting as the team leader and technical architect for a proprietary service creation and integration platform for both the IT and Telecom industries in addition to a in-house big data real-time analytics solution. He is always fascinated by SOA, middleware services and mobile development. Byron is co-founder and Executive Editor at Java Code Geeks.\n\n# Generate random Integer within given range\n\nIn this example we shall show you how to generate a random Integer within a given range, using `random()` method of Math. The class Math contains methods for performing basic numeric operations such as the elementary exponential, logarithm, square root, and trigonometric functions. To generate a random Integer within a given range one should perform the following steps:\n\n• Use `random()` method of Math to get a double value with a positive sign, greater than or equal to 0.0 and less than 1.0.\n• Multiply the result to a number. For example, multiply the result to 100. The maximum of this is 100 and the minimum 0.\n• You can also add a number to the result. For example add 50 to the result. Now the range is between 50 and 150,\n\nas described in the code snippet below.\n\n```package com.javacodegeeks.snippets.core;\n\npublic class RandomIntWithinGivenRange {\n\npublic static void main(String args[]) {\n\n// This example will return a random integer\n// in the range [-50,50]\nint random1 = (int)(Math.random()*100)-50;\nSystem.out.println(\"Value 1 = \" + random1);\n\n// This example will return a random integer\n// in the range [50,150]\nint random2 = (int)(Math.random()*100)+50;\nSystem.out.println(\"Value 2 = \" + random2);\n}\n}\n```\n\nOutput:\n\n``````Value 1 = -43\nValue 2 = 111```\n```\n\nThis was an example of how to generate a random Integer within a given range in Java.\n\n# Do you want to know how to develop your skillset to become a Java Rockstar?\n\n## Subscribe to our newsletter to start Rocking right now!\n\n### and many more ....", null, "" ]

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[ "#", null, "Reasoning Questions: Ranking And Position Set 12\n\nHello Aspirants. Welcome to Online Reasoning Section with explanation in AffairsCloud.com. Here we are creating Best question samples from Ranking and Position with explanation, which is common for competitive exams!!!\n\nQuestions Penned by Yogit\n\n(Q1 – Q5) Answer the following questions based on the arrangement given below:-\nY W % 2 @ C N 3 A S M 9 \\$ ! D * T 4 # B L + 7 ? 5 G F 6\n\n1. How many such symbols are there in the arrangement which are immediately followed by letter and immediately preceded by a number ?\na) one\nb) two\nc) three\nd) more than three\ne) None of these\nExplanation :  [email protected] and 4#B\n\n2. If the numbers immediately after the symbols get their value doubled. Then find the sum of all such numbers.\na) 24\nb) 28\nc) 32\nd) 34\ne) None of these\nExplanation :\n2, 7 and 5. So the sum of these numbers = 4+14+10 = 28\n\n3. How many such numbers are there which are immediately followed by letter but not immediately preceded by the symbol ?\na) one\nb) two\nc) three\nd) more than three\ne) None of these\nExplanation : N3A\n\n4. If all the symbols from the above expression is removed then which one is 3rd to the left of 7th from the left end.\na) %\nb) C\nc) N\nd) 3\ne) None of these\nAnswer – b) C Explanation :\nAfter removing the symbols, A is 7th from the left and C is 3rd to the left of A\n\n5. How many such vowels are there which is immediately followed by consonants and immediately preceded by number?\na) More than three\nb) one\nc) two\nd) three\ne) None of these\nExplanation : 3AS\n\n(Q6 – Q10) Answer the following questions based on the arrangement given below:-\nG % A Y H D 7 L @ S 1 E X 3 J F \\$ Z M 0 \\$ R P 9 U 5 !\n\n1. What will come in place of question mark?\nGA!  %Y5  AHU  ?\na) DY9\nb) YDP\nc) YD9\nd) YD0\ne) None of these\nExplanation\nSeries is +2 and then last letter from the right end and so on\n\n2. If every second number/letter/symbol from the right replace with the days of the week starting from Monday, then which letter will replace Thursday?\na) 0\nb) M\nc) \\$\nd) 9\ne) None of these\nExplanation :\n5 = Monday, 9 = Tuesday, R = Wednesday and 0 = Thursday\n\n3. If the order of the last 10 characters get reversed then what will be the seventh to the right of fourteenth form the left end.\na) P\nb) R\nc) Z\nd) \\$\ne) None of these\nExplanation :\n14th from the left = 3, now the order of the last 10 get reversed so, \\$ will be 7th to the right of 3.\n\n4. If all the numbers are dropped from the above sequence, then which element is eleventh from the right end?\na) E\nb) 3\nc) X\nd) \\$\ne) None of these\nExplanation :\nAfter removing numbers, eleventh from the right end = X\n\n5. How many such vowels are there in the above arrangement each of which is immediately followed by a number?\na) More than three\nb) one\nc) two\nd) three\ne) None of these" ]

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[ "# Defining exponentiation on the integers\n\nIf one defines the integers as equivalence classes of pairs of natural numbers, there is a (canonical?) way to define addition and multiplication for the integers based on addition and multiplication for the natural numbers, but I’m not sure if the same may be said for exponentiation.\n\nOne could define exponentiation for the integers similar to how it can be defined for the natural numbers, by applying the recursion principle to multiplication, but I’m wondering if there is an approach that “agrees in spirit” with how addition and multiplication on $\\mathbf{Z}$ can be defined in terms of those operations in $\\mathbf{N}$.\n\n• of course you could. use the biomial theorem. Jul 1, 2015 at 2:40\n• I'm not sure I follow. Jul 1, 2015 at 2:44\n\nIt follows from the binomia theorem, how exponentiation has to be declared for a pair: $$(a,b)^n=(a-b)^n=\\sum(-1)^k \\binom{n}{k} a^kb^{n-k}=\\sum_{k \\; \\text{even}}\\binom{n}{k} a^kb^{n-k} - \\sum_{k \\;\\text{odd}}\\binom{n}{k}a^kb^{n-k}=(\\sum_{k \\; \\text{even}} \\binom{n}{k}a^kb^{n-k}, \\sum_{k \\;\\text{odd}}\\binom{n}{k}a^kb^{n-k})$$ This is exponentiation for natural number exponents.\nFor nonnatural exponents exponentiation does not always result in a natural number. But one can define $(a,b)^{(c,d)}$ is the solution $x$ of $$(a,b)^c=x(a,b)^d$$ if one exists." ]

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[ "# 37486 (number)\n\n37,486 (thirty-seven thousand four hundred eighty-six) is an even five-digits composite number following 37485 and preceding 37487. In scientific notation, it is written as 3.7486 × 104. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 18,742 positive integers (up to 37486) that are relatively prime to 37486.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 5\n• Sum of Digits 28\n• Digital Root 1\n\n## Name\n\nShort name 37 thousand 486 thirty-seven thousand four hundred eighty-six\n\n## Notation\n\nScientific notation 3.7486 × 104 37.486 × 103\n\n## Prime Factorization of 37486\n\nPrime Factorization 2 × 18743\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 37486 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 37,486 is 2 × 18743. Since it has a total of 2 prime factors, 37,486 is a composite number.\n\n## Divisors of 37486\n\n1, 2, 18743, 37486\n\n4 divisors\n\n Even divisors 2 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 56232 Sum of all the positive divisors of n s(n) 18746 Sum of the proper positive divisors of n A(n) 14058 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 193.613 Returns the nth root of the product of n divisors H(n) 2.66652 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 37,486 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 37,486) is 56,232, the average is 14,058.\n\n## Other Arithmetic Functions (n = 37486)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 18742 Total number of positive integers not greater than n that are coprime to n λ(n) 18742 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3964 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 18,742 positive integers (less than 37,486) that are coprime with 37,486. And there are approximately 3,964 prime numbers less than or equal to 37,486.\n\n## Divisibility of 37486\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 2 1 4 1 6 1\n\nThe number 37,486 is divisible by 2.\n\n## Classification of 37486\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Square Free\n\n## Base conversion (37486)\n\nBase System Value\n2 Binary 1001001001101110\n3 Ternary 1220102101\n4 Quaternary 21021232\n5 Quinary 2144421\n6 Senary 445314\n8 Octal 111156\n10 Decimal 37486\n12 Duodecimal 1983a\n20 Vigesimal 4de6\n36 Base36 sxa\n\n## Basic calculations (n = 37486)\n\n### Multiplication\n\nn×y\n n×2 74972 112458 149944 187430\n\n### Division\n\nn÷y\n n÷2 18743 12495.3 9371.5 7497.2\n\n### Exponentiation\n\nny\n n2 1405200196 52675334547256 1974587590838438416 74019390430169702462176\n\n### Nth Root\n\ny√n\n 2√n 193.613 33.4675 13.9145 8.21815\n\n## 37486 as geometric shapes\n\n### Circle\n\n Diameter 74972 235531 4.41457e+09\n\n### Sphere\n\n Volume 2.20646e+14 1.76583e+10 235531\n\n### Square\n\nLength = n\n Perimeter 149944 1.4052e+09 53013.2\n\n### Cube\n\nLength = n\n Surface area 8.4312e+09 5.26753e+13 64927.7\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 112458 6.0847e+08 32463.8\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.43388e+09 6.20785e+12 30607.2\n\n## Cryptographic Hash Functions\n\nmd5 a2f32c0381843529ca285ed87f5d1982 286b8d8cb228989f8957dce054f94ce154c757a8 fa7ab5eba46dd3013f5b8212ccf631b808eeb99ecf11a9df45d6edb61c687179 e4ab7fe64623a6c1217d13d0eb24fdb7a100f3716cb27209b9ac8afef5a6f3a7d00575897f57cca0064869a6068ef951efa02890c554d4b34bbe27cefc528fac c5842c6effba7e182b6cfcb9eb3935150330a3d7" ]

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[ "# [R-meta] Variance inflation factors (VIF) calculations applicable to a rma.mv model\n\nMartineau, Roger Roger@M@rtine@u @ending from AGR@GC@CA\nThu May 31 21:31:27 CEST 2018\n\n```Dear list members,\n\nI have the following multilevel model. I want to calculate variance inflation factors (VIF) among Xs to test collinearity.\n\nDoes someone know how to get VIF from a rma.mv model ?\n\n> model19 <- rma.mv(MPY, SEM^2, data=dataset,\n+ mods = ~\n+ X1 +\n+ X2 +\n+ X3 +\n+ X4 +\n+ X5,\n+ random = ~ 1|experiment/study/ID,\n+ method = \"REML\", digits=4)\n> robust(model19, cluster= dataset\\$experiment)\n\nI have this VIF function that I can use for a lmer model (see below) but it doesn’t work for the rma.mv model.\n\n# VIF for lmer model\nvif.mer <- function (fit) {\nv <- vcov(fit)\nnam <- names(fixef(fit))\nns <- sum(1 * (nam == \"Intercept\" | nam == \"(Intercept)\"))\nif (ns > 0) {\nv <- v[-(1:ns), -(1:ns), drop = FALSE]\nnam <- nam[-(1:ns)]\n}\nd <- diag(v)^0.5\nv <- diag(solve(v/(d %o% d)))\nnames(v) <- nam\nv\n}\n\nBest,\n\nRoger ☺\n\nRoger Martineau, mv Ph.D.\nCentre de recherche et de développement\nsur le bovin laitier et le porc\nAgriculture et agroalimentaire Canada/Agriculture and Agri-Food Canada\nTéléphone/Telephone: 819-780-7319\nTélécopieur/Facsimile: 819-564-5507\n2000, Rue Collège / 2000, College Street\nSherbrooke (Québec) J1M 0C8\nCanada\nroger.martineau using agr.gc.ca<mailto:roger.martineau using agr.gc.ca>\n\n[[alternative HTML version deleted]]\n\n```\n\nMore information about the R-sig-meta-analysis mailing list" ]

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[ "due Wednesday, November 3rd by 10pm (Waterloo local time)\nIn order to receive full credits, you must show all the work leading to your solution. You may of\ncourse work together with your classmates, but you must write up the solutions independently.\n\nProblems\n$# 1$ Let $E \\subseteq \\mathbb{R}^{d}$ be a non-empty compact set. The following questions are independent.\n(a) (4 pts) Prove that for all $\\varepsilon>0$ there exist $N \\in \\mathbb{N}$ and $x_{1}, \\cdots, x_{N} \\in E$ such that $E \\subseteq \\cup_{j=1}^{N} B_{\\varepsilon}\\left(x_{j}\\right)$\n(b) (4 pts) Suppose $\\left{F_{i}\\right}_{i \\in I}$ is an arbitrary collection of closed subsets of $\\mathbb{R}^{d}$ such that $F_{i} \\subseteq E$ for all $i \\in I$ and that $F_{i_{1}} \\cap \\cdots \\cap F_{i_{N}} \\neq \\emptyset$ for any $N \\in \\mathbb{N}$ and $i_{1}, \\cdots, i_{N} \\in I$. Prove that $\\cap_{i \\in I} F_{i} \\neq \\emptyset$\n\n#2 Prove that the following sets are compact.\n(a) $(5 \\mathrm{pts}) E={0} \\cup\\left(\\cup_{n=1}^{\\infty} \\overline{B_{\\frac{1}{n}}\\left(a_{n}\\right)}\\right) \\subseteq \\mathbb{R}^{d}$, where $\\left(a_{n}\\right){\\mathbb{N}}$ is a given sequence in $\\mathbb{R}^{d}$ converging to 0 . (b) $(7$ pts $) E={0} \\cup\\left{\\frac{1}{n} \\mid n \\in \\mathbb{N}\\right} \\cup\\left{\\frac{1}{n}+\\frac{1}{m} \\mid n, m \\in \\mathbb{N}\\right} \\subseteq \\mathbb{R}$.\n\n$# 3$ (10 pts) Let $E, K$ be disjoint subsets of $\\mathbb{R}^{d}$, with $E$ closed and $K$ compact. Prove that there exists $\\alpha>0$ such that $|x-y| \\geq \\alpha$ for all $x \\in E$ and $y \\in K .$ (First prove that for all $y \\in K$ there exists $r{y}>0$ such that $B_{r_{y}}(y) \\cap E=\\emptyset$.)\n\n$# 4$ (8 pts) Let $E$ be an uncountable subset of $\\mathbb{R}^{d}$. Prove that there exists $x \\in E$ such that $B_{\\delta}(x) \\cap E$ is uncountable for all $\\delta>0$.\n\n### 5 (12 pts) Let $E$ be a subset of $\\mathbb{R}^{d}$. Prove that the following two statements are equivalent.\n\n(i) $E$ is compact.\n(ii) For any infinite subset $F \\subseteq E$, there exists $x \\in E$ such that $B_{\\delta}(x) \\cap F$ is infinite for all $\\delta>0$" ]

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[ "# time efficient key value store for fast lookup\n\nLet's state a collection of key/values.\n\nKey is an unsigned integer (0 - 2^32-1). Hense there are comparable. Value is few bytes fixed size.\n\nThere are 4M items in the collection. Keys are not evenly spread in the key space i.e there is no way to predict hole size between two consecutive keys.\n\nThis collection is created once and never modify after that, even if not prohibited.\n\nThe aim is to lookup a value by its key. But if key is not found then the previous one (as in integer natural order) must be returned.\n\nI'm looking for a data-structure that can be mmap on disk which would be reasonably space efficient but most importantly very time efficient for the lookup function.\n\nThere is no known pattern on how the keys are queried.\n\n• You can use a hash table. – Yuval Filmus Jun 26 '17 at 22:01\n• Not if you need to find the \"previous one in integer order\", you can't. – Pseudonym Jun 28 '17 at 6:11\n\nMake a table with 2^22 entries to lookup the highest 22 bits of the key. Each entry is responsible for one value on average (but may contain up to 1024).\n\nEntry #i in the table, which is responsible for 0 to 1024 values with keys from 1024i to 1024i + 1023, contains the following: 1. The number of keys in that range. 2. If there are 0 keys, the value of the next lower available key. If there is 1 key, the lower 10 bit of the key in that range, and its value. If there are two or more keys, an index j into a second table.\n\nIn the second table, we store the lower 10 bit of the key, plus the value, for every key where there are two or more keys in the range from 1024i to 1024i + 1023. For values in the second table, we use binary search.\n\nThe minimum possible size for such any such data structure is $\\log_2{ 2^{32} \\choose 4\\times 10^6} \\approx 4.6\\times10^7$ bits, or around 5.5MB.\n\nAn array of 4 million 32-bit numbers is only 15.25MB. So even relatively space-inefficient data structures should fit comfortably in RAM.\n\nAny rank/select dictionary will work. Briefly, given a set $S$ of values, we can define the operations as:\n\n$$\\hbox{rank}(x) = \\left| \\{ y \\in S\\ : y < x\\} \\right|$$ $$\\hbox{select}(i) = \\min \\{ x : \\hbox{rank}(x) = i \\}$$\n\nIntuitively, $\\hbox{rank}(x)$ is the number of elements in the set less than x, and $\\hbox{select}(i)$ is the ith smallest element in S.\n\nAs an example of a simple implementation, if you store the set as a sorted array of integers, then select is just an array lookup, and rank can be implemented with binary search.\n\nGiven those two queries, the value that you're looking for is $\\hbox{select}(\\hbox{rank}(x) - 1)$. However, it should be possible to take pretty much any rank/select data structure (e.g. these ones) and implement this query directly.\n\nYou may also want to consider some variant of a van Emde Boas tree. But to be honest, a B-tree (with careful implementation of the leaf nodes) might do the job.\n\nSince the data structure is only created once, you can simply use an array ordered by the key. Each element in the array contains the key-value pair. Sorting the array is O(n*log(n)), which will be fast for 4e6 items and only needs to be done once. You can then use binary search (which is O(log2(n) or ~22) on the array to find the key. If not found, simply return the prior element in the array.\n\nThis will be very fast and will fit even more easily in RAM because there are no pointers. It also has the advantage of being extremely simple and easy to write. If the data needs to be persisted, you just write the array to disk, use mmap if you want, etc.\n\nIf you have only 2^32 numbers, just make a lookup table on disk. If the number of value bytes (=n) is low enough, then just copy the values for the non existent keys from the last valid one. Otherwise introduce another indirection where you link from that lookup table to our list of 4M values.\n\nYou can calculate the access point for the value exactly: key * n Bytes. Lookup speed is O(1) for existing and non existing keys.\n\nDisk consumption is 2^32 * n Bytes for value.\n\n• That would take 36GB of disk (9 bytes value) and only 0.1% would be actually used. This will never be mapped in memory so reads will be slow. This is the worst solution ever. – Setop Jul 2 '17 at 21:39\n• Like your honest comment :) Beside that, instead of using 9 bytes, you can use only 2 bytes to point to your 4m values. That would give you 2^33 + 9*4m bytes = 8 GB. That is something that can hang in memory. It really depends how much time, resources, money and scalability issues you have, if you want to make your algorithm more complicated or not... – CFrei Jul 5 '17 at 10:10\n• 2 bytes only allow to store 64k values, not 4m. It won't fit. – Setop Jul 6 '17 at 6:27\n• Yeah, sorry, you need 22 bits/3 bytes for 4 mio entries, I miss calced... well. was a try. – CFrei Jul 6 '17 at 20:36" ]

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[ "# Convert Imperial Units To Meters\n\nTo convert imperial units to meters you can use our multi-unit converter to select the unit to convert and the amount to be converted by selecting the unit from the drop down and entering the amount in the box below. Additionally you can convert from meters by entering the amount in the meters box.\n\n Amount Unit Meters inches  feet  yards  furlongs  miles  chains  links  rods  poles  perches  hands  fathoms  cables  natuical m = m\n\n## Imperial Units Background\n\nImperial units are a system of units defined by the Weight and Measures Act of 1824 in the United Kingdom, these were later refined in 1959 (as were US Customary units hence the use of the term Imperial Units in the US as the 1959 revision eradicated a number of the differences between the two systems).\n\nUS Customary units were based on English units at the time of US independence and hence had many slight differences in definitions from imperial units.\n\nImperial units were used across the British Empire and hence most English speaking countries used this system at one time or another, though all have now adopted the metric system, with only Canada still legally allowing the use of imperial units, although they have officially adopted metric units.\n\n## 1 Meter Expressed In Imperial Units\n\n• 1 meter = 39.3700787 inches\n• 1 meter = 9.8425197 hands\n• 1 meter = 4.9709695 links\n• 1 meter = 3.2808399 feet\n• 1 meter = 1.0936133 yards\n• 1 meter = 0.5468066 fathoms\n• 1 meter = 0.1988388 rods/poles/perches\n• 1 meter = 0.0497097 chains\n• 1 meter = 0.0053961 cables\n• 1 meter = 0.0049709 furlongs\n• 1 meter = 0.0006214 miles\n• 1 meter = 0.0005400 nautical miles\n\n## Imperial Units Expressed As Meters\n\n• 1 inch = 0.0254 meters\n• 1 hand = 0.1016 meters\n• 1 link = 0.201168 meters\n• 1 foot = 0.3048 meters\n• 1 yard = 0.9144 meters\n• 1 fathom = 1.8288002 meters\n• 1 rod/pole/perch = 5.0291995 meters\n• 1 chain = 20.1167981 meters\n• 1 cable = 185.3190 meters\n• 1 furlong = 201.1708 meters\n• 1 mile = 1,609.344 meters\n• 1 nautical mile = 1,852 meters" ]

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[ "#!/usr/bin/perl # # Author: M.Fuerstenau, Oce Printing Systems GmbH, Poing, Germany # # Purpose: # # c2n.cgi is a filter/wrapper for the frontend of Cacti to include it in Nagios. # # Prerequisites: # - guest account in cacti must be enabled # - CGI.pm # - libwww-perl # - a database user how is able to read all tables in the Cacti database # # Changes: # # 15.05.2008 - Version 1.0 # # 24.09.2008 - Version 1.1 # - renamed from cacti.cgi to c2n.cgi # - c2n.cgi can now handle multiple installations of Cacti with seperate databases. # # 11.10.2008 - Version 1.2 # - Minor code revision - removed some nested if/else # - Some minor bugs using the calendar fixed # - Allow/added the zoom function where possible # - Added some lines in the interface filter to remove unnessesary grey lines. # # 15.01.2009 - Version 1.3 # - Minor bug fix - /cacti was hardcoded in 47 lines (oh yeah - the good old copy & paste) # Therefore defining another URI in global.php than /cacti would cause porblems. # Replaced by the variable \\$CACTI_URI. Enter your URI here. # Thanks to Thomas Wollner for detecting this bug. use strict; use CGI qw/:cgi-lib/; use LWP::Simple; use DBI; use Date::Calc qw(:all); use File::Basename; use vars qw(\\$DOC \\$DATE_TO_EPOCH \\$DSN \\$DATABASE \\$DBHOST \\$DBPORT \\$DBH \\$ID @ROW \\$STH \\$SQLSTATEMENT); use vars qw(\\$SERVER \\$HOST \\$DBPASS \\$DBLOGIN \\$DATE_FROM_EPOCH \\$DATE_TO \\$DATE_FROM \\$SCRIPT_NAME \\$SCRIPT_NAME_BASE); use vars qw(%formParameter \\$URL2FETCH \\$QUERY_PARMS \\$NAGIOS); use vars qw(\\$CACTI_INSTANCE \\$CALLING_SERVER %URL_PART1 %DBHOST %DBLOGIN %DBPASS %DATABASE %DBPORT); # Main part ------------------------------------------------------------------------ &ReadParse(*formParameter); \\$HOST = \\$formParameter{'host'}; # \\$HOST is the name of the host you are # asking performance graphs for. \\$NAGIOS = \\$formParameter{'nagios'}; # \\$NAGIOS is a variable identifying only # the part of the script which has to be used \\$CACTI_INSTANCE = \\$formParameter{'cacti_instance'}; # This variable is only a value for # identifying the server and the database # cacti is running \\$CALLING_SERVER = \\$ENV{'SERVER_NAME'}; # \\$CALLING_SERVER is the calling host. \\$SCRIPT_NAME = \\$ENV{'SCRIPT_NAME'}; \\$SCRIPT_NAME_BASE = basename(\\$SCRIPT_NAME); \\$QUERY_PARMS = \\$ENV{'QUERY_STRING'}; # The complete query string \\$CACTI_URI = \"cacti\"; # The URI of Cacti as defined in Cacti # in global.php # This is needed for the reverse proxy %URL_PART1 = ( 0 => \"\\$CALLING_SERVER\", 1 => \"\\$CALLING_SERVER/ci1\" ); %DBHOST = ( 0 => \"mysqlsrv\", 1 => \"cacti_probe\" ); %DBLOGIN = ( 0 => \"stat4nag\", 1 => \"stat4nag\" ); %DBPASS = ( 0 => \"stat\", 1 => \"stat\" ); %DATABASE = ( 0 => \"cacti\", 1 => \"cacti\" ); %DBPORT = ( 0 => \"3306\", 1 => \"3306\" ); \\$SERVER = \\$URL_PART1{\\$CACTI_INSTANCE}; \\$DBHOST = \\$DBHOST{\\$CACTI_INSTANCE}; \\$DBLOGIN = \\$DBLOGIN{\\$CACTI_INSTANCE}; \\$DBPASS = \\$DBPASS{\\$CACTI_INSTANCE}; \\$DATABASE = \\$DATABASE{\\$CACTI_INSTANCE}; \\$DBPORT = \\$DBPORT{\\$CACTI_INSTANCE}; # First call - Main status page # \\$NAGIOS is an identifier for the page wanted if (\\$NAGIOS eq 1) { get_host_id(); print \"Content-type: text/html\\n\\n\"; \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph_view.php?action=preview&host_id=\\$ID&graph_template_id=0&filter=\"; \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; # Removing all unwanted scrap, rewriting URL/URI and so on using som regular expressions \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/href=\\'graph.php\\?/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&/g; \\$DOC =~ s/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/href=\\'cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/g; \\$DOC =~ s///isog; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'/href=\\'http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?/g; \\$DOC =~ s///isog; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/Presets:<\\/strong>//iso; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s///iso; \\$DOC =~ s/< href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/isog; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/href=\\\"include/href=\\\"\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/href=\\\"images/href=\\\"\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/nagios=3\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph_view.php\\&action=preview\\&host_id=\\$ID\\&graph_template_id=0\\&page=/isog; \\$DOC =~ s/\n/\n/; print \"\\$DOC\"; exit; } if (\\$NAGIOS eq 2) { print \"Content-type: text/html\\n\\n\"; \\$QUERY_PARMS =~ s/^.*function=//; \\$QUERY_PARMS =~ s/php\\&?/php\\?/; \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/\\$QUERY_PARMS\"; \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; # Removing all unwanted scrap, rewriting URL/URI and so on using som regular expressions \\$DOC =~ s///isog; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'graph.php\\?/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&/g; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'/href=\\'http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?/g; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"\\http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/href=\\\"include/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/href=\\\"images/href=\\\"\\http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/nagios=3\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph_view.php\\&action=preview\\&host_id=\\$ID\\&graph_template_id=0\\&page=/isog; print \"\\$DOC\"; exit; } if (\\$NAGIOS eq 3) { my \\$PAGE=\\$formParameter{'page'}; my \\$HOSTID=\\$formParameter{'host_id'}; print \"Content-type: text/html\\n\\n\"; \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph_view.php?action=preview&host_id=\\$HOSTID&graph_template_id=0&page=\\$PAGE\"; \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/href=\\'graph.php\\?/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&/g; \\$DOC =~ s/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/href=\\'cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/g; \\$DOC =~ s///isog; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'/href=\\'http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?/g; \\$DOC =~ s///isog; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/Presets:<\\/strong>//iso; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s///iso; \\$DOC =~ s/< href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/isog; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/href=\\\"include/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/.*?<\\/form>//iso; \\$DOC =~ s/graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/nagios=3\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph_view.php\\&action=preview\\&host_id=\\$HOSTID\\&graph_template_id=0\\&page=/isog; \\$DOC =~ s/Previous<\\/a>/Previous<\\/a>/iso; print \"\\$DOC\"; exit; } if (\\$NAGIOS eq 4) { my \\$PAGE=\\$formParameter{'page'}; my \\$HOSTID=\\$formParameter{'host_id'}; \\$DATE_FROM = \\$formParameter{'date_from'}; \\$DATE_TO = \\$formParameter{'date_to'}; \\$DATE_FROM_EPOCH = \\$formParameter{'date_from_epoch'}; \\$DATE_TO_EPOCH = \\$formParameter{'date_to_epoch'}; print \"Content-type: text/html\\n\\n\"; \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph_view.php?action=preview&host_id=\\$HOSTID&graph_template_id=0&page=\\$PAGE\"; \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/href=\\'graph.php\\?/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&/g; \\$DOC =~ s/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/href=\\'cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/g; \\$DOC =~ s///isog; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'/href=\\'http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?/g; \\$DOC =~ s///isog; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/Presets:<\\/strong>//iso; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s///iso; \\$DOC =~ s/title=\\'Graph Begin Timestamp\\' size=\\'14\\' value=\\'.*?\\'/title=\\'Graph Begin Timestamp\\' size=\\'14\\' value=\\'\\$DATE_FROM\\'/g; \\$DOC =~ s/title=\\'Graph End Timestamp\\' size=\\'14\\' value=\\'.*?\\'/title=\\'Graph End Timestamp\\' size=\\'14\\' value=\\'\\$DATE_TO\\'/g; \\$DOC =~ s/graph_start=.*?&/graph_start=\\$DATE_FROM_EPOCH&/g; \\$DOC =~ s/graph_end=.*?\\/'/graph_end=\\$DATE_TO_EPOCH\\'/g; \\$DOC =~ s/< href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/isog; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/href=\\\"include/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/.*?<\\/form>//iso; \\$DOC =~ s/\\$SCRIPT_NAME_BASE\\?graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/\\$SCRIPT_NAME_BASE\\?nagios=4\\&cacti_instance=\\$CACTI_INSTANCE\\&date_from=\\$DATE_FROM&date_to=\\$DATE_TO\\&date_from_epoch=\\$DATE_FROM_EPOCH\\&date_to_epoch=\\$DATE_TO_EPOCH\\&function=graph_view.php&action=preview\\&host_id=\\$HOSTID\\&graph_template_id=0\\&page=/isog; \\$DOC =~ s/graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/\\$SCRIPT_NAME_BASE\\?nagios=4\\&date_from=\\$DATE_FROM&date_to=\\$DATE_TO\\&date_from_epoch=\\$DATE_FROM_EPOCH\\&date_to_epoch=\\$DATE_TO_EPOCH\\&function=graph_view.php&action=preview\\&host_id=\\$HOSTID\\&graph_template_id=0\\&page=/isog; print \"\\$DOC\"; exit; } else { my \\$CAL_SET=\\$formParameter{'cal_select'}; if (\\$CAL_SET eq \"\") { \\$CAL_SET=0; } if (\\$CAL_SET eq 0) { my \\$LOCAL_GRAPH_ID=\\$formParameter{'local_graph_id'}; my \\$RRA_ID=\\$formParameter{'rra_id'}; my \\$GRAPH_START=\\$formParameter{'graph_start'}; my \\$GRAPH_END=\\$formParameter{'graph_end'}; print \"Content-type: text/html\\n\\n\"; # The rra-id determines which type/graphic has to be zoomed if (\\$RRA_ID eq 0) { \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph.php?action=zoom&local_graph_id=\\$LOCAL_GRAPH_ID&rra_id=0&graph_start=\\$GRAPH_START&graph_end=\\$GRAPH_END\"; } else { \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph.php?action=zoom&local_graph_id=\\$LOCAL_GRAPH_ID&rra_id=\\$RRA_ID&view_type=\"; } \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/\n//iso; \\$DOC =~ s/var cURLBase = \\\"graph.php/var cURLBase = \\\"http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php/g; print \"\\$DOC\"; exit; } else { print \"Content-type: text/html\\n\\n\"; # Now it is a little bit tricky. The calendar function. All submission of parameters # is done by get. Unfortuntely using the calendar a part is being send using post. # Therefore some paramters are available in \\$QUERY_PARMS but cannot be splitted up by # ReadParse and must be splitted up manually below. \\$HOST = \\$QUERY_PARMS; \\$HOST =~ s/^host=//; \\$HOST =~ s/&.*\\$//; \\$CACTI_INSTANCE = \\$QUERY_PARMS; \\$CACTI_INSTANCE =~ s/^.*cacti_instance=//; # And another phenomenon I haven't solved. Although these vars have # been defined as global in the beginning they are not set here. # Therefore I had to set them seperately here. Bullhsit. \\$DBHOST = \\$DBHOST{\\$CACTI_INSTANCE}; \\$SERVER = \\$URL_PART1{\\$CACTI_INSTANCE}; \\$DBLOGIN = \\$DBLOGIN{\\$CACTI_INSTANCE}; \\$DBPASS = \\$DBPASS{\\$CACTI_INSTANCE}; \\$DATABASE = \\$DATABASE{\\$CACTI_INSTANCE}; \\$DBPORT = \\$DBPORT{\\$CACTI_INSTANCE}; get_host_id(); \\$DATE_FROM = \\$formParameter{'date1'}; \\$DATE_TO = \\$formParameter{'date2'}; \\$DATE_FROM_EPOCH = conv_date(\\$DATE_FROM); \\$DATE_TO_EPOCH = conv_date(\\$DATE_TO); \\$URL2FETCH = \"http://\\$SERVER/\\$CACTI_URI/graph_view.php?action=preview&host_id=\\$ID&graph_template_id=0&filter=\"; \\$DOC = LWP::Simple::get \\$URL2FETCH; die \"Couldn't get it!\" unless defined \\$DOC; # Removing all unwanted scrap, rewriting URL/URI and so on using som regular expressions \\$DOC =~ s/style=\\'background-color: #f5f5f5; border: 1px solid #bbbbbb;\\'//iso; \\$DOC =~ s/href=\\'graph.php\\?/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&/g; \\$DOC =~ s/href=\\'nagios=2\\&cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/href=\\'cacti_instance=\\$CACTI_INSTANCE\\&function=graph.php\\&action=zoom/g; \\$DOC =~ s///isog; \\$DOC =~ s///isog; \\$DOC =~ s/href=\\'/href=\\'http:\\/\\/\\$CALLING_SERVER\\$SCRIPT_NAME\\?/g; \\$DOC =~ s///isog; \\$DOC =~ s/src=\\'/src=\\'http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\//g; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/table>//iso; \\$DOC =~ s/\n.*?<\\/table>//iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/<< Previous<\\/strong>/  <\\/strong>/g; \\$DOC =~ s/Next >><\\/strong>/  <\\/strong>/g; \\$DOC =~ s///iso; \\$DOC =~ s///iso; \\$DOC =~ s/Presets:<\\/strong>//iso; \\$DOC =~ s/\n.*?<\\/td>//iso; \\$DOC =~ s/.*?<\\/td>//iso; \\$DOC =~ s///iso; \\$DOC =~ s/title=\\'Graph Begin Timestamp\\' size=\\'14\\' value=\\'.*?\\'/title=\\'Graph Begin Timestamp\\' size=\\'14\\' value=\\'\\$DATE_FROM\\'/g; \\$DOC =~ s/title=\\'Graph End Timestamp\\' size=\\'14\\' value=\\'.*?\\'/title=\\'Graph End Timestamp\\' size=\\'14\\' value=\\'\\$DATE_TO\\'/g; \\$DOC =~ s/graph_start=.*?&/graph_start=\\$DATE_FROM_EPOCH&/g; \\$DOC =~ s/graph_end=.*?\\'/graph_end=\\$DATE_TO_EPOCH\\'/g; \\$DOC =~ s/src=\\\"include/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/src=\\\"images/src=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/href=\\\"include/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/include/g; \\$DOC =~ s/href=\\\"images/href=\\\"http:\\/\\/\\$SERVER\\/\\$CACTI_URI\\/images/g; \\$DOC =~ s/\\$SCRIPT_NAME_BASE\\?graph_view.php\\?action=preview\\&graph_template_id=0\\&page=/\\$SCRIPT_NAME_BASE\\?nagios=4\\&cacti_instance=\\$CACTI_INSTANCE\\&date_from=\\$DATE_FROM&date_to=\\$DATE_TO\\&date_from_epoch=\\$DATE_FROM_EPOCH\\&date_to_epoch=\\$DATE_TO_EPOCH\\&function=graph_view.php&action=preview\\&host_id=\\$ID\\&graph_template_id=0\\&page=/isog; \\$DOC =~ s//\n/; print \"\\$DOC\"; exit; } } sub error { print \"Content-type: text/html\\n\\n\"; print \"\\n\"; print \"\\n\"; print \"\\n\"; print \"Performance Data Error\\n\"; print \"\\n\"; print \"\\n\"; print \"\\n\"; print \"\\n\"; print \"\\n\"; print \"\\n\"; print \"\n\\n\"; print \"\n\\n\"; print \"\n\\n\"; print \"\n\\n\"; print \"\n\\n\"; print \"\n\\n\"; print \"\n\n# Sorry - no performance data for this hostavailable\n\n\\n\"; print \"\\n\"; print \"\\n\"; } sub get_host_id { # Connect to the database ---------------------------------------------------------- \\$DSN = \"DBI:mysql:database=\\$DATABASE;host=\\$DBHOST;port=\\$DBPORT\"; \\$DBH = DBI->connect(\\$DSN, \\$DBLOGIN, \\$DBPASS, { PrintError => 0, ### Don't report errors via warn( ) RaiseError => 1 ### Do report errors via die( ) } ); # Constuct a SQL statement ---------------------------------------------------------- # get the id for the host delivered by nagios \\$SQLSTATEMENT = \"SELECT id FROM host where description = \\\"\\$HOST\\\"\"; # ----- Prepare a SQL statement for execution \\$STH = \\$DBH->prepare( \\$SQLSTATEMENT ); # ----- Execute the statement in the database \\$STH->execute(); # ----- Retrieve the returned rows of data while ( @ROW = \\$STH->fetchrow_array( ) ) { # Put the hashed password in the var \\$ID = \\$ROW; } \\$DBH->disconnect; if (\\$ID eq \"\") { error(); exit 1; } } sub conv_date { my \\$DATE2CONV = \\$_; my \\$YEAR = \"\\$DATE2CONV\"; \\$YEAR =~ s/-.*//; my \\$MONTH = \"\\$DATE2CONV\"; \\$MONTH =~ s/^.*?-//; \\$MONTH =~ s/-.*//; my \\$DAY = \"\\$DATE2CONV\"; \\$DAY =~ s/^.*?-//; \\$DAY =~ s/^.*?-//; \\$DAY =~ s/ .*//; my \\$HOUR = \"\\$DATE2CONV\"; \\$HOUR =~ s/^.*? //; \\$HOUR =~ s/...\\$//; my \\$MIN = \"\\$DATE2CONV\"; \\$MIN =~ s/^.*?://; \\$DATE2CONV = Mktime(\\$YEAR,\\$MONTH,\\$DAY, \\$HOUR,\\$MIN,0); return \\$DATE2CONV; }" ]

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[ "1. Home\n2. Log-linear modelling\n\n# Log-linear modelling\n\nSelect menu: Stats | Regression Analysis | Log-linear Models\n\nLog-linear modelling is appropriate for data consisting of counts, which are assumed to come from Poisson distributions.\n\nStats | Regression Analysis | Log-linear Models.\n2. Fill in the fields as required then click Run.\n\nYou can set additional Options then after running, you can save the results by clicking Save.", null, "The probabilities of the distributions are studied by fitting linear models on the logarithmic scale; that is, in the terminology of generalized linear models, there is a logarithmic link function between the expected counts and the combination of linear effects. This link implies that the underlying model is multiplicative, so the effect of a treatment, for example, will be to produce a proportionate increase (or decrease) in the expected counts rather than to add (or subtract) a constant amount.\n\n## Response variate\n\nThe name of the variate containing the observed counts needs to be entered into the Response variate field, and can be selected from those in the Available data list.\n\n## Maximal model\n\nThe Maximal model lets you specify the most complicated model that you are likely to want to consider. It may be left blank, but this may lead to subsequent inability to compare models because of data that are missing for some explanatory variates but not for others.\n\n## Model to be fitted\n\nThe model to be fitted is specified by entering a model formula into the Model to be fitted field. The formula can involve both variates and factors which can be selected from the Available data field, and operators from the Operators field.\n\n## Censored data\n\nWhen the option Censored are censored option is ticked, you can set a censoring value in upper bound (Direction: Right) or lower bound (Direction: Left) field at which values are censored. Right censoring means that you are certain the value is at least as great as the bound, but may be higher. This may occur when it is not possible to count above a particular value. Left censoring means that you are certain the value is at most as large as the bound, but may be lower. This may occur if you can not detect less than a given count.\n\n## Action buttons\n\n Run Run the analysis. Cancel Close the dialog without further changes. Options Opens a dialog where additional options and settings can be specified. Defaults Reset to the default settings. Clicking the right mouse on this button produces a shortcut menu where you can choose to set the options using the currently stored defaults or the Genstat default settings. Save Opens a dialog where you can save results from the analysis. Predict Lets you calculate predictions based on the current regression model. Change model This changes the current model by adding or dropping explanatory terms, thus allowing a sequence of models to be fitted and assessed. If you specified a maximal model, all new terms must have appeared in that model. If you did not specify a maximal model and a term is introduced with a missing value for a unit previously used in the regression, the model sequence is interrupted and information will be available only for the current model (excluding that unit) and the new model. Further output Lets you display additional results and graphical output from the analysis.\n\n## Action Icons", null, "Pin Controls whether to keep the dialog open when you click Run. When the pin is down", null, "the dialog will remain open, otherwise when the pin is up", null, "the dialog will close.", null, "Restore Restore names into edit fields and default settings.", null, "Clear Clear all fields and list boxes.", null, "Help Open the Help topic for this dialog." ]

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[ "Is 4058 a prime number? What are the divisors of 4058?\n\n## Parity of 4 058\n\n4 058 is an even number, because it is evenly divisible by 2: 4 058 / 2 = 2 029.\n\nFind out more:\n\n## Is 4 058 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 4 058 is about 63.702.\n\nThus, the square root of 4 058 is not an integer, and therefore 4 058 is not a square number.\n\n## What is the square number of 4 058?\n\nThe square of a number (here 4 058) is the result of the product of this number (4 058) by itself (i.e., 4 058 × 4 058); the square of 4 058 is sometimes called \"raising 4 058 to the power 2\", or \"4 058 squared\".\n\nThe square of 4 058 is 16 467 364 because 4 058 × 4 058 = 4 0582 = 16 467 364.\n\nAs a consequence, 4 058 is the square root of 16 467 364.\n\n## Number of digits of 4 058\n\n4 058 is a number with 4 digits.\n\n## What are the multiples of 4 058?\n\nThe multiples of 4 058 are all integers evenly divisible by 4 058, that is all numbers such that the remainder of the division by 4 058 is zero. There are infinitely many multiples of 4 058. The smallest multiples of 4 058 are:\n\n## Numbers near 4 058\n\n### Nearest numbers from 4 058\n\nFind out whether some integer is a prime number" ]

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[ "", null, "", null, "", null, "", null, "", null, "Home Search   What's New Index Books Links Q & A Newsletter Banners   Feedback Tip Jar", null, "", null, "", null, "", null, "", null, "MSDN Visual Basic Community", null, "", null, "", null, "Title Graph a function with parameters entered by the user using a convenient scale graph, equation, scale Graphics\n\nUse a PictureBox's Scale properties to make a convenient scale. Read the parameter values from TextBoxes. Then use a Do loop to increment the independent variable that determines the function values.\n\n```Private Sub cmdDraw_Click()\nConst PI = 3.14159265\nDim A As Single\nDim B As Single\nDim theta As Single\nDim dtheta As Single\n\nA = CSng(txtA.Text)\nB = CSng(txtB.Text)\npicGraph.ScaleLeft = 0\npicGraph.ScaleWidth = 4 * PI\npicGraph.ScaleTop = 2\npicGraph.ScaleHeight = -4\npicGraph.AutoRedraw = True\n\ntheta = 0\npicGraph.Cls\npicGraph.CurrentX = theta\npicGraph.CurrentY = Sin(A * theta) + Sin(B * theta)\n\ndtheta = 4 * PI / picGraph.ScaleX(picGraph.ScaleWidth, _\npicGraph.ScaleMode, vbPixels)\nDo While theta <= 4 * PI + dtheta / 2\npicGraph.Line -(theta, Sin(A * theta) + Sin(B * _\ntheta))\ntheta = theta + dtheta\nLoop\nEnd Sub```", null, "", null, "Copyright © 1997-2010 Rocky Mountain Computer Consulting, Inc.   All rights reserved.", null, "", null, "Updated", null, "" ]

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[ "", null, "# Question: How Many Square Feet Will 1000 Bricks Cover?\n\n## How much does brick cost per 1000?\n\nBricks are usually sold in large quantities: 1,000 bricks cost between \\$550 on average.\n\nHomeowners typically pay between \\$340 and \\$850 for 1,000 bricks but could pay anywhere from \\$250 to \\$3,730..\n\n## How many bricks do I need for 144 square feet?\n\nDivide the total square inches into 144, which is the total number of inches in a square foot. If the brick is 32 square inches, for instance, the result is 4.5 bricks per square foot (144 divided by 32).\n\n## How do I calculate sq ft?\n\nTo find square feet, multiply the length measurement in feet by the width measurement in feet. This yields a product called the area, which is expressed in square feet (or square inches if you are calculating a much smaller space, such as a dollhouse).\n\n## How many bricks can a bricklayer lay in a day?\n\n600 bricksWorking at average speeds one good bricklayer might lay 600 bricks in the day. So 1200 bricks between the gang will amount to 20m² of single skin face brickwork (60 bricks per m²).\n\n## How do you calculate brick coverage?\n\nFor a singular layer brick wall, multiply the length of the wall by the height to get the area. Multiply that area by 60 to get the number of bricks you should need, then add 10% for wastage. That’s the short answer and assumes ‘standard’ brick and mortar sizes. It can also vary based on the type of structure.\n\n## How many square feet is a 10×10 wall?\n\nArea of the floor or ceiling: Multiply the length by the width (10 feet x 12 feet = 120 square feet of area). Area of a wall: Multiply the width of the wall by its height. So one of the walls is 80 square feet (10 feet wide x 8 feet high) and the other is 96 square feet (12 feet x 8 feet).\n\n## How many bricks do I need for a 4 inch wall?\n\nBrick calculation for 4 inch brick wall Brick wall calculation for single brick wall or half brick wide wall or 4 inch brick wall requires 55 bricks per square metre.\n\n## How many bricks do I need for a 100 square foot patio?\n\nIn this case, 144 divided by 32 is 4 1/2 pavers per square foot. Multiply that number by the area square footage; if you have 100 square feet to fill, you need at least 450 pavers.\n\n## How many bricks are in a square foot?\n\n7 bricksA wall built with a standard modular brick will require 7 bricks per square foot to complete the project. Projects using different brick sizes will require a different amount.\n\n## How many bags of cement do I need for 1000 bricks?\n\n3 bags cementTo lay 1000 bricks = 3 bags cement + 0.6 cu. m. sand.\n\n## How heavy is a brick?\n\nabout 5 poundsYou can expect an average brick weight to be about 5 pounds (2.27 kg) for a standard red clay brick. The standard size measures to be 8-inch by 2 1/4-inch by 4-inch. Bricks are used as a building material for a variety of projects such as walls, fireplaces, patios, and walkways.\n\n## How many bricks do I need for 200 square feet?\n\nAbout seven modular thin brick are needed to cover a square foot, so 200 square feet X 7 thin brick per square foot = 1,400 thin brick to cover the wall.\n\n## How many bricks do I need for a 10×10 patio?\n\n450 bricksFor our example, this means you’ll need 450 bricks to cover your 10’x10′ patio space.\n\n## What size is a 200 square foot room?\n\nA 200 square foot space is about the size of a one-car garage. In other words, the average car could fit into 200 square foot place with a small amount of wiggle room. Imagine a space smaller than a bedroom for 100 square feet.\n\n## How many sq ft is 20×20?\n\n400 square feetSo if a area is 20 foot wide by 20 foot long, 20 x 20 = 400 square feet." ]

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[ "Solutions by everydaycalculation.com\n\n1st number: 1 2/6, 2nd number: 42/84\n\n8/6 + 42/84 is 11/6.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 6 and 84 is 84\n2. For the 1st fraction, since 6 × 14 = 84,\n8/6 = 8 × 14/6 × 14 = 112/84\n3. Likewise, for the 2nd fraction, since 84 × 1 = 84,\n42/84 = 42 × 1/84 × 1 = 42/84\n112/84 + 42/84 = 112 + 42/84 = 154/84\n5. 154/84 simplified gives 11/6\n6. So, 8/6 + 42/84 = 11/6\nIn mixed form: 15/6\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "## Concept#\n\n• Adaptive sampling algorithms usually employ empirical policies, and they are not based on any mathematical decission process.\n\n• We descrive adaptive sampling in terms of a multi-armed bandit problem to develop a novel adaptive sampling algorithm, Adaptive Bandit [ref], providing strong fundamentals to tackle the exploration-exploitation dilemma faced in adaptive sampling.\n\n• Adaptive Bandit is framed into a reinforcement-learning based framework, using an action-value function and an upper confidence bound selection algorithm, improving adaptive sampling’s performance and versatility when faced against different free energy landscape.\n\n• Discretized conformational states are defined as actions, and each action has an associated reward distribution. When an action is picked, the algorithm computes the associated reward for that action, based on MSM free energy estimations, and applies a policy to select the next action.\n\n• AdaptiveBandit relies on the UCB1 algorithm to optimize the action-picking policy, defining an upper confidence bound for each action based on the number of times the agent has picked that action and the total number of actions taken\n\n$a_t = argmax_{a\\in\\mathcal{A}}\\left[{Q_t(a) + c\\sqrt{\\frac{\\ln{t}}{N_t(a)}}}\\right]$\n\nA. Pérez, P. Herrera-Nieto, S. Doerr and G. De Fabritiis, AdaptiveBandit: A multi-armed bandit framework for adaptive sampling in molecular simulations. arXiv preprint 2020; arXiv:2002.12582.\n\nAuer P. Using confidence bounds for exploitation-exploration trade-offs. Journal of Machine Learning Research. 2002; 3(Nov):397-422.\n\n## Getting started#\n\nThis tutorial will show you how to properly set up an AdaptiveBandit project, highlighting the main differences with respect to the standard adaptive sampling. As an example, we will perform some folding simulations using the chicken villin headpice (PDB: 2F4K).\n\nLet’s start by importing HTMD and the AdaptiveBandit class:\n\nfrom htmd.ui import *\n\n\nAdaptiveBandit uses the same project structure as adaptive sampling, with each simulation being associated to a single directory with all the files to run it.\n\nTo begin, get the starting generators here. You can also download the data using wget -O gen.tar.gz https://ndownloader.figshare.com/files/25767026. You will have to uncompress that tar.gz file and allow execution in all run.sh files.\n\nfor file in glob('./generators/*/run.sh'):\nos.chmod(file, 0o755)\n\n\nThese generators contain prepared unfolded structures of villin, which we want to simulate long enough to reach the folded native structure.\n\nWe start our AdaptiveBandit project in the same way as with adaptive sampling, by defining the queue used for simulations.\n\nqueue = LocalGPUQueue()\n\nab = AdaptiveBandit()\nab.app = queue\n\n\nThen, we define the nmin, nmax and nframes to set the maximum amount of simulated frames\n\nab.nmin=0\nab.nmax=2\nab.nframes = 1000000\n\n\nAnd we choose the projection and clustering method used to construct a Markov model at each epoch\n\nab.clustmethod = MiniBatchKMeans\nab.projection = MetricSelfDistance('protein and name CA')\n\n\nUp until now, the setup is exactly the same as with AdaptiveMD. However, AdaptiveBandit has an additional parameter, which sets the $$c$$ parameter from the UCB1 equation:\n\nab.exploration = 0.01\n\n\nAdditionally, AdaptiveBandit accepts a goal function as an input that will be used to initialize our action-value estimates. In this example, we will use the contacts goal function defined in the previous tutorial to initialize the $$Q(a)$$ values. The goal_init parameter sets an $$N_t(a)$$ initial value proportional to the max frames per cluster at the end of the run, which represents the statistical certainty we give to the goal function.\n\nref = Molecule('2F4K')\n\ndef contactGoal(mol, crystal):\ncrystalCO = MetricSelfDistance('protein and name CA', periodic=None,\nmetric='contacts',\nthreshold=10).project(crystal).squeeze()\nproj = MetricSelfDistance('protein and name CA',\nmetric='contacts',\nthreshold=10).project(mol)\n# How many crystal contacts are seen?\nco_score = np.sum(proj[:, crystalCO] == 1, axis=1).astype(float)\nco_score = co_score / np.sum(crystalCO)\nreturn co_score\n\nab.goalfunction = (contactGoal, (ref,))\nab.goal_init = 0.3\n\n\nAnd now, we just need to launch our AdaptiveBandit run:\n\nab.run()" ]

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[ "# Granular Monotonic Binning in SAS\n\nIn the post (https://statcompute.wordpress.com/2017/06/15/finer-monotonic-binning-based-on-isotonic-regression), it is shown how to do a finer monotonic binning with isotonic regression in R.\n\nBelow is a SAS macro implementing the monotonic binning with the same idea of isotonic regression. This macro is more efficient than the one shown in (https://statcompute.wordpress.com/2012/06/10/a-sas-macro-implementing-monotonic-woe-transformation-in-scorecard-development) without iterative binning and is also able to significantly increase the binning granularity.\n\n`%macro monobin(data = , y = , x = );`\n`options mprint mlogic;`\n`data _data_ (keep = _x _y);`\n`  ``set` `&data;`\n`  ``where` `&y ``in` `(0, 1) ``and` `&x ~= .;`\n`  ``_y = &y;`\n`  ``_x = &x;`\n`run;`\n`proc transreg data = _last_ noprint;`\n`  ``model identity(_y) = monotone(_x);`\n`  ``output` `out` `= _tmp1 tip = _t;`\n`run;`\n`proc summary data = _last_ nway;`\n`  ``class _t_x;`\n`  ``output` `out` `= _data_ (``drop` `= _freq_ _type_) mean(_y) = _rate;`\n`run;`\n`proc sort data = _last_;`\n`  ``by` `_rate;`\n`run;`\n`data _tmp2;`\n`  ``set` `_last_;`\n`  ``by` `_rate;`\n`  ``_idx = _n_;`\n`  ``if _rate = 0 ``then` `_idx = _idx + 1;`\n`  ``if _rate = 1 ``then` `_idx = _idx - 1;`\n`run;`\n`  `\n`proc sql noprint;`\n`create` `table`\n`  ``_tmp3 ``as`\n`select`\n`  ``a.*,`\n`  ``b._idx`\n`from`\n`  ``_tmp1 ``as` `a ``inner` `join` `_tmp2 ``as` `b`\n`on`\n`  ``a._t_x = b._t_x;`\n`  `\n`create` `table`\n`  ``_tmp4 ``as`\n`select`\n`  ``a._idx,`\n`  ``min``(a._x)                                               ``as` `_min_x,`\n`  ``max``(a._x)                                               ``as` `_max_x,`\n`  ``sum``(a._y)                                               ``as` `_bads,`\n`  ``count``(a._y)                                             ``as` `_freq,`\n`  ``mean(a._y)                                              ``as` `_rate,`\n`  ``sum``(a._y) / b.bads                                      ``as` `_bpct,`\n`  ``sum``(1 - a._y) / (b.freq - b.bads)                       ``as` `_gpct,`\n`  ``log(calculated _bpct / calculated _gpct)                ``as` `_woe,`\n`  ``(calculated _bpct - calculated _gpct) * calculated _woe ``as` `_iv`\n`from`\n`  ``_tmp3 ``as` `a, (``select` `count``(*) ``as` `freq, ``sum``(_y) ``as` `bads ``from` `_tmp3) ``as` `b`\n`group` `by`\n`  ``a._idx;`\n`quit;`\n`title ``\"Monotonic WoE Binning for %upcase(%trim(&x))\"``;`\n`proc print data = _last_ label noobs;`\n`  ``var _min_x _max_x _bads _freq _rate _woe _iv;`\n`  ``label`\n`    ``_min_x = ``\"Lower\"`\n`    ``_max_x = ``\"Upper\"`\n`    ``_bads  = ``\"#Bads\"`\n`    ``_freq  = ``\"#Freq\"`\n`    ``_rate  = ``\"BadRate\"`\n`    ``_woe   = ``\"WoE\"`\n`    ``_iv    = ``\"IV\"``;`\n`  ``sum` `_bads _freq _iv;`\n`run;`\n`title;`\n`%mend monobin;`\nBelow is the sample output for LTV, showing an identical binning scheme to the one generated by the R isobin() function.", null, "更多相关文章请看 https://statcompute.wordpress.com/category/scorecard/" ]

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[ "The in the last term means that the second terms of the binomial factors must each contain y. As the chart on the right shows you $$-2 \\cdot -2$$ is positive 4 ...so we do have to consider these negative factors. In general, we can use the following steps to factor a quadratic of the form. Remove Common Factors if possible 2. \\\\ Substitute that factor pair into two binomials . a = 1 ax 2 + bx + c. 1. (If you need help factoring trinomials when $$a \\ne 1$$, then go here. Thanks to the SQA and authors for making the excellent resources below freely available. Find the product ac.. 2. The resulting factors will … Factoring quadratic equations worksheet. List the factors of the FACTORING QUADRATIC TRINOMIALS Example 2 : X2 - 8X + 15 Step 2 Factor the first term which is x2 (x )(x ) Step 4 Check the middle term (x - 5)(x - 3) -5x multiply -5 and x + -3x multiply -3 and x -8x Add the 2 terms. start color #11accd, a, end color #11accd, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #aa87ff, c, end color #aa87ff. Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Recent Articles. \\\\ Factor Trinomials of the Form x2 + bxy + cy2 Sometimes you’ll need to factor trinomials of the form with two variables, such as The first term,, is the product of the first terms of the binomial factors,. Since 1 and 4 add up to 5 and multiply together to get 4, we can factor it like: (x+1) (x+4) In other words, we will use this approach whenever the coefficient in front of x2 is 1. Find the factors of the product whose sum is the middle term of the trinomial. two factors of ac that add to give b, 3. $$, Write down the factor pairs of$$ \\red 6 $$, Identify which factor pair from the previous step sum up to$$ \\blue{-5} $$. Once a trinomial has been factorised, you may be asked to: Solve the trinomial (find the roots) Sketch the parabola. Quadratics – Worksheets. Factoring of quadratic polynomials (second-degree polynomials) is done by “un-FOILing,” which means we start with the result of a FOIL problem and work backwards to find the two binomial factors. The standard format for the quadratic equation is: ax2 + bx + … ), Identify a, b and c in the trinomial ax2 + bx + c,$$ In fact, this is not even a trinomial because there are 2 terms, It's always easier to understand a new concept by looking at a specific example so you might want scroll down and do that first. A more complex situation is factoring trinomials when the leading coefficient is not one. In order to solve the quadratic equation ax 2 + bx + c = 0 by factorization, the following steps are used: a = 1 If you are factoring a quadratic like x^2+5x+4 you want to find two numbers that. Starting Out 1 Set up your expression. Factor Trinomials using the “ac” Method. We can factor trinomials by first looking for factors that are common (that is the GCF) Example: Factor the following trinomials: Find the Greatest Common Factor - GCF. \\blue { b = 5} \\\\ A trinomial expression takes the form: \\ [a {x^2} + bx + c\\] To factorise a trinomial expression, put it back into a pair of brackets. Instead of multiplying two binomials to get a trinomial, you will write the trinomial as a product of two binomials. $$, Write down all factor pairs of$$ \\red 6 $$, Identify which factor pair from the previous step sum up to$$ \\blue 5$$, Factor the trinomial below$$ x^2 - 2x -3 $$, Identify a, b and c in the trinomial$$ax^2 + bx + c$$,$$ The trinomials on the left have the same constants 1, −3, −10 but different arguments. \\\\ The Procedure. $$\\text{Examples of Quadratic Trinomials}$$, $$\\red { \\text{Non }}\\text{-Examples of Quadratic Trinomials}$$, this is not a quadratic trinomial because there is an exponent that is $$\\red { \\text{ greater than 2} }$$, this is not a quadratic trinomial because there is not exponent of 2. In short, it is a quadratic expression with … a = 1 \\\\ Factoring trinomials when a is equal to 1 Factoring trinomials is the inverse of multiplying two binomials. Group terms with common factors. ... Factoring trinomials. Just follow these easy steps. \\\\ \\blue{ b = -2 } Further, we can say x+6=0 and x+2=0 and x =-6,-2 thereby are the roots Add up to 5. Check to see if the constant in either the first … \\blue{ b - 5} 5. Factoring quadratics by grouping. We can then pull out the GCF by using the distributive property in reverse. \\blue { b = -2} \\\\ $$, Write down all factor pairs of$$ \\red 3 $$, Identify which factor pair from the previous step sum up to$$ \\blue {4 } $$, Substitute that factor pair into two binomials, Factor the trinomial below$$ x^ 2 + 5x + 6 $$,$$ \\\\ James W. Brennan, Summary of Steps to Factor Quadratic Trinomials, Split Real World Math Horror Stories from Real encounters, Identify a, $$\\blue b$$ , and $$\\red c$$ in the trinomial $$ax^2 + \\blue bx + \\red c$$, Write down all factor pairs of $$\\red c$$, Identify which factor pair from the previous step. (Note: since c is negative we only need to think about pairs that have 1 negative factor and 1 positive factor. Solution : In the quadratic expression above, the coefficient of x 2 is 1. (Note: since $$\\red 4$$ is positive we only need to think about pairs that are either both positive or both negative. Video Tutorial of Factoring a Trinomial . ax 2 + hx + kx + c. 4. Another way to factor trinomials of the form $$ax^2+bx+c$$ is the “ac” method. Find two numbers whose product is $$12$$ and whose sum is $$-7\\text{. Video transcript. Factoring trinomials is easiest when the leading coefficient (the coefficient on the squared term) is one. \\\\ \\\\ Solving quadratic equations by factoring is all about writing the quadratic function as a product of two binomials functions of one degree each. \\red{ c = -15} a = 1 No need to guess and check. \\blue { b =4 } In Example A.57 we factor quadratic trinomials in which one or more of the coefficients is negative. That is the only difference between them. 4x+2=2 (x+6) Simplify Example A Worked example to illustrate how the factoring calculator Works: An algebra calculator that finds the roots to a quadratic equation of the form ax2 + bx + c = 0 , Write down all factor pairs of \\red {-3 } (yes, the negative sign matters! There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. ©1998-2002 Introduction to Physics. Example A.57. Factoring Polynomials Chapter Sections § 13.1 The Greatest Common Factor Factors Factors (either numbers or polynomials) When an integer is written as a product of integers, each of the integers in the product is a factor of the original number. Interactive simulation the most controversial math riddle ever! Factor the trinomial below x^2 - 2x - 15, Identify a, b and c in the trinomial ax^2 + bx + c , Factoring Special Cases 1 Check for prime numbers. List the factors of the \\red { c= -3} Factor x2 – 5x + 6 If c is \"minus\", then the factors will be of alternating signs; that is, one will be \"plus\" and one will be \"minus\". When given a trinomial, or a quadratic, it can be useful for purposes of canceling and simplifying to factor it. Summary of Steps to Factor Quadratic Trinomials 1. This page will focus on quadratic trinomials. If b is \"minus\", then the larger of the two factors is \"minus\". Factoring a quadratic equation can be defined as the process of breaking the equation into the product of its factors. Group the two pairs of terms that have common factors and simplify. If b is \"plus\", then the larger of the two factors is \"plus\". Factoring simple quadratics review. A Quadratic Trinomial. \\(\\displaystyle x^2-7x+12$$ $$\\displaystyle x^2-x-12$$ Solution. In other words, we can also say that factorization is the reverse of multiplying out. Test all the possible Find constant term, 3. Try the entered exercise, or type in your own exercise. ax^2 + bx + c. Where a, b, and c are all numbers. Rewrite the trinomial as four – term expressions by replacing the middle term by the sum factor. 3. When factoring trinomials, the first step would be to try to find the greatest common factor (GCF). In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. To factor a trinomial in the form x 2 + bx + c, find two integers, r and s, whose product is c and whose sum is b. Rewrite the trinomial as x 2 + rx + sx + c and then use grouping and the distributive property to factor the polynomial. 2. Problem 1 : Factor : x 2 + 6x + 5. 1. : \\red{ c = 3} Factoring Trinomials Calculator is a free online tool that displays the factors of given trinomial. Factoring using AC Method 1. (The only difference being that a quadratic trinomial has a degree of 2.) 4. Factoring Trinomials in the form x 2 + bx + c . the middle term into two terms, using the numbers found in step, 4. This formula only works when $$a = 1$$ . First, what is a quadratic trinomial? A trinomial is a polynomial with 3 terms.. We’ve seen already seen factorising into single brackets, but this time we will be factorising quadratics into double brackets. Factoring Trinomials, a = 1. You can use the Mathway widget below to practice factoring expression that are quadratic in form. In other words, there must be an exponent of '2' and that exponent must be the greatest exponent. \\\\ ), Identify which factor pair from the previous step sums up to $$\\blue { -2}$$, Substitute that factor pair into two binomials, Factor the trinomial below $$x^2 - 5x + 6$$, Identify a, b and c in the trinomial $$ax^2+ bx + c$$, $$How to factor expressions. Practice: Factoring quadratics with a common factor.$$, Write down the factor pairs of $$\\red{ -15}$$ Remember a negative times a positive is a negative. \\red {c = 6} Quadratics are algebraic expressions that include the term, x^2, in the general form, . factoring ax^2 + bx + c when \"a\" greater than 1. coefficient of the x2 term, 2. The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. Factoring Quadratic Polynomials Worksheet - Problems. Formula For Factoring Trinomials (when $$a = 1$$ ) Rewrite the quadratic as. Factor. Group the four terms into two pairs, copyright BYJU’S online factoring trinomials calculator tool makes the calculation faster, and it displays the factors of a trinomial in a fraction of seconds. Find the product of the leading term and the last term. a x 2 + b x + c. \\blueD ax^2+\\goldD bx+\\purpleC c ax2 +bx +c. binomials you can make from these factors, 2. With the quadratic equation in this form: Step 1: Find two numbers that multiply to give ac (in other words a times c), and add to give b. Remember a negative times a negative is a positive. \\red { c = 4 } Multiply together to get 4. Factorising an expression is to write it as a product of its factors. Then click the button to compare your answer to Mathway's. The degree of a quadratic trinomial must be '2'. \\blue { b = 5} a = 1 \\red{ c = 6} Split Whenever a quadratic has constants 3, 2, −1, then for any argument, the factoring will be (3 times the argument − 1) (argument + 1). Factorising Quadratics. Find two numbers h and k such that. Identify which factor pair from the previous step sums up to $$\\blue b$$, If you'd like, you can check your work by multiplying the two binomials and verify that you get the original trinomial, Factor the following trinomial: $$x^2 + 4x + 3$$, Identify a, b and c in the trinomial $$ax^2 + bx + c$$, $$Next lesson. (The “ac” method is sometimes called the grouping method.) \\\\ Please use the below for … Note: For the rest of this page, 'factoring trinomials' will refer to factoring 'quadratic trinomials'.$$, Write down all factors of $$\\red c$$ which multiply to $$\\red { \\fbox {4}}$$. ), Identify which factor pair from the previous step sums up to $$\\blue{-2}$$. the middle term into two terms, using the numbers found in step. hk = ac (h and k are factors of the product of the coefficient of x 2 and the constant term)AND h + k = b (h and k add to give the coefficient of x)3. Solver. Factoring a quadratic is like un-doing the “FOIL” process. Given a general quadratic trinomial. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. For example the trinomail quadratic,can we written as (x+6) (x+2)=0, where (x+2) and (x+6) are the binomial terms each of degree 1. 2. If the sum of the terms is the middle term in the given quadratic trinomial then the factors are correct. a = 1 (Click \"Tap to view steps\" to be taken directly to the Mathway site for a paid upgrade.) This means that factoring a quadratic expression is the process of taking a trinomial and turning it into multiplication of two binomials - basically FOIL backwards. Coefficient is not one type in your own exercise quadratic trinomials in which one or more of binomial. To try to find the factors of the x2 term, 3, but this time we use., but this time we will be factorising quadratics into double brackets \\! ( GCF ) that factorization is the “ ac ” method. video I want to the. Coefficient in front of x2 is 1 is often called a quadratic, it be... A more complex situation is factoring trinomials when equal to 1 factoring when. This approach whenever the coefficient on the squared term ) is the of... Is like un-doing the “ ac ” method is sometimes called the grouping method. thanks to the widget... Possible binomials you can use the Mathway site for a paid upgrade. then click the button to compare answer... \\Blued ax^2+\\goldD bx+\\purpleC c ax2 +bx +c of this page, 'factoring trinomials ' ' will refer to 'quadratic... Calculator is a negative times a positive methods: common factor, difference two! By factoring is all about writing the quadratic expression above, the coefficient of trinomial... Examples of factoring quadratic trinomials a second degree polynomial, which is often called quadratic. ( the “ ac ” method. ( click Tap to view ''! Whose sum is the inverse of multiplying two binomials term into two terms using. The possible binomials you can make from these factors, 2. seen into. Have the same constants 1, −3, −10 but different arguments Tap to steps! Is factoring trinomials in which one or more of the form words, must! The possible binomials you factoring quadratic trinomials make from these factors, 2. + 5 type in your own exercise '... Widget below to practice factoring expression that are quadratic in form the coefficient of x 2 + bx c.. Common factor, difference of two squares, trinomial/quadratic expression and completing the.... Trinomials, the first step would be to try to find the greatest exponent factor it factors of the term... Like x^2+5x+4 you want to do a bunch of examples of factoring a second degree polynomial, which is called! Bunch of examples of factoring a quadratic by using the numbers found in step 4. Quadratic is like un-doing the “ ac ” method is sometimes called the grouping.! Ax^2+\\Goldd bx+\\purpleC c ax2 +bx +c try to find the greatest common factor, difference of two,! A x 2 + 6x + 5 numbers found in step, 4 are quadratic in form ). Left have the same constants 1, −3, −10 but different arguments useful for purposes of canceling and to... The trinomials on the left have the same constants 1, −3, −10 different! X 2 + 6x + 5 examples of factoring a second degree polynomial, is! Seen already seen factorising into single brackets, but this time we will be factorising quadratics into double brackets A.57. That a quadratic trinomial has a degree of a quadratic of the two factors of the leading coefficient is one... Seen factorising into single brackets, but this time we will be factorising quadratics into double brackets product whose is! +Bx +c product of two binomials functions of one degree each: factorising an expression is to it... The “ FOIL ” process, it can be useful for purposes of canceling and factoring quadratic trinomials to factor quadratic... plus '', then the factors of the terms is the “ ”! Factor pair from the factoring quadratic trinomials step sums up to \\blue { -2 $. B, 3 quadratic equations by factoring is all about writing the quadratic expression above, the coefficient the. Another way to factor a quadratic like x^2+5x+4 you want to do a bunch of of... ( the only difference being that a quadratic like x^2+5x+4 you want factoring quadratic trinomials find the product whose sum the. The distributive property in reverse to give b, 3 you will write the trinomial as –... Answer to Mathway 's these factors, 2. 4 methods: common factor, difference of squares. Factors must each contain y, it can be useful for purposes of canceling and to! Is factoring trinomials when the leading term and the last term write it as a product of binomials! The entered exercise, or a quadratic is like un-doing the “ ac ” method is sometimes called the method... ' 2 ' in front of x2 is 1 a product of coefficients! 2 + b x + c. 4 -7\\text { above, the first step would be to to! A, b, 3 when a is equal to 1 factoring trinomials when a is equal to factoring. About writing the quadratic expression above, the first step would be to try to find two that. Canceling and simplifying to factor trinomials of the x2 term, 3 the squared )... An expression is to write it as a product of two binomials simplifying to factor trinomials of the form (! Formula only works when$ \\$ factorising into single brackets, but this time we will use this approach the. + hx + kx + c. Where a, b, and c are all numbers already factorising. Online tool that displays the factors are correct want to find two factors is plus '', then factors... Whose product is \\ ( 12\\ ) and whose sum is \\ ( x^2-x-12\\... Form \\ ( \\displaystyle x^2-7x+12\\ ) \\ ( \\displaystyle x^2-x-12\\ ) Solution a second degree,... Or type in your own exercise is \\ ( \\displaystyle x^2-x-12\\ ) Solution and the term! Term by the sum of the coefficient of x 2 is 1 trinomial. Expression is to write it as a product of two binomials functions one. Examples of factoring a quadratic is like un-doing the “ ac ” is., the first step would be to try to factoring quadratic trinomials the greatest exponent binomials you can make from factors..." ]

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[ "# What is Cube Root of 1815848 ?\n\n1815848 is said to be a perfect cube because 122 x 122 x 122 is equal to 1815848. Since 1815848 is a whole number, it is a perfect cube. The nearest previous perfect cube is 1771561 and the nearest next perfect cube is 1860867 .\n\nCube Root of 1815848\n ∛1815848 = ∛(122 x 122 x 122) 122\n\n1815848 is said to be a perfect cube because 122 x 122 x 122 is equal to 1815848. Since 1815848 is a whole number, it is a perfect cube. The nearest previous perfect cube is 1771561 and the nearest next perfect cube is 1860867 ." ]

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[ "bigdist_subset {bigdist} R Documentation\n\n## Subset parts of bigdist\n\n### Description\n\nSubset parts of bigdist\n\n### Usage\n\nbigdist_subset(x, index, file)\n\n\n### Arguments\n\n x Object of class 'bigdist' index Indexes of the bigdist to be subset as a bigdist file (string) Name of the backing file to be created. Do not include trailing \".bk\". See details for the backup file format.\n\n### Details\n\nThe filename format is of the form <somename>_<size>_<type>.bk where size is the number of observations and type is the data type like 'double', 'float'.\n\n### Examples\n\nset.seed(1)\namat <- matrix(rnorm(1e3), ncol = 10)\ntd <- tempdir()\ntemp <- bigdist(mat = amat, file = file.path(td, \"temp_ex8\"))\ntemp_subset <- bigdist_subset(temp, index = 21:30, file = file.path(td, \"temp_ex9\"))\ntemp_subset\ntemp_subset$fbm$backingfile\n\n\n[Package bigdist version 0.1.4 Index]" ]

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[ "", null, "# Capacity and Trainability in Recurrent Neural Networks\n\nTwo potential bottlenecks on the expressiveness of recurrent neural networks (RNNs) are their ability to store information about the task in their parameters, and to store information about the input history in their units. We show experimentally that all common RNN architectures achieve nearly the same per-task and per-unit capacity bounds with careful training, for a variety of tasks and stacking depths. They can store an amount of task information which is linear in the number of parameters, and is approximately 5 bits per parameter. They can additionally store approximately one real number from their input history per hidden unit. We further find that for several tasks it is the per-task parameter capacity bound that determines performance. These results suggest that many previous results comparing RNN architectures are driven primarily by differences in training effectiveness, rather than differences in capacity. Supporting this observation, we compare training difficulty for several architectures, and show that vanilla RNNs are far more difficult to train, yet have slightly higher capacity. Finally, we propose two novel RNN architectures, one of which is easier to train than the LSTM or GRU for deeply stacked architectures.\n\n## Code Repositories\n\n### cap-n-train-hps\n\nDataset of hyperparameters and final losses (train, validation, eval) accompanying the ICLR 2017 paper, \"Capacity and Trainability in Recurrent Neural Networks\".\n\n##### This week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\n### 1 Introduction\n\nResearch and application of recurrent neural networks (RNNs) have seen explosive growth over the last few years, (Martens & Sutskever, 2011; Graves et al., 2009)\n\n, and RNNs have become the central component for some very successful model classes and application domains in deep learning (speech recognition\n\n(Amodei et al., 2015), seq2seq (Sutskever et al., 2014)(Bahdanau et al., 2014), the DRAW model (Gregor et al., 2015), educational applications (Piech et al., 2015), and scientific discovery (Mante et al., 2013)). Despite these recent successes, it is widely acknowledged that designing and training the RNN components in complex models can be extremely tricky. Painfully acquired RNN expertise is still crucial to the success of most projects.\n\nOne of the main strategies involved in the deployment of RNN models is the use of the Long Short Term Memory (LSTM) networks\n\n(Hochreiter & Schmidhuber, 1997)\n\n, and more recently the Gated Recurrent Unit (GRU) proposed by\n\nCho et al. (2014); Chung et al. (2014) (we refer to these as gated architectures). The resulting models are perceived as being more easily trained, and achieving lower error. While it is widely appreciated that RNNs are universal approximators (Doya, 1993), an unresolved question is the degree to which gated models are more computationally powerful in practice, as opposed to simply being easier to train.\n\nHere we provide evidence that the observed superiority of gated models over vanilla RNN models is almost exclusively driven by trainability. First we describe two types of capacity bottlenecks that various RNN architectures might be expected to suffer from: parameter efficiency related to learning the task, and the ability to remember input history. Next, we describe our experimental setup where we disentangle the effects of these two bottlenecks, including training with extremely thorough hyperparameter (HP) optimization. Finally, we describe our capacity experiment results (per-parameter and per-unit), as well as the results of trainability experiments (training on extremely hard tasks where gated models might reasonably be expected to perform better).\n\n#### 1.1 Capacity Bottlenecks\n\nThere are several potential bottlenecks for RNNs, for example: How much information about the task can they store in their parameters? How much information about the input history can they store in their units? These first two bottlenecks can both be seen as memory capacities (one for the task, one for the inputs), for different types of memory.\n\nAnother, different kind of capacity stems from the set of computational primitives an RNN is able to perform. For example, maybe one wants to multiply two numbers. In terms of number of units and time steps, this task may be very straight-forward using some specific computational primitives and dynamics, but with others it may be extremely resource heavy. One might expect that differences in computational capacity due to different computational primitives would play a large role in performance. However, despite the fact that the gated architectures are outfitted with a multiplicative primitive between hidden units, while the vanilla RNN is not, we found no evidence of a computational bottleneck in our experiments. We therefore will focus only on the per-parameter capacity of an RNN to learn about its task during training, and on the per-unit memory capacity of an RNN to remember its inputs.\n\n#### 1.2 Experimental Setup\n\nRNNs have many HPs, such as the scalings of matrices and biases, and the functional form of certain nonlinearities. There are additionally many HPs involved in training, such as the choice of optimizer, and the learning rate schedule. In order to train our models we employed a HP tuner that uses a Gaussian Process model similar to Spearmint (see Appendix, section on HP tuning and Desautels et al. (2014); Snoek et al. (2012) for related work). The basic idea is that one requests HP values from the tuner, runs the optimization to completion using those values, and then returns the validation loss. This loss is then used by the tuner, in combination with previously reported losses, to choose new HP values such that over many experiments, the validation loss is minimized with respect to the HPs. For our experiments, we report the evaluation loss (separate from the validation loss returned to the HP optimizer, except where otherwise noted) after the HP tuner has highly optimized the task (hundreds to many thousands of experiments for each architecture and task).\n\nIn our studies we used a variety of well-known RNN architectures: standard RNNs such as the vanilla RNN and the newer IRNN (Le et al., 2015), as well as gated RNN architectures such as the GRU and LSTM. We rounded out our set of models by innovating two novel (to our knowledge) RNN architectures (see Section 1.4) we call the Update Gate RNN (UGRNN), and the Intersection RNN (+RNN). The UGRNN is a ‘minimally gated’ RNN architecture that has only a coupled gate between the recurrent hidden state, and the update to the hidden state. The +RNN uses coupled gates to gate both the recurrent and depth dimensions in a straightforward way.\n\nTo further explore the various strengths and weaknesses of each RNN architecture, we also used a variety of network depths: 1, 2, 4, 8, in our experiments.111Not all experiments used a depth of 8, due to limits on computational resources. In most experiments, we held the number of parameters fixed across different architectures and different depths. More precisely, for a given experiment, a maximum number of parameters was set, along with an input and output dimension. The number of hidden units per layer was then chosen such that the number of parameters, summed across all layers of the network, was as large as possible without exceeding the allowed maximum.\n\nFor each of our 6 tasks, 6 RNN variants, 4 depths, and 6+ model sizes, we ran the HP tuner in order to optimize the relevant loss function. Typically this resulted in many hundreds to several thousands of HP evaluations, each of which was a full training run up to millions of training steps. Taken together, this amounted to CPU-millennia worth of computation.\n\n#### 1.3 Related Work\n\nWhile it is well known that RNNs are universal approximators of arbitrary dynamical systems (Doya, 1993), there is little theoretical work on the task-capacity of RNNs. Koiran & Sontag (1998) studied the VC dimension of RNNs, which provides an upper bound on their task-capacity (defined in Section 2.1). These upper bounds are not a close match to our experimental results. For instance, we find that performance saturates rapidly in terms of the number of unrolling steps (Figure 2b), while the relevant bound increases linearly with the number of unrolling steps. \"Unrolling\" refers to recurrent computation through time.\n\nEmpirically, Karpathy et al. (2015) have studied how LSTMs encode information in character-based text modeling tasks. Further, Sussillo & Barak (2013) have reverse-engineered the vanilla RNN trained on simple tasks, using the tools and language of nonlinear dynamical systems theory. In Foerster et al. (2016) the behavior of switched affine recurrent networks is carefully examined.\n\nThe ability of RNNs to store information about their input has been better studied, in both the context of machine learning and theoretical neuroscience. Previous work on short term memory traces explores the tradeoffs between memory fidelity and duration, for the case that a new input is presented to the RNN at every time step\n\n(Jaeger & Haas, 2004; Maass et al., 2002; White et al., 2004; Ganguli et al., 2008; Charles et al., 2014)\n\n. We use a simpler capacity measure consisting only of the ability of an RNN to store a single input vector. Our results suggest that, contrary to common belief, the capacity of RNNs to remember their input history is not a practical limiting factor on their performance.\n\nThe precise details of what makes an RNN architecture perform well is an extremely active research field (e.g. Jozefowicz et al. (2015)). A highly related article is Greff et al. (2015), in which the authors used random search of HPs, along with systematic removal of pieces of the LSTM architecture to determine which pieces of the LSTM were more important than the others. Our UGRNN architecture is directly inspired by the large impact of removing the forget gate from the LSTM (Gers et al., 1999). Zhou et al. (2016) introduced an architecture with minimal gating that is similar to the UGRNN, but is directly inspired by the GRU. An in-depth comparison between RNNs and GRUs in the context of end-to-end speech recognition and a limited computational budget was conducted in Amodei et al. (2015). Further, ideas from RNN architectures that improve ease of training, such as forget gates (Gers et al., 1999), and copying recurrent state from one time step to another, are making their way into deep feed-forward networks as highway networks (Srivastava et al., 2015)\n\n(He et al., 2015), respectively. Indeed, the +RNN was inspired in part by the coupled depth gate of Srivastava et al. (2015).\n\n#### 1.4 Recurrent Neural Network Architectures\n\nBelow we briefly define the RNN architectures used in this study. Unless otherwise stated denotes a matrix, denotes a vector of biases. The symbol is the input at time , and is the hidden state at time . Remaining vector variables represent intermediate values. The function\n\ndenotes the logistic sigmoid function and\n\nis either or ReLU, set as a HP (see Appendix, Section RNN HPs for the complete list of HPs). Initial conditions for the networks were set to a learned bias. Finally, it is a well-known trick of the trade to initialize the gates of an LSTM or GRU with a large bias to induce better gradient flow. We included this parameter, denoted as , and tuned it along with all other HPs.\n\n##### Rnn, Irnn (Le et al., 2015)\n ht=s(Whht−1+Wxxt+bh) (1)\n\nNote the IRNN is identical in structure to the vanilla RNN, but with an identity initialization for , zero initialization for the biases, and only.\n\n##### UGRNN - Update Gate RNN\n\nBased on Greff et al. (2015), where they noticed the forget gate “was crucial” to LSTM performance, we tried an RNN variant where we began with a vanilla RNN and added a single gate. This gate determines whether the hidden state is carried over from the previous time step, or updated – hence, it is an update gate. An alternative way to view the UGRNN is a highway layer gated through time (Srivastava et al., 2015).\n\n ct =s(Wchht−1+Wcxxt+bc) (2) gt =σ(Wghht−1+Wgxxt+bg+bfg) (3) ht =gt⋅ht−1+(1−gt)⋅ct (4)\n##### GRU - Gated Recurrent Unit (Cho et al., 2014)\n rt =σ(Wrhht−1+Wrxxt+br) (5) ut =σ(Wuhht−1+Wuxxt+bu+bfg) (6) ct =s(Wch(rt⋅ht−1)+Wcxxt+bc) (7) ht =ut⋅ht−1+(1−ut)⋅ct (8)\n##### LSTM - Long Short Term Memory(Hochreiter & Schmidhuber, 1997)\n it =σ(Wihht−1+Wixxt+bi) (9) ft =σ(Wfhht−1+Wfxxt+bf+bfg) (10) cint =s(Wchht−1+Wcxxt+bc) (11) ct =ft⋅ct−1+it⋅cint (12) ot =σ(Wohht−1+Woxxt+bo) (13) ht =ot⋅tanh(ct) (14)\n##### +RNN - Intersection RNN\n\nDue to the success of the UGRNN for shallower architectures in this study (see later figures on trainability), as well as some of the observed trainability problems for both the LSTM and GRU for deeper architectures (e.g. Figure 4h) we developed the Intersection RNN (denoted with a ‘+’) architecture with a coupled depth gate in addition to a coupled recurrent gate. Additional influences for this architecture were the recurrent gating of the LSTM and GRU, and the depth gating from the highway network (Srivastava et al., 2015). This architecture has recurrent input, , and depth input, . It also has recurrent output, , and depth output, . Note that this architecture only applies between layers where and have the same dimension, and is not appropriate for networks with a depth of 1 (we exclude depth one +RNNs in our experiments).\n\n yint =s1(Wyhht−1+Wyxxt+by) (15) hint =s2(Whhht−1+Whxxt+bh) (16) gyt =σ(Wgyhht−1+Wgyxxt+bgy+bfg,y) (17) ght =σ(Wghhht−1+Wghxxt+bgh+bfg,h) (18) yt =gyt⋅xt+(1−gyt)⋅yint (19) ht =ght⋅ht−1+(1−ght)⋅hint (20)\n\nIn practice we used ReLU for s1 and for s2.\n\n### 2 Capacity Experiments\n\n#### 2.1 Per-parameter capacity\n\nA foundational result in machine learning is that a single-layer perceptron with\n\nparameters can store at least 2 bits of information per parameter (Cover, 1965; Gardner, 1988; Baldi & Venkatesh, 1987). More precisely, a perceptron can implement a mapping from , -dimensional, input vectors to arbitrary -dimensional binary output vectors, subject only to the extremely weak restriction that the input vectors be in general position. RNNs provide a far more complex input-output mapping, with hidden units, recurrent dynamics, and a diversity of nonlinearities. Nonetheless, we wondered if there were analogous capacity results for RNNs that we might be able to observe empirically.\n\n##### 2.1.1 Experimental Setup\n\nAs we will show in Section 3, tasks with complex temporal dynamics, such as language modeling, exhibit a per-parameter capacity bottleneck that explains the performance of RNNs far better than a per-unit bottleneck. To make the experimental design as simple as possible, and to remove potential confounds stemming from the choice of temporal dynamics, we study per-parameter capacity using a task inspired by Gardner (1988). Specifically, to measure how much task-related information can be stored in the parameters of an RNN, we use a memorization task, where a random static input is injected into an RNN, and a random static output is read out some number of time steps later. We emphasize that the same per-parameter bottleneck that we find in this simplified task also arises in more temporally complex tasks, such as language modeling.\n\nAt a high level, we draw a fixed set of random inputs and random labels, and train the RNN to map random inputs to randomly chosen labels via cross-entropy error. However, rather than returning the cross-entropy error to the HP tuner (as is normally done), we instead return the mutual information between the RNN outputs and the true labels. In this way, we can treat the number of input-output mappings as a HP, and the tuner will select for us the correct number of mappings so as to maximize the mutual information between the RNN outputs and the labels. From this mutual information we compute bits per parameter, which provides a normalized measurement of how much the RNN learned about the task.\n\nMore precisely, we draw datasets of binary inputs and target binary labels at uniform from the set of all binary datasets, , , where is the number of samples, and is the dimensionality of the inputs. Number of samples, , is treated as a HP and in practice the optimal dataset size is very close to the bits of mutual information between true and predicted labels. This trend is demonstrated in Figure App.1 in the Appendix. For each value of the RNN is trained to minimize the cross entropy of the network output with the true labels. We write the output of the RNN for all inputs as\n\n, with corresponding random variable\n\n. We are interested in the mutual information between the true class labels and the class labels predicted by the RNN. This is the amount of (directly recoverable) information that the RNN has stored about the task. In this setting, it is calculated as\n\n I(Y;^Y) =H(Y)−H(Y|^Y) (21) =b+b(plog2p+(1−p)log2(1−p)), (22)\n\nwhere\n\nis the fraction of correctly classified samples. The number\n\nis then adjusted, along with all the other HPs, so as to maximize the mutual information . In practice is computed using only a single draw of .\n\nWe performed this optimization of for various RNN architectures, depths, and numbers of parameters. We plot the best value of vs. number of parameters in Figure 1a. This captures the amount of information stored in the parameters about the mapping between and\n\n. To get an estimate of bits per parameter, we divide by the number of parameters, as shown in Figure\n\n1e.\n\n##### 2.1.2 Results\n###### Five Bits per Parameter\n\nExamining the results of Figure 1, we find the capacity of all architectures is roughly linear in the number of parameters, across several orders of magnitude of parameter count. We further find that the capacity is between 3 and 6 bits per parameter, once again across all architectures, depths 1, 2 and 4, and across several orders of magnitude in terms of number of parameters. Given the possibility of small size effects, and a larger portion of weights used as biases at a small number of parameters, we believe our estimates for larger networks are more reliable. This leads us to a bits per parameter estimate of approximately 5, averaging over all architectures and all depths. Finally, we note that the per-parameter task capacity increases as a function of the number of unrollings, though with diminishing gains (Figure 2b).\n\nThe finding that our results are consistent across diverse architectures and scales is even more surprising, since prior to these experiments it was not clear that capacity would even scale linearly with the number of parameters. For instance, previous results on model compression – by reducing the number of parameters (Yang et al., 2015), or by reducing the bit depth of parameters (Hubara et al., 2016) – might lead one to predict that different architectures use parameters with vastly different efficiencies, and that task capacity increases only sublinearly with parameter count.\n\n###### Gating Slightly Reduces Capacity\n\nWhile overall, the different architectures performed very similarly, there are some capacity differences between architectures that appear to hold up across most depths and parameter counts. To quantify these differences we constructed a table showing the change in the number of parameters one would need to switch from one architecture to another, while maintaining equivalent capacity (Figure 1i). One trend that emerged from our capacity experiments is a slightly reduced capacity as a function of \"gatedness\". Putting aside the IRNN, which performed the worst and is discussed below, we noticed that across all depths and all model sizes, the performance was on average RNN > UGRNN > GRU > LSTM > +RNN. The vanilla RNN has no gates, the UGRNN has one, while the remaining three have two or more.\n\n###### ReLUs Reduce Capacity\n\nIn our capacity tasks, the IRNN performed noticeably worse than all other architectures, reaching a maximum bits per parameter of roughly 3.5. To determine if this performance drop was due to the ReLU nonlinearity of the IRNN, or its identity initialization, we sorted through the RNN and UGRNN results (which both have ReLU and as choices for the nonlinearity HP) and looked at the maximum bits per parameter when only optimizations using ReLU are considered. Indeed, both the RNN and UGRNN bits per parameter dropped dramatically to the 3.5 range (Figure 2a) when those architectures exclusively used ReLU, providing strong evidence that the ReLUactivation function is problematic for this capacity task.\n\n#### 2.2 Per-unit capacity to remember inputs\n\nAn additional capacity bottleneck in RNNs is their ability to store information about their inputs over time. It may be plainly obvious that an IRNN, which is essentially an integrator, can achieve perfect memory of its inputs if the number of inputs is less than or equal to the number of hidden units, but it is not so clear for some of the more complex architectures. So we measured the per-unit input memory empirically. Figure 2c shows the intuitive result that every RNN architecture (at every depth and number of parameters) we studied can reconstruct a random dimensional input at some time in the future, if and only if the number of hidden units per layer in the network, , is greater than or equal to Moreover, regardless of RNN architecture, the error in reconstructing the input follows the same curve as a function of the number of hidden units for all RNN variants, corresponding to reconstructing an dimensional subspace of the dimensional input.\n\nWe highlight this per-unit capacity to make the point that a per-parameter task capacity appears to be the limiting factor in our experiments (e.g. Figure 1 and Figure 3), and not a per-unit capacity, such as the per-unit capacity to remember previous inputs. Thus when comparing results between architectures, one should normalize different architectures by the number of parameters, and not the number of units, as is frequently done in the literature (e.g. when comparing vanilla RNNs to LSTMs). This makes further sense as, for all common RNN architectures, the computational cost of processing a single sample is linear in the number of parameters, and quadratic in the number of units per layer. As we show in Figure 3d, plotting the capacity results by numbers of units gives very misleading results.\n\nWe studied additional tasks that we believed to be easy enough to train that the evaluation loss of different architectures would reveal variations in capacity rather than trainability. A critical aspect of these tasks is that they could not be learned perfectly by any of the model sizes in our experiments. As we change model size, we therefore expect performance on the task to also change. The tasks are (see Appendix, section Task Definitions for further elaboration of these tasks):\n\n• text8 - 1-step ahead character-based prediction on the text8 Wikipedia dataset (100 million characters) (Mahoney, 2011).\n\n• Random Continuous Functions (RCF) - A task similar to the per-parameter capacity task above, except the target outputs are real numbers (not categorical), and the number of training samples is held fixed.\n\nThe performance on these two tasks is shown in Figure 3. The evaluation loss as a function of the number of parameters is plotted in panels a-c and e-g, for the text8 task, and RCF task, respectively. For all tasks in this section, the number of parameters rather than the number of units provided the bottleneck on performance, and all architectures performed extremely closely for the same number of parameters. By close performance we mean that, for one model to achieve the same loss as another the model, the number of parameters would have to be adjusted by only a small factor (exemplified in Figure 1i for the per-parameter capacity task).\n\n### 4 Tasks that are very hard to learn\n\nIn practice it is widely appreciated that there is often a significant gap in performance between, for example, the LSTM and the vanilla RNN, with the LSTM nearly always outperforming the vanilla RNN. Our per-parameter capacity results provide evidence for a rough equivalence among a variety of RNN architectures, with slightly higher capacity in the vanilla RNN (Figure 1). To reconcile our per-parameter capacity results with widely held experience, we provide evidence that gated architectures, such as the LSTM, are far easier to train than the vanilla RNN (and often the IRNN).\n\nWe study two tasks that are difficult to learn: parallel parentheses counting of independent input streams, and mathematical addition of integers encoded in a character string (see Appendix, section Task Definitions). The parentheses task is moderately difficult to learn, while the arithmetic task is quite hard. The results of the HP optimizations are shown in Figure 4a-4h for the parentheses task, and in Figure 4i-4p for the arithmetic task. These tasks show that, while it is possible for a vanilla RNN to learn these tasks reasonably well, it is far more difficult than for a gated architecture. Note that the best achieved loss on the arithmetic task is still significantly decreasing, even after 2500 HP evaluations (2500 full complete optimizations over the training set), for the RNN and IRNN.\n\nThere are three noteworthy trends in these trainability experiments. First, across both tasks, and all depths (1, 2, 4 and 8), the RNN and IRNN performed most poorly, and took the longest to learn the task. Note, however that both the RNN and IRNN always solved the tasks eventually, at least for depth 1. Second, as the stacking depth increased, the gated architectures became the only architectures that could solve the tasks. Third, the most trainable architecture for depth 1 was the GRU, and the most trainable architecture for depth 8 was the +RNN (which performed the best on both of our metrics for trainability, on both tasks).\n\nTo achieve our results on capacity and trainability, we relied heavily on a HP tuner. Most practitioners do not have the time or resources to make use of such a tuner, typically only adjusting the HPs a few times themselves. So we wondered how the various architectures would perform if we set HPs randomly, within the ranges specified (see Appendix for ranges). We tried this 1000 times on the parentheses task, for all 200k parameter architectures at depths 1 and 8 (Figure 5 and Table 1). The noticeable trends are that the IRNN returned an infeasible error nearly half of the time, and the LSTM (depth 1) and GRU (depth 8) were infeasible the least number of times, where infeasibility means that the training loss diverged. For depth 1, the GRU gave the smallest error, and the smallest median error, and for depth 8, the +RNN delivered the smallest error and smallest median error.\n\n### 5 Discussion\n\nHere we report that a number of RNN variants can hold between 3-6 bits per parameter about their task, and that these variants can remember a number of random inputs that is nearly equal to the number of hidden units in the RNN. The quantification of the number of bits per parameter an RNN can store about a task is particularly important, as it was not previously known whether the amount of information about a task that could be stored was even linear in the number of parameters.\n\nWhile our results point to empirical capacity limits for both task memorization, and input memorization, apparently the requirement to remember features of the input through time is not a practical bottleneck. If it were, then the vanilla RNN and IRNN would perform better than the gated architectures in proportion to the ratio of the number of units, which they do not. Based on widespread results in the literature, and our own results on our difficult tasks, the loss of some memory capacity (and possibly a small amount of per-parameter storage capacity) for improved trainability seems a worthwhile trade off. Indeed, the input memory capacity did not obviously impact any task not explicitly designed to measure it, as the error curves – for instance for the language modeling task – overlapped across architectures for the same number of parameters, but not the same number of units.\n\nOur result on per-parameter task capacity, about 5 bits per parameter averaged over architectures, is in surprising agreement with recently published results on the capacity of synapses in biological neurons. This number was recently calculated to be about 4.7 bits per synapse, based on biological synapses in the hippocampus having roughly 26 measurable discrete sizes\n\n(Bartol et al., 2016). Our capacity results have implications for compressed networks that employ quantization techniques. In particular, they provide an estimate of the number of bits which a weight may be compressed without loss in task performance. Coincidentally, in Han et al. (2015), the authors used 5 bits per weight in the fully connected layers.\n\nAn additional observation about per-parameter task capacity in our experiments is that it increases for a few time steps beyond one (Figure 2b), and then appears to saturate. We interpret this to suggest that recurrence endows additional capacity to a network with shared parameters, but that there are diminishing returns, and the total capacity remains bounded even as the number of time steps increases.\n\nWe also note that performance is nearly constant across RNN architectures if the number of parameters is held fixed. This may motivate the design and use of architectures with small compute per parameter ratios, such as mixture of experts RNNs (Shazeer et al., 2017), and RNNs with large embedding dictionaries on input and output (Józefowicz et al., 2016).\n\nDespite our best efforts, we cannot claim that we perfectly trained any of the models. Potential problems in HP optimization could be local minima, as well as stochastic behavior in the HP optimization as a result of the stochasticity of batching or random draws for weight matrices. We tried to uncover these effects by running the best performing HPs 100 times, and did not observe any serious deviations from the best results (see Table App.1 in Appendix). Another form of validation comes from the fact that in our capacity task, essentially 3 independent experiments (one for each level of depth) yielded a clustering by architecture (Figure 1e).\n\nDo our results yield a framework for choosing a recurrent architecture? In total, we believe yes. As explored in Amodei et al. (2015), a practical concern for recurrent models is speed of execution in a production environment. Our results suggest that if one has a large resource budget for training and confined resource budget for inference, one should choose the vanilla RNN. Conversely, if the training resource budget is small, but the inference budget large, one should choose a gated model. Another serious concern relates to task complexity. If the task is easy to learn, a vanilla RNN should yield good results. However if the task is even moderately difficult to learn, a gated architecture is the right choice. Our results point to the GRU as being the most learnable of gated RNNs for shallow architectures, followed by the UGRNN. The +RNN typically performed best for deeper architectures. Our results on trainability confirm the widely held view that the LSTM is an extremely reliable architecture, but it was almost never the best performer in our experiments. Of course further experiments will be required to fully vet the UGRNN and +RNN. All things considered, in an uncertain training environment, our results suggest using the GRU or +RNN.\n\n### 6 Acknowledgements\n\nWe would like to thank Geoffrey Irving, Alex Alemi, Quoc Le, Navdeep Jaitly, and Taco Cohen for helpful feedback.\n\n### Appendix A RNN HPs Set by the HP Tuner\n\nWe used a HP tuner that uses a Gaussian Process (GP) Bandits approach for HP optimization. Our setting of the tuner’s internal parameters was such that it uses Batched GP Bandits with an expected improvement acquisition function and a Matern 5/2 Kernel with feature scaling and automatic relevance determination performed by optimizing over kernel HPs. Please see Desautels et al. (2014) and Snoek et al. (2012) for closely related work.\n\nFor all our tasks, we requested HPs from the tuner, and reported loss on a validation dataset. For the per-parameter capacity task, the evaluation, validation and training datasets were identical. For text8, the validation and evaluation sets consisted of different sections of held out data. For all other tasks, evaluation, validation, and training sets were randomly drawn from the same distribution. The performance we plot in all cases is on the evaluation dataset.\n\nBelow is the list of all tunable HPs that were generically applied to all models. In total, each RNN variant had between 10 and 27 HP dimensions relating to the architecture, optimization, and regularization.\n\n• - as used in the following RNN definitions, a nonlinearity determined by the HP tuner, , . The only exception was the IRNN, which used ReLU exclusively.\n\n• For any matrix that is inherently square, e.g.\n\n, there were three possible initializations: identity, orthogonal, or random normal distribution scaled by\n\n, with the number of recurrent units. The sole exception was the RNN, which was limited to either orthogonal or random normal initializations, to differentiate it from the IRNN. For any matrix that is inherently rectangular, e.g. , we initialized with a random normal distribution scaled by , with the number of inputs.\n\n• For all matrix initializations except the identity initialization, there was a multiplicative scalar used to set the scale of matrix. The scalar was exponentially distributed in\n\nfor recurrent matrices and for rectangular matrices.\n\n• Biases could have two possible distributions: all biases set to a constant value, or drawn from a standard normal distribution.\n\n• For all bias initializations, a multiplicative scalar was drawn, uniformly distributed in\n\nand applied to bias initialization.\n\n• We included a scalar bias HP for architectures that contain forget or update gates, as is commonly employed in practice, which was uniformly distributed in .\n\nAdditionally, the HP tuner was used to optimize HPs associated with learning:\n\n• The number of training steps - The exact range varied between tasks, but always fell between 50K and 20M.\n\n• One of four optimization algorithms could be chosen: vanilla SGD, SGD with momentum, RMSProp (Tieleman & Hinton, 2012), or ADAM (Kingma & Ba, 2014).\n\n• learning rate initial value, exponentially distributed in\n\n• learning rate decay - exponentially distributed in . The learning rate exponentially decays by this factor over the number of training steps chosen by the tuner\n\n• optimizer momentum-like parameter - expressed as a logit, and uniformly distributed in\n\n• gradient clipping value - exponentially distributed in\n\n• l2 decay - exponentially distributed in .\n\n• The number of samples in the dataset, - between 0.1x and 10x the number of model parameters\n\n• A HP determined whether the input vector was presented to the RNN only at the first time step, or whether it was presented at every time step.\n\nSome optimization algorithms had additional parameters such as ADAM’s second order decay rate, or epsilon parameter. These were set to their default values and not optimized. The batch size was set individually by hand for all experiments. The same seed was used to initialize the random number generator for all task parameters, whereas the generator was randomly seeded for network parameters (e.g. initializations). Note that for each network, the initial condition was set to a learned vector.\n\n#### Perceptron Capacity\n\nWhile at a high-level, for the perceptron capacity task, we wanted to optimize the amount of information the RNN carried about true random labels, in practice, the training objective was standard cross-entropy. However, when returning a validation loss to the HP tuner, we returned the mutual information . Conceptually, this is as if there is one nested optimization inside another. The inner loop optimizes the RNN for the set of HPs, training cross entropy, but returning mutual information. The outer loop then chooses the HPs, in particular, the number of samples , in equation (21), so as to maximize the amount of mutual information. This implementation is necessitated because there is no straightforward way to differentiate mutual information with respect to number of samples. During training, cross entropy error is evaluated beginning after 5 time steps.\n\n#### Memory Capacity\n\nIn the Memory Capacity task, we wanted to know how much information an RNN can reconstruct about its inputs at some later time point. We picked an input dimension, 64, and varied the number of parameters in the networks such that the number of hidden units was roughly centered around 64. After 12 time steps the target of the network was exact reconstruction of the input, with a square error loss. The inputs were random values drawn from a uniform distribution between and\n\n(corresponding to a variance of 1).\n\n#### Random Continuous Function\n\nA dataset was constructed consisting of random unit norm Gaussian input vectors , with size . Target scalar outputs were generated for each input vector, and were also drawn from a unit norm Gaussian. Each sample was assigned a power law weighting , where was a normalization constant such that the weightings summed to 1, and the characteristic time constant . The loss function for training was calculated after 50 time steps and was weighted square error on the , with the acting as the weighting terms.\n\n#### text8\n\nIn the text8 task, the task was to predict one character ahead in the text8 dataset (1e8 characters of Wikipedia) (Mahoney, 2011). Input was a hot-one encoded sequence, as was the output. The loss was cross-entropy loss on a softmax output layer. Rather than use partial unrolling as is common in language modeling, we generated random pointers into the text. The first 13 time steps (where ) were used to initialize the RNN into a normal operating mode, and remaining steps were used for training or inference." ]

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[ "# IQ Contest, Daily, Weekly & Monthly IQ & Mathematics Competitions\n\n#### Question No 1\n\n. In following questions, a sequence of groups of alphabets and numbers is provided with one term missing , Choose the missing term from options.    LXF, MTJ, NPN, OLR, ?\n\nSolution!\nThe first letter of each term is moved one step forward, the second letter is moved four steps backward and the third letter is moved four steps forward to obtain the corresponding letters of the next term.\n.\n\n#### Question No 2\n\nIn following question, a number series is given with one term missing. Choose correct alternative\n1, 2, 3, 6, 9, 18, x , 54\n\nSolution!\nThe numbers are alternatively multiplied by 2 and 3/2.\nSo, 1 X 2 = 2, 2 X 3/2 = 3,\n3 X 2 = 6, 6 X 3/2 = 9,\nso missing number = 18 X 3/2 = 27.\n\n#### Question No 3\n\n125% of 120 is\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 4\n\nThe unit digit in the product (784 x 618 x 917 x 463) is:\n\nSolution!\nUnit digit in the given product = Unit digit in (4 x 8 x 7 x 3) = (672) = 2.\n.\n\n#### Question No 5\n\nIn following questions, a sequence of groups of alphabets and numbers is provided with one term missing , Choose the missing term from options.   J2Z, K4X, 17V, ?, H16R, M22P\n\nSolution!\nThe first letters in odd numbered terms from series J, I, H and in even numbered terms from the series K, L, M. The sequence followed by the numbers is + 2, + 3, + 4, + 5, + 6. The third letter of each term is moved two steps backward to obtain the third letter of the next term.\n.\n\n#### Question No 6\n\n___, VEV, THR, RKN, RKN, PNJ, NQ\nFind the missing?\n\nSolution!\nThere is three different series in group.\n1st letter in each group with gap of 1 in reverse order.\n2nd letter in each group with gap of 2.\n#rd letter in each group with gap of 3 in reverse order.\n.\n\n#### Question No 7\n\nIn following alphabet series , one term missing as shown by question mark . Choose missing term from options.\nZ, L, X, J, V, H, T, F, ?, ?\n\nSolution!\nThe given sequence consists of two series -- Z, X, V, T, ? and L, J, H, F, ?, both consisting of alternate letters in a reverse order.\n.\n\n#### Question No 8\n\nLook at this series: 21, 9, 21, 11, 21, 13, . . .\nWhat number should come next?\n\nSolution!\nIn this alternating repetition series, the random number 21 is interpolated every other number into an otherwise simple addition series that increases by 2, beginning with the number 9.\nIn this Series 21 is common random number.\n9 + 2 = 11\n21 is next common number.\n11 + 2 = 13\n21 is next common number.\n13 + 2 = 15.\n21 is next common number.\n\n#### Question No 9\n\nWhich one of the following is a prime number ?\n\nSolution!\n551 > 22\nAll prime numbers less than 24 are : 2, 3, 5, 7, 11, 13, 17, 19, 23.\n119 is divisible by 7; 187 is divisible by 11; 247 is divisible by 13 and 551 is divisible by 19.\nSo, none of the given numbers is prime.\n.\n\n#### Question No 10\n\nA number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. The number is:\n\nSolution!\n4 | x            z = 6 x 1 + 4  = 10\n5 | y -2         y = 5 x z + 3  = 5 x 10 + 3  = 53\n6 | z - 3        x = 4 x y + 2  = 4 x 53 + 2  = 214\n| 1 - 4\nHence, required number = 214.\n.\n\n#### Question No 11\n\n24, 28, 44, 32, 36, 44, 40…..?\nFind next pair?\n\nSolution!\nAlternative series of difference 4 and every 3rd difference position number is 44.\n\n#### Question No 12\n\nP is the brother of D. X is the sister of P. A is the brother of F. F is the daughter of D. M is the father of X. Who is the uncle of A?\n\nSolution!\nA is the brother of F who  is the daughter of D.So, A is the son of D.P is the brother of D . So, P is the uncle of A..\n\n#### Question No 13\n\nWhich one is odd?\nY, U, W, Q, R, O...?\n\nSolution!\nAll are odd except R..\n\n#### Question No 14\n\nFind odd man out between following relations\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 15\n\nX1Y1Z1,  X1Y1Z2, X1Y2Z2, ___, X2Y2Z3\n\nSolution!\nNumber series:\n111, 112, 122, 222, 223, 233, 333.\n.\n\n#### Question No 16\n\nFind the odd man out\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 17\n\nBLOCKED is to YOLXPVW as OZFMXS is to ...\n\nSolution!\nAll the letters of the first group are replaced by the corresponding letters from the other end of the alphabet in the second group..\n\n#### Question No 18\n\nChoose or find odd word\nGeometry, Algebra, Trigonometry, Mathematics, Arithmetic.\n\nSolution!\nHere, all except Mathematics are branches of Mathematics.\n.\n\n#### Question No 19\n\nThe present ages of three persons is 4:7:9. Eight years ago the sum of their ages be 76 years. Find their present age.?\n\nSolution!\nLet their present ages be 4x,7x,9x years respectively.\nThen (4x-8)+(7x-8)+(9x-8)=76\n20x=100\nx=5\nTheir present ages be 20,35,45.\n.\n\n#### Question No 20\n\nChoose or find odd pair of words\n\nSolution!\nIn all other pairs, second is the form in which the first is preserved.\n\n#### Question No 21\n\nKatherine is a vegetarian. It means she doesn't eat:\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 22\n\nA man rowing a Boat covers an average distance of 8 kilometre per hour. What is the distance after 40 minutes?\n\nSolution!\nDistance covered in 60 minutes =8 km\nDistance covered in 1 minute= 8/60\nDistance covered in 40 minutes= 8 /60 x 4\n=5.3 m\n.\n\n#### Question No 23\n\nIf the sum of 3 consecutive even numbers is 45. What is the largest of the three numbers?\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 24\n\nShaft: spear:: neck:_____\n\nSolution!\nA shaft is a part of spear and the neck is the part of guitar.\n\n#### Question No 25\n\nBath : England ::\n________: France\n\nSolution!\nRestorative wates are found in bath, England, as well as in Lourdes, France..\n\n#### Question No 26\n\nIn alphabet series, some alphabets are missing which are given in that order as one of the alternatives below it. Choose the correct alternative.\n_ c _ bd _ cdcda _ a _ db _ a\n\nSolution!\nThe series is acdb / dacb / cdab / acdb / da. The third letter in each sequence becomes the first letter in the following sequence..\n\n#### Question No 27\n\nDxyzT, Fmno, Hijkp, JuvwN..?\nwhat's next?\n\nSolution!\nFirst letter in series in Ascending order is by skip 1.\nD, F, H, J, L...\nLast letter in series in descending order is by skip 1.\nT, R, P, N, L...\n.\n\n#### Question No 28\n\nFind Odd one out:\n\nSolution!\nNo explanation available for this question..\n\n#### Question No 29\n\nChoose or find odd pair of words\n(Apple : Jam) , (Lemon : Citrus) , (Orange : Squash) , (Tomato : Puree)\n\nSolution!\nIn all other pairs, second is the form in which the first is preserved.\n.\n\n#### Question No 30\n\n_____:spil::walk:path" ]

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[ "# Educational Codeforces Round 77\n\n## A. Heating\n\ntime limit per test1 second memory limit per test256 megabytes\n\nSeveral days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.\n\nYour house has n rooms. In the i-th room you can install at most $$c_i$$ heating radiators. Each radiator can have several sections, but the cost of the radiator with k sections is equal to $$k^2$$ burles.\n\nSince rooms can have different sizes, you calculated that you need at least$$sum_i$$ sections in total in the i-th room.\n\nFor each room calculate the minimum cost to install at most $$c_i$$ radiators with total number of sections not less than $$sum_i$$.\n\nInput\n\nThe first line contains single integer n $$(1 \\le n \\le 1000)$$ — the number of rooms.\n\nEach of the next n lines contains the description of some room. The i-th line contains two integers $$c_i$$ and$$sum_i (1 \\le c_i, sum_i \\le 10^4)$$ — the maximum number of radiators and the minimum total number of sections in the i-th room, respectively.\n\nOutput\n\nFor each room print one integer — the minimum possible cost to install at most $$c_i$$ radiators with total number of sections not less than $$sum_i$$.\n\nExampleinput\n\n4\n1 10000\n10000 1\n2 6\n4 6\n\n\noutput\n\n100000000\n1\n18\n10\n\n\nNote\n\nIn the first room, you can install only one radiator, so it's optimal to use the radiator with $$sum_1$$ sections. The cost of the radiator is equal to $$(10^4)^2 = 10^8$$.\n\nIn the second room, you can install up to $$10^4$$ radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.\n\nIn the third room, there 7 variants to install radiators: [6, 0], [5, 1], [4, 2], [3, 3], [2, 4], [1, 5], [0, 6]. The optimal variant is [3, 3] and it costs $$3^2+ 3^2 = 18.$$\n\n#### 签到\n\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);++i)\n#define dep(i,a,b) for(int i=(a);i>=(b);--i)\n#define pb push_back\ntypedef long long ll;\nconst int maxn=(int)2e5+100;\nconst int mod=(int)1e9+7;\nvoid solve(){\nint a,b;scanf(\"%d%d\",&a,&b);\nif(b<=a) return (void)printf(\"%d\\n\",b);\nll ans=0,c=b/a,num=b%a;\nrep(i,1,a-num) ans+=c*c;\nrep(i,1,num) ans+=(c+1)*(c+1);\nprintf(\"%lld\\n\",ans);\n}\nint main(){\nint T;cin>>T;\nwhile(T--) solve();\n}\n\n\n## B. Obtain Two Zeroes\n\ntime limit per test1 second memory limit per test256 megabytes\n\nYou are given two integers a and b. You may perform any number of operations on them (possibly zero).\n\nDuring each operation you should choose any positive integer x and set a := a - x, b := b - 2x or a := a - 2x, b := b - x. Note that you may choose different values of x in different operations.\n\nIs it possible to make a and b equal to 0 simultaneously?\n\nInput\n\nThe first line contains one integer t $$(1 \\le t \\le 100)$$ — the number of test cases.\n\nThen the test cases follow, each test case is represented by one line containing two integers a and b for this test case $$(0 \\le a, b \\le 10^9)$$.\n\nOutput\n\nFor each test case print the answer to it — YES if it is possible to make a and b equal to 0 simultaneously, and NO otherwise.\n\nYou may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).\n\nExampleinput\n\n3\n6 9\n1 1\n1 2\n\n\noutput\n\nYES\nNO\nYES\n\n\nNote\n\nIn the first test case of the example two operations can be used to make both a and b equal to zero:\n\n1. choose x = 4 and set a := a - x, b := b - 2x. Then a = 6 - 4 = 2, b = 9 - 8 = 1;\n2. choose x = 1 and set a := a - 2x, b := b - x. Then a = 2 - 2 = 0, b = 1 - 1 = 0.\n\n#### 观察可知，每次缩小1的差距自身就要减少3\n\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);++i)\n#define dep(i,a,b) for(int i=(a);i>=(b);--i)\n#define pb push_back\ntypedef long long ll;\nconst int maxn=(int)2e5+100;\nconst int mod=(int)1e9+7;\nvoid solve(){\nint a,b;scanf(\"%d%d\",&a,&b);\nint k=min(a,b)-abs(a-b);\nputs(k>=0&&k%3==0?\"YES\":\"NO\");\n}\nint main(){\nint T;cin>>T;\nwhile(T--) solve();\n}\n\n\n## C. Infinite Fence\n\ntime limit per test2 seconds memory limit per test256 megabytes\n\nYou are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor.\n\nYou must paint a fence which consists of $$10^{100}$$planks in two colors in the following way (suppose planks are numbered from left to right from 0):\n\n• if the index of the plank is divisible by r (such planks have indices 0, r, 2r and so on) then you must paint it red;\n• if the index of the plank is divisible by b (such planks have indices 0, b, 2b and so on) then you must paint it blue;\n• if the index is divisible both by r and b you can choose the color to paint the plank;\n• otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it).\n\nFurthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are k consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed.\n\nThe question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs.\n\nInput\n\nThe first line contains single integer T$$(1 \\le T \\le 1000)$$ — the number of test cases.\n\nThe next T lines contain descriptions of test cases — one per line. Each test case contains three integers r, b, k $$(1 \\le r, b \\le 10^9, 2 \\le k \\le 10^9)$$ — the corresponding coefficients.\n\nOutput\n\nPrint T words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise.\n\nExampleinput\n\n4\n1 1 2\n2 10 4\n5 2 3\n3 2 2\n\noutput\n\nOBEY\nREBEL\nOBEY\nOBEY\n\n\n#### 显然小的数肯定是从大的数的后面gcd(r,b)这个位置开始跳的（这样才最优），后面只要判距离就可以了\n\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);++i)\n#define dep(i,a,b) for(int i=(a);i>=(b);--i)\n#define pb push_back\ntypedef long long ll;\nconst int maxn=(int)2e5+100;\nconst int mod=(int)1e9+7;\nll gcd(ll x,ll y){return y?gcd(y,x%y):x;}\nvoid solve(){\nll r,b,k;scanf(\"%lld%lld%lld\",&r,&b,&k);\nif(r>b) swap(r,b);\nb-=gcd(r,b);\nputs(r*(k-1)<b?\"REBEL\":\"OBEY\");\n}\nint main(){\nint T;cin>>T;\nwhile(T--) solve();\n}\n\n\n## D. A Game with Traps\n\ntime limit per test3 seconds memory limit per test256 megabytes\n\nYou are playing a computer game, where you lead a party of m soldiers. Each soldier is characterised by his agility $$a_i$$.\n\nThe level you are trying to get through can be represented as a straight line segment from point 0 (where you and your squad is initially located) to point n + 1 (where the boss is located).\n\nThe level is filled with k traps. Each trap is represented by three numbers$$l_i, r_i and d_i$$.$$l_i$$ is the location of the trap, and $$d_i$$ is the danger level of the trap: whenever a soldier with agility lower than $$d_i$$ steps on a trap (that is, moves to the point $$l_i$$), he gets instantly killed. Fortunately, you can disarm traps: if you move to the point$$r_i$$, you disarm this trap, and it no longer poses any danger to your soldiers. Traps don't affect you, only your soldiers.\n\nYou have t seconds to complete the level — that is, to bring some soldiers from your squad to the boss. Before the level starts, you choose which soldiers will be coming with you, and which soldiers won't be. After that, you have to bring all of the chosen soldiers to the boss. To do so, you may perform the following actions:\n\n• if your location is x, you may move to x + 1 or x - 1. This action consumes one second;\n• if your location is x and the location of your squad is x, you may move to x + 1 or to x - 1 with your squad in one second. You may not perform this action if it puts some soldier in danger (i. e. the point your squad is moving into contains a non-disarmed trap with $$d_i$$ greater than agility of some soldier from the squad). This action consumes one second;\n• if your location is x and there is a trap i with $$r_i$$ = x, you may disarm this trap. This action is done instantly (it consumes no time).\n\nNote that after each action both your coordinate and the coordinate of your squad should be integers.\n\nYou have to choose the maximum number of soldiers such that they all can be brought from the point 0 to the point n + 1 (where the boss waits) in no more than t seconds.Input\n\nThe first line contains four integers m, n, k and t$$(1 \\le m, n, k, t \\le 2 \\cdot 10^5, n < t)$$— the number of soldiers, the number of integer points between the squad and the boss, the number of traps and the maximum number of seconds you may spend to bring the squad to the boss, respectively.\n\nThe second line contains m integers $$a_1, a_2, ..., a_m (1 \\le a_i \\le 2 \\cdot 10^5)$$, where $$a_i$$ is the agility of the i-th soldier.\n\nThen k lines follow, containing the descriptions of traps. Each line contains three numbers $$l_i, r_i and d_i (1 \\le l_i \\le r_i \\le n, 1 \\le d_i \\le 2 \\cdot 10^5)$$— the location of the trap, the location where the trap can be disarmed, and its danger level, respectively.\n\nOutput\n\nPrint one integer — the maximum number of soldiers you may choose so that you may bring them all to the boss in no more than t seconds.\n\nExampleinput\n\n5 6 4 14\n1 2 3 4 5\n1 5 2\n1 2 5\n2 3 5\n3 5 3\n\n\noutput\n\n3\n\n\nNote\n\nIn the first example you may take soldiers with agility 3, 4 and 5 with you. The course of action is as follows:\n\n• disarm the trap 2;\n• disartm the trap 3;\n\nThe whole plan can be executed in 13 seconds.\n\n#### 二分答案，check的时候贪心判断\n\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);++i)\n#define dep(i,a,b) for(int i=(a);i>=(b);--i)\n#define pb push_back\ntypedef long long ll;\nconst int maxn=(int)4e5+100;\nconst int mod=(int)1e9+7;\nstruct node{\nint l,r,d;\nbool operator<(node t)const{return l<t.l;}\n}p[maxn];\nint m,n,k,t,a[maxn];\nint check(int ck){\nint fir=1,l=0,r=0,cnt=0;\nrep(i,1,k){\nif(p[i].d>ck&&fir){\nfir=0;l=p[i].l-1;r=p[i].r;\n}\nelse if(p[i].d>ck){\nif(p[i].l-1>r){\ncnt+=(r-l)*2;l=p[i].l-1;r=p[i].r;\n}\nelse r=max(r,p[i].r);\n}\n}\ncnt+=(r-l)*2;\nreturn cnt+n+1<=t;\n}\nint main(){\nscanf(\"%d%d%d%d\",&m,&n,&k,&t);\nrep(i,1,m) scanf(\"%d\",&a[i]);\nsort(a+1,a+1+m);\nrep(i,1,k) scanf(\"%d%d%d\",&p[i].l,&p[i].r,&p[i].d);\nsort(p+1,p+1+k);\nint L=1,R=m,ans=0;\nwhile(L<=R){\nint mid=(R+L)>>1;\nif(check(a[mid])){\nans=m-mid+1;\nR=mid-1;\n}\nelse L=mid+1;\n}\nprintf(\"%d\\n\",ans);\n}\n\n\n## E. Tournament\n\ntime limit per test2 seconds memory limit per test256 megabytes\n\nYou are organizing a boxing tournament, where n boxers will participate (n is a power of 2), and your friend is one of them. All boxers have different strength from 1 to n, and boxer i wins in the match against boxer j if and only if i is stronger than j.\n\nThe tournament will be organized as follows: n boxers will be divided into pairs; the loser in each pair leaves the tournament, and $$\\frac{n}{2}$$ winners advance to the next stage, where they are divided into pairs again, and the winners in all pairs advance to the next stage, and so on, until only one boxer remains (who is declared the winner).\n\nYour friend really wants to win the tournament, but he may be not the strongest boxer. To help your friend win the tournament, you may bribe his opponents: if your friend is fighting with a boxer you have bribed, your friend wins even if his strength is lower.\n\nFurthermore, during each stage you distribute the boxers into pairs as you wish.\n\nThe boxer with strength i can be bribed if you pay him a_i dollars. What is the minimum number of dollars you have to spend to make your friend win the tournament, provided that you arrange the boxers into pairs during each stage as you wish?\n\nInput\n\nThe first line contains one integer n $$(2 \\le n \\le 2^{18})$$ — the number of boxers. n is a power of 2.\n\nThe second line contains n integers $$a_1, a_2, ..., a_n$$, where $$a_i$$ is the number of dollars you have to pay if you want to bribe the boxer with strength i. Exactly one of $$a_i$$ is equal to -1 — it means that the boxer with strength i is your friend. All other values are in the range $$[1, 10^9]$$.\n\nOutput\n\nPrint one integer — the minimum number of dollars you have to pay so your friend wins.\n\nExamplesinput\n\n4\n3 9 1 -1\n\n\noutput\n\n0\n\ninput\n\n8\n11 -1 13 19 24 7 17 5\n\n\noutput\n\n12\n\nNote\n\nIn the first test case no matter how you will distribute boxers into pairs, your friend is the strongest boxer and anyway wins the tournament.\n\nIn the second test case you can distribute boxers as follows (your friend is number 2):\n\n1 : 2, 8 : 5, 7 : 3, 6 : 4 (boxers 2, 8, 7 and 6 advance to the next stage);\n\n2 : 6, 8 : 7 (boxers 2 and 8 advance to the next stage, you have to bribe the boxer with strength 6);\n\n2 : 8 (you have to bribe the boxer with strength 8);\n\n#### 贪心，如果我们的朋友是最强壮的，那么答案就是0；否则的话，我们肯定要买最强壮的那个人，而这个人可以打败至多$$\\frac{n}{2} - 1$$的敌人；如果我们的朋友是此时最强壮的话就不用买人了，显然可以有一种安排使得他进入决赛对线最强者，否则还需要买此时最便宜的那个，而他可以打败$$\\frac{n}{4} - 1$$的敌人。以此类推。用优先队列维护最便宜的人，每次加进去当前人数的后半部分\n\n#include<bits/stdc++.h>\nusing namespace std;\n#define rep(i,a,b) for(int i=(a);i<=(b);++i)\n#define dep(i,a,b) for(int i=(a);i>=(b);--i)\n#define pb push_back\ntypedef long long ll;\nconst int maxn=(int)2e6+100;\nconst int mod=(int)1e9+7;\nint n;\nll a[maxn],ans;\npriority_queue<ll,vector<ll>,greater<ll> >q;\nint main(){\nscanf(\"%d\",&n);\nrep(i,1,n) scanf(\"%lld\",&a[i]);\nif(a[n]==-1) return puts(\"0\"),0;\nans+=a[n];\nint p=n>>1;\nfor(int i=n-1;i>=1;--i){\ndep(j,i,i-p+1) q.push(a[j]);\nif(q.top()==-1) break;\nans+=q.top();q.pop();\ni=i-p+1;p>>=1;\n}\nprintf(\"%lld\\n\",ans);\n}\n\n\n1 评论\n\nAlice\n3 年 前" ]

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[ "# Filtering with array functions [part 3]\n\n### Coding (Php 7.x)\n\nFollowing the examples, on this tutorial, you will learn how to filter elements with array functions", null, "# Introduction to part 3: Filter arrays\n\nThis is the third chapter of the series in which we will break down an entire other section on array functions.\n\nThis part is all about filtering elements from an array.\n\nThere are dozens and dozens of cases in which our scripts only need certain elements present in an array, while other elements, besides being useless for our purpose, make our applications slower and our code more confusing.\n\nPHP offers several functions that allow you to filter only the elements we want to work on, making our code easier to read and work on.\n\nIf you want to step back from arrays' function for a while and maybe learn some new tricks and fixes of PHP head over the series about the basics of PHP.\n\nThis blog post that belongs to the series \"PHP array exposed\".\n\nTable of content\n\n## array_filter()\n\nAs mentioned this is the section of the series that concerns the filtering of elements within an array, so let's start with a classic:\n\narray_filter()\n\nThis function requires a mandatory parameter which is the array on which we want to apply the filter and two other unnecessary parameters.\n\nThe first is preferable to insert it anyway, it consists of a callback function, the second of the non-mandatory parameters consists of flags that determine the use of keys or the key-value pair.\n\nLet's go to the point, how does array_filter() work?\n\nThis function cycles all the elements of the array as the first parm and passes them to the array function used as a callback, one after the other.\n\nIf, after evaluating the element, the function returns true the value is returned inside the array if not filtered out.\n\nNote that the array keys are preserved.\n\nIf you have already seen some episodes of this series, you will already know that I use simple references to understand.\n\nIn this case, instead, I will shamelessly steal an example from the PHP.net manual,\n\nThe reason is simple, I believe that the proposed example is quite difficult for a beginner and I am taking this opportunity to explain it, in simple words.\n\nHere is the example proposed:\n\n```function odd(\\$var)\n{\n// returns whether the input integer is odd\nreturn(\\$var & 1);\n}\n\nfunction even(\\$var)\n{\n// returns whether the input integer is even\nreturn(!(\\$var & 1));\n}\n\n\\$array1 = array(\"a\"=>1, \"b\"=>2, \"c\"=>3, \"d\"=>4, \"e\"=>5);\n\\$array2 = array(6, 7, 8, 9, 10, 11, 12);\n\necho \"Odd :\\n\";\nprint_r(array_filter(\\$array1, \"odd\"));\necho \"Even:\\n\";\nprint_r(array_filter(\\$array2, \"even\"));\n\nOdd :\nArray\n(\n[a] => 1\n[c] => 3\n[e] => 5\n)\nEven:\nArray\n(\n => 6\n => 8\n => 10\n => 12\n)```\n\nIn the example above the function array_filter() takes an array as a parameter and invokes a function (second parameter), the function, in turn, evaluates the element passed as an attribute and returns a Boolean value.\n\nNow, what is this?\n\nreturn(\\$var & 1);\n\nIt's a AND operation in BITWISE.\n\nAll even numbers have the least significant bit set to 0, whereas all odd numbers have the least significant bit set to 1.\n\nSo it simply \"ANDs\" two numbers together.\n\n3 for instance, it is represented as 11 in binary. 11 & 01 = 01 => true, so it is odd.\n\n2 instead, it is represented as 10 in binary. 10 & 01 = 00 => false, so it is even.\n\nI have already described the BITWISE operators during an article on the basics of PHP.\nIt will surely increase your skills.\n\n## array_reduce()\n\nWhich makes it one of the least used features I've ever seen in my career.\n\narray_reduce() needs a callback function which makes it more complicated than average, the more difficult part is that the callback requires two parameters to work.\n\nLet's take a look,\n\nWhat this function does is to Iteratively reduce the array to a single value using a callback function.\n\nWhat does it mean?\n\nIn simple terms, this function takes an array and, after processing the elements according to a callback, returns a single variable from it.\n\nAs you can imagine this function takes at least 2 parameters, the first is the array to be processed, the second is the function callback that does the job.\n\nA third parameter concerns the initial value from which we want to start in our callback, however, it is not mandatory to use it.\n\n```function addComma(\\$e1,\\$e2)\n{\nreturn \\$e1 . \", \" . \\$e2;\n}\n\\$dwarfs = array('Doc', 'Grumpy', 'Happy', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy');\n\n// Doc, Grumpy, Happy, Sleepy, Dopey, Bashful, Sneezy\n\n// Dwarfs, Doc, Grumpy, Happy, Sleepy, Dopey, Bashful, Sneezy```", null, "## array_intersec()\n\nWe have gone from a fairly complicated function to understand to a very simple one.\n\narray_intersec() is in fact very simple, being one of the first available functions since it was first published in PHP 4.0.1.\n\nIts task is to filter the elements of an array, indicated as the first mandatory parameter and commonly called a master array, against one or more selected arrays as subsequent parameters.\n\nIf an element is present in the master array and is also present in any of the others then it will be returned to the resulting array.\n\n```\\$dwarfs1 = array('Doc', 'Grumpy', 'Happy', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy');\n\\$dwarfs2 = array('Doc', 'Dopey');\n\\$dwarfs3 = array('Doc', 'Grumpy', 'Dopey', 'Bashful', 'Sneezy');\n\\$result=array_intersect(\\$dwarfs1, \\$dwarfs2, \\$dwarfs3);\nprint_r(\\$result);\nArray ( => 'Doc', => 'Dopey' )```\n\nNote that, to be treated as equal elements, the value of the comparison === between them must be in true.\n\n## array_intersect_assoc()\n\nThis function came shortly after the previous one, and like the previous one, it inherits a very simple syntax to understand.\n\narray_intersect_assoc() takes a minimum of 2 arrays as parameters, filters the first array, called master, with the others.\n\nif both keys and values are identical in all the arrays provided, then they will be added to the array returned by the function.\n\n```\\$dwarfs = array(\n'first' => 'Grumpy',\n'second' => 'Happy',\n'third' => 'Sleepy',\n'fourth' => 'Dopey',\n'fifth' => 'Bashful',\n'sixth' => 'Sneezy',\n'seventh' => 'Doc'\n);\n\\$dwarfs2 = array(\n'first' => 'Grumpy',\n0 => 'Happy',\n1 => 'Dopey',\n2 => 'Bashful',\n'sixth' => 'Sneezy',\n);\n\\$dwarfs3 = array(\n'first' => 'Grumpy',\n0 => 'Dopey',\n'sixth' => 'Sneezy',\n);\n\\$result=array_intersect_assoc(\\$dwarfs1, \\$dwarfs2, \\$dwarfs3);\nprint_r(\\$result);\nArray ( ['first' ] => 'Grumpy', ['sixth'] => 'Sneezy')```\n\nAs you can see the filtering returned only 2 of all elements within the arrays.\n\nIn all three arrays there is the pair ['first'] => 'Grumpy' and ['sixth'] => 'Sneezy'.\n\nYou can also note that in all arrays there is the value ‘Dopey’, but it has different keys so it is not evaluated.\n\n## array_diff_assoc()\n\nAnother relative of the functions explained above and another function that is part of the 'array filters' category.\n\narray_uintersect() works in a way very similar to array_intersect() but differs in the fact that it has a callback function, indicated as the last parameter, that compares arrays' elements provided.\n\n```function comparison(\\$a,\\$b)\n{\nreturn \\$a ⇔ \\$b;\n}\n\n\\$dwarfs1 = array('Doc', 'Grumpy', 'Sneezy');\n\\$dwarfs2 = array('Doc', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy');\n\nprint_r(array_uintersect(\\$dwarfs1, \\$dwarfs2, \"comparison\"));\nArray ( => 'Doc', => 'Sneezy')```\n\nNote that from PHP 5 the language provides the function array_intersect_assoc(), it works similarly to array_intersect() but it compares both keys and values.\n\nThe family of array_intersect and array_diff is quite big, counting 10 elements in total, they are all very similar both in working methods and syntax so here is the list and a few little hints if you decide to dive into them:\n\n• array_intersect,\n• array_ intersect_ assoc,\n• array_ intersect_ key,\n• array_ intersect_ uassoc,\n• array_ intersect_ ukey,\n• array_ diff\n• array_ diff_ assoc,\n• array_ diff_ key,\n• array_ diff_ uassoc,\n• array_ diff_ ukey,\n\nLegend:\nIntersect: Computes the intersections;\nDiff: Computes the difference;\nAssoc: evaluates both keys and values;\nUassoc: the callback function is used for the indices comparison;\nUkey: evaluates the keys;\n\n## max()\n\nIt will surely happen that during your career, you will have to at least once look for the maximum value between a series of variables or between the elements within an array,\n\nPHP 4 has provided you with its own programmers a feature to make this search much easier.\n\nthe max() function takes an array as a parameter or two or more variables that can be of different types and compare each of them returning the maximum value.\n\nYou can also evaluate different types of variables and they will be evaluated using standard comparison rules.\n\n```\\$maxNumber = max(2, 3, 1, 6, 7);\n// 7\n\\$maxArrayElement = max([2, 4, 5]);\n// 5\n\n\\$maxDwarfs = ['Doc', 'Grumpy', 'Happy', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy'];\n// 'Doc'\n\n// The string 'hello' when compared to an int is treated as 0\n\\$maxIntString = max(1, 'Doc');\n// 1\n\n// Multiple arrays compares the elements from left to right 3 < 4\n\\$maxArray = max([1, 2, 3], [1, 2, 4]);\n// [1, 2, 4]\n\n// The value 0 is evaluated as false whereas TRUE is evaluates as 1\n\\$maxIntBool = max(0, TRUE);\n// TRUE```\n\n## min()\n\nthe min() function finds the smallest value among a set of values.\n\nbasically, if you invert all that is written in the part regarding the max() function you will get the result of min().\n\n## count() & sizeof()\n\nIf you are working with an element that is an array or an object that implements the built-in interface of PHP Countable, you can use the array count() function.\n\nThis function works for both arrays and objects.\n\ncount() takes 2 values as a parameter, the first and the variable to be counted, the second parameter, which is not mandatory, consists of a flag that defines the counting method.\n\nThe flag can have two values: COUNT_NORMAL and COUNT_RECURSIVE, the second is useful in the case of associative arrays.\n\n```\\$dwarfs = ['Doc', 'Grumpy', 'Happy', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy'];\nvar_dump(count(\\$dwarfs));\n// int(7)\n\n\\$disney = [\n\"dwarfs\" => ['Doc', 'Grumpy', 'Happy', 'Sleepy', 'Dopey', 'Bashful', 'Sneezy'],\n\"originals\" =>[\"Mickey Mouse\",\"Pete\", \"Goofy\", \"Minnie Mouse\", \"Pluto\"]\n];\n// Normal count\nvar_dump(count(\\$disney));\n// int(2)\n\n// Recursive count\nvar_dump(count(\\$disney, COUNT_RECURSIVE));\n// int(14)```\n\nSome more about the count function on the official manual\n\n# Conclusion\n\nAs you have seen, the portfolio of array functions that PHP has provided us with, and continues to give us, is incredibly broad,\n\nIn this section of the series we have seen extremely simple functions, for example, the pair min() and max(), and others that go deeper in detail like array_ intersect_ uassoc().\n\nThe good news is that there is no job and no open vacancy that requires you to know all these functions by heart,\n\nAnother good news is that, now that you understand them, you can mark this post on your browser and return whenever you want.\n\nAfter discussing the filters, in the next chapter of the series, you will go to see another part that constitutes the foundation of PHP functions.\n\nHow to order items inside.\n\nIf you have learned something new today and have not already done so, subscribe to the newsletter in order to be notified when the next chapter is published.\n\nIf you like this content and the next set of examples on arrays functions, stay tuned and subscribe to the newsletter so you will be notified.\n\nIf you want to learn more about PHP array functions take a look at the tutorials already published which are part of this series by clicking on the links below.\n\n***", null, "### Learn to code, gain a new skill, get a new job\n\n#### Whatever your goal — Treehouse will get you there\n\nThey are currently making a free 4-month offer (valued at \\$ 100).\n\nHave a look at it!.\n\nIf you like this content and you are hungry for some more join the Facebook's community in which we share info and news just like this one!\n\n### Other posts that might interest you\n\nCoding (Php 7.x) Mar 25, 2019\n\n#### Array function in PHP [Part 2]\n\nCoding (Php 7.x) Apr 26, 2019\n\n#### Array functions [sorting elements]\n\nCoding (Php 7.x) May 7, 2019\n\n#### Beyond popular array functions in php [with examples]", null, "" ]

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[ "# Area distortion under certain classes of quasiconformal mappings\n\n## Abstract\n\nIn this paper we study the hyperbolic and Euclidean area distortion of measurable sets under some classes of K-quasiconformal mappings from the upper half-plane and the unit disk onto themselves, respectively.\n\n## 1 Introduction\n\nLet $$f:\\Omega\\to \\mathbb {C}$$ be an ACL (absolute continuous on lines) homeomorphism in a domain $$\\Omega\\subset \\mathbb {C}$$ that preserves orientation. If f satisfies\n\n$$D_{f}= \\frac{\\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert }{\\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert } \\leq K \\quad \\mbox{a.e.}$$\n\nfor some $$K\\geq1$$, then f is a K-quasiconformal mapping, where\n\n$$f_{z} = \\frac{1}{2} ( f_{x} -if_{y} ) \\quad \\mbox{and}\\quad f _{\\overline{z}} = \\frac{1}{2} ( f_{x} +if_{y} ),$$\n\nand $$D_{f}$$ is called the dilatation of f.\n\nIn 1956 Beurling and Alhfors solved the boundary value problem for quasiconformal mappings . If $$M\\geq1$$, they gave an explicit formula for the extension of an M-quasisymmetric function $$h:\\mathbb {R}\\to \\mathbb {R}$$ to a quasiconformal mapping $$f=u+iv$$ from $$\\mathbb {H}$$ onto itself, where $$\\mathbb {H}$$ denotes the upper half-plane. The mapping f is called the Beurling-Alhfors extension of h. In particular f satisfies (see )\n\n$$\\frac{1}{C y^{2}}\\leq\\frac{J_{f}(z)}{v^{2}}\\leq\\frac{C}{y^{2}},$$\n\nwhere $$J_{f}$$ denotes the Jacobian of f and $$C=C(K)>0$$ depends on $$K=K(M)$$, the maximal dilatation of f. Thus, for each measurable subset E of $$\\mathbb {H}$$, it holds that\n\n$$\\frac{A_{\\mathcal{H}}(E)}{C}\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr) \\leq C A_{ \\mathcal{H}}(E),$$\n\nwhere $$A_{\\mathcal{H}}(\\cdot)$$ denotes the hyperbolic area in the half-plane $$\\mathbb {H}$$.\n\nIn 1994 Astala proved that if f is a K-quasiconformal mapping from the unit disk $$\\mathbb {D}$$ onto itself, normalized by $$f(0)=0$$, and if E is any measurable subset of the unit disk, then $$A_{e}(f(E))\\leqslant a(K)A_{e}(E)^{1/K}$$, where $$A_{e}(\\cdot)$$ denotes the Euclidean area and $$a(K)\\to1$$ when $$K\\to1^{+}$$.\n\nIn 1998 Reséndis and Porter obtained some results about area distortion under quasiconformal mappings on the unit disk $$\\mathbb {D}$$ onto itself with respect to the hyperbolic measure. They also showed the existence of explodable sets; this kind of sets has bounded hyperbolic area, but under a specific quasiconformal mapping its image has infinite hyperbolic area.\n\nIn recent years harmonic quasiconformal mappings have been extensively studied, see and the papers cited therein. The following two recent results are very close to the results presented in this paper.\n\nIn 2007 Knežević and Mateljević proved the following Schwarz-Pick type distortion theorem.\n\n### Theorem 1\n\nLet f be a K-quasiconformal harmonic mapping from the unit disk $$\\mathbb {D}$$ onto itself, then\n\n$$\\frac{1}{1+k}\\frac{1-\\vert f(z)\\vert ^{2}}{1-\\vert z\\vert ^{2}}\\leq\\bigl\\vert f_{z}(z)\\bigr\\vert \\leq \\frac{1}{1-k}\\frac{1-\\vert f(z)\\vert ^{2}}{1-\\vert z\\vert ^{2}},$$\n\nwhere k is defined by $$k:=\\frac{K-1}{K+1}$$.\n\nIn 2012 Min Chen and Xingdi Chen studied the class of $$(K,K')$$ quasiconformal maps from $$\\mathbb {H}$$ onto itself, and they obtained the following result about area distortion of harmonic mappings.\n\n### Theorem 2\n\nLet $$f(z)=u(z)+iv(z)$$ be a harmonic mapping from $$\\mathbb {H}$$ onto itself and continuous on $$\\mathbb {H}\\cup \\mathbb {R}$$ with $$f(\\infty)=\\infty$$. In particular, f has the form $$f(z)=f(x+iy)=u(x,y)+icy$$ for some $$c>0$$ (see ). If f is K-quasiconformal and $$E\\subset \\mathbb {H}$$ is any measurable set, then\n\n1. (i)\n\n$$A_{e}(f(E))\\leq c^{2}KA_{e}(E)$$,\n\n2. (ii)\n\n$$A_{\\mathcal{H}}(f(E))\\leq K A_{\\mathcal{H}}(E)$$,\n\n3. (iii)\n\n$$0< u_{x}\\leq\\frac{c(K+1+\\sqrt{(K+1)^{2}-4})}{2}$$,\n\n4. (iv)\n\n$$\\vert u_{y} \\vert \\leq c\\sqrt{(K+1)\\frac{K+1+ \\sqrt{(K+1)^{2}-4}}{2}}$$.\n\nIn this paper we use the following hyperbolic density definitions:\n\n$$\\frac{2\\vert dz\\vert }{1-\\vert z\\vert ^{2}}\\quad \\mbox{and}\\quad \\frac{ \\vert dw\\vert }{\\operatorname {Im}w}$$\n\nfor the unit disk $$\\mathbb {D}$$ and the upper half-plane $$\\mathbb {H}$$, respectively, see and . We denote also by $$A_{\\mathcal{H}}$$ the hyperbolic area in the unit disk $$\\mathbb {D}$$.\n\n## 2 Results and discussion\n\nOur purpose in this article is to continue the study of the hyperbolic area distortion under K-quasiconformal mappings from the upper half-plane $$\\mathbb {H}$$ onto itself or from the unit disk $$\\mathbb {D}$$ onto itself. Due to the existence of explodable sets (see ), we study some particular classes of quasiconformal mappings. First we apply the result of Knežević and Mateljević to estimate the hyperbolic area distortion under harmonic quasiconformal mappings from $$\\mathbb {D}$$ onto itself, and we remove the hypothesis of harmonicity in Theorem 2.\n\nAdditionally, we generalize even more the class studied by Chen and Chen in . More precisely, the main result of this paper is the following.\n\n### Theorem 3\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself such that f maps a family of horocyclics with a common tangent point onto a family of horocyclics. Then, for each measurable set $$E\\subset \\mathbb {H}$$, the following inequalities hold:\n\n$$\\frac{1}{K^{9}} A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K ^{9} A_{\\mathcal{H}}(E).$$\n\nThese bounds are asymptotically sharp when $$K\\to1^{+}$$.\n\nAdditionally, we obtain some results about radial and angular quasiconformal mappings. Motivated by the generalization mentioned above, we finally describe a set that contains the region of values of the partial derivatives of K-quasiconformal mappings.\n\n### 2.1 Harmonic quasiconformal mappings\n\nIn this part we use Theorem 1 to estimate the hyperbolic area distortion under quasiconformal harmonic mappings from the unit disk onto itself and analyze the hyperbolic and Euclidean area distortion under quasiconformal mappings $$f(z)=f(x+iy)=u(x,y)+icy$$, with $$c>0$$, from $$\\mathbb {H}$$ onto itself, without the hypothesis of harmonicity of f (see ), and we sketch the proof of items (i) and (ii) of Theorem 2 as a corollary of this result and get better bounds than the ones obtained in items (iii) and (iv) in the same theorem. See and for results in hyperbolic geometry.\n\n### Theorem 4\n\nLet f be a K-quasiconformal harmonic mapping from the unit disk $$\\mathbb {D}$$ onto itself. If $$E\\subset \\mathbb {D}$$ is a measurable set, then\n\n$$\\frac{1}{K}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq \\biggl( \\frac{K+1}{2} \\biggr) ^{2}A_{\\mathcal{H}}(E).$$\n\nThese bounds are asymptotically sharp when $$K\\to1^{+}$$.\n\n### Proof\n\nLet $$w=f(z)=u(z)+iv(z)$$, $$z\\in \\mathbb {D}$$. By hypothesis and Theorem 1, the mapping f satisfies\n\n\\begin{aligned} \\biggl( \\frac{1}{1+k}\\frac{1-\\vert f(z)\\vert ^{2}}{1-\\vert z\\vert ^{2}} \\biggr) ^{2} \\leq& \\bigl\\vert f_{z}(z)\\bigr\\vert ^{2} \\\\ \\leq& \\biggl( \\frac{1}{1-k}\\frac {1-\\vert f(z)\\vert ^{2}}{1-\\vert z\\vert ^{2}} \\biggr) ^{2}. \\end{aligned}\n\nMoreover, the Jacobian $$J_{f}$$ of f satisfies\n\n$$\\bigl(1-k^{2}\\bigr)\\vert f_{z}\\vert ^{2}\\leq J_{f}=\\vert f_{z}\\vert ^{2}-\\vert f_{\\overline{z}}\\vert ^{2} \\leq \\vert f_{z}\\vert ^{2} .$$\n\nThus, for each measurable set $$E\\subset \\mathbb {D}$$, the following equalities hold:\n\n\\begin{aligned}& A_{\\mathcal{H}}f(E)= \\int_{f(E)}\\frac{du\\,dv}{(1-\\vert w\\vert ^{2})^{2}}= \\int _{E}\\frac{J _{f}(z)}{(1-\\vert f(z)\\vert ^{2})^{2}}\\,dx\\,dy, \\\\& \\frac{1-k^{2}}{(1+k)^{2}}A_{\\mathcal{H}}(E)\\leq \\int_{E} \\frac{J_{f}(z)\\,dx\\,dy}{(1-\\vert f(z)\\vert ^{2})^{2}}\\leq\\frac{1}{(1-k)^{2}}A _{\\mathcal{H}}(E) \\end{aligned}\n\nand the result follows immediately. □\n\n### Theorem 5\n\nLet f be a K-quasiconformal mapping from the upper half-plane $$\\mathbb {H}$$ onto itself given by $$f(z)=f(x+iy)=u(x,y)+icy$$, with $$c>0$$. If there exists $$M>0$$ such that $$\\vert f_{z}(z)\\vert -\\vert f_{\\overline{z}}(z)\\vert \\leq M$$ a.e., then for any measurable subset E of $$\\mathbb {H}$$ the following inequalities hold:\n\n1. (i)\n\n$$A_{e}(f(E))\\leq M^{2}K A_{e}(E)$$,\n\n2. (ii)\n\n$$A_{\\mathcal{H}}(f(E))\\leq\\frac{M^{2}K}{c^{2}}A_{ \\mathcal{H}}(E)$$.\n\n### Proof\n\nSince f is a K-quasiconformal mapping, f satisfies $$\\vert f_{z}\\vert + \\vert f_{\\overline{z}}\\vert \\leq K (\\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert )$$ a.e. in $$\\mathbb {H}$$, and by hypothesis we get $$J_{f}=(\\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert ) (\\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert )\\leq KM^{2}$$ a.e. in $$\\mathbb {H}$$. Then, for any measurable set $$E\\subset \\mathbb {H}$$,\n\n$$A_{e}\\bigl(f(E)\\bigr)= \\int_{f(E)}du \\,dv= \\int_{E}J_{f}\\,dx \\,dy\\leq KM^{2}A_{e}(E)$$\n\nand for the second result we have\n\n$$A_{\\mathcal{H}}\\bigl(f(E)\\bigr)= \\int_{f(E)}\\frac{du \\,dv}{(\\operatorname {Im}w)^{2}}= \\int _{E}\\frac{J_{f}\\,dx \\,dy}{(cy)^{2}}\\leq\\frac{M^{2}K}{c^{2}}A_{ \\mathcal{H}}(E).$$\n\n□\n\nIf we additionally suppose that f is a harmonic mapping, then there exists a holomorphic function $$g:\\mathbb {H}\\to \\mathbb {C}$$ such that $$f(z)= \\operatorname {Re}g(z) +icy$$. Thus\n\n$$\\vert f_{z}\\vert =\\frac{1}{2} \\bigl\\vert g'(z)+c\\bigr\\vert \\quad \\mbox{and} \\quad \\vert f_{\\overline{z}}\\vert = \\frac{1}{2} \\bigl\\vert g'(z)-c\\bigr\\vert .$$\n\nSince\n\n$$\\frac{\\vert f_{\\overline{z}}\\vert }{\\vert f_{z}\\vert } \\leq k \\quad \\mbox{or equivalently}\\quad \\bigl\\vert g'(z)-c\\bigr\\vert ^{2} \\leq k^{2} \\bigl\\vert g'(z) +c\\bigr\\vert ^{2},$$\n\nwe obtain that $$g'(z)$$ belongs to the circle D with center $$(\\frac{1+k^{2}}{1-k^{2}}c,0)$$ and radius $$\\frac{2ck}{1-k^{2}}$$. Hence, for each $$w\\in D$$, we get the estimations\n\n$$\\vert w+c\\vert -\\vert w-c\\vert \\leq2c\\quad \\mbox{and}\\quad \\vert w+c\\vert + \\vert w-c\\vert \\leq2Kc.$$\n\nIn particular we have\n\n$$\\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert \\leq c\\quad \\mbox{and}\\quad \\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert \\leq Kc.$$\n\nSo we obtain Theorem 2 as a corollary of Theorem 5.\n\nWe note that Theorem 5 gives some information about the partial derivatives, in specific we obtain that if f satisfies the hypothesis of the last theorem, then $$f_{z}$$ and $$f_{\\overline{z}}$$ are bounded. The previous sketch gives us also some idea to study quasiconformal mappings of the form $$f(x+iy)=u(x,y)+iv(y)$$.\n\n### 2.2 Quasiconformal mappings $$f(x+iy)=u(x,y)+iv(y)$$\n\nWe now generalize the class studied by Cheng and Chen (see ) in two directions. First, we will show that it is possible to avoid the harmonic hypothesis, and second, we will prove that the class of K-quasiconformal mappings given by $$f(x+iy)=u(x,y)+iv(y)$$ is a family that accepts asymptotically sharp bi-bounds for the area distortion.\n\nLet $$1\\leq K<\\infty$$ and $$\\Omega\\subset \\mathbb {C}$$ be a domain. Suppose that $$f:\\Omega\\to \\mathbb {C}$$ is a K-quasiconformal mapping given by\n\n$$f(x+iy)=u(x,y)+iv(x,y).$$\n(1)\n\nThen f satisfies\n\n$$\\biggl\\vert \\frac{f_{\\overline{z}}}{f_{z}} \\biggr\\vert \\leq k, \\quad \\mbox{a.e.}\\quad \\mbox{or equivalently}\\quad \\biggl\\vert \\frac{f_{ \\overline{z}}}{f_{z}} \\biggr\\vert ^{2}\\leq k^{2},\\quad \\mbox{a.e.}$$\n(2)\n\nInequality (2) is satisfied if and only if\n\n$$(u_{x}-v_{y}) ^{2}+(v_{x}+u_{y})^{2} \\leq k ^{2} \\bigl((u_{x}+v_{y})^{2}+(v _{x}-u_{y})^{2}\\bigr)\\quad \\mbox{a.e.}$$\n\nor equivalently\n\n$$u_{x}^{2}+u_{y}^{2}+v_{x}^{2}+v_{y}^{2}- \\frac{1+k^{2}}{1-k^{2}}2u _{x}v_{y}+\\frac{1+k^{2}}{1-k^{2}}2u_{y}v_{x} \\leq0 \\quad \\mbox{a.e.}$$\n\nDefine\n\n$$\\alpha=\\alpha(k):=\\frac{1+k^{2}}{1-k^{2}}\\geq1.$$\n(3)\n\nThen f satisfies inequality (2) if and only if\n\n$$u_{x}^{2}+u_{y}^{2}+v_{x}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}+2\\alpha u _{y}v_{x}\\leq0 \\quad \\mbox{a.e.}$$\n(4)\n\nFrom now on each expression that involves partial derivatives will be true almost everywhere (a.e.) and Ω will denote a domain of the complex plane $$\\mathbb {C}$$.\n\nIn this part we focus on K-quasiconformal mappings f from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. In particular f can be extended homeomorphically to $$\\overline{\\mathbb {H}}$$, $$u(x,y)$$ is ACL and $$v(y)$$ is absolutely continuous. We know that f satisfies\n\n$$u_{x}^{2}+u_{y}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}\\leq0\\quad \\mbox{a.e.}$$\n(5)\n\nDespite the fact that v depends only on the variable y, we write $$v_{y}$$ instead of $$v'$$ to emphasize the dependence on y. The next result gives the principal characteristics of the mapping f, and most of them are consequences of its quasiconformal properties (the rest are easy to prove).\n\n### Proposition 1\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. Then\n\n• The function $$y\\mapsto v(y)$$ is a homeomorphism from $$[0,\\infty) \\to[0,\\infty)$$ that is absolutely continuous and differentiable a.e.\n\n• For almost every $$y\\in[0,\\infty)$$, the function $$x\\mapsto u(x,y)$$ is a homeomorphism from $$\\mathbb {R}$$ onto itself that is absolutely continuous and differentiable a.e.\n\n• The K-quasiconformal inverse mapping $$f^{-1}:\\mathbb {H}\\to \\mathbb {H}$$ has the same form as f, that is, $$f^{-1}(x+iy)=w(x,y)+iv^{-1}(y)$$.\n\n• Let $$y_{0}\\in(0,\\infty)$$ be fixed. The function $$h:\\mathbb {H}\\to \\mathbb {H}$$ defined by $$h(x+iy)=u(x,y+y_{0})+i[v(y+y_{0})-v(y_{0})]$$ is a K-quasiconformal mapping. In particular each function $$x\\mapsto u(x,y _{0})$$ is quasisymmetric.\n\n• If $$g:\\mathbb {H}\\to \\mathbb {H}$$ is a $$K'$$-quasiconformal mapping given by $$g(x+iy)=l(x,y)+iw(y)$$, then $$f\\circ g$$ is a $$KK'$$-quasiconformal mapping of the same form.\n\n• The mapping f leaves invariant the family of horocyclics with tangential point at infinity.\n\nWe study inequality (5) in more detail. To get this, we complete in (5) the square in $$v_{y}$$, so we obtain\n\n$$(u_{x}-\\alpha v_{y})^{2} + u_{y}^{2} \\leq v_{y}^{2}\\bigl(\\alpha^{2}-1\\bigr)\\quad \\mbox{a.e.}$$\n\nThis inequality defines a circle a.e., thus $$u_{x}$$ and $$v_{y}$$ satisfy in particular\n\n$$\\alpha v_{y}-v_{y}\\sqrt{\\alpha^{2}-1}\\leq u_{x}\\leq\\alpha v_{y}+ v _{y}\\sqrt{ \\alpha^{2}-1}\\quad \\mbox{a.e.}$$\n\nand\n\n$$-v_{y}\\sqrt{\\alpha^{2}-1}\\leq u_{y}\\leq v_{y} \\sqrt{\\alpha^{2}-1}\\quad \\mbox{a.e.}$$\n\nIn fact, the circle is a subset of the square described by the previous inequalities. Observe that\n\n$$K=\\alpha+\\sqrt{\\alpha^{2}-1}$$\n(6)\n\nand\n\n$$\\frac{1}{K}=\\alpha-\\sqrt{\\alpha^{2}-1},$$\n\nwhere K is the maximal dilatation of f. Let $$C= \\sqrt{\\alpha^{2}-1}$$. Then $$0\\leq C$$. With this notation the last inequalities can be written as follows:\n\n$$\\frac{v_{y}}{K}\\leq u_{x}\\leq K v_{y} \\quad \\mbox{a.e.}$$\n(7)\n\nand\n\n$$-C v_{y}\\leq u_{y}\\leq Cv_{y} \\quad \\mbox{a.e.}$$\n(8)\n\nGiven $$0\\leq x$$, we integrate (7) on the interval $$[0,x]$$\n\n$$\\int_{0}^{x}\\frac{v_{y}(y)}{K}\\,dt\\leq \\int_{0}^{x}u_{x}(t,y)\\,dt\\leq \\int_{0}^{x} Kv_{y}(y)\\,dt.$$\n\nIf we choose any fixed $$y\\in(0,\\infty)$$ such that $$u(x,y)$$ is absolutely continuous with respect to x, then we obtain\n\n$$\\frac{x}{K}v_{y}(y)+u(0,y)\\leq u(x,y)\\leq Kx v_{y}(y)+u(0,y)$$\n(9)\n\nfor each $$x\\in \\mathbb {R}$$ and almost every $$y\\in(0,\\infty)$$. Using the left-hand side of the last inequality, we get\n\n$$\\limsup_{y\\to0^{+}} \\biggl[ \\frac{v_{y}(y)x}{K}+u(0,y) \\biggr] \\leq \\limsup_{y\\to0^{+}}u(x,y),$$\n\nand since $$u(x,y)$$ is continuous, we obtain\n\n$$\\frac{x}{K}\\limsup_{y\\to0^{+}}v_{y}(y)\\leq u(x,0)-u(0,0)< \\infty.$$\n\nFor this reason, $$\\limsup_{y\\to0^{+}} v_{y}(y)$$ exists and consequently $$\\liminf_{y\\to0^{+}}v_{y}(y)$$ exists too. We define $$v_{y}^{+}(0):= \\limsup_{y\\to0^{+}}v_{y}(y)$$ and $$v_{y}^{-}(0):=\\liminf_{y\\to0^{+}}v _{y}(y)$$. With this notation, we obtain from (9)\n\n$$\\frac{v_{y}^{+} (0)x}{K}+u(0,0)\\leq u(x,0)\\leq K v_{y}^{-}(0)x+u(0,0).$$\n(10)\n\nOn the other hand, we choose any fixed $$x\\in[0,\\infty)$$ such that $$u(x,y)$$ is absolutely continuous with respect to y, and we integrate (8) on the interval $$[0,y]$$. So\n\n$$\\int_{0}^{y} -Cv_{y}(t)\\,dt\\leq \\int_{0}^{y} u_{y}(x,t)\\,dt\\leq \\int_{0} ^{y} Cv_{y}(t)\\,dt$$\n\nand, since $$v(y)$$ is absolutely continuous, we obtain\n\n$$-Cv(y)+u(x,0)\\leq u(x,y)\\leq Cv(y)+u(x,0)$$\n\nfor $$y\\in[0,\\infty)$$ and almost every $$x\\in[0,\\infty)$$. By an argument of continuity of the mapping f and density, we have\n\n$$-Cv(y)+u(x,0)\\leq u(x,y)\\leq Cv(y)+u(x,0)$$\n(11)\n\nfor all $$(x,y)\\in[0,\\infty)\\times[0,\\infty)$$. Setting $$x=0$$ in the previous inequality, we get\n\n$$-Cv(y)+u(0,0)\\leq u(0,y)\\leq Cv(y)+u(0,0).$$\n(12)\n\nThus, combining (9) and (12), we have\n\n$$\\frac{v_{y}(y)x}{K}-Cv(y)+u(0,0)\\leq u(x,y)\\leq Kxv_{y}(y)+Cv(y)+u(0,0)$$\n\nfor each $$x\\in \\mathbb {R}$$ and almost every $$y\\in(0,\\infty)$$. In the same way we use (10) and (11) to obtain\n\n$$\\frac{x}{K}v_{y}^{+}(0)-Cv(y)+u(0,0)\\leq u(x,y)\\leq Kxv_{y}^{-}(0)+Cv(y)+u(0,0).$$\n\nWe combine the left- and right-hand sides of the previous inequalities to get\n\n$$\\frac{x}{K}v_{y}(y)-Cv(y)+u(0,0)\\leq u(x,y)\\leq Kxv_{y}^{-}(0)+Cv(y)+u(0,0)$$\n(13)\n\nand\n\n$$\\frac{x}{K}v_{y}^{+}(0)-Cv(y)+u(0,0) \\leq u(x,y)\\leq Kxv_{y}(y)+Cv(y)+u(0,0)$$\n(14)\n\nfor each $$x\\in \\mathbb {R}$$ and almost every $$y\\in(0,\\infty)$$. Since the left- and right-hand sides of inequalities (13) and (14) represent linear equations in the variable x, we compare their slopes and the fact that $$x\\geq0$$ to conclude\n\n$$\\frac{v_{y}(y)}{K}\\leq Kv_{y}^{-}(0)\\quad \\mbox{and}\\quad \\frac{v_{y}^{+}(0)}{K}\\leq Kv_{y}(y)$$\n\nfor each $$x\\in \\mathbb {R}$$ and almost every $$y\\in(0,\\infty)$$. Hence\n\n$$\\frac{ v_{y}^{+}(0)}{K^{2}}\\leq v_{y}(y)\\leq K^{2} v_{y}^{-}(0)$$\n\nfor each $$x\\in \\mathbb {R}$$ and almost every $$y\\in(0,\\infty)$$. We recall that $$v(y)$$ is absolutely continuous, and we integrate the above inequalities on the interval $$[0,y]$$\n\n$$\\frac{1}{K^{2}} \\int_{0}^{y}v_{y}^{+}(0)\\,dt\\leq \\int_{0}^{y}v_{y}(t)\\,dt \\leq K^{2} \\int_{0}^{y}v_{y}^{-}(0)\\,dt$$\n\nto get\n\n$$\\frac{1}{K^{2}}v_{y}^{+}(0)y\\leq v(y)\\leq K^{2}v_{y}^{-}(0)y.$$\n(15)\n\nIn particular $$0< v_{y}^{-}(0)\\leq v_{y}^{+}(0)<\\infty$$ and v has right Dini’s derivatives at 0. If $$x<0$$, we obtain the same relation as (15) and have the next result.\n\n### Theorem 6\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. Then $$v_{y}^{*}(0)= \\limsup_{y\\to0^{+}}v_{y}(y)$$ and $$v_{y}^{-}(0)= \\liminf_{y\\to0^{+}}v_{y}(y)$$ are finite, and the partial derivatives of f satisfy the following inequalities:\n\n1. 1.\n\n$$\\frac{1}{K^{2}}v_{y}^{+}(0)\\leq v_{y}(y)\\leq K^{2}v_{y}^{-}(0)$$ for almost every $$y\\in(0,\\infty)$$.\n\n2. 2.\n\n$$\\frac{1}{K^{3}} v_{y}^{+}(0)\\leq u_{x}(x,y)\\leq K^{3} v_{y}^{-}(0)$$ for almost every $$x+i y\\in \\mathbb {H}$$.\n\n3. 3.\n\n$$\\vert u_{y}(x,y) \\vert \\leq\\frac{K(K^{2}-1)}{2} v_{y}^{-}(0)$$ for almost every $$x+i y\\in \\mathbb {H}$$.\n\nIn particular by Proposition  6 the partial derivatives of f belong to some truncate solid cone.\n\nWe can combine the previous result with items (iii) and (iv) of Theorem 2 to obtain the following result.\n\n### Corollary 1\n\nLet f be a harmonic K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself with $$f(\\infty)=\\infty$$. In particular there exists $$c> 0$$ such that $$f(x+iy)=u(x,y)+icy$$. Then\n\n$$\\frac{c}{K^{3}}\\leq u_{x} \\leq A(K)c$$\n\nand\n\n$$\\vert u_{y} \\vert \\leq B(K)c$$\n\nwith\n\n\\begin{aligned} &A(K)= \\textstyle\\begin{cases} K^{3} & \\textit{if } K\\in[1,a], \\\\ \\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2} &\\textit{if } K\\in[a,\\infty), \\end{cases}\\displaystyle \\\\ &B(K)= \\textstyle\\begin{cases} \\frac{K(K^{2}-1)}{2} & \\textit{if } K\\in[1,b], \\\\ \\sqrt{(K+1)\\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2}} &\\textit{if } K \\in[b,\\infty), \\end{cases}\\displaystyle \\end{aligned}\n\nwhere $$a=1. 12373\\ldots$$ and $$b=1. 95371\\ldots$$ are the solutions of the equations $$K^{3}= \\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2}$$ and $$\\frac{K(K^{2}-1)}{2}= \\sqrt{(K+1)\\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2}}$$, respectively.\n\n### Proof\n\nThe left-hand side of the first inequality is immediate from item 1 of Theorem 6. We now consider the right-hand side estimations of $$u_{x}$$ in item 2 of Theorem 6, with $$v_{y}^{-}(0)=v_{y}^{+}(0)=c$$ and item (iii) of Theorem 2. Then\n\n$$u_{x} \\leq\\min \\biggl\\{ cK^{3}, \\frac{c(K+1+\\sqrt{(K+1)^{2}-4})}{2} \\biggr\\} .$$\n\nWe consider the equation\n\n$$K^{3}= \\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2}$$\n\nand obtain the value of a by solving $$K^{6} -K^{4} -K^{3} +1=0$$ that has only two real solutions $$K=1$$ and $$K=1. 12373\\ldots=a$$.\n\nWe now consider the estimations of $$\\vert u_{y}\\vert$$ in item 3 of Theorem 6 and item (iv) of Theorem 2. Thus\n\n$$\\vert u_{y}\\vert \\leq\\min \\biggl\\{ \\frac{cK(K^{2}-1)}{2}, c \\sqrt{(K+1)\\frac{K+1+ \\sqrt{(K+1)^{2}-4}}{2}} \\biggr\\} .$$\n\nThe value of b is given by the real solution of\n\n$$\\frac{K(K^{2}-1)}{2}=\\sqrt{(K+1) \\frac{K+1+\\sqrt{(K+1)^{2}-4}}{2}}$$\n\nthat can be reduced to the equation\n\n$$\\biggl( \\frac{K+1}{2} \\biggr) \\bigl(K^{4}-2K^{3}+K^{2}-2 \\bigr)= \\sqrt{(K+1)^{2}-4},$$\n\nand the real solution of this equation is $$K=1.95371\\ldots=b$$. This value is obtained through the equation\n\n$$\\biggl( \\frac{(K+1)^{2}}{4} \\biggr) \\bigl(K^{4}-2K^{3}+K^{2}-2 \\bigr)^{2}=(K+1)^{2}-4$$\n\ndiscarding the value $$K=1.55739\\ldots$$ that obviously is not a solution of the original equation. □\n\nThe last corollary shows that these estimations are asymptotically sharp when $$K\\to1^{+}$$ improving the result obtained in .\n\n### Theorem 7\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. Then\n\n1. 1.\n\nThere exists $$M>0$$ such that $$\\vert f_{z}\\vert - \\vert f_{\\overline{z}}\\vert \\leq M$$ and $$\\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert \\leq KM$$ a.e.\n\n2. 2.\n\nThe mapping f is Lipschitz in $$\\mathbb {H}$$.\n\n3. 3.\n\nThe mapping f is hyperbolically Lipschitz in $$\\mathbb {H}$$.\n\n### Proof\n\n1. 1.\n\nSince\n\n\\begin{aligned} \\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert &= \\frac{1}{2}\\sqrt{(u_{x}+v_{y})^{2}+u _{y}^{2}}-\\frac{1}{2}\\sqrt{(u_{x}-v_{y})^{2}+u_{y}^{2}} \\\\ &\\leq\\frac{1}{2}\\sqrt{2\\bigl(u_{x}^{2}+v_{y}^{2}+u_{y}^{2} \\bigr)} \\quad \\mbox{a.e.} \\end{aligned}\n\nWe estimate the last expression using Theorem 6 to obtain\n\n\\begin{aligned}& \\frac{1}{2}\\sqrt{2\\bigl(u_{x}^{2}+v_{y}^{2}+u_{y}^{2} \\bigr)} \\\\& \\quad \\leq\\frac{\\sqrt{2}}{2}\\sqrt{\\bigl(K^{3} v_{y}^{-}(0)\\bigr)^{2}+\\bigl(K^{2} v _{y}^{-}(0)\\bigr)^{2}+ \\biggl( \\frac{K^{2}-1}{2}Kv_{y}^{-}(0) \\biggr) ^{2}} \\\\& \\quad =\\frac{K^{2}v_{y}^{-}(0)}{\\sqrt{2}}\\sqrt{K^{2}+1+ \\biggl( \\frac{K ^{2}-1}{2K} \\biggr) ^{2}} \\\\& \\quad =\\frac{K^{2}v_{y}^{-}(0)}{\\sqrt{2}}\\sqrt{\\frac{5K^{4}+2K^{2}+1}{4K ^{2}}} \\\\& \\quad =\\frac{Kv_{y}^{-}(0)}{2\\sqrt{2}}\\sqrt{5K^{4}+2K^{2}+1} \\quad \\mbox{a.e.} \\end{aligned}\n\nThus we choose $$M=\\frac{Kv_{y}^{-}(0)}{2\\sqrt{2}}\\sqrt{5K^{4}+2K ^{2}+1}$$.\n\n2. 2.\n\nLet $$z_{1}, z_{2}\\in \\mathbb {H}$$ and l be the Euclidean segment that joints $$z_{1}$$ with $$z_{2}$$. Then\n\n\\begin{aligned} \\bigl\\vert f(z_{1})-f(z_{2})\\bigr\\vert \\leq& \\int_{f(l)}\\vert df\\vert \\\\ =& \\int_{l}\\bigl(\\vert f_{z}\\vert + \\vert f_{\\overline{z}}\\vert \\bigr)\\vert dz\\vert \\\\ \\leq& MK\\vert z_{1}-z_{2}\\vert . \\end{aligned}\n3. 3.\n\nLet $$z_{1}, z_{2}\\in \\mathbb {H}$$ and l be the hyperbolic segment that joints $$z_{1}$$ with $$z_{2}$$. Then\n\n\\begin{aligned} d_{\\mathcal{H}} \\bigl( f(z_{1}),f(z_{2}) \\bigr) &\\leq \\int_{f(l)} \\frac{\\vert dw\\vert }{\\operatorname {Im}w}= \\int_{l}\\frac{\\vert df\\vert }{v} \\\\ &\\leq \\int_{l} \\bigl(\\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert \\bigr)\\frac{\\vert dz\\vert }{v}\\leq MK \\int _{l}\\frac{\\vert dz\\vert }{v}\\leq LMKd_{\\mathcal{H}}(z_{1},z_{2}), \\end{aligned}\n\nwhere $$L=\\frac{K^{2}}{v_{y}^{+}(0)}$$ and $$d_{\\mathcal{H}}$$ denotes the hyperbolic metric. □\n\nNow we can extend even more the result obtained by Min Chen and Xindy Chen .\n\n### Theorem 8\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. Then, for each measurable set $$E\\subset \\mathbb {H}$$,\n\n1. 1.\n\n$$\\frac{(v_{y}^{+}(0))^{2}}{K^{5}} A_{e}(E)\\leq A_{e}(f(E))\\leq K ^{5}(v_{y}^{-}(0))^{2} A_{e}(E)$$.\n\n2. 2.\n\n$$\\frac{1}{K^{9}} ( \\frac {v_{y}^{+}(0)}{v_{y}^{-}(0)} ) ^{2} A _{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}(f(E))\\leq K^{9} ( \\frac{v _{y}^{-}(0)}{v_{y}^{+}(0)} ) ^{2} A_{\\mathcal{H}}(E)$$.\n\nSince $$v_{y}^{-}(0)\\leq v_{y}^{+}(0)$$, we have\n\n$$\\frac{1}{K^{9}} A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K ^{9} A_{\\mathcal{H}}(E),$$\n\nand these inequalities are asymptotically sharp when $$K\\to1^{+}$$.\n\n### Proof\n\nLet $$E\\subset \\mathbb {H}$$ be a measurable set. The Jacobian of f is $$J_{f}=u_{x}v_{y}$$. By (7) and item 1 of Theorem 6,\n\n$$\\frac{(v_{y}^{+}(0))^{2}}{K^{5}}\\leq J_{f}\\leq K^{5} \\bigl(v_{y}^{-}(0)\\bigr)^{2}\\quad \\mbox{a.e}.$$\n\nThe Euclidean area of $$f(E)$$ is\n\n$$\\int_{E}J_{f}\\,dx\\,dy=A_{e}\\bigl(f(E)\\bigr)$$\n\nand in consequence\n\n$$\\frac{(v_{y}^{+}(0))^{2}}{K^{5}} A_{e}(E)\\leq A_{e}\\bigl(f(E)\\bigr)\\leq K^{5}\\bigl(v _{y}^{-}(0)\\bigr)^{2} A_{e}(E).$$\n\nOn the other hand, for the hyperbolic area only, we see that from (15) it follows\n\n$$\\frac{1}{K^{9}} \\biggl( \\frac{v_{y}^{+}(0)}{v_{y}^{-}(0)} \\biggr) ^{2} A _{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K^{9} \\biggl( \\frac{v _{y}^{-}(0)}{v_{y}^{+}(0)} \\biggr) ^{2} A_{\\mathcal{H}}(E).$$\n\n□\n\nNow, we can prove Theorem 3.\n\n### Proof of Theorem 3\n\nIf $$f:\\mathbb {H}\\to \\mathbb {H}$$ is a K-quasiconformal mapping that leaves invariant the family of horocyclics with common tangent point at ∞, then $$f(x+iy)=u(x,y)+iv(y)$$ and is exactly Theorem 8.\n\nWe now suppose that $$f:\\mathbb {H}\\to \\mathbb {H}$$ is a K-quasiconformal mapping such that f maps horocyclics with tangential point at $$x_{0}\\in \\mathbb {R}$$ onto horocyclics with tangential point at $$x_{1}\\in \\mathbb {R}$$. For $$i=0, 1$$, let $$H_{i}$$ be a horocyclic with tangential point at $$x_{i}$$ and $$H_{\\infty}^{i}\\subset \\mathbb {H}$$ be a horocyclic with tangential point at infinity. Then there exist Möbius transformations $$T_{i}$$ such that $$T_{i}(H_{i})=H_{\\infty}^{i}$$. We define $$\\hat{f}:H_{\\infty}^{0}\\to H_{\\infty}^{1}$$ by $$\\hat{f}(z)=(T _{1}\\circ f\\circ T_{0}^{-1})(z)$$, then is K-quasiconformal and can be written in the form $$\\hat{f}(x+iy)=u(x,y)+iv(y)$$. By Theorem 8 we have that if $$E\\subset \\mathbb {H}$$ is a measurable set, then satisfies\n\n$$\\frac{1}{K^{9}} A_{\\mathcal{H}}\\bigl(T_{0}(E)\\bigr)\\leq A_{\\mathcal{H}}\\bigl( \\hat{f}\\bigl(T_{0}(E)\\bigr)\\bigr)\\leq K^{9} A_{\\mathcal{H}}\\bigl(T_{0}(E)\\bigr).$$\n\nSince the hyperbolic area in $$\\mathbb {H}$$ is invariant under the Möbius transformations $$T_{i}$$, the result follows immediately. The other case is similar. □\n\n### Corollary 2\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself given by $$f(x+iy)=u(x,y)+iv(y)$$. Then, for each measurable set $$E\\subset \\mathbb {H}$$,\n\n1. (i)\n\nIf v is differentiable at 0, then\n\n$$\\frac{(v'_{y}(0))^{2}}{K^{9}} A_{e}(E)\\leq A_{e}\\bigl(f(E)\\bigr)\\leq K^{9} \\bigl(v'_{y}(0)\\bigr)^{2} A_{e}(E).$$\n2. (ii)\n\nIf v is continuously differentiable in a neighborhood of 0,\n\n$$\\frac{(v'_{y}(0))^{2}}{K^{5}} A_{e}(E)\\leq A_{e}\\bigl(f(E)\\bigr)\\leq K^{5} \\bigl(v'_{y}(0)\\bigr)^{2} A_{e}(E).$$\n\nThese inequalities are asymptotically sharp when $$K\\to1^{+}$$.\n\n### Proof\n\nIf v is differentiable at 0, then from (15) we get $$\\frac{v_{y}^{-}(0)}{K^{2}}\\leq v_{y}'(0)\\leq K^{2} v_{y}^{+}(0)$$. If v is continuously differentiable in a neighborhood of 0, then $$v_{y}^{-}(0)=v_{y}'(0)=v_{y}^{+}(0)$$. We prove the corollary applying item 1 of Theorem 8. □\n\nThe following examples of quasiconformal mappings are not harmonic, thus we are generalizing the results obtained in .\n\n### Example 1\n\n1. 1.\n\nLet $$f:\\mathbb {H}\\to \\mathbb {H}$$ given by\n\n$$f(x+iy)=x+\\frac{i}{2} \\bigl( 2y\\arctan y+2\\pi y-\\ln\\bigl(1+y^{2} \\bigr) \\bigr).$$\n\nThen f is a $$\\frac{3\\pi}{2}$$-quasiconformal mapping with\n\n$$v_{y}^{+}(0)=v_{y}^{-}(0)=v'(0)= \\pi.$$\n\nThus, for each measurable set $$E\\subset \\mathbb {H}$$, we have\n\n$$\\frac{32}{243\\pi^{3}}A_{e}(E)\\leq A_{e}\\bigl(f(E)\\bigr)\\leq \\frac{243\\pi^{7}}{32}A_{e}(E)$$\n\nand\n\n$$\\frac{512}{19{,}683\\pi^{9}}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr) \\leq \\frac{19{,}683\\pi^{9}}{512}A_{\\mathcal{H}}(E).$$\n2. 2.\n\nLet $$f:\\mathbb {H}\\to \\mathbb {H}$$ given by $$f(x+iy)=2x+\\sin(x+y)+iy$$. Then f is a $$\\frac{11 +\\sqrt{85}}{6}$$-quasiconformal mapping with\n\n$$v_{y}^{+}(0)=v_{y}^{-}(0)=v'(0)=1.$$\n\nIt is enough to observe that the dilatation of f is given by\n\n\\begin{aligned}& D_{f}(x,y) \\\\& \\quad =\\frac{\\sqrt{\\cos^{2}(x+y)+(3+\\cos(x+y))^{2}}+\\sqrt{\\cos^{2}(x+y)+(1+ \\cos(x+y))^{2}}}{\\sqrt{\\cos^{2}(x+y)+(3+\\cos(x+y))^{2}}-\\sqrt{ \\cos^{2}(x+y)+(1+\\cos(x+y))^{2}}} \\end{aligned}\n\nand the function $$D_{f}(x,a-x)$$ depends only on a, more precisely\n\n$$D_{f}(x,a-x)=\\frac{\\sqrt{\\cos^{2}a+(3+\\cos a)^{2}}+\\sqrt{\\cos ^{2}a+(1+\\cos a)^{2}}}{\\sqrt{\\cos^{2}a+(3+\\cos a)^{2}}-\\sqrt{\\cos ^{2}a+(1+\\cos a)^{2}}}.$$\n\nThe maximum of $$D_{f}(x,a-x)$$ is attained at the critical points $$a=2l\\pi$$, $$l=0, \\pm1, \\pm2,\\ldots$$ and the maximal dilatation of f is\n\n$$\\frac{\\sqrt{17}+\\sqrt{5}}{\\sqrt{17}-\\sqrt{5}}=\\frac{11+ \\sqrt{85}}{6}.$$\n\nThus, for each measurable set $$E\\subset \\mathbb {H}$$, we have\n\n$$\\biggl( \\frac{6}{11+\\sqrt{85}} \\biggr) ^{5}A_{e}(E)\\leq A_{e}\\bigl(f(E)\\bigr) \\leq \\biggl( \\frac{11+\\sqrt{85}}{6} \\biggr) ^{5}A_{e}(E)$$\n\nand\n\n$$\\biggl( \\frac{6}{11+\\sqrt{85}} \\biggr) ^{9}A_{\\mathcal{H}}(E)\\leq A _{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq \\biggl( \\frac{11+\\sqrt{85}}{6} \\biggr) ^{9}A _{\\mathcal{H}}(E).$$\n\n### 2.3 Angular and radial quasiconformal mappings\n\nIn this part we obtain the results of area distortion for radial and angular mappings. In the case of angular mappings, we use the hyperbolic model of the unit disk $$\\mathbb{D}$$.\n\n### Proposition 2\n\nLet $$f:\\Omega\\to \\mathbb {C}$$ be an ACL mapping. If $$f(re^{i\\theta})= u(re ^{i\\theta})+i v(re^{i\\theta})$$, then for a.e. in Ω\n\n\\begin{aligned} &4\\vert f_{z}\\vert ^{2} = \\biggl( u_{r} + \\frac{v_{\\theta}}{r} \\biggr) ^{2}+ \\biggl( v_{r} - \\frac{u_{\\theta}}{r} \\biggr) ^{2}, \\end{aligned}\n(16)\n\\begin{aligned} &4\\vert f_{\\overline{z}}\\vert ^{2}= \\biggl( u_{r} - \\frac{v_{\\theta}}{r} \\biggr) ^{2}+ \\biggl( v_{r} + \\frac{u_{\\theta}}{r} \\biggr) ^{2} \\end{aligned}\n(17)\n\nand the Jacobian of f is\n\n$$J_{f} =\\frac{1}{r}(u_{r}v_{\\theta}-u_{\\theta}v_{r}).$$\n(18)\n\nA mapping $$f:\\mathbb {H}\\to \\mathbb {H}$$ is said to be radial at $$x\\in \\mathbb {R}$$ if f leaves invariant all Euclidean rays in $$\\mathbb {H}$$ that meet at x.\n\nLet $$f:\\mathbb {H}\\to \\mathbb {H}$$ be a radial mapping at x. Since hyperbolic area is invariant under horizontal translations, we can assume that the point $$x\\in \\mathbb {R}$$, where the Euclidean rays meet, is $$x=0$$. If f is a radial mapping, then f can be written in polar coordinates $$(r,\\theta)$$ as $$f(z)=f(re^{i\\theta})=\\rho(r,\\theta)e^{i\\theta}$$, with $$\\rho(r, \\theta):(0,\\infty]\\times(0,\\pi)\\to(0,\\infty)$$ if $$z=re^{i\\theta }$$.\n\n### Lemma 1\n\nLet f be an ACL mapping from $$\\mathbb {H}$$ onto itself. Suppose that f is a radial mapping at 0. Then its Jacobian is $$J_{f}=\\frac{\\rho\\rho _{r}}{r}$$ a.e. If f preserves orientation, then $$\\rho_{r}> 0$$ a.e.\n\n### Proof\n\nSince $$f(z)=f(re^{i\\theta})=\\rho(r,\\theta)e^{i\\theta}$$, then $$u(r,\\theta)=\\rho(r,\\theta)\\cos\\theta$$ and $$v(r,\\theta)=\\rho(r, \\theta)\\sin\\theta$$, and the proof is immediate from (18). □\n\n### Proposition 3\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself. Suppose that f is a radial mapping at 0. Then the function ρ satisfies the following:\n\n1. 1.\n\nFor $$1\\leq r<\\infty$$,\n\n$$r^{\\frac{1}{K}}\\leq\\frac{\\rho(r,\\theta)}{\\rho(1,\\theta)}\\leq r ^{K};$$\n2. 2.\n\nFor $$0< r<1$$,\n\n$$r^{K}\\leq\\frac{\\rho(r,\\theta)}{\\rho(1,\\theta)}\\leq r^{ \\frac{1}{K}}.$$\n\n### Proof\n\nWe first prove that the function $$(0,\\infty)\\ni r\\mapsto\\ln\\rho(r, \\theta)$$ is absolutely continuous for almost every $$\\theta\\in(0, \\pi)$$. It is enough to prove that for every $$M>1$$, the function $$[\\frac{1}{M},M]\\ni r\\mapsto\\ln\\rho(r,\\theta)$$ is absolutely continuous for almost every $$\\theta\\in(0,\\pi)$$. Let $$\\Omega=\\{z=x+i \\theta\\in \\mathbb {C}: (x,\\theta)\\in(-\\infty,\\infty)\\times(0, \\pi)\\}$$. Then the mapping $$\\log\\circ f\\circ\\exp: \\Omega\\to \\Omega$$ is K-quasiconformal. Thus the function $$(-\\infty,\\infty) \\ni x\\mapsto\\ln\\rho(e^{x},\\theta)$$ is absolutely continuous for almost every $$\\theta\\in(0,\\pi)$$. Let $$\\varepsilon>0$$. There exists $$\\delta>0$$ such that for every finite collection of disjoint intervals $$(a_{j},b_{j})\\subset \\mathbb {R}$$, $$j=1, 2, \\ldots, n$$, with $$\\sum_{j=1} ^{n} (b_{j} -a_{j}) < \\delta$$, then\n\n$$\\sum_{j=1}^{n} \\bigl( \\ln\\rho \\bigl(e^{b_{j}},\\theta\\bigr)-\\ln\\rho\\bigl(e^{a_{j}}, \\theta\\bigr) \\bigr) < \\varepsilon.$$\n\nSince lnr is absolutely continuous on $$[\\frac{1}{M},M]$$, there exists $$\\delta'>0$$ such that for every finite collection of disjoint intervals $$(c_{l},d_{l})\\subset[\\frac{1}{M},M]$$, $$l=1, 2, \\ldots, m$$, with $$\\sum_{l=1}^{m} (d_{l} -c_{l}) < \\delta'$$, then\n\n$$\\sum_{l=1}^{m} (\\ln d_{l}-\\ln c_{l})< \\delta,$$\n\nand by the previous inequality\n\n$$\\sum_{l=1}^{m} \\bigl( \\ln \\rho(d_{l},\\theta)-\\ln\\rho(c_{l},\\theta) \\bigr) < \\varepsilon.$$\n\nIf $$f(z)=\\rho(r,\\theta)e^{i\\theta}$$, from (16) and (17) we get\n\n$$\\vert f_{z}\\vert ^{2}=\\frac{1}{4} \\biggl( \\rho^{2}_{r}+2\\frac{\\rho_{r}\\rho }{r}+\\frac{ \\rho^{2}}{r^{2}}+ \\frac{\\rho^{2}_{\\theta}}{r^{2}} \\biggr)\\quad \\mbox{a.e.}$$\n(19)\n\nand\n\n$$\\vert f_{\\overline{z}}\\vert ^{2}=\\frac{1}{4} \\biggl( \\rho_{r}^{2}-\\frac{2\\rho _{r} \\rho}{r}+ \\frac{\\rho^{2}}{r^{2}}+\\frac{\\rho_{\\theta}^{2}}{r ^{2}} \\biggr)\\quad \\mbox{a.e.}$$\n(20)\n\nBy (2), (19) and (20) we have\n\n$$\\frac{1}{4} \\biggl[ \\biggl( \\rho_{r}-\\frac{\\rho}{r} \\biggr) ^{2}+\\frac{ \\rho^{2}_{\\theta}}{r^{2}} \\biggr] \\leq k^{2} \\frac{1}{4} \\biggl[ \\biggl( \\rho _{r}+\\frac{\\rho}{r} \\biggr) ^{2}+\\frac{\\rho^{2}_{\\theta }}{r^{2}} \\biggr]\\quad \\mbox{a.e.}$$\n\nor equivalently\n\n$$\\frac{2(r^{2}\\rho_{r}^{2}+\\rho^{2}+\\rho_{\\theta}^{2})}{4r\\rho_{r} \\rho}\\leq\\frac{k^{2}+1}{1 -k^{2}}=\\alpha\\quad \\mbox{a.e.}$$\n\nThen\n\n$$\\frac{1}{2} \\biggl[ \\frac{r\\rho_{r}}{\\rho}+\\frac{\\rho}{r\\rho _{r}} \\biggr] \\leq \\alpha\\quad \\mbox{a.e.}$$\n\nand thus\n\n$$\\biggl( \\frac{r\\rho_{r}}{\\rho} \\biggr) ^{2}-2\\alpha \\biggl( \\frac {r\\rho _{r}}{\\rho} \\biggr) +1\\leq0 \\quad \\mbox{a.e.}$$\n\nSolving this inequality, we obtain\n\n$$\\frac{1}{Kr}\\leq\\frac{\\rho_{r}}{\\rho}\\leq\\frac{K}{r}\\quad \\mbox{a.e.}$$\n(21)\n\nor equivalently\n\n$$\\frac{1}{Kr}\\leq\\frac{\\partial}{\\partial r}\\ln\\rho\\leq \\frac{K}{r}\\quad \\mbox{a.e.}$$\n\nWe choose any fixed $$\\theta\\in(0,\\pi)$$ such that $$\\ln\\rho(r, \\theta)$$ is absolutely continuous on r, and we integrate the previous inequality on the interval $$[1,R]$$ to get\n\n$$\\int_{1}^{R}\\frac{1}{Kr}\\,dr\\leq \\int_{1}^{R}\\frac{\\partial }{\\partial r}\\ln\\rho \\,dr\\leq \\int_{1}^{R}\\frac{K}{r}\\,dr.$$\n\nThus\n\n$$\\frac{1}{K} \\ln r \\bigl\\vert _{1}^{R}\\leq\\ln \\rho(r,\\theta)\\bigr\\vert _{1}^{R} \\leq K \\ln r \\vert _{1}^{R} \\quad \\mbox{for almost every }\\theta\\in(0, \\pi)\\mbox{ and }R\\in[1,\\infty).$$\n\nBy an argument of continuity of f and density, we finally obtain\n\n$$R^{\\frac{1}{K}}\\leq\\frac{\\rho(R,\\theta)}{\\rho(1,\\theta)}\\leq R ^{K}\\quad \\mbox{for all }(R, \\theta)\\in[1,\\infty)\\times(0,\\pi).$$\n\nIn a similar way, if we suppose that $$0< R<1$$, then\n\n$$R^{K}\\leq\\frac{\\rho(R,\\theta)}{\\rho(1,\\theta)}\\leq R^{ \\frac{1}{K}}\\quad \\mbox{for all }(R, \\theta)\\in(0,1)\\times(0, \\pi).$$\n\n□\n\n### Theorem 9\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {H}$$ onto itself that leaves invariant each ray in $$\\mathbb {H}$$ that meets a real base point. If $$E\\subset \\mathbb {H}$$ is a measurable set, then\n\n$$\\frac{1}{K}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K A_{ \\mathcal{H}}(E).$$\n\nThese inequalities are asymptotically sharp when $$K\\to1^{+}$$.\n\n### Proof\n\nWe can suppose that the base point is 0 since hyperbolic area is invariant under horizontal translations. Let $$E\\subset \\mathbb {H}$$ be a measurable set and Ê denote the set E in polar coordinates. If $$f:\\mathbb {H}\\to \\mathbb {H}$$ is given by $$f(z)=\\rho(r,\\theta)e ^{i\\theta}$$, then\n\n$$A_{\\mathcal{H}}\\bigl(f(E)\\bigr)= \\iint_{f(E)}\\frac{du \\,dv}{(\\operatorname {Im}w)^{2}}= \\iint _{E}\\frac{ J_{f}\\,dx \\,dy}{(\\operatorname {Im}f(z))^{2}}= \\iint_{\\widehat{E}}\\frac{ \\rho_{r}\\,dr \\,d\\theta}{\\rho\\sin^{2}\\theta}.$$\n\nBy (21) we have\n\n$$\\frac{1}{Kr}\\leq\\frac{\\rho_{r}}{\\rho}\\leq\\frac{K}{r} \\quad \\mbox{a.e.}$$\n\nthen\n\n$$\\frac{1}{K} \\iint_{\\widehat{E}}\\frac{r \\,dr \\,d\\theta}{r^{2}\\sin^{2} \\theta}\\leq \\iint_{\\widehat{E}}\\frac{ \\rho_{r}\\,dr \\,d\\theta}{\\rho \\sin^{2}\\theta}\\leq K \\iint_{\\widehat{E}}\\frac{r \\,dr \\,d\\theta}{r^{2} \\sin^{2}\\theta}$$\n\nor equivalently, in rectangular coordinates,\n\n$$\\frac{1}{K} \\iint_{E}\\frac{ dx \\,dy}{y^{2}}\\leq A_{\\mathcal{H}}\\bigl(f(E) \\bigr) \\leq K \\iint_{E}\\frac{ dx \\,dy}{y^{2}},$$\n\nthat is,\n\n$$\\frac{1}{K}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K A_{ \\mathcal{H}}(E).$$\n\n□\n\n### Example 2\n\nLet $$f:\\mathbb {H}\\to \\mathbb {H}$$ be the $$\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}$$-quasiconformal mapping which is radial at 0 given by\n\n$$f\\bigl(re^{i\\theta}\\bigr)=r \\biggl( \\frac{\\theta^{2}}{2}+\\theta+1 \\biggr) e^{i \\theta}.$$\n\nThen, for all measurable set $$E\\subset \\mathbb {H}$$, the mapping f satisfies\n\n$$\\frac{\\sqrt{2}-1}{\\sqrt{2}+1}A_{\\mathcal{H}}(E)\\leq A_{ \\mathcal{H}}\\bigl(f(E)\\bigr)\\leq \\frac{\\sqrt{2}+1}{\\sqrt{2}-1} A_{ \\mathcal{H}}(E).$$\n\nIn the following case we consider the hyperbolic model in the unit disk $$\\mathbb {D}$$.\n\nA mapping $$f:\\mathbb {D}\\to \\mathbb {D}$$ is said to be angular at $$0\\in \\mathbb {D}$$ if f leaves invariant each circle in $$\\mathbb {D}$$ with center at 0.\n\nAn angular mapping f at 0 can be written as $$f(z)= f(re^{i\\theta })=r e ^{i\\varphi(r,\\theta)}$$, where $$\\varphi:[0,1)\\times[0,2 \\pi]\\to \\mathbb {R}$$.\n\n### Lemma 2\n\nLet f be an ACL mapping from $$\\mathbb {D}$$ onto itself. Suppose that f is angular at 0. Then its Jacobian is $$J_{f}=\\varphi_{\\theta}$$. If f preserves orientation, then $$\\varphi_{\\theta}> 0$$ a.e.\n\n### Proof\n\nSince $$f(z)=f(re^{i\\theta})=re^{i\\varphi(r,\\theta)}$$, then $$u(r,\\theta)=r\\cos\\varphi(r,\\theta)$$ and $$v(r,\\theta)=r\\sin \\varphi(r,\\theta)$$, and the proof is immediate from (18). □\n\n### Proposition 4\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {D}$$ onto itself which is angular at 0. Then\n\n$$\\frac{1}{K}\\leq\\varphi_{\\theta}\\leq K \\quad \\textit{a.e. in }[0,1) \\times[0,2\\pi].$$\n(22)\n\n### Proof\n\nIf $$f(z)=f(re^{i\\theta})=re^{i\\varphi(r,\\theta)}$$, from (16) and (17) we get\n\n\\begin{aligned}& \\begin{aligned} &4\\vert f_{z}\\vert ^{2} =(1+\\varphi_{\\theta})^{2} +r^{2} \\varphi_{r}^{2}\\quad \\mbox{a.e. } \\\\ &4\\vert f_{\\overline{z}}\\vert ^{2} =(1-\\varphi_{\\theta})^{2} +r^{2} \\varphi _{r}^{2}\\quad \\mbox{a.e. } \\end{aligned} \\end{aligned}\n(23)\n\nSince\n\n$$\\frac{(1-\\varphi_{\\theta})^{2} }{(1+\\varphi_{\\theta})^{2} }\\leq \\frac{ \\vert f_{\\overline{z}}\\vert ^{2}}{\\vert f_{z}\\vert ^{2}}\\leq k^{2} \\quad \\mbox{a.e.},$$\n\nwe get the result. □\n\n### Corollary 3\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {D}$$ onto itself. Suppose that f is angular at 0. Then, for each $$0< \\theta_{1}<\\theta _{2}<2\\pi$$ and $$r\\in(0,1)$$, the following holds:\n\n$$\\frac{\\theta_{2}-\\theta_{1}}{K}\\leq\\varphi(r,\\theta_{2}) -\\varphi (r, \\theta_{1})\\leq K (\\theta_{2}-\\theta_{1}).$$\n\n### Proof\n\nIt is immediate integrating (22) and applying the continuity of $$\\varphi(r,\\theta)$$ and density. Thus, it is enough to prove that $$\\theta\\mapsto\\varphi(r,\\theta)$$, $$\\theta\\in(0,2 \\pi)$$, is absolutely continuous for almost every $$r\\in(0,1)$$.\n\nLet $$\\Omega_{1}=\\{z=x+i\\theta\\in \\mathbb {C}: -\\infty< x <0 \\mbox{ and } 0< \\theta<2\\pi\\}$$ and $$\\Omega_{2}=\\{z=x+i\\theta\\in \\mathbb {C}: -\\infty< x <0 \\mbox{ and } \\varphi(r,0)< \\theta<2\\pi+ \\varphi(r,0)\\}$$. Then the mapping $$\\log\\circ f\\circ\\exp: \\Omega _{1} \\to\\Omega_{2}$$ is K-quasiconformal. Thus the function $$(0,2\\pi)\\ni\\theta\\mapsto\\varphi(e^{x},\\theta)$$ is absolutely continuous for almost every $$x\\in(-\\infty,0)$$, or equivalently $$\\theta\\mapsto\\varphi(r,\\theta)$$, $$\\theta\\in(0,2\\pi)$$, is absolutely continuous for almost every $$r\\in(0,1)$$. □\n\n### Theorem 10\n\nLet f be a K-quasiconformal mapping from $$\\mathbb {D}$$ onto itself which is angular at 0. If $$E\\subset \\mathbb {H}$$ is a measurable set, then\n\n$$\\frac{1}{K}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr)\\leq K A_{ \\mathcal{H}}(E).$$\n\nThese inequalities are asymptotically sharp when $$K\\to1^{+}$$.\n\n### Proof\n\nLet $$E\\subset \\mathbb {D}$$ be a measurable set and Ê denote the set E in polar coordinates. If $$f:\\mathbb {D}\\to \\mathbb {D}$$ is given as before by $$f(z)=re^{i\\varphi(r,\\theta)}$$, then\n\n$$A_{\\mathcal{H}}\\bigl(f(E)\\bigr)= \\iint_{f(E)} \\frac{4\\,du\\,dv}{(1-\\vert w\\vert ^{2})^{2}}= \\iint_{E}\\frac{ 4J_{f}\\,dx \\,dy}{(1-\\vert f(z)\\vert ^{2})^{2}}= \\iint_{\\widehat{E}}\\frac{ 4 \\varphi_{\\theta} r \\,dr \\,d\\theta}{(1- r^{2})^{2}}.$$\n\nBy (22) we have\n\n$$\\frac{1}{K} \\iint_{\\widehat{E}}\\frac{4r \\,dr\\,d\\theta}{(1- r^{2})^{2}} \\leq \\iint_{\\widehat{E}}\\frac{ 4\\varphi_{\\theta} r \\,dr \\,d\\theta }{(1- r^{2})^{2}}\\leq K \\iint_{\\widehat{E}}\\frac{4r \\,dr\\,d\\theta}{(1- r ^{2})^{2}}$$\n\nor equivalently, in rectangular coordinates,\n\n$$\\frac{1}{K} \\iint_{E}\\frac{ 4\\,dx \\,dy}{(1-\\vert z\\vert ^{2})^{2}}\\leq A _{\\mathcal{H}}\\bigl(f(E) \\bigr)\\leq K \\iint_{E}\\frac{ 4\\,dx \\,dy}{(1-\\vert z\\vert ^{2})^{2}},$$\n\nand this concludes the proof. □\n\n### Example 3\n\nLet $$f:\\mathbb {D}\\to \\mathbb {D}$$ be a mapping given by $$f(re^{i\\theta})=re^{i \\varphi(r,\\theta)}$$, where $$\\varphi(r,\\theta)= \\theta e^{(\\theta -2\\pi)r}$$. Then by (23)\n\n\\begin{aligned} \\frac{\\vert f_{z}\\vert +\\vert f_{\\overline{z}}\\vert }{\\vert f_{z}\\vert -\\vert f_{\\overline{z}}\\vert } & \\leq\\frac{4\\vert f_{z}\\vert ^{2}}{\\vert f_{z}\\vert ^{2}-\\vert f_{\\overline{z}}\\vert ^{2}} \\\\ & =\\frac{e ^{2r(\\theta-2\\pi)}r^{2}\\theta^{2}(\\theta-2\\pi)^{2} +(1+ e^{r( \\theta-2\\pi)}(1+r\\theta))^{2}}{(1+r\\theta)e^{r(\\theta-2\\pi)}} \\\\ &\\leq e^{r(\\theta-2\\pi)}r^{2}\\theta^{2}(\\theta-2 \\pi)^{2}+\\frac{1}{e ^{r(\\theta-2\\pi)}}+2+(1+r\\theta)e^{r(\\theta-2\\pi)} \\\\ &\\leq\\pi^{4}+e^{2\\pi r}+2+(1+2\\pi) \\\\ &\\leq\\pi^{4}+2\\pi+3+e^{2\\pi}. \\end{aligned}\n\nThen f is a $$\\pi^{4}+2\\pi+3+e^{2\\pi}$$-quasiconformal mapping. Thus, for each measurable set E of the unit disk $$\\mathbb {D}$$, the following inequalities hold:\n\n$$\\frac{1}{\\pi^{4}+2\\pi+3+e^{2\\pi}} A_{\\mathcal{H}} (E)\\leq A_{ \\mathcal{H}} \\bigl(f(E)\\bigr) \\leq\\bigl(\\pi^{4}+2\\pi+3+e^{2\\pi}\\bigr) A_{\\mathcal{H}} (E).$$\n\nNumerical evidence says that the maximal dilatation of f can be $$e^{2\\pi}$$.\n\nThe following example shows that the result of Theorem 10 can not be generalized to radial mappings at 0.\n\n### Example 4\n\nLet $$K\\geq1$$. Let $$f, g:\\mathbb {D}\\to \\mathbb {D}$$ be the K-quasiconformal mappings given by $$f(re^{i\\theta}) = r^{\\frac{1}{K}} e^{i\\theta}$$ and $$g(re^{i\\theta}) = r^{K} e^{i\\theta}$$. For each $$r\\in(0,1)$$, define $$E_{r}=\\{ z\\in \\mathbb {D}: \\vert z\\vert < r \\}$$. Then\n\n$$A_{\\mathcal{H}}(E_{r})=\\frac{4\\pi r^{2}}{1-r^{2}} , \\qquad A_{\\mathcal{H}} \\bigl(f(E _{r})\\bigr)=\\frac{4\\pi r^{\\frac{2}{K}}}{1-r^{\\frac{2}{K}}} ,\\qquad A_{ \\mathcal{H}}\\bigl( g(E_{r})\\bigr)=\\frac{4\\pi r^{2K}}{1-r^{2K}}.$$\n\nThen if $$K>1$$, there is not $$C>0$$ such that\n\n$$A_{\\mathcal{H}} \\bigl(f(E_{r})\\bigr)\\leq C A_{\\mathcal{H}}(E_{r}) \\quad \\mbox{or}\\quad \\frac{ A_{\\mathcal{H}}(E_{r})}{C} \\leq A_{ \\mathcal{H}} \\bigl(g(E_{r}) \\bigr)\\quad \\mbox{for all }r\\in(0,1).$$\n\n### 2.4 The Beurling-Ahlfors extension\n\nUsing the Beurling-Ahlfors (BA) extension, we give explicit examples of quasiconformal mappings of the form $$f(x,y)=u(x,y)+iv(x,y)$$ and their associated bi-bounds for the hyperbolic area distortion.\n\nMore precisely, let $$h:\\mathbb {R}\\to \\mathbb {R}$$ be an increasing homeomorphism and define its Beurling-Ahlfors extension $$f:\\overline{\\mathbb {R}}\\to \\overline{\\mathbb {R}}$$ by $$f(x+iy)=u(x,y)+iv(x,y)$$, where\n\n$$u(x,y)=\\frac{1}{2y} \\int_{-y}^{y}h(x+t)\\,dt,\\qquad v(x,y)=\\frac{1}{2y} \\int_{0}^{y} \\bigl( h(x+t)-h(x-t) \\bigr)\\,dt.$$\n\nLet $$M\\geq1$$. An increasing homeomorphism $$h:\\mathbb {R}\\to \\mathbb {R}$$ is M-quasisymmetric if\n\n$$\\frac{1}{M}\\leq\\frac{h(x+t)-h(x)}{h(x)-h(x-t)}\\leq M$$\n\nfor all $$x\\in \\mathbb {R}$$ and $$t>0$$. It is well known that its BA-extension is a $$K=K(M)\\geq1$$ quasiconformal mapping, even more; Ahlfors proved in that this extension is a quasi-isometry, that is, there exists $$0< C=C(K)<\\infty$$ such that\n\n$$\\frac{1}{Cy^{2}}\\leq\\frac{J(z)}{v^{2}}\\leq\\frac{C}{y^{2}}\\quad \\mbox{a.e.}$$\n\n### Example 5\n\nLet $$h(x)=x$$. Then h is a 1-quasisymmetric function and its BA-extension is the 2-quasiconformal and harmonic mapping defined by $$f(x+iy)=x+\\frac{iy}{2}$$. Moreover, for each measurable set $$E\\subset \\mathbb {H}$$,\n\n$$A_{e}\\bigl(f(E)\\bigr)=\\frac{1}{2}A_{e}(E) \\quad \\mbox{and}\\quad A_{\\mathcal{H}}\\bigl(f(E)\\bigr)=2A _{\\mathcal{H}}(E).$$\n\n### Example 6\n\nLet $$g(x)=x^{3}$$. Then g is a $$7+4\\sqrt{3}$$-quasisymmetric function, and its BA-extension is the $$20. 7872\\ldots$$ quasiconformal mapping $$f(x+iy)=x^{3}+xy^{2}+\\frac{i}{4}(6x^{2}y+y^{3})$$. In particular $$f(x,y)$$ does not have bounded derivatives. Moreover,\n\n$$\\frac{y^{2} J(x+iy)}{v(x,y)^{2}}= \\frac{12(6 x^{4} - 3 x^{2} y^{2} + y^{4})}{(6 x^{2} + y^{2})^{2}}.$$\n\nSetting $$y=cx$$ in the right-hand side, we get for $$x\\neq0$$\n\n$$r(c)= \\frac{12 (6 - 3 c^{2} + c^{4})}{(6 + c^{2})^{2}}$$\n\nand it is easy to see that\n\n$$\\frac{3}{4} \\leq r(c)< 12\\quad \\mbox{for each }c\\in \\mathbb {R}.$$\n\nThus, for each measurable set $$E\\subset \\mathbb {H}$$, we have\n\n$$\\frac{3}{4}A_{\\mathcal{H}}(E)\\leq A_{\\mathcal{H}}\\bigl(f(E)\\bigr) \\leq12 A_{ \\mathcal{H}}(E).$$\n\nLet\n\n$$E=\\biggl\\{ (x,y)\\in \\mathbb {H}: x\\in[1,\\infty), 0< y\\leq\\frac{1}{x^{2}}\\biggr\\} .$$\n\nWe have that $$A_{e}(E)=1$$ and, since $$J_{f}(x+iy)=\\frac{3}{4}(6x^{4}-3x ^{2}y^{2}+y^{4})$$, it holds $$A_{e}(f(E))=\\infty$$. Thus f explodes Euclidean area.\n\n### Example 7\n\nLet $$g(x+iy)=2x+\\sin(x+y)+iy$$. Then g is a $$\\frac{11+\\sqrt{85}}{6}$$-quasiconformal mapping from $$\\mathbb {H}$$ onto itself. Thus the function $$h(x)=2x+\\sin x$$ is $$2. 2\\ldots$$ -quasisymmetric and its BA-extension is $$3. 2\\ldots$$ -quasiconformal, given by\n\n$$f(x+iy)=2x + \\frac{\\sin x\\sin y}{y}+ \\frac{i}{y} \\biggl( y^{2} + \\cos x -\\frac{\\cos(x+ y)}{2} -\\frac{\\cos(x-y)}{2} \\biggr).$$\n\nIn a forthcoming paper we study more deeply quasi-isometric properties of the BA extension.\n\n### 2.5 A set that contains the region of values of the partial derivatives of K-quasiconformal mappings\n\nIn this part we study some particular forms of the mapping f in (1). First, suppose that f is a K-quasiconformal mapping given by $$f(x+iy)=u(x)+iv(y)$$. Then, by (4), its partial derivatives satisfy the inequality\n\n$$u_{x}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}\\leq0\\quad \\mbox{a.e.}$$\n(24)\n\nSince $$\\alpha\\geq1$$, the discriminant of $$u_{x}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}$$ is non-negative and (24) defines the interior of an angular region with the identification $$u_{x} \\sim x- \\mathrm{axis}$$ and $$v_{y} \\sim y-\\mathrm{axis}$$. Thus we have proved.\n\n### Theorem 11\n\nLet $$1\\leq K <\\infty$$. If $$f:\\Omega\\to \\mathbb {C}$$ is a K-quasiconformal mapping given by $$f(x+iy)=u(x)+iv(y)$$, then its partial derivatives belong to one of the angular regions defined by (24).\n\n### Proof\n\nThe proof follows from the fact that the Jacobian of f is always positive. □\n\nIf f is a K-quasiconformal mapping given by $$f(x+iy)=u(x,y)+iv(y)$$, then (4) reduces to\n\n$$u_{x}^{2}+u_{y}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}\\leq0 \\quad \\mbox{a.e.}$$\n(25)\n\nInequality (25) suggests studying the quadratic form $$Q(x,y,w)=x^{2}+y^{2}+w^{2}-2\\alpha xw$$, whose associated symmetric matrix is\n\n$$N= \\left( \\begin{matrix}{} 1 & 0 & -\\alpha \\\\ 0 & 1 & 0 \\\\ -\\alpha& 0 & 1 \\end{matrix}\\right) .$$\n\n### Proposition 5\n\nThere exists an invertible matrix P such that $$P^{-1}NP=D$$, where\n\n$$D= \\left( \\begin{matrix}{} 1-\\alpha& 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1+\\alpha \\end{matrix}\\right) .$$\n\n### Proof\n\nThe eigenvalues of N are $$\\lambda_{1}=1-\\alpha$$, $$\\lambda_{2}=1$$ and $$\\lambda_{3}=1+\\alpha$$ with eigenvectors $$(1,0,1)$$, $$(0,1,0)$$ and $$(1,0,-1)$$, respectively. After normalization we obtain the basis $$\\mathcal{B}:=\\{(\\frac{1}{\\sqrt{2}},0,\\frac{1}{\\sqrt{2}}), (0,1,0), (\\frac{1}{\\sqrt{2}},0,-\\frac{1}{\\sqrt{2}})\\}$$. Set\n\n$$P= \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ 0 & 1 & 0 \\\\ \\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} \\end{matrix}\\right) =P^{-1}.$$\n\nA simple calculus ends the proof. □\n\n### Corollary 4\n\nThe quadratic form $$\\widehat{Q}(\\widehat{x},\\widehat{y},\\widehat{w})=(1- \\alpha)\\widehat{x}^{2}+\\widehat{y}^{2}+(1+\\alpha)\\widehat{w}^{2}$$ represents the quadratic form $$Q(x,y,w)$$ with basis $$\\mathcal{B}$$, where\n\n$$\\left( \\begin{matrix}{} x \\\\ y \\\\ w \\end{matrix}\\right) = \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ 0 & 1 & 0 \\\\ \\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} \\end{matrix}\\right) \\left( \\begin{matrix}{} \\widehat{x} \\\\ \\widehat{y} \\\\ \\widehat{w} \\end{matrix}\\right) .$$\n\nIn particular $$Q(x,y,w)\\leq0$$ if and only if $$\\widehat{Q}( \\widehat{x},\\widehat{y},\\widehat{w})\\leq0$$.\n\nThe solution of $$\\widehat{Q}(\\widehat{x},\\widehat{y},\\widehat {w})=0$$ is a double cone if $$\\alpha>1$$. In fact, $$\\widehat{Q}(\\widehat{x}, \\widehat{y},\\widehat{w})=0$$ if and only if\n\n$$(\\alpha-1)\\widehat{x}^{2}=\\widehat{y}^{2}+(1+\\alpha) \\widehat{w}^{2}$$\n\nor equivalently\n\n$$\\widehat{x}^{2}=\\frac{\\widehat{y}^{2}}{(\\alpha-1)}+\\frac{(1+\\alpha)}{( \\alpha-1)} \\widehat{w}^{2}$$\n\nand this is the equation of a double elliptic cone in $$\\mathbb {R}^{3}$$ in the basis $$\\mathcal{B}$$.\n\n### Proposition 6\n\nLet $$1< K <\\infty$$. If $$f:\\Omega\\to \\mathbb {C}$$ is a K-quasiconformal mapping given by $$f(x+iy)=u(x,y)+iv(y)$$, then its partial derivatives $$u_{x}$$, $$u_{y}$$ and $$v_{y}$$ belong to one branch of the elliptic cone (25).\n\n### Proof\n\nAs we saw f is K-quasiconformal if and only if $$u_{x}$$, $$u_{y}$$ and $$v_{y}$$ satisfy $$Q(u_{x},u_{y},v_{y})\\leq0$$ that describes the solid cone (25). As f preserves orientation, then $$J_{f}=u _{x}v_{y}>0$$ a.e. Since $$v_{y}>0$$ a.e., then necessarily $$u_{x}>0$$ a.e., and the result follows. □\n\nWe do not study the case $$f(x+iy)=u(x,y)+iv(x)$$ because this kind of mapping is not a homeomorphism from $$\\mathbb {H}$$ onto itself and in consequence is not quasiconformal.\n\nWe consider now the general case, that is, a quasiconformal mapping $$f:\\Omega\\subset \\mathbb {C}\\to \\mathbb {C}$$ given by $$f(x+iy)=u(x,y)+iv(x,y)$$, then by (4) we have that\n\n$$u_{x}^{2}+u_{y}^{2}+v_{x}^{2}+v_{y}^{2}-2 \\alpha u_{x}v_{y}+2\\alpha v _{x}u_{y}\\leq0\\quad \\mbox{a.e.}$$\n\nIn this case we study the quadratic form $$Q(x,y,z,w)=x^{2}+y^{2}+z ^{2}+w^{2}-2\\alpha xw+2\\alpha yz$$ with the associated symmetric matrix\n\n$$N= \\left( \\begin{matrix}{} 1 & 0 & 0 & -\\alpha \\\\ 0 & 1 & \\alpha& 0 \\\\ 0 & \\alpha& 1 & 0 \\\\ -\\alpha& 0 & 0 & 1 \\end{matrix}\\right) .$$\n\n### Proposition 7\n\nThere exists an invertible matrix P such that $$P^{-1}NP=D$$, where\n\n$$D= \\left( \\begin{matrix}{} 1-\\alpha& 0 & 0 & 0 \\\\ 0 & 1-\\alpha& 0 & 0 \\\\ 0 & 0 & 1+\\alpha& 0 \\\\ 0 & 0 & 0 & 1+\\alpha \\end{matrix}\\right) .$$\n\n### Proof\n\nThe characteristic polynomial of the matrix N is $$(1-\\lambda)^{4}- \\alpha^{2}(1-\\lambda)^{2}-\\alpha^{2}(1-\\lambda)^{2}+\\alpha^{4}$$ with eigenvalues $$\\lambda_{1}=1+\\alpha$$, $$\\lambda_{2}=1-\\alpha$$, and both with multiplicity two. The eigenvectors of $$\\lambda_{1}$$ are $$(1,0,0,-1)$$ and $$(0,1,1,0)$$ and for $$\\lambda_{2}$$ are $$(1,0,0,1)$$ and $$(0,1,-1,0)$$. After normalization we obtain the matrix\n\n$$P= \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ 0 & -\\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} & 0 \\end{matrix}\\right)$$\n\nwith inverse\n\n$$P^{-1}= \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}} & 0 & 0 & \\frac{1}{\\sqrt{2}} \\\\ 0 & \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} & 0 \\\\ \\frac{1}{\\sqrt{2}} & 0 & 0 & -\\frac{1}{\\sqrt{2}} \\\\ 0 & \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 0 \\end{matrix}\\right) .$$\n\nThus\n\n$$P^{-1}NP=D.$$\n\n□\n\n### Corollary 5\n\n$$\\widehat{Q}(\\widehat{x},\\widehat{y},\\widehat{z},\\widehat{w})=(1- \\alpha) \\widehat{x}^{2}+(1-\\alpha)\\widehat{y}^{2}+(1+\\alpha) \\widehat{z}^{2}+(1+\\alpha)\\widehat{w}^{2}$$\n\n$$Q(x,y,z,w)=x^{2}+y^{2}+z^{2}+w^{2}-2 \\alpha xw+2\\alpha yz$$\n\nin the basis\n\n$$\\mathcal{C}= \\biggl\\{ \\biggl(\\frac{1}{\\sqrt{2}},0,0,-\\frac{1}{\\sqrt {2}} \\biggr),\\biggl(0,\\frac{1}{ \\sqrt{2}},\\frac{1}{\\sqrt{2}},0\\biggr),\\biggl( \\frac{1}{\\sqrt{2}},0,0,\\frac{1}{ \\sqrt{2}}\\biggr),\\biggl(0,\\frac{1}{\\sqrt{2}},- \\frac{1}{\\sqrt{2}},0\\biggr) \\biggr\\} ,$$\n\nwhere\n\n$$\\left( \\begin{matrix}{} x \\\\ y \\\\ z \\\\ w \\end{matrix}\\right) = \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} & 0 \\\\ 0 & \\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ 0 & -\\frac{1}{\\sqrt{2}} & 0 & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & 0 & -\\frac{1}{\\sqrt{2}} & 0 \\end{matrix}\\right) \\left( \\begin{matrix}{} \\widehat{x} \\\\ \\widehat{y} \\\\ \\widehat{z} \\\\ \\widehat{w} \\end{matrix}\\right) .$$\n\nIn particular $$Q(x,y,z,w)\\leq0$$ if and only if $$\\widehat{Q}( \\widehat{x},\\widehat{y},\\widehat{z},\\widehat{w})\\leq0$$.\n\n### Proof\n\nWe have the relations\n\n$$\\left( \\begin{matrix}{} x \\\\ y \\\\ z \\\\ w \\end{matrix}\\right) = \\left( \\begin{matrix}{} \\frac{1}{\\sqrt{2}}(\\widehat{x}+\\widehat{z}) \\\\ \\frac{1}{\\sqrt{2}}(\\widehat{y}+\\widehat{w}) \\\\ \\frac{1}{\\sqrt{2}}(\\widehat{w}-\\widehat{y}) \\\\ \\frac{1}{\\sqrt{2}}(\\widehat{x}-\\widehat{z}) \\end{matrix}\\right) .$$\n\nThus\n\n\\begin{aligned} Q(x,y,z,w) =&x^{2}+y^{2}+z^{2}+w^{2}-2 \\alpha xw+2\\alpha yz \\\\ =&\\frac{1}{2} ( \\widehat{x}+\\widehat{z} ) ^{2}+ \\frac{1}{2}( \\widehat{y}+\\widehat{w})^{2}+\\frac{1}{2}( \\widehat{w}-\\widehat{y})^{2}+ \\frac{1}{2} ( \\widehat{x}- \\widehat{z} ) ^{2} \\\\ &{}-2\\alpha \\biggl[ \\frac{1}{\\sqrt{2}} ( \\widehat{x}+\\widehat {z} ) \\biggr] \\biggl[ \\frac{1}{\\sqrt{2}} ( \\widehat{x}-\\widehat{z} ) \\biggr] +2 \\alpha \\biggl[ \\frac{1}{\\sqrt{2}} ( \\widehat{w}+\\widehat {y} ) \\biggr] \\biggl[ \\frac{1}{\\sqrt{2}} ( \\widehat{w}-\\widehat{y} ) \\biggr] \\\\ =&\\frac{1}{2}\\bigl[\\widehat{x}^{2}+2\\widehat{x}\\widehat{z}+ \\widehat{z} ^{2}\\bigr]+\\frac{1}{2}\\bigl[\\widehat{y}^{2}+2 \\widehat{y}\\widehat{w}+\\widehat{w} ^{2}\\bigr] \\\\ &{}+\\frac{1}{2}\\bigl[\\widehat{x}^{2}-2\\widehat{x}\\widehat{z}+ \\widehat{z} ^{2}\\bigr]+\\frac{1}{2}\\bigl[\\widehat{w}^{2}-2 \\widehat{w}\\widehat{y}+\\widehat{y} ^{2}\\bigr] \\\\ &{}-\\alpha\\bigl[\\widehat{x}^{2}-\\widehat{z}^{2}\\bigr]+\\alpha \\bigl[\\widehat{w}^{2}- \\widehat{y}^{2}\\bigr] \\\\ =&(1-\\alpha)\\widehat{x}^{2}+(1-\\alpha)\\widehat{y}^{2}+(1+ \\alpha) \\widehat{z}^{2}+(1+\\alpha)\\widehat{w}^{2} \\\\ =&\\widehat{Q}(\\widehat{x},\\widehat{y},\\widehat{z},\\widehat{w}); \\end{aligned}\n\nand this means that $$Q(x,y,z,w)\\leq0$$ if and only if $$\\widehat{Q}( \\widehat{x},\\widehat{y},\\widehat{z},\\widehat{w})\\leq0$$. □\n\n### Proposition 8\n\nLet $$1\\leq K <\\infty$$. Let $$f:\\Omega\\to \\mathbb {C}$$ be a K-quasiconformal mapping given by $$f(x+iy)=u(x,y)+iv(x,y)$$. Then its partial derivatives $$u_{x}$$, $$u_{y}$$ and $$v_{x}$$, $$v_{y}$$ belong to the solid bounded by $$Q(x,y,z,w)= \\widehat{Q}(\\widehat{x},\\widehat{y},\\widehat{z}, \\widehat{w})=0$$.\n\n## 3 Conclusions\n\nThe classes of mappings introduced in this paper have precise geometrical meaning, in particular, the class of quasiconformal mappings $$f(z)=u(x,y) +i v(y)$$; see, for example, Proposition 1.\n\nAs it is showed in Theorems 3, 6, 8, 9, 10 and Corollaries 1 and 2, we have obtained left and right asymptotic bounds for the hyperbolic or Euclidean area distortion. In some previous results only right bounds were known. Moreover, the examples showed that the different classes of mappings defined in the paper are not empty.\n\nWe do not know whether the branch of the elliptic cone (25), mentioned in Proposition 6, coincides or not with the region of variation of the partial derivatives of quasiconformal mappings $$f(z)=u(x,y)+iv(y)$$.\n\n## References\n\n1. Beurling, A, Ahlfors, LV: The boundary correspondence under quasiconformal mappings. Acta Math. 96, 125-142 (1956)\n\n2. Ahlfors, LV: Quasiconformal reflections. Acta Math. 109, 291-301 (1963)\n\n3. Astala, K: Area distortion of quasiconformal mappings. Acta Math. 173, 37-60 (1994)\n\n4. Porter, RM, Reséndis, LF: Quasiconformally explodable sets. Complex Var. Theory Appl. 36, 379-392 (1998)\n\n5. Emerenko, A, Hamilton, DH: On the area distortion by quasiconformal mappings. Proc. Am. Math. Soc. 123, 2793-2797 (1995)\n\n6. Chen, X, Qian, T: Estimate of hyperbolically partial derivatives of ρ-harmonic quasiconformal mappings and its applications. Complex Var. Elliptic Equ. 60, 875-892 (2015)\n\n7. Dongmian, F, Xinzhong, H: Harmonic K-quasiconformal mappings from the unit disk onto half planes. Bull. Malays. Math. Soc. 39(1), 339-347 (2016)\n\n8. Kalaj, D, Mateljevi, M: Quasiconformal harmonic mappings and generalizations. In: Proceedings of the ICM2010 Satellite Conference International Workshop on Harmonic and Quasiconformal Mappings (HQM2010), vol. 18, pp. 239-260 (2010)\n\n9. Partyka, D, Sakan, K: On a asymptotically sharp variant of Heinz’s inequality. Ann. Acad. Sci. Fenn., Math. 30, 167-182 (2005)\n\n10. Knežević, M, Mateljević, M: On the quasi-isometries of harmonic quasiconformal mappings. J. Math. Anal. Appl. 334(1), 404-413 (2007)\n\n11. Chen, M, Chen, X: $$(K,K')$$-quasiconformal harmonic mappings of the upper half plane onto itself. Ann. Acad. Sci. Fenn., Math. 37, 265-276 (2012)\n\n12. Axler, S Bourdon, P, Ramey, W: Harmonic Function Theory, pp. 1-259. Springer, New York (2001)\n\n13. Anderson, J: Hyperbolic Geometry, pp. 1-230. Springer, London (2003)\n\n14. Beardon, A: The Geometry of Discrete Groups, pp. 1-338. Springer, New York (1983)\n\n## Acknowledgements\n\nWe are very grateful to the referees for valuable suggestions which have improved both the presentation and contents of this paper.\n\nThis research was partially supported by CONACYT and Programa de Análisis Matemático de la Universidad Autónoma Metropolitana-Azcapotzalco.\n\nSome examples in this paper were calculated using Wolfram Mathematica, supported by Universidad Autónoma Metropolitana-Azcapotzalco.\n\nThe material of this paper belongs to the theory of plane quasiconformal mappings. The omnipresence of double inequalities associated to fundamental facts is well known in this theory, for example:\n\n(a) modulus of normalized quasiconformal mappings from the unit disk onto itself;\n\n(b) quasi-isometry of the Beurling-Ahlfors extension;\n\n(c) bi-Lipschitz inequalities for quasiconformal harmonic mappings; etc.\n\nIn this paper we study the hyperbolic or Euclidean area distortion under certain classes of quasiconformal mappings, defined in the upper half-plane or the unit disk. We found left and right bounds for the mentioned distortions. In particular, we are generalizing the results obtained for harmonic quasiconformal mappings from the upper half-plane onto itself (see Chen and Chen $$(K,K')$$ -quasiconformal harmonic mappings of the upper half plane onto itself Ann. Aca. Scien. Fen, pp.265-276 (2012)). We also give a straightforward application of a recent result to this topic (see Knežević and Mateljević On the quasi-isometries of harmonic quasiconformal mappings Journal of Mathematical Analysis and Applications, pp.404-413 (2007)). Of course the generality of results is very desirable in mathematics; however, some particular cases have their own interest.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Lino F Reséndis O.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors’ contributions\n\nAll the authors contributed equally and significantly in writing this paper. All the authors read and approved the final manuscript.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Rights and permissions", null, "" ]

[ null, "https://journalofinequalitiesandapplications.springeropen.com/track/article/10.1186/s13660-017-1481-1", null ]

[ "Search a number\nBaseRepresentation\nbin100000100111…\n…111011101011\n3121002111101002\n4200213323223\n54142132141\n6503145215\n7132456305\noct40477353\n917074332\n108552171\n114911411\n122a4520b\n131a05874\n1411c8975\n15b3de9b\nhex827eeb\n\n8552171 has 4 divisors (see below), whose sum is σ = 8570952. Its totient is φ = 8533392.\n\nThe previous prime is 8552161. The next prime is 8552177. The reversal of 8552171 is 1712558.\n\nIt is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 1712558 = 2856279.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 8552171 - 26 = 8552107 is a prime.\n\nIt is an Ulam number.\n\nIt is a Duffinian number.\n\nIt is not an unprimeable number, because it can be changed into a prime (8552177) by changing a digit.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 8690 + ... + 9623.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (2142738).\n\nAlmost surely, 28552171 is an apocalyptic number.\n\n8552171 is a deficient number, since it is larger than the sum of its proper divisors (18781).\n\n8552171 is a wasteful number, since it uses less digits than its factorization.\n\n8552171 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 18780.\n\nThe product of its digits is 2800, while the sum is 29.\n\nThe square root of 8552171 is about 2924.4095130470. The cubic root of 8552171 is about 204.4994410587.\n\nThe spelling of 8552171 in words is \"eight million, five hundred fifty-two thousand, one hundred seventy-one\".\n\nDivisors: 1 467 18313 8552171" ]

[ null ]

[ "", null, "# Double.toString(double) sometimes produces incorrect results\n\nXMLWordPrintable\n\n#### Details\n\n• Type:", null, "CSR\n• Status: Closed\n• Priority:", null, "P3\n• Resolution: Approved\n• Fix Version/s:\n• Component/s:\n• Labels:\nNone\n• Subcomponent:\n• Compatibility Kind:\nbehavioral\n• Compatibility Risk:\nminimal\n• Compatibility Risk Description:\nHide\nThe current specification does not always uniquely specify the resulting string. In addition, there are many documented anomalies in the implementation (JDK-4511638) that are still present. Thus, the risk of encountering slightly different results is latent since ever and can potentially manifest at each release of the platform: any correction, even without a spec modification, would introduce slight incompatibilities anyway.\n\nThe proposed specification aims at uniquely determine the resulting value. When accompanied by a conforming implementation, it should help removing the uncertainties associated with the current, more vague spec.\n\nThe behavior changes will also be visible in certain other parts of the platform, including java.math (JDK-8205592).\nShow\nThe current specification does not always uniquely specify the resulting string. In addition, there are many documented anomalies in the implementation ( JDK-4511638 ) that are still present. Thus, the risk of encountering slightly different results is latent since ever and can potentially manifest at each release of the platform: any correction, even without a spec modification, would introduce slight incompatibilities anyway. The proposed specification aims at uniquely determine the resulting value. When accompanied by a conforming implementation, it should help removing the uncertainties associated with the current, more vague spec. The behavior changes will also be visible in certain other parts of the platform, including java.math ( JDK-8205592 ).\n• Interface Kind:\nJava API\n• Scope:\nSE\n\n## Summary\n\nModify the specification (Javadoc) of Double::toString(double) and Float::toString(float) to ensure a uniquely determined resulting string value in all cases.\n\n## Problem\n\nThe current Javadoc specifications of the mentioned methods are somehow vague in what the resulting strings should be.\n\nOn the one hand, a strict reading leads to believe that the digits in the resulting string are drawn, from left to right, from the exact value of the argument, until the number represented by the string is near enough to the argument as to round to it according to the default IEEE 754 round-to-closest mode.\n\nOn the other hand, a more lenient interpretation of the Javadoc and the observed behavior both lead to the conclusion that the digits appearing in the result are those of an unspecified number that also rounds to the argument. While the spec makes it clear that it must be a shortest one that possibly rounds to the argument, sometimes there are more choices. The spec says nothing in these cases. In summary, it is not always clear from which number the digits are eventually drawn. In absence of a more specific description of this number, the result is not always uniquely determined and different implementations are thus allowed to return different strings.\n\n## Solution\n\nSpecify that the conversion is split in two separate stages. The first selects a unique, well specified decimal number that represents the argument and that meets the properties listed in the specification section below. The second stage then format this decimal number as a string, as specified below.\n\nThe current and the proposed specs, while different in wording, determine exactly the same resulting strings for the vast majority of cases. Their results, however, might differ where the current one is not specific enough.\n\n## Specification\n\nDouble::toString(double):\n\n`````` /**\n* Returns a string representation of the {@code double}\n* argument. All characters mentioned below are ASCII characters.\n* <ul>\n* <li>If the argument is NaN, the result is the string\n* \"{@code NaN}\".\n* <li>Otherwise, the result is a string that represents the sign and\n* magnitude (absolute value) of the argument. If the sign is negative,\n* the first character of the result is '{@code -}'\n* ({@code '\\u005Cu002D'}); if the sign is positive, no sign character\n* appears in the result. As for the magnitude <i>m</i>:\n* <ul>\n* <li>If <i>m</i> is infinity, it is represented by the characters\n* {@code \"Infinity\"}; thus, positive infinity produces the result\n* {@code \"Infinity\"} and negative infinity produces the result\n* {@code \"-Infinity\"}.\n*\n* <li>If <i>m</i> is zero, it is represented by the characters\n* {@code \"0.0\"}; thus, negative zero produces the result\n* {@code \"-0.0\"} and positive zero produces the result\n* {@code \"0.0\"}.\n*\n* <li> Otherwise <i>m</i> is positive and finite.\n* It is converted to a string in two stages:\n* <ul>\n* <li> <em>Selection of a decimal</em>:\n* A well-defined decimal <i>d</i><sub><i>m</i></sub>\n* is selected to represent <i>m</i>.\n* This decimal is (almost always) the <em>shortest</em> one that\n* rounds to <i>m</i> according to the round to nearest\n* rounding policy of IEEE 754 floating-point arithmetic.\n* <li> <em>Formatting as a string</em>:\n* The decimal <i>d</i><sub><i>m</i></sub> is formatted as a string,\n* either in plain or in computerized scientific notation,\n* depending on its value.\n* </ul>\n* </ul>\n* </ul>\n*\n* <p>A <em>decimal</em> is a number of the form\n* <i>s</i>&times;10<sup><i>i</i></sup>\n* for some (unique) integers <i>s</i> &gt; 0 and <i>i</i> such that\n* <i>s</i> is not a multiple of 10.\n* These integers are the <em>significand</em> and\n* the <em>exponent</em>, respectively, of the decimal.\n* The <em>length</em> of the decimal is the (unique)\n* positive integer <i>n</i> meeting\n* 10<sup><i>n</i>-1</sup> &le; <i>s</i> &lt; 10<sup><i>n</i></sup>.\n*\n* <p>The decimal <i>d</i><sub><i>m</i></sub> for a finite positive <i>m</i>\n* is defined as follows:\n* <ul>\n* <li>Let <i>R</i> be the set of all decimals that round to <i>m</i>\n* according to the usual <em>round to nearest</em> rounding policy of\n* IEEE 754 floating-point arithmetic.\n* <li>Let <i>p</i> be the minimal length over all decimals in <i>R</i>.\n* <li>When <i>p</i> &ge; 2, let <i>T</i> be the set of all decimals\n* in <i>R</i> with length <i>p</i>.\n* Otherwise, let <i>T</i> be the set of all decimals\n* in <i>R</i> with length 1 or 2.\n* <li>Define <i>d</i><sub><i>m</i></sub> as the decimal in <i>T</i>\n* that is closest to <i>m</i>.\n* Or if there are two such decimals in <i>T</i>,\n* select the one with the even significand.\n* </ul>\n*\n* <p>The (uniquely) selected decimal <i>d</i><sub><i>m</i></sub>\n* is then formatted.\n* Let <i>s</i>, <i>i</i> and <i>n</i> be the significand, exponent and\n* length of <i>d</i><sub><i>m</i></sub>, respectively.\n* Further, let <i>e</i> = <i>n</i> + <i>i</i> - 1 and let\n* <i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub>\n* be the usual decimal expansion of <i>s</i>.\n* Note that <i>s</i><sub>1</sub> &ne; 0\n* and <i>s</i><sub><i>n</i></sub> &ne; 0.\n* Below, the decimal point <code>.</code> is {@code '\\u005Cu002E'}\n* and the exponent indicator <code>E</code> is {@code '\\u005Cu0045'}.\n* <ul>\n* <li>Case -3 &le; <i>e</i> &lt; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <code>0.0</code>&hellip;<code>0</code><!--\n* --><i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub>,\n* where there are exactly -(<i>n</i> + <i>i</i>) zeroes between\n* the decimal point and <i>s</i><sub>1</sub>.\n* For example, 123 &times; 10<sup>-4</sup> is formatted as\n* {@code 0.0123}.\n* <li>Case 0 &le; <i>e</i> &lt; 7:\n* <ul>\n* <li>Subcase <i>i</i> &ge; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub><!--\n* --><code>0</code>&hellip;<code>0.0</code>,\n* where there are exactly <i>i</i> zeroes\n* between <i>s</i><sub><i>n</i></sub> and the decimal point.\n* For example, 123 &times; 10<sup>2</sup> is formatted as\n* {@code 12300.0}.\n* <li>Subcase <i>i</i> &lt; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub>&hellip;<!--\n* --><i>s</i><sub><i>n</i>+<i>i</i></sub><code>.</code><!--\n* --><i>s</i><sub><i>n</i>+<i>i</i>+1</sub>&hellip;<!--\n* --><i>s</i><sub><i>n</i></sub>,\n* where there are exactly -<i>i</i> digits to the right of\n* the decimal point.\n* For example, 123 &times; 10<sup>-1</sup> is formatted as\n* {@code 12.3}.\n* </ul>\n* <li>Case <i>e</i> &lt; -3 or <i>e</i> &ge; 7:\n* computerized scientific notation is used to format\n* <i>d</i><sub><i>m</i></sub>.\n* Here <i>e</i> is formatted as by {@link Integer#toString(int)}.\n* <ul>\n* <li>Subcase <i>n</i> = 1:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub><code>.0E</code><i>e</i>.\n* For example, 1 &times; 10<sup>23</sup> is formatted as\n* {@code 1.0E23}.\n* <li>Subcase <i>n</i> &gt; 1:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub><code>.</code><i>s</i><sub>2</sub><!--\n* -->&hellip;<i>s</i><sub><i>n</i></sub><code>E</code><i>e</i>.\n* For example, 123 &times; 10<sup>-21</sup> is formatted as\n* {@code 1.23E-19}.\n* </ul>\n* </ul>\n*\n* <p>To create localized string representations of a floating-point\n* value, use subclasses of {@link java.text.NumberFormat}.\n*\n* @param d the {@code double} to be converted.\n* @return a string representation of the argument.\n*/\npublic static String toString(double d) {}``````\n\nFloat::toString(float):\n\n`````` /**\n* Returns a string representation of the {@code float}\n* argument. All characters mentioned below are ASCII characters.\n* <ul>\n* <li>If the argument is NaN, the result is the string\n* \"{@code NaN}\".\n* <li>Otherwise, the result is a string that represents the sign and\n* magnitude (absolute value) of the argument. If the sign is\n* negative, the first character of the result is\n* '{@code -}' ({@code '\\u005Cu002D'}); if the sign is\n* positive, no sign character appears in the result. As for\n* the magnitude <i>m</i>:\n* <ul>\n* <li>If <i>m</i> is infinity, it is represented by the characters\n* {@code \"Infinity\"}; thus, positive infinity produces\n* the result {@code \"Infinity\"} and negative infinity\n* produces the result {@code \"-Infinity\"}.\n* <li>If <i>m</i> is zero, it is represented by the characters\n* {@code \"0.0\"}; thus, negative zero produces the result\n* {@code \"-0.0\"} and positive zero produces the result\n* {@code \"0.0\"}.\n*\n* <li> Otherwise <i>m</i> is positive and finite.\n* It is converted to a string in two stages:\n* <ul>\n* <li> <em>Selection of a decimal</em>:\n* A well-defined decimal <i>d</i><sub><i>m</i></sub>\n* is selected to represent <i>m</i>.\n* This decimal is (almost always) the <em>shortest</em> one that\n* rounds to <i>m</i> according to the round to nearest\n* rounding policy of IEEE 754 floating-point arithmetic.\n* <li> <em>Formatting as a string</em>:\n* The decimal <i>d</i><sub><i>m</i></sub> is formatted as a string,\n* either in plain or in computerized scientific notation,\n* depending on its value.\n* </ul>\n* </ul>\n* </ul>\n*\n* <p>A <em>decimal</em> is a number of the form\n* <i>s</i>&times;10<sup><i>i</i></sup>\n* for some (unique) integers <i>s</i> &gt; 0 and <i>i</i> such that\n* <i>s</i> is not a multiple of 10.\n* These integers are the <em>significand</em> and\n* the <em>exponent</em>, respectively, of the decimal.\n* The <em>length</em> of the decimal is the (unique)\n* positive integer <i>n</i> meeting\n* 10<sup><i>n</i>-1</sup> &le; <i>s</i> &lt; 10<sup><i>n</i></sup>.\n*\n* <p>The decimal <i>d</i><sub><i>m</i></sub> for a finite positive <i>m</i>\n* is defined as follows:\n* <ul>\n* <li>Let <i>R</i> be the set of all decimals that round to <i>m</i>\n* according to the usual <em>round to nearest</em> rounding policy of\n* IEEE 754 floating-point arithmetic.\n* <li>Let <i>p</i> be the minimal length over all decimals in <i>R</i>.\n* <li>When <i>p</i> &ge; 2, let <i>T</i> be the set of all decimals\n* in <i>R</i> with length <i>p</i>.\n* Otherwise, let <i>T</i> be the set of all decimals\n* in <i>R</i> with length 1 or 2.\n* <li>Define <i>d</i><sub><i>m</i></sub> as the decimal in <i>T</i>\n* that is closest to <i>m</i>.\n* Or if there are two such decimals in <i>T</i>,\n* select the one with the even significand.\n* </ul>\n*\n* <p>The (uniquely) selected decimal <i>d</i><sub><i>m</i></sub>\n* is then formatted.\n* Let <i>s</i>, <i>i</i> and <i>n</i> be the significand, exponent and\n* length of <i>d</i><sub><i>m</i></sub>, respectively.\n* Further, let <i>e</i> = <i>n</i> + <i>i</i> - 1 and let\n* <i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub>\n* be the usual decimal expansion of <i>s</i>.\n* Note that <i>s</i><sub>1</sub> &ne; 0\n* and <i>s</i><sub><i>n</i></sub> &ne; 0.\n* Below, the decimal point <code>.</code> is {@code '\\u005Cu002E'}\n* and the exponent indicator <code>E</code> is {@code '\\u005Cu0045'}.\n* <ul>\n* <li>Case -3 &le; <i>e</i> &lt; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <code>0.0</code>&hellip;<code>0</code><!--\n* --><i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub>,\n* where there are exactly -(<i>n</i> + <i>i</i>) zeroes between\n* the decimal point and <i>s</i><sub>1</sub>.\n* For example, 123 &times; 10<sup>-4</sup> is formatted as\n* {@code 0.0123}.\n* <li>Case 0 &le; <i>e</i> &lt; 7:\n* <ul>\n* <li>Subcase <i>i</i> &ge; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub>&hellip;<i>s</i><sub><i>n</i></sub><!--\n* --><code>0</code>&hellip;<code>0.0</code>,\n* where there are exactly <i>i</i> zeroes\n* between <i>s</i><sub><i>n</i></sub> and the decimal point.\n* For example, 123 &times; 10<sup>2</sup> is formatted as\n* {@code 12300.0}.\n* <li>Subcase <i>i</i> &lt; 0:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub>&hellip;<!--\n* --><i>s</i><sub><i>n</i>+<i>i</i></sub><code>.</code><!--\n* --><i>s</i><sub><i>n</i>+<i>i</i>+1</sub>&hellip;<!--\n* --><i>s</i><sub><i>n</i></sub>,\n* where there are exactly -<i>i</i> digits to the right of\n* the decimal point.\n* For example, 123 &times; 10<sup>-1</sup> is formatted as\n* {@code 12.3}.\n* </ul>\n* <li>Case <i>e</i> &lt; -3 or <i>e</i> &ge; 7:\n* computerized scientific notation is used to format\n* <i>d</i><sub><i>m</i></sub>.\n* Here <i>e</i> is formatted as by {@link Integer#toString(int)}.\n* <ul>\n* <li>Subcase <i>n</i> = 1:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub><code>.0E</code><i>e</i>.\n* For example, 1 &times; 10<sup>23</sup> is formatted as\n* {@code 1.0E23}.\n* <li>Subcase <i>n</i> &gt; 1:\n* <i>d</i><sub><i>m</i></sub> is formatted as\n* <i>s</i><sub>1</sub><code>.</code><i>s</i><sub>2</sub><!--\n* -->&hellip;<i>s</i><sub><i>n</i></sub><code>E</code><i>e</i>.\n* For example, 123 &times; 10<sup>-21</sup> is formatted as\n* {@code 1.23E-19}.\n* </ul>\n* </ul>\n*\n* <p>To create localized string representations of a floating-point\n* value, use subclasses of {@link java.text.NumberFormat}.\n*\n* @param f the float to be converted.\n* @return a string representation of the argument.\n*/\npublic static String toString(float f) {}``````\n\n#### Attachments\n\n1. Double.diff\n8 kB\n2. Float.diff\n8 kB\n3. specdiff-4511638.00.zip\n171 kB\n4. specdiff-4511638.01.zip\n193 kB\n\n#### People\n\nAssignee:", null, "Raffaello Giulietti\nReporter:", null, "Bill Strathearn (Inactive)\nReviewed By:\nBrian Burkhalter, Guy Steele" ]

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[ "## CBSE Class 12 Maths Question Bank and Important Questions Chapter wise\n\nCBSE Question Bank for Class 12 Maths contains all the important questions Chapter wise. Just after understanding a topic students need to attempt critical questions of that topic to be able to make sure that they understand the topic fully and are ready from examination point of view. Thus Question bank help students to practice and understand every particular chapter of Maths totality and also get well prepared for final exams.\n\n#### Total Papers : 57\n\nClass 12 Maths Marks Distribution\nUnits Marks\nRelations and Functions 8\nAlgebra 10\nCalculus 35\nVectors and Three - Dimensional Geometry 14\nLinear Programming 5\nProbability 8\nTotal 80\nInternal Assessment 20\n\nMaths Topics to be covered for Class 12 Science\n\n• Relations and Functions\n• Inverse Trigonometric Function\n• Matrices,Determinants\n• Continuity and Differentiability\n• Applications of Derivatives\n• Integrals\n• Applications of the Integrals\n• Differential Equations\n• Vectors\n• Three Dimensional Geometry\n• Linear Programming\n• Probability\n\nStructure of CBSE Maths Sample Paper for Class 12 Science is\n\nType of Question Marks per Question Total No. of Questions Total Marks\nObjective Type Questions 1 20 20\nShort Answer Type Questions 2 6 12\nLong Answer Type Question - 1 4 6 24\nLong Answer Type Question - 2 6 4 24\nTotal 36 80\n\nFor Preparation of exams students can also check out other resource material\n\nCBSE Class 12 Maths Sample Papers\n\nCBSE Class 12 Maths Worksheets\n\nCBSE Class 12 Maths Question Papers\n\nCBSE Class 12 Maths Test Papers\n\nCBSE Class 12 Maths Revision Notes\n\nImportance of Question Bank for Exam Preparation?\n\nThere are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows\n\n1. Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.\n2. CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.\n3. Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.\n4. Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.\nUpload Papers & Earn 50 Points" ]

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[ "(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);" ]

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[ "• Counting by numbers other than one\n• Even and Odd Numbers\n• Counting by Twos with Even Numbers\n• Next Even Number when Counting by Twos\n• Counting by Twos with Odd Numbers\n• Next Odd Number when Counting by Twos\n• Counting by Threes\n• Next Number when Counting by Threes\n• Counting by Fours\n• Next Number when Counting by Fours\n• Counting by Fives\n• Next Number when Counting by Fives\n• Counting by Tens\n• Next Number when Counting by Tens\n• Addition of Five Digit Numbers\n• Relationship between Addition and Subtraction\n• Subtraction\n• Subtracting with Mental Math\n• Subtraction with Pencil and Paper\n• Rounding and Estimation\n• Rounding to the nearest ten\n• Rounding to the nearest hundred\n• Rounding Numbers to the Nearest Thousand\n• Subtraction using Estimation\n• Front End Estimation of Sums\n• Front End Estimation of Differences\n• Multiplication Facts\n• Multiplication Facts for Zero to Three\n• Multiplication Facts for Four to Six\n• Multiplication Facts for Seven to Nine\n• Multiplication Practice for Zero to Nine\n• Multiplication Facts for Ten to Twelve\n• Multiplication Practice for Zero to Twelve\n• Multiplication\n• Multiplying Tens\n• Multiplication without Renaming 2 digit\n• Multiplication of Two Digit Numbers\n• Multiplying Hundreds\n• Multiplication without Renaming 3 digit\n• Multiplication with Renaming 3 digit\n• Multiplication of Three Numbers\n• MULTIPLICATION OF MORE THAN 2 digit NUMBERS\n• Relationship of Multiplication and Division\n• Inverse Relationship between Multiplication and Division\n• Inverse Relationship between Division and Multiplication\n• Division Facts\n• Relationship of Subtraction to Division\n• Division Facts for Zero to Three\n• Division Facts for Four to Six\n• Division Facts for Seven to Nine\n• Division Practice for One to Nine\n• Division Facts for Ten to Twelve\n• Division Practice for One to Twelve\n• Division\n• Dividing Tens\n• Dividing Without a Remainder\n• Dividing Hundreds\n•  Fractions\n• Comparing Fractions with Like Denominators\n• Comparing Fractions with Unlike Denominators\n• Identifying Equivalent Fractions\n• Comparing Fractions with Different Numerators and Denominators\n• Subtraction of Fractions\n• multiplication and division of fractions\n• Decimals\n• Decimals - Tenths\n• Decimals - Hundredths\n• Decimals -  Thousandths\n• Subtraction of Decimals\n• MULTIPLICATION OF DECIMALS\n• Consumer Math\n• Converting Coin Values\n• Change From a Purchase\n• Coins For Change\n• Measurement - Time\n• Addition of Hours and Minutes\n• Converting Minutes to Hours\n• Converting Hours to Minutes\n• Time Periods\n• Days in Months\n• Months\n• Time - Convert weeks to days\n• Time - Convert weeks and days to days\n• Time - Add Days and Weeks\n• Time - Convert hours to days and hours\n• Time - Convert days to hours\n• Time - Convert days and hours to hours\n• Time - Add Days and hours\n• Time - Convert hours to minutes\n• Time - Add hours and minutes\n• Time - Convert minutes to seconds\n• Time - Convert minutes and seconds to seconds\n• Time - Add Minutes and seconds\n• Time - Convert days to weeks and days\n• Time - Convert seconds to minutes and seconds\n• Time zones Continental United States" ]

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[ "Prealgebra 2e\n\n# 9.2Solve Money Applications\n\nPrealgebra 2e9.2 Solve Money Applications\n\n### Learning Objectives\n\nBy the end of this section, you will be able to:\n• Solve coin word problems\n• Solve ticket and stamp word problems\nBe Prepared 9.4\n\nBefore you get started, take this readiness quiz.\n\nMultiply: $14(0.25).14(0.25).$\nIf you missed this problem, review Example 5.15.\n\nBe Prepared 9.5\n\nSimplify: $100(0.2+0.05n).100(0.2+0.05n).$\nIf you missed this problem, review Example 7.22.\n\nBe Prepared 9.6\n\nSolve: $0.25x+0.10(x+4)=2.50.25x+0.10(x+4)=2.5$\nIf you missed this problem, review Example 8.44.\n\n### Solve Coin Word Problems\n\nImagine taking a handful of coins from your pocket or purse and placing them on your desk. How would you determine the value of that pile of coins?\n\nIf you can form a step-by-step plan for finding the total value of the coins, it will help you as you begin solving coin word problems.\n\nOne way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, you would add the total value of each pile.\n\nFigure 9.4 To determine the total value of a stack of nickels, multiply the number of nickels times the value of one nickel.(Credit: Darren Hester via ppdigital)\n\nHow would you determine the value of each pile? Think about the dime pile—how much is it worth? If you count the number of dimes, you'll know how many you have—the number of dimes.\n\nBut this does not tell you the value of all the dimes. Say you counted $1717$ dimes, how much are they worth? Each dime is worth $0.100.10$—that is the value of one dime. To find the total value of the pile of $1717$ dimes, multiply $1717$ by $0.100.10$ to get $1.70.1.70.$ This is the total value of all $1717$ dimes.\n\n$17·0.10=1.70number·value=total value17·0.10=1.70number·value=total value$\n\n### Finding the Total Value for Coins of the Same Type\n\nFor coins of the same type, the total value can be found as follows:\n\n$number·value=total valuenumber·value=total value$\n\nwhere number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.\n\nYou could continue this process for each type of coin, and then you would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.\n\nLet's look at a specific case. Suppose there are $1414$ quarters, $1717$ dimes, $2121$ nickels, and $3939$ pennies. We'll make a table to organize the information – the type of coin, the number of each, and the value.\n\nType $NumberNumber$ $Value ()Value ()$ $Total Value ()Total Value ()$\nQuarters $1414$ $0.250.25$ $3.503.50$\nDimes $1717$ $0.100.10$ $1.701.70$\nNickels $2121$ $0.050.05$ $1.051.05$\nPennies $3939$ $0.010.01$ $0.390.39$\n$6.646.64$\nTable 9.1\n\nThe total value of all the coins is $6.64.6.64.$ Notice how Table 9.1 helped us organize all the information. Let's see how this method is used to solve a coin word problem.\n\n### Example 9.11\n\nAdalberto has $2.252.25$ in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?\n\nTry It 9.21\n\nMichaela has $2.052.05$ in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?\n\nTry It 9.22\n\nLiliana has $2.102.10$ in nickels and quarters in her backpack. She has $1212$ more nickels than quarters. How many coins of each type does she have?\n\n### How To\n\n#### Solve a coin word problem.\n\n1. Step 1. Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.\n2. Step 2. Identify what you are looking for.\n3. Step 3. Name what you are looking for. Choose a variable to represent that quantity.\n• Use variable expressions to represent the number of each type of coin and write them in the table.\n• Multiply the number times the value to get the total value of each type of coin.\n4. Step 4. Translate into an equation. Write the equation by adding the total values of all the types of coins.\n5. Step 5. Solve the equation using good algebra techniques.\n6. Step 6. Check the answer in the problem and make sure it makes sense.\n7. Step 7. Answer the question with a complete sentence.\n\nYou may find it helpful to put all the numbers into the table to make sure they check.\n\nType Number Value ($) Total Value Table 9.2 ### Example 9.12 Maria has $2.432.43$ in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have? Try It 9.23 Sumanta has $4.204.20$ in nickels and dimes in her desk drawer. She has twice as many nickels as dimes. How many coins of each type does she have? Try It 9.24 Alison has three times as many dimes as quarters in her purse. She has $9.359.35$ altogether. How many coins of each type does she have? In the next example, we'll show only the completed table—make sure you understand how to fill it in step by step. ### Example 9.13 Danny has $2.142.14$ worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have? Try It 9.25 Jesse has $6.556.55$ worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have? Try It 9.26 Elaine has $7.007.00$ in dimes and nickels in her coin jar. The number of dimes that Elaine has is seven less than three times the number of nickels. How many of each coin does Elaine have? ### Solve Ticket and Stamp Word Problems The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values; so we can organize the information in tables much like we did for coin problems. ### Example 9.14 At a school concert, the total value of tickets sold was $1,506.1,506.$ Student tickets sold for $66$ each and adult tickets sold for $99$ each. The number of adult tickets sold was $55$ less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold? Try It 9.27 The first day of a water polo tournament, the total value of tickets sold was $17,610.17,610.$ One-day passes sold for $2020$ and tournament passes sold for $30.30.$ The number of tournament passes sold was $3737$ more than the number of day passes sold. How many day passes and how many tournament passes were sold? Try It 9.28 At the movie theater, the total value of tickets sold was $2,612.50.2,612.50.$ Adult tickets sold for $1010$ each and senior/child tickets sold for $7.507.50$ each. The number of senior/child tickets sold was $2525$ less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold? Now we'll do one where we fill in the table all at once. ### Example 9.15 Monica paid $10.4410.44$ for stamps she needed to mail the invitations to her sister's baby shower. The number of $49-cent49-cent$ stamps was four more than twice the number of $8-cent8-cent$ stamps. How many $49-cent49-cent$ stamps and how many $8-cent8-cent$ stamps did Monica buy? Try It 9.29 Eric paid $16.6416.64$ for stamps so he could mail thank you notes for his wedding gifts. The number of $49-cent49-cent$ stamps was eight more than twice the number of $8-cent8-cent$ stamps. How many $49-cent49-cent$ stamps and how many $8-cent8-cent$ stamps did Eric buy? Try It 9.30 Kailee paid $14.8414.84$ for stamps. The number of $49-cent49-cent$ stamps was four less than three times the number of $21-cent21-cent$ stamps. How many $49-cent49-cent$ stamps and how many $21-cent21-cent$ stamps did Kailee buy? ### Section 9.2 Exercises #### Practice Makes Perfect Solve Coin Word Problems In the following exercises, solve the coin word problems. 51. Jaime has $2.602.60$ in dimes and nickels. The number of dimes is $1414$ more than the number of nickels. How many of each coin does he have? 52. Lee has $1.751.75$ in dimes and nickels. The number of nickels is $1111$ more than the number of dimes. How many of each coin does he have? 53. Ngo has a collection of dimes and quarters with a total value of $3.50.3.50.$ The number of dimes is $77$ more than the number of quarters. How many of each coin does he have? 54. Connor has a collection of dimes and quarters with a total value of $6.30.6.30.$ The number of dimes is $1414$ more than the number of quarters. How many of each coin does he have? 55. Carolyn has $2.552.55$ in her purse in nickels and dimes. The number of nickels is $99$ less than three times the number of dimes. Find the number of each type of coin. 56. Julio has $2.752.75$ in his pocket in nickels and dimes. The number of dimes is $1010$ less than twice the number of nickels. Find the number of each type of coin. 57. Chi has $11.3011.30$ in dimes and quarters. The number of dimes is $33$ more than three times the number of quarters. How many dimes and nickels does Chi have? 58. Tyler has $9.709.70$ in dimes and quarters. The number of quarters is $88$ more than four times the number of dimes. How many of each coin does he have? 59. A cash box of $11$ and $55$ bills is worth $45.45.$ The number of $11$ bills is $33$ more than the number of $55$ bills. How many of each bill does it contain? 60. Joe's wallet contains $11$ and $55$ bills worth $47.47.$ The number of $11$ bills is $55$ more than the number of $55$ bills. How many of each bill does he have? 61. In a cash drawer there is $125125$ in $55$ and $1010$ bills. The number of $1010$ bills is twice the number of $55$ bills. How many of each are in the drawer? 62. John has $175175$ in $55$ and $1010$ bills in his drawer. The number of $55$ bills is three times the number of $1010$ bills. How many of each are in the drawer? 63. Mukul has $3.753.75$ in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket? 64. Vina has $4.704.70$ in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse? Solve Ticket and Stamp Word Problems In the following exercises, solve the ticket and stamp word problems. 65. The play took in $550550$ one night. The number of$8 adult tickets was $1010$ less than twice the number of $55$ child tickets. How many of each ticket were sold?\n\n66.\n\nIf the number of $88$ child tickets is seventeen less than three times the number of $1212$ adult tickets and the theater took in $584,584,$ how many of each ticket were sold?\n\n67.\n\nThe movie theater took in $1,2201,220$ one Monday night. The number of $77$ child tickets was ten more than twice the number of $99$ adult tickets. How many of each were sold?\n\n68.\n\nThe ball game took in $1,3401,340$ one Saturday. The number of $1212$ adult tickets was $1515$ more than twice the number of $55$ child tickets. How many of each were sold?\n\n69.\n\nJulie went to the post office and bought both $0.490.49$ stamps and $0.340.34$ postcards for her office's bills She spent $62.60.62.60.$ The number of stamps was $2020$ more than twice the number of postcards. How many of each did she buy?\n\n70.\n\nBefore he left for college out of state, Jason went to the post office and bought both $0.490.49$ stamps and $0.340.34$ postcards and spent $12.52.12.52.$ The number of stamps was $44$ more than twice the number of postcards. How many of each did he buy?\n\n71.\n\nMaria spent $16.8016.80$ at the post office. She bought three times as many $0.490.49$ stamps as $0.210.21$ stamps. How many of each did she buy?\n\n72.\n\nHector spent $43.4043.40$ at the post office. He bought four times as many $0.490.49$ stamps as $0.210.21$ stamps. How many of each did he buy?\n\n73.\n\nHilda has $210210$ worth of $1010$ and $1212$ stock shares. The numbers of $1010$ shares is $55$ more than twice the number of $1212$ shares. How many of each does she have?\n\n74.\n\nMario invested $475475$ in $4545$ and $2525$ stock shares. The number of $2525$ shares was $55$ less than three times the number of $4545$ shares. How many of each type of share did he buy?\n\n#### Everyday Math\n\n75.\n\nParent Volunteer As the treasurer of her daughter's Girl Scout troop, Laney collected money for some girls and adults to go to a $3-day3-day$ camp. Each girl paid $7575$ and each adult paid $30.30.$ The total amount of money collected for camp was $765.765.$ If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?\n\n76.\n\nParent Volunteer Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school year. She didn't remember how many boys had paid the $2424$ full-year registration fee and how many had paid a $1616$ partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If $400400$ was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?\n\n#### Writing Exercises\n\n77.\n\nSuppose you have $66$ quarters, $99$ dimes, and $44$ pennies. Explain how you find the total value of all the coins.\n\n78.\n\nDo you find it helpful to use a table when solving coin problems? Why or why not?\n\n79.\n\nIn the table used to solve coin problems, one column is labeled “number” and another column is labeled ‘“value.” What is the difference between the number and the value?\n\n80.\n\nWhat similarities and differences did you see between solving the coin problems and the ticket and stamp problems?\n\n#### Self Check\n\nAfter completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.\n\nAfter reviewing this checklist, what will you do to become confident for all objectives?" ]

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[ "Input impedance of an op Amp\n\nI noticed that the input impedance of an op Amp is extremely high. Why is that so?\n\n• It's part of the definition of what a opamp is. The ideal opamp has infinite input impedance. Jun 13 '12 at 15:59\n\nIt's one of the rules. For an ideal opamp goes that\n\n• input impedance is infinite (input current is zero)\n• gain is infinite\n• output impedance is zero\n• input offset voltage is zero\n\nIf these requirements aren't met several basic opamp circuits wouldn't work. Take for instance the inverting amplifier.", null, "The transfer function is\n\n$\\mathrm{V_{OUT} = - \\dfrac{R_f}{R_{IN}} \\cdot V_{IN}}$\n\nas derived in this answer. The proof relies on the infinite input impedance, but you can't explain the transfer function based on both inputs equal, because that's not a property of the opamp! So-called proofs that start from the fact that the inverting input is at ground are invalid.\n\nNote that FET input opamps do better than their BJT input counterparts. The former will have pA input current, whereas for the latter this may be several $\\mu$A.\n\nOpamps for everyone\n\n• If the op amp and all other circuits were ideal in all regards except for input impedance, the op amp's input impedance would not affect the behavior of that circuit unless it has a zero-ESR capacitive component; if nothing else would ever cause any potential to appear across the op amp's inputs, it wouldn't matter whether there existed any means for current to flow between them. Of course, if the op amp or other components are non-ideal, adding input impedance may affect the way those non-ideal components behave. Jun 13 '12 at 15:55\n• @supercat - suppose both resistors 1k, input impedance to -12V: 1k. Vin +1V. Then instead of Vout = -1V you'd get Vout = +11V. Pull KCL from the library and do the calculation. Jun 13 '12 at 16:02\n• Input impedance to a non-ground voltage potential would be a problem, of course. If the non-inverting input doesn't budge from ground, however, inverting-input impedance to either ground or to the non-inverting input would be non-factors in an otherwise-ideal op amp used with ideal components. Jun 13 '12 at 16:05\n• A number of practical op amp designs have historically had a relatively low and non-linear impedance between the two inputs, and some still do (some, for example, specify that current may flow between the inputs if the potential between them exceeds a fraction of a volt). Whether or not this is a problem will depend upon how the various non-ideal characteristics of the op amp and surrounding components interact. Jun 13 '12 at 16:08\n• @supercat - But \"ground\" doesn't mean anything to an opamp if it's fed from +/-12V. Look at the schematic of any opamp, you won't find a net that's ground. Impedance is always to either of the supplies. You would be right that an impedance to ground, and the pin grounded, wouldn't see any current, but that's not the case. Jun 13 '12 at 16:10\n\nA very high input impedance gets us closer to an ideal op-amp. The characteristics of an ideal op-amp are:\n\n• Infinite bandwidth\n• Infinite gain\n• Infinite input resistance\n\nThe ideal op-amp exists because using it as a basis for analysis provides several worthwhile shortcuts that simplify the math involved. The infinite input resistance is important because it ensures that no current goes into the op-amp. This simplifies the analysis of feedback op-amp circuits.\n\nAlso, high input impedance is desirable in most circumstances. It allows a signal with very weak drive to be correctly read by the op-amp and amplified. If it had low input impedance the op-amp would draw down the voltage of the weak signal and not properly amplify it.\n\nThe higher the input impedance, the less likely that the opamp itself will effect the input signal.\n\nThink of input and output impedance as a voltage divider made from two resistors. The input impedance of the opamp is the lower resistor, while the output impedance of whatever device is feeding the opamp is the high resistor.\n\nThe best case is when the output impedance is super low while the input impedance is super high. In this case the voltage divider is barely bringing the voltage down.\n\nWorst case is when the output impedance is super high and the input impedance is super low. Then the signal might be divided down to 1/100th of the original voltage, or worse!\n\nSo, it's better if the input to the opamp has the highest impedance possible.\n\nThere are cases where you might want a lower input impedance. But in these cases you usually just put a load resistor on the input signal and don't rely on the opamp to do that for you. Your load resistor (or terminating resistor, or whatever) will have higher tolerances than what the opamp will give you normally.\n\nThe answers are all good, but there is a more basic and more important concept behind that, in part explained by David.\n\nThe Op-amp, at least the most common type of op-amp, is a voltage-input, voltage-output component: it means that it takes a voltage as input, does some operation and spits out another voltage.\n\nReading a voltage requires attaching an instrument to the voltage source, as you would do with a voltmeter: this instrument has to be put in parallel (has to read the same voltage) with the source itself, and will see a current on it that depends on the input resistance of the instrument itself, in this case the op-amp. The current will also flow on the output resistance of the source, as shown in the picture:", null, "The current on the resistor will cause a drop on the input voltage to the op-amp, that won't be anymore equal to the source: we don't want that.\n\nSo if the op-amp has an infinite resistance, the current on the input will be zero and the read voltage will be Vg, as desired. Of course that's an ideal condition, but the higher the better.\n\n• I think this is a valid answer, if there is something wrong please point it Jun 14 '12 at 7:01\n• Downvoter had posted a comment, but for some reason deleted it. I don't recall his motive. Jun 14 '12 at 16:59" ]

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[ "# mAth rounD 15.5\n\n16 -15 就是四舍五入啊\n\nMath.round(11.5)==12 Math.round(-11.5)==-11 round方法返回与参数最接近的长整数,参数加1/2后,求其floor\n\nMath.round方法并不是我们常说的四舍五入,而是四舍六入五看偶,就是说小数点第一位小于五舍去,大于五进一,等于五则要看按照四舍五入方法得到的值是偶数还是奇数,是偶数则进一,奇数则舍去.\n\nMath.round()是四舍五入的函数,所以Math.round(11.5)是12, Math.round(-11.5)是-11\n\njava最开始还是基于C语言的,所以它也拥有C语言的特点,当一个浮点数取整时,如果没有特殊的转换直接取值的话,那么它只取整数,如int a = (int) 15.5; // 直接写int a = 15.5;是会报错的哦System.out.println(a); //最终的结果是15,而不是\n\nmath类中提供了三个与取整有关的方法:ceil、floor、round,这些方法的作用与它们的英文名称的含义相对应,例如,ceil的英文意义是天花板,该方法就表示向上取整,所以,math.ceil(11.3)的结果为12,math.ceil(-11.3)的结果是-11;floor的英文\n\nround()方法你可以这样理解:就是括号内的数+0.5之后,向下取值,比如:round(3.4)就是3.4+0.5=3.9,向下取值是3,所以round(3.4)=3; 那么round(-10.5)就是-10.5+0.5=-10,向下取值就是-10,所以round(-10.5)=-10\n\n1. 首先 Math 没有round(double,int)方法2. 如果是Math.round(11.4) = 11" ]

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[ "# What are the boundaries of x and y if 2x - 3y >= 9 and - x - 4y >= 8??\n\nJan 20, 2018\n\n$x \\ge \\frac{37}{25}$\n$y \\ge \\frac{25}{11}$.\n\n#### Explanation:\n\n$2 x - 3 y \\ge 9$\n\n$\\left(- x - 4 y \\ge 8\\right) \\cdot 2 = - 2 x - 8 y \\ge 16$\n\nadd $2 x - 3 y \\ge 9$\n+ $- 2 x - 8 y \\ge 16$\n\nYou get $11 y \\ge 25$\nSo, $y \\ge \\frac{25}{11}$.\nYou plug in $\\frac{25}{11}$ into one of the equation and solve for x.\n$2 x - 3 \\left(\\frac{25}{11}\\right) \\ge 9$\n$2 x \\ge \\frac{74}{25}$\n$x \\ge \\frac{37}{25}$" ]

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[ "# 230\n\nThis number is a composite.", null, "230 = 2 x 23 x (2 + 3) or (2 x 23 x 5) is the only number with prime factors produced simply from its first two digits, which happen to be the first two primes. [Blunden]", null, "230 is the first number such that it and the next number are both the product of three distinct primes (230 = 2*5*23 and 231 = 3*7*11). [Axoy]" ]

[ null, "https://primes.utm.edu/gifs/check.gif", null, "https://primes.utm.edu/gifs/check.gif", null ]

[ "# Description\n\nAllows for the conditional execution of some statements. It takes the general form:\n\n`IF expression THEN|ELSE statement`\n\nWhere:\n\n• The expression evaluates to a value of Boolean TRUE or FALSE.\n• If the expression is TRUE, the statements defined by the THEN clause will execute (if present).\n• If the expression is FALSE, the statements defined by the ELSE clause, if any will execute.\n• The THEN and ELSE clauses may take two different forms being single and multiple line statements.\n\nThe simplest form of either clause is of the form:\n\n`IF A THEN CRT A`\n\nor\n\n`IF A ELSE CRT A`\n\nHowever, the END keyword may be used to expand the clauses to enclose multiple lines of code as so:\n\n```     IF A THEN\nA = A*6\nCRT A\nEND ELSE\nA = 76\nCRT A\nEND```\n\nIt is possible to combine the single and multi-line versions of either clause to make complex combinations of the command. For reasons of readability it is suggested that where both clauses are present for an IF statement that the same form of each clause is coded.\n\nIF statements can be nested within either clause to any number of levels.\n\nAn example of use is as:\n\n```0001     CRT \"Are you sure (Y/N) \":" ]

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[ "Short-term trading is a difficult task due to fluctuating demand and supply in the stock market. These demands and supply are reflected in stock prices. The stock prices may be predicted using technical indicators. Most of the existing literature considered the limited technical indicators to measure short-term prices. We have considered 82 different combinations of technical indicators to predict the stock prices. We evaluate RICARDO PLC prediction models with Modular Neural Network (DNN Layer) and Lasso Regression1,2,3,4 and conclude that the LON:RCDO stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold LON:RCDO stock.\n\nKeywords: LON:RCDO, RICARDO PLC, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.\n\n## Key Points\n\n1. How can neural networks improve predictions?\n2. Can neural networks predict stock market?\n3. Can statistics predict the future?", null, "## LON:RCDO Target Price Prediction Modeling Methodology\n\nStock market trading is an activity in which investors need fast and accurate information to make effective decisions. Since many stocks are traded on a stock exchange, numerous factors influence the decision-making process. Moreover, the behaviour of stock prices is uncertain and hard to predict. For these reasons, stock price prediction is an important process and a challenging one. We consider RICARDO PLC Stock Decision Process with Lasso Regression where A is the set of discrete actions of LON:RCDO stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Lasso Regression)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Modular Neural Network (DNN Layer)) X S(n):→ (n+1 year) $\\stackrel{\\to }{S}=\\left({s}_{1},{s}_{2},{s}_{3}\\right)$\n\nn:Time series to forecast\n\np:Price signals of LON:RCDO stock\n\nj:Nash equilibria\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## LON:RCDO Stock Forecast (Buy or Sell) for (n+1 year)\n\nSample Set: Neural Network\nStock/Index: LON:RCDO RICARDO PLC\nTime series to forecast n: 05 Oct 2022 for (n+1 year)\n\nAccording to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold LON:RCDO stock.\n\nX axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Yellow to Green): *Technical Analysis%\n\n## Conclusions\n\nRICARDO PLC assigned short-term B2 & long-term B2 forecasted stock rating. We evaluate the prediction models Modular Neural Network (DNN Layer) with Lasso Regression1,2,3,4 and conclude that the LON:RCDO stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold LON:RCDO stock.\n\n### Financial State Forecast for LON:RCDO Stock Options & Futures\n\nRating Short-Term Long-Term Senior\nOutlook*B2B2\nOperational Risk 7031\nMarket Risk3155\nTechnical Analysis6970\nFundamental Analysis4651\nRisk Unsystematic6343\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 89 out of 100 with 794 signals.\n\n## References\n\n1. Chernozhukov V, Escanciano JC, Ichimura H, Newey WK. 2016b. Locally robust semiparametric estimation. arXiv:1608.00033 [math.ST]\n2. Li L, Chen S, Kleban J, Gupta A. 2014. Counterfactual estimation and optimization of click metrics for search engines: a case study. In Proceedings of the 24th International Conference on the World Wide Web, pp. 929–34. New York: ACM\n3. Li L, Chu W, Langford J, Moon T, Wang X. 2012. An unbiased offline evaluation of contextual bandit algo- rithms with generalized linear models. In Proceedings of 4th ACM International Conference on Web Search and Data Mining, pp. 297–306. New York: ACM\n4. M. Benaim, J. Hofbauer, and S. Sorin. Stochastic approximations and differential inclusions, Part II: Appli- cations. Mathematics of Operations Research, 31(4):673–695, 2006\n5. Arjovsky M, Bottou L. 2017. Towards principled methods for training generative adversarial networks. arXiv:1701.04862 [stat.ML]\n6. Canova, F. B. E. Hansen (1995), \"Are seasonal patterns constant over time? A test for seasonal stability,\" Journal of Business and Economic Statistics, 13, 237–252.\n7. Bai J, Ng S. 2017. Principal components and regularized estimation of factor models. arXiv:1708.08137 [stat.ME]\nFrequently Asked QuestionsQ: What is the prediction methodology for LON:RCDO stock?\nA: LON:RCDO stock prediction methodology: We evaluate the prediction models Modular Neural Network (DNN Layer) and Lasso Regression\nQ: Is LON:RCDO stock a buy or sell?\nA: The dominant strategy among neural network is to Hold LON:RCDO Stock.\nQ: Is RICARDO PLC stock a good investment?\nA: The consensus rating for RICARDO PLC is Hold and assigned short-term B2 & long-term B2 forecasted stock rating.\nQ: What is the consensus rating of LON:RCDO stock?\nA: The consensus rating for LON:RCDO is Hold.\nQ: What is the prediction period for LON:RCDO stock?\nA: The prediction period for LON:RCDO is (n+1 year)" ]

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[ "Instruction\n1\nThe pace of inflation is used to measure the intensity of inflationary processes, their dynamics along with the index of inflation. It characterizes the rate of depreciation of money and purchasing power for a certain period. The rate of inflation expresses the increase in the average price level in percentage to their face value at the beginning of the study period.\n2\nAt the conclusion of long-term contracts it is necessary to determine the projected annual pace of inflation and the growth of its index. It is necessary to calculate the expected average monthly growth rate of inflation. This information can be taken in the published forecasts of economic and social development of the country in the coming period. The results of these forecasts that become the basis of subsequent factor analysis of inflation in the financial activities of the company.\n3\nThe projected annual rate of inflation (TIG) is calculated according to the formula:\nTIG = (1+Tim) n – 1, where:\nTim - the expected average rate of inflation in the upcoming period\nn – the degree to which you want to build the number (1+Tim) equal to the number of months in the forecast period. If the period is a year, then n = 12.\n4\nIn this formula you can calculate not only the predicted annual rate of inflation, but the value of this parameter for any forthcoming period. For example, if a calculation covers a period of two years, the degree n, which erected a number (1+Tim) will be equal to 24.\n5\nThe rate of inflation can be used to calculate the predicted annual index of inflation (IIG):\nIIG = 1 + TIG or:\nIIG = (1 + Tim)n.\n6\nThe pace of inflation is used to construct real interest rates, taking into account the factor of inflation. For this calculation, use the projected nominal interest rate in the financial market, which is usually reflected in the prices of futures and options contracts on the stock exchange. According to the Fisher Model, the real interest rate Ip is calculated by the formula:\nIp= (I – TI) / (1 + TI), where:\nI – nominal interest rate (actual or projected for a certain period." ]

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[ "Back to Kemp Acoustics Home", null, "", null, "", null, "", null, "Next: Pressure radiation from a Up: The piston approximation Previous: The piston approximation   Contents\n\n## Pressure radiation from a piston terminated in an infinite baffle\n\nConsider a rigid piston in a rigid infinite baffle as shown in figure 3.1. The piston vibrates uniformly with a sinusoidal velocity of amplitude", null, "normal to the baffle.", null, "In order to calculate the behaviour of this system, we split the piston into infinitesimal simple source elements and sum the resulting pressure fields. A piston surface element of area", null, "is present at", null, ". This surface element oscillates with a velocity amplitude of", null, "normal to the baffle and acts as a simple source of spherical pressure waves. These are represented on the diagram by a hemispherical shell, with the acoustic pressure at a distance", null, "from the source element given by", null, "(3.2)\n\nwhere", null, "is the simple source strength and a", null, "time factor is assumed throughout. The part", null, "is known as the Green's function and implies that the pressure oscillates sinusoidally in space with wavelength", null, "and with an amplitude that dies as", null, ". Integrating (3.2) over", null, ", the surface of the whole piston, we get the total pressure field due to the sum of all the source elements that make up the piston.", null, "(3.3)\n\nNote that the integrand is singular (tends to infinity) as", null, "tends to zero. This problem must be addressed before numerical integration is possible.\n\nBack to Kemp Acoustics Home", null, "", null, "", null, "", null, "Next: Pressure radiation from a Up: The piston approximation Previous: The piston approximation   Contents\nJonathan Kemp 2003-03-24" ]

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[ "# Algebra of complex numbers(Complex Number 1 NCERT)\n\n0\n856\n\nAlgebra of complex numbers\n\nIn this Topic, we will see the algebra of complex numbers.\n\nLet z1 = a + ib and z2 = c + id be any two complex numbers. Then, the sum z1 + z2 is defined as follows:\nz1 + z2 = (a + c) + i (b + d), which is again a complex number.\nFor example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8\n\nThe addition of complex numbers satisfy the following properties:\n\n• The closure lawThe sum of two complex numbers is a complex number, i.e., z1 + z2 is a complex number for all complex numbers z1 and z2.\n• The commutative lawFor any two complex numbers z1 and z2, z1 + z2 = z2 + z1\n• The associative lawFor any three complex numbers z1, z2, z3, (z1 + z2) + z3 = z1 + (z2 + z3).\n• The existence of additive identityThere exists the complex number 0 + i 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number z, z + 0 = z.\n• The existence of additive inverseTo every complex number z = a + ib, we have the complex number – a + i(– b) (denoted as – z), called the additive inverse or negative of z. We observe that z + (–z) = 0 (the additive identity).\n\nDifference of two complex numbers\nGiven any two complex numbers z1 and z2, the difference z1 – z2 is defined as follows:\nz1 – z2 = z1 + (– z2).\nFor example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4i\nand (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i\n\nMultiplication of two complex numbers\nLet z1 = a + ib and z2 = c + id be any two complex numbers. Then, the product z1 z2 is defined as follows:\nz1 z2 = (ac – bd) + i(ad + bc)\nFor example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28\n\nThe multiplication of complex numbers possesses the following properties, which we state without proofs.\n\n• The closure lawThe product of two complex numbers is a complex number, i.e., z1z2 is a complex number for all complex numbers z1 and z2.\n• The commutative lawFor any two complex numbers z1 and z2, z1z2 = z2z1\n• The associative lawFor any three complex numbers z1, z2, z3, (z1z2)z3 = z1(z2z3).\n• The existence of multiplicative identityThere exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, for every complex number z..\n• The existence of multiplicative inverseFor every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number (a/a2 + b2) + i (-b/a2 + b2) (denoted by 1/z or z-1 ), called the multiplicative inverse of z such that z • 1/z = 1 (the multiplicative identity).\n• The distributive lawFor any three complex numbers z1, z2, z3,\n(a) z1 (z2 + z3) = z1 z2 + z1 z3\n(b) (z1 + z2) z3 = z1 z3 + z2 z3\n\nDivision of two complex numbers\nGiven any two complex numbers z1 and z2, where z2 ≠ 0, the quotient z1 / z1 is defined by\nz1 / z2 = z1 • 1/z2\nFor example, let z1 = 6 + 3i and z2 = 2 – i , Then", null, "Power of i  we know that", null, "The square roots of a negative real number\nNote that i2 = –1 and ( – i)2 = i2 = – 1\nTherefore, the square roots of – 1 are i, – i. However, by the symbol √− , we would mean i only. Now, we can see that i and –i both are the solutions of the equation x2 + 1 = 0 or x2 = –1.\n\nSimilarly (√3 i)2 = (√3)2 i2 = 3(-1) = -3\n(√3 i)2 = (√3)2 i2 = -3\nTherefore, the square roots of –3 are √3 i and −√3i .\nAgain, the symbol √−3 is meant to represent √3 i only, i.e., √-3 = √3 i .\nGenerally, if a is a positive real number, √−a = √ a √−1 = √a i ,\nWe already know that √a × √b = √ab for all positive real number a and b. This result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0? Let us examine.\n\nNote that\ni2 = √−1 √−1 √(−1)(−1) (by assuming √a × √b = √ab for all real numbers)\n= √1 = 1, which is a contradiction to the fact that i = − .\n\nTherefore,√a × √b ≠ √ab if both a and b are negative real numbers.\nFurther, if any of a and b is zero, then, clearly, √a × √b = √ab =0.\n\nIdentities\nWe prove the following identity\n(z1 + z2)2 = z12 + z22 + 2z1z2 , for all complex numbers z1 and z2\nProof We have,\n(z1 + z2)2 = (z1 + z2) (z1 + z2),\n= (z1 + z2) z1 + (z1 + z2) z2 (Distributive law)\n= z12 + z2z1 + z1z2 + z22 (Distributive law)\n= z12 + z1z2 + z1z2 + z12 (Commutative law of multiplicatoin)\n= z12 + z22 + 2z1z2\n\nSimilarly, we can prove the following identities:\n(i) (z1 – z2)2 = z12 – 2z1z2 + z22\n(ii) (z1 + z2)3 = z13 + 3z12z2 + 3z1z22 + z23\n(iii) (z1 – z2)3 = z13 – 3z12z2 + 3z1z22 – z23\n(iv) z12 – z22 = (z1 + z2) (z1 – z2\nIn fact, many other identities which are true for all real numbers, can be proved to be true for all complex numbers.\n\nExample Express the following in the form of a + bi:", null, "SHARE" ]

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[ "# Free Probability 03 Practice Test - 9th Grade\n\nWhen a coin is tossed, what is the probability of occurrence of \"head\" as the outcome?\n\nA. 14\nB. 35\nC. 13\nD. 12\n\n#### SOLUTION\n\nSolution : D\n\nWhen you flip a coin, there are two outcomes either head or tail. So the probability of occurrence of the head is the same as the probability of occurrence of the tail that is 12\n\nThe range of probability is\n\nA.\n\n-1 to 1 (inclusive)\n\nB.\n\n0 to 1 (inclusive)\n\nC.\n\n1 to 2 (inclusive)\n\nD.\n\n-1 to 0 (inclusive)\n\n#### SOLUTION\n\nSolution : B\n\n.\n\nA coin was tossed 1500 times and out of these, 100 times, the coin stuck somewhere so the result was not counted. The frequency of occurrence of a head was 700. Find the probability of occurrence of a tail.\n\nA.\n\n0.46\n\nB.\n\n0.5\n\nC.\n\n0.3\n\nD.\n\n1\n\n#### SOLUTION\n\nSolution : B\n\nA coin was tossed 1500 times but 100 times result was not counted.\n\nSo, the total number of times result counted = 1400\n\nNumber of times head comes = 700\n\nNumber of times tail comes = 1400 - 700 = 700\n\nProbability of occurrence of tail =Number of times tail comesTotal number of times result counted\n\nProbability of occurrence of tail =7001400\n\nProbability of occurrence of tail = 0.5\n\nOutcome2356Frequency10141620\n\nA dice was thrown 80 times and the frequency of occurrence of 2, 3, 5 and 6 is given in the table. What is the probability of occurrence of 1 or 4?\n\nA.\n\n1\n\nB.\n\n0.67\n\nC.\n\n0.25\n\nD.\n\n14\n\n#### SOLUTION\n\nSolution : C and D\n\nTotal frequency of occurrence of 2, 3, 5 and 6 = 60.\n\nFrequency of occurrence of 1 or 4 = 80 - 60 = 20.\n\nProbability of occurrence of 1 or 4Frequency of occurrence of 1 or 4Total number of times die is thrown\n\nSo, probability of occurrence of 1 or 4 = 2080 = 14 = 0.25\n\nA goldsmith can choose between 3 metals A, B or C to make an artifact. What is the probability that he chooses C?\n\nA.\n\n13\n\nB.\n\n0\n\nC.\n\n1\n\nD.\n\n23\n\n#### SOLUTION\n\nSolution : A\n\nSince the goldsmith can choose amongst A, B and C, he has 3 possibilities of which the favorable outcome is choosing C. Therefore, probability that he chooses C is 13.\n\nBagChocolates112219318422511\n\n5 bags had some amount of chocolates in them, as shown in the table. The probability of occurrence of more than 18 chocolates in a bag if a bag is chosen at random will be ___\n\n#### SOLUTION\n\nSolution :\n\nOut of the givne 5 bags, bag 2 and bag 4 have more than 18 chocolates.\nProbability(E) =number of favorable eventstotal number of events\nProbablity of a bag containing more than 18 chocolates =number of bags containing more than 18 chocolatestotal number of bags=25=0.4\n\nIf an alphabet is picked from the word MISSIMMIPPI what is the probability that an alphabet 'M' is picked?\n\nA.\n\n111\n\nB.\n\n311\n\nC.\n\n811\n\nD.\n\n1\n\n#### SOLUTION\n\nSolution : B\n\nTotal number of alphabets in the word MISSIMMIPPI = 11\n\nNumber of times M comes = 3\n\nProbability of picking M = Number of times M comesTotal number of alphabets in the word MISSIMMIPPI\n\nSo, probability of its occurrence will be 311.\n\nA coin was flipped 100 times. What is the maximum probability of occurrence of tail?\n\nA.\n\n0.5\n\nB.\n\n1\n\nC.\n\n0\n\nD.\n\n0.75\n\n#### SOLUTION\n\nSolution : B\n\nA coin is tossed 100 times and consider the case when you only get tail 100 times and no head. So, in all the trials you will have a tail.\n\nProbability of getting a tail = Number of times tails appearNumber of times a coin is tossed\n\nProbability of getting a tail​ = 100100=1\n\nWhat will be the probability that two friends have the same birthday date in a normal year ?\n\nA. 173\nB. 1365\nC. 7365\nD. 12365\n\n#### SOLUTION\n\nSolution : B\n\nNumber of days in a year = 365\nTherefore, total possible outcomes = 365\n\nIf they have birthday on same day, then number of favourable outcomes =1\n\nTherefore, required probability =1365\n\nA die is rolled twice. Find the probability that 5 will come up both the times.\n\nA. 136\nB. 1836\nC. 636\nD. 436\n\n#### SOLUTION\n\nSolution : A\n\nNumber of all possible outcomes = (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5) and (6,6).\n\nThe number of all possible outcomes = 36.\n\nThere is only one case (5,5) when 5 comes up both the times.\n\nProbability that 5 will come up both the times=Number of times 5 comes up both the timesNumber of all possible outcomes\n\nP (5 will come up both the times) =136" ]

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[ "# Constrained Triangle\n\nGeometry Level 5", null, "We have three parallel lines $l, m$ and $n$ and an equilateral triangle $CDE$ with its vertices on these lines as shown in figure.\n\nThe distance between the lines $l$ and $m$ is 5 and between $m$ and $n$ is 2.\n\nThe side length of triangle can be expressed as $A \\sqrt{B}$, where $A$ and $B$ are positive integers with $B$ square-free. Find $A + B$.\n\n×" ]

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[ "# Statistical hypothesis testing\n\nThis is what people want:\n\n1. State a hypothesis.\n2. Gather data.\n3. Find out the probability that the hypothesis is true, given the data.\n\nThis is not how statistical hypothesis testing works. It's more like this:\n\n1. State a hypothesis.\n2. Gather data.\n3. Reason about the likeliness of data assuming that the hypothesis is true.\n1. If the data is likely given the hypothesis, we fail to reject the hypothesis.\n2. If the data is unlikely given the hypothesis, we reject the hypothesis.\n\nHow do we measure the likeliness or “unlikeliness” of data? Using a probability. And a threshold, which has been fixed to .05. A century of controversy ensues.\n\nSome teaching materials about hypothesis testing seem to jump directly to “computing test statistics”, say the z-score, and getting the results as quick as possible, with no indication as to why the z-score is used or why the test has a “right tail” or “two tails”. Here I will try to show the reasoning behind every step.\n\nLet's work on a simple example of the kind we can find everywhere. We read that the average height of people in the world is 150 cm. We want to challenge this assumption. So we state a hypothesis:\n\n$$H_0: \\mu = 150$$\n\nThis is the null hypothesis, the status quo, which we are trying to challenge. The alternative hypothesis could be that people are actually taller, so:\n\n$$H_1: \\mu > 150$$\n\nNow gather data from $n=1000$ people… and get a sample average $\\bar{x}$.\n\nHow would we interpret $\\bar{x}$? Let's say $\\bar{x} = 160$. Do we immediately conclude that $H_1$ is true? What about $\\bar{x} = 200$, that would be overwhelming evidence, right? But how about $\\bar{x} = 151$, does that support $H_1$, or is it only due to chance that we got a value slightly greater than $150$? Or $\\bar{x} = 149$, for that matter: one centimeter should be within the expected variation from a random sample, shouldn't it? Hey, even $\\bar{x} = 220$ doesn't allow jumping to conclusions, what if we were really unlucky and sampled from a country with lots of really tall people?\n\nNo matter the value, there is a chance of making a wrong inference from $\\bar{x}$. Namely:\n\n1. We could get $\\bar{x} > 150$ even if $H_0$ is true.\n2. We could get $\\bar{x} \\leq 150$ even if $H_0$ is not true.\n\nThese scenarios would be surprising, but not implausible. We can quantify how probable these scenarios are. For instance, let's say we get $\\bar{x} = 160$. So we are really, really tempted to conclude that we should reject $H_0$ once for all. But… could it be due to chance? What are the odds of getting $\\bar{x} = 160$ or worse (i.e. greater) if $H_0$ is true?\n\n\\begin{aligned} P(\\bar{x} \\geq 160 | H_0 \\: true) &= P(\\bar{x} \\geq 160 | \\mu = 150) \\\\ &= P\\left(\\frac{\\bar{x}-\\mu}{\\sigma/\\sqrt{n}} \\geq \\frac{160-\\mu}{\\sigma/\\sqrt{n}} \\middle| \\mu = 150\\right) \\\\ &= P\\left(Z \\geq \\frac{160-\\mu}{\\sigma/\\sqrt{n}} \\middle| \\mu = 150\\right) \\end{aligned}\n\nIn line 2 we substract the mean and divide by the sample standard deviation to form a quantity $Z$ whose distribution is $N(0,1)$. This is simply a fact about the normal distribution (of course, we are assuming that heights are normal here…). For the sake of simplicity, assume the standard deviation $\\sigma=30$ is known. So:\n\n\\begin{aligned} P(\\bar{x} \\geq 160 | H_0 \\: true) &= P\\left(Z \\geq \\frac{160-\\mu}{\\sigma/\\sqrt{n}} \\middle| \\mu = 150\\right) \\\\ &= P\\left(Z \\geq \\frac{160-150}{30/\\sqrt{1000}}\\right) \\\\ &= P(Z \\geq 1.05) \\\\ &= 0.15 \\end{aligned}\n\nYay, we have a result! The probability of getting a sample such as the observed, assuming $H_0$ is true, equals $0.15$!\n\nBut… how does this value help us in deciding whether we should reject $H_0$? Well $0.15$ sounds like a considerable value, close to the probability of getting any given number by rolling a die. Maybe we shouldn't take chances and decide that, maybe, $H_0$ is true? (or rather, fail to reject $H_0$. You know, because assuming $H_0$ is true would explain the data). We would certainly fail to reject $H_0$ if we had obtained, say, $0.99$ (“if the data is very likely when $H_0$ is true, then surely $H_0$ holds, huh?”). And if we had obtained $0.00001$, we would be thinking that the data is unlikely when $H_0$ is true, so we would reject $H_0$, right?\n\nThis $0.15$ value we obtained is called the p-value. Some statistician guy said: if the p-value is less than $0.05$, reject $H_0$. Otherwise, fail to reject it.\n\nThis threshold $\\alpha=0.05$ is called the significance level of the test. It quantifies how likely we are to mistakenly reject $H_0$.\n\n## The fallacy\n\n(will write about Jacob Cohen's paper “The Earth is Round (p .05)”).", null, "" ]

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[ "×\n×\n\n# If A is an n n skew-symmetric matrix and n is odd, prove that det(A) = 0", null, "ISBN: 9780321964670 380\n\n## Solution for problem 59 Chapter 3.2\n\nDifferential Equations | 4th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Differential Equations | 4th Edition\n\n4 5 1 333 Reviews\n18\n3\nProblem 59\n\nIf A is an n n skew-symmetric matrix and n is odd, prove that det(A) = 0.\n\nStep-by-Step Solution:\nStep 1 of 3\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780321964670\n\nThis full solution covers the following key subjects: . This expansive textbook survival guide covers 91 chapters, and 2967 solutions. This textbook survival guide was created for the textbook: Differential Equations, edition: 4. The answer to “If A is an n n skew-symmetric matrix and n is odd, prove that det(A) = 0.” is broken down into a number of easy to follow steps, and 17 words. The full step-by-step solution to problem: 59 from chapter: 3.2 was answered by , our top Math solution expert on 03/13/18, 06:45PM. Differential Equations was written by and is associated to the ISBN: 9780321964670. Since the solution to 59 from 3.2 chapter was answered, more than 219 students have viewed the full step-by-step answer.\n\nUnlock Textbook Solution" ]

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[ "Welcome to the Airtable Community! If you're new here, check out our Getting Started area to get the most out of your community experience.\n\n# Question about nested if(and()) function\n\nTopic Labels: Formulas\n1333 4\ncancel\nShowing results for\nDid you mean:", null, "", null, "4 - Data Explorer\n\nHi, I’m trying to figure out why I’m experiencing issues with my nested IF(AND()) function on Airtable. When I limit the fields in AND() to just two conditions, it works, but anything else and Airtable won’t let me use the formula at all.\n\nWhat works\n\nIF(AND({Person}= “1”, {A}= “Yes”), “Approved”,\n\nIF(AND({Person}= “2”, {B}= “Yes”), “Approved”,\n\nIF(AND({Person}= “3”, {C}= “Yes”), “Approved”,\n\nIF(AND({Person}= “4”, {D}= “Yes”), “Approved”,\n\nIF({Quote} = “0.00”\n\n, “Temporary approval”, “Unapproved”)))))\n\nWhat doesn’t\n\nIF(AND({Person}= “1”, {A}= “Yes”), “Approved”,\n\nIF(AND({Person}= “2”, {B}= “Yes”), “Approved”,\n\nIF(AND({Person}= “3”, {C}= “Yes”), “Approved”,\n\nIF(AND({Person}= “4”, {D}= “Yes”), “Approved”,\n\nIF(AND({Person}= “5”, {E}= “Yes”, {F}= “Yes”), “Approved”,\n\nIF(AND({Person}= “6”, {E}= “Yes”, {F}= “Yes”, {G}= “Yes”), “Approved”,\n\nIF({Quote} = “0.00”\n\n, “Temporary approval”, “Unapproved” )))))))\n\n4 Replies 4", null, "", null, "16 - Uranus\n\nI don’t see anything wrong from a syntax point. AND() does not have a limit on how many clauses it can accept (and if it does, it isn’t two). What types of fields are E, F, and G? There could be an errant comma or quote that I’m not seeing. It also could be the trailing space after “Unapproved” before the closing parenthesis.\n\nAlso your formula can be simplified by using an OR since so many outcomes result in “Approved”\n\n``````IF(\nOR(\nAND({Person}=\"1\", {A}=\"Yes\"),\nAND({Person}=\"2\", {B}=\"Yes\"),\nAND({Person}=\"3\", {C}=\"Yes\"),\nAND({Person}=\"4\", {D}=\"Yes\"),\nAND({Person}=\"5\", {E}=\"Yes\", {F}=\"Yes\"),\nAND({Person}=\"6\", {E}=\"Yes\", {F}=\"Yes\", {G}=\"Yes\")\n),\n\"Approved\",\nIF({Quote}=\"0.00\", \"Temporary approval\", \"Unapproved\")\n)\n``````", null, "", null, "4 - Data Explorer\n\nThank you for your answer! I added the trailing space to differentiate between different lines of the formula better but didn’t have that syntax when I used it in Airtable.\n\nThe Person fields are text with a colon, possibly an ampersand, and spaces (e.g. {Person} = “C&I: 5000”); the alphabetical fields are names ({A} for example could be {John Smith}).\n\nBased on your comments I’ve edited my formula to the following with all 8 approval person types but still have received the error: “Sorry, there was a problem saving this field. Invalid formula. Please check your formula text.”\n\nI would appreciate any advice you have!\n\nIF(\n\nOR(\n\nAND({Person}= “1”, {A}= “Yes”),\n\nAND({Person}= “2”, {B}= “Yes”),\n\nAND({Person }= “3”, {C}= “Yes”),\n\nAND({Person}= “4”, {D}= “Yes”),\n\nAND({Person} = “5”, {E} = “Yes”, {F} = “Yes”, {G} = “Yes”),\n\nAND({Person} = “6”, {E} = “Yes”, {F} = “Yes”),\n\nAND({Person} = “7”, {E} = “Yes”),\n\nAND({Person} = “8”, {H} = “Yes”),\n\n“Approved”, IF({Quote} = “0.00”, “Temporary approval”, “Unapproved”)))", null, "", null, "16 - Uranus\n\nYou’re missing a closing parenthesis for the OR, or its in the wrong spot. The should be no comma after your last AND, there should be the closing parenthesis for the OR.", null, "", null, "4 - Data Explorer\n\nI appreciate this and have moved my closing parenthesis for OR–I am somehow still experiencing an error but will do some more troubleshooting on my end before I ask a follow up question.", null, "" ]

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[ "### Translate\n\nCentrifugal pump Calculations\n\n### Formulas & Calculations\n\nHow to Calculate...\n\n• Hydraulic power calculation\n• Pump Shaft Power\n• Electrical Power\n• Specific Speed\n• Centrifugal Pump Calculations Energy loss in discharge valve throttling\n• NPSHa - Net Positive Suction Head Available\n• Head ↔ Pressure\n• Volumetric flow ↔ Mass flow\n• Break Power Calculation\n\n# Formulas\n\nHead is net work done for unit weight of water.\n\nH = Head ; Unit - Meter\nP = Water Pressure ; Unit - kPa, psi\nΓ = Specific weight of Fluid ; Unit - kN/m³\nV = Water Velocity ; Unit - m/s\ng = Gravitional force ; Unit = m/s² Value = 9.8m/s²\nZ = Elevation to be lift\n\nHorizontal pump - Datum is a horizontal plane from center line of shaft\n\nVertical pump - Datum is horizontal plane of impeller eye.\n\nSee calculations\n\n### Pump Capacity ( Volume flow rate of pump)\n\nCapacity is the amount of water pumped per unit time.\n\nCapacity Q = A V\n\nA = Area of pipe\nV = Volume flow rate\n\nSee calculations\n\n### Hydraulic power calculation\n\n⍴ = densityGravitional force of fluid ; Unit = kg/m³\ng = Gravitional force ; Unit = m/s² Value = 9.8m/s²\nH = Total Head = Hd - Hs; Unit = m\nHd = Discharge head\nHs = Suction head\nQ = Discharge flow ; Unit = m³/hour\n\nSee calculations\n\n### Pump Shaft Power\n\nηpump = Pump efficiency\n\nMaximum shaft power\n\nMmax = ɑ1 ⍴ g H Q / η0\n\nSee calculations\n\n### Electrical Power\n\nηmotor = Motor efficiency\n\nSee calculations" ]

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[ "Ounces To Grams\n\n# 901 oz to g901 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 901 ounce to grams?\n\n 901 oz * 28.349523125 g = 25542.9203356 g 1 oz\nA common question is How many ounce in 901 gram? And the answer is 31.7818397166 oz in 901 g. Likewise the question how many gram in 901 ounce has the answer of 25542.9203356 g in 901 oz.\n\n## How much are 901 ounces in grams?\n\n901 ounces equal 25542.9203356 grams (901oz = 25542.9203356g). Converting 901 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 901 oz to g.\n\n## Convert 901 oz to common mass\n\nUnitMass\nMicrogram25542920335.6 µg\nMilligram25542920.3356 mg\nGram25542.9203356 g\nOunce901.0 oz\nPound56.3125 lbs\nKilogram25.5429203356 kg\nStone4.0223214286 st\nUS ton0.02815625 ton\nTonne0.0255429203 t\nImperial ton0.0251395089 Long tons\n\n## What is 901 ounces in g?\n\nTo convert 901 oz to g multiply the mass in ounces by 28.349523125. The 901 oz in g formula is [g] = 901 * 28.349523125. Thus, for 901 ounces in gram we get 25542.9203356 g.\n\n## 901 Ounce Conversion Table", null, "## Alternative spelling\n\n901 Ounces to Grams, 901 oz to Grams, 901 oz in Grams, 901 Ounces in g, 901 oz to g, 901 Ounce to Grams, 901 Ounce in Grams, 901 Ounce in Gram, 901 Ounce to g, 901 Ounce in g," ]

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[ "# When do we need to separate areas while calculating definite integrals?\n\nMany times when the function has range $f(x) \\lt 0$ and also $f(x) > 0$ and we want to calculate an area of the function we need to separate the area in two parts. Thus, we will calculate the whole area as: $$S = \\big{\\lvert} S_1\\big\\rvert +\\big\\lvert S_2\\big\\rvert$$\n\nThis is because the definite integral of $S_1$ or the one of $S_2$ could be negative.\n\nHowever, in many cases we don't. For example, with $$f(x) = x + \\sin^2x$$", null, "Here in this image we don't need to separate the areas for calculating for example the area between $-\\pi\\le x \\le \\pi$. But why?\n\nCould you give examples where we need to separate the areas and sum them up with absolute values, and examples where we don't need.\n\n• What do you mean \"here\" we don't? WolframAlpha does not mention areas anywhere ... – Hagen von Eitzen Apr 9 '16 at 15:24\n• Do you know odd,even functions we split up areas if by doing so we get even and odd function which would give resultant area as $0$ – Archis Welankar Apr 9 '16 at 15:27\n• @HagenvonEitzen I mean here by the image of the function. – Pichi Wuana Apr 9 '16 at 15:28\n• @HagenvonEitzen I edited the question. – Pichi Wuana Apr 9 '16 at 15:30\n• Basically you don't need any extra tinkering with the function if you can just compute a primitive. – Captain Lama Apr 9 '16 at 15:31" ]

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[ "# Pressure of air inside a glass\n\n• B\n• Pushoam\n\n#### Pushoam\n\nTL;DR Summary\nHow does putting a cover affect the pressure inside a glass filled with air?\nLet's consider an uncovered glass. Air particles are present in the glass.", null, "$$P_1 = P_a$$ $$P_2 =P_1 +\\rho gh = P_a +\\rho g h$$where ##P_A## is atmospheric pressuere and ##\\rho ## is air density.", null, "Now, if I cover the glass with a plastic card, then what is ## P_1##?\n$$P_2 =P_1 +\\rho gh$$\n1) ## P_1 ## is pressure due to motion of air particles and the air particles near the cover interact with the cover and its speed may change and hence ##P_1## may be less or more than ##P_a##.\n\n2) Following three forces are acting on the cover:", null, "a) force due to pressure ##P_1## of air particles\nb) normal force N due to glass walls\nc) cover's weight W\n\nApplying Newton's first law gives,\n$$P_1 A+ N = W$$ $$P_1 A = W - N$$\nNow, since normal force is self-adjustable, let's take a light plastic card such that N = 0. Hence, in this case ##P_1 A = W ##.\nFor a plastic card with mass 20g and area 20cm2, ## P_1 = 10 Pa## which is lower than the atmospheric pressure.\nSo, the conclusion is: putting a cover reduces the pressure of air inside the glass. Is this correct?\n\nSo, the conclusion is: putting a cover reduces the pressure of air inside the glass. Is this correct?\nNo.\nThere will be a hydrostatic difference in pressure due to the thickness of the card, but that difference will be overcome by the density of the card, which is resting on the glass.\n\n•", null, "russ_watters\nNo.\nThere will be a hydrostatic difference in pressure due to the thickness of the card, but that difference will be overcome by the density of the card, which is resting on the glass.\nAnd/or the card will bend down/inward due to its own weight and the resulting pressure in the glass will be higher than atmospheric.\n\nDefine Pa as the atmospheric pressure at the top edge, inside the glass.\nThe following four forces are acting on a cover of thickness; t\na) force due to pressure of air from below; A·Pa\nb) force due to pressure of air from above; A·(Pa - ρ·g·t)\nc) cover's weight; W\nd) normal force upwards due to glass wall; N\nN + A·Pa = W + A·(Pa - ρ·g·t)\nN = W - ρ·g·t\nρ·g·t\nis the buoyancy of the card" ]

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[ "", null, "counting degenerate dimensions", null, "• Question\n\n•", null, "Hi,\n\nIn my fact table I have two degenerate dimension (OrderNo, ItemNo). The grain is at billing level, so 1 OrderNo can have many rows, same with the ItemNo.\n\nQuestion, how do I best represent these as measures in my cube (OrderNo Count and ItemNo count). I can do a distinct count to create both measures, however I'm trying to avoid distinct counts, one reason this will create two new measure groups, so I assume processing and query performance will decrease (my fact table is roughly 200million rows).\n\nThese are important measures for me because many questions are about about the production... how many orders last month, how many orders for supplier x and so on...\n\nAre there alternatives to distinct counts?\n\nTuesday, April 16, 2013 9:46 AM\n\n•", null, "Hi Toro,\n\nOK.  This could be because you have filtered on an attribute of the [Dim Order] dimension.  In this case it would be fixed by adding Existing to the NonEmpty() function ie try this.  Also, probably safest to add the measure you are using for NonEmpty.\n\ncount(nonempty(Existing [Dim Order].[ORDER_ID].[ORDER_ID].members,Measures.Orders))\n\nRichard\n\n• Edited by Monday, April 22, 2013 12:49 AM correction\n• Marked as answer by Thursday, April 25, 2013 9:09 AM\nMonday, April 22, 2013 12:48 AM\n\nAll replies\n\n•", null, "Hi Toro, If I understand correctly, each Order can have multiple Items.  ie. Item belongs to an Order?  If this is the case, I would be thinking of having one dimension with OrderNo, ItemNo as the composite key.  Then, any time you wanted to know how many Orders there were, or how many Items in an Order(s) it is just a matter of counting the nonempty() members of the dimension.\n\nRegarding performance of this design, it will somewhat depend on the size of the OrderItem dimension.  However, a query that is asking for the number of items in an Order will perform very well as ssas will restrict the count nonempty to the OrderItems for that particular Order.  Similarly, queries that filter on attributes of Order or Item will run faster than non filtered queries as ssas will scan a smaller number of OrderItem members.  For this reason, you might want to make OrderMonth etc a properties of the Order.\n\nHope that helps,\n\nRichard\n\nThursday, April 18, 2013 4:04 AM\n•", null, "Hi Richard,\n\nThanks for the reply. Ok I have tried your suggestion, however maybe I'm missing something...\n\nI have created the dimension and a calculated member as follows (count(nonempty([Dim Order].[ORDER_ID].members)))\n\nWhen browsing the cube, the calculated member returns the correct total, however when I introduce any other dimensions to slice the data, the calculated member continues to returns the full total\n\ne.g.\n\ntotal orders = 100 (OK)\n\nintroduce dimension date\n\n2012 orders = 100 (not ok, should be 40)\n\n2013 orders = 100 (not ok, should be 60)\n\nthanks\n\nThursday, April 18, 2013 10:05 AM\n•", null, "Hi Toro,\n\nOK.  This could be because you have filtered on an attribute of the [Dim Order] dimension.  In this case it would be fixed by adding Existing to the NonEmpty() function ie try this.  Also, probably safest to add the measure you are using for NonEmpty.\n\ncount(nonempty(Existing [Dim Order].[ORDER_ID].[ORDER_ID].members,Measures.Orders))\n\nRichard\n\n• Edited by Monday, April 22, 2013 12:49 AM correction\n• Marked as answer by Thursday, April 25, 2013 9:09 AM\nMonday, April 22, 2013 12:48 AM" ]

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[ "# Weight Loss Calculator\n\nAugust 14, 2011\n\nThere are many people who are looking for a calculator that will give them details of the calories that they must take to lose. Here is a simple calculator that will help them in calculating the calories they must take to lose weight. The calculator also helps people calculate the calories they must reduce for faster weight loss.\n\n``` .wlc_table<br /> {</p> <p> }<br /> .wlc_table label<br /> {<br /> float:left;<br /> width:100px;<br /> }<br /> ```\n\n` `\n\n``` Your age: Your gender: MaleFemale Your height: ft & in OR cm Your weight: lbs OR kg Daily activity level: No sport/exerciseLight activity (sport 1-3 times per week)Moderate activity (sport 3-5 times per week)High activity (everyday exercise)Extreme activity (professional athlete) How much weight you wish to lose? lbs OR kg How much time do you have? days ```\n\n``` <br /> function validateCalculator(frm)<br /> {<br /> var age=frm.age.value;<br /> if(isNaN(age) || age<6 || age > 125 || age==\"\")<br /> {<br /> alert(\"Please enter your age, numbers only.\");<br /> frm.age.focus();<br /> return false;<br /> }</p> <p> var height_ft=frm.height_ft.value;<br /> if(isNaN(height_ft) || height_ft<0) height_ft=\"\"; var height_in=frm.height_in.value; if(isNaN(height_in) || height_in<0) height_in=\"\"; var height_cm=frm.height_cm.value; if(isNaN(height_cm) || height_cm<0) height_cm=\"\"; if(height_ft==\"\" &#038;&#038; height_cm==\"\" &#038;&#038; height_in==\"\") { alert(\"Please enter your height, numbers only\"); return false; } var weight_lb=frm.weight_lb.value; if(isNaN(weight_lb) || weight_lb<0) weight_lb=\"\"; var weight_kg=frm.weight_kg.value; if(isNaN(weight_kg) || weight_kg<0) weight_kg=\"\"; if(weight_kg==\"\" &#038;&#038; weight_lb==\"\") { alert(\"Please enter your weight, numbers only.\"); return false; } var lose_lb=frm.lose_lb.value; if(isNaN(lose_lb) || lose_lb<0) lose_lb=\"\"; var lose_kg=frm.lose_kg.value; if(isNaN(lose_kg) || lose_kg<0) lose_kg=\"\"; if(lose_kg==\"\" &#038;&#038; lose_lb==\"\") { alert(\"Please enter how much weight you want to lose, numbers only.\"); return false; } var days=frm.days.value; if(isNaN(days) || days<0 || days==\"\") { alert(\"Please enter how many days you have to reach the goal, numbers only.\"); frm.days.focus(); return false; } } function calculateHeight(fld) { if(fld.name==\"height_in\" || fld.name==\"height_ft\") { // calculate height in inches if(isNaN(fld.form.height_in.value) || fld.form.height_in.value==\"\") inches=0; else inches=fld.form.height_in.value; if(isNaN(fld.form.height_ft.value) || fld.form.height_ft.value==\"\") feet=0; else feet=fld.form.height_ft.value; inches=parseInt(parseInt(feet*12) + parseInt(inches)); h=Math.round(inches*2.54); fld.form.height_cm.value=h; } else { // turn cm into feets and inches if(isNaN(fld.value) || fld.value==\"\") cm=0; else cm=fld.value; totalInches=Math.round(cm/2.54); inches=totalInches%12; feet=(totalInches-inches)/12; fld.form.height_ft.value=feet; fld.form.height_in.value=inches; } } function calculateWeight(fld) { if(fld.name==\"weight_lb\" || fld.name==\"lose_lb\") { // calculate in kg if(isNaN(fld.value) || fld.value==\"\") w=0; else w=fld.value; wKg=Math.round(w*0.453*10)/10; if(fld.name==\"weight_lb\") fld.form.weight_kg.value=wKg; else fld.form.lose_kg.value=wKg; } else { // calculate in lbs if(isNaN(fld.value) || fld.value==\"\") w=0; else w=fld.value; wP=Math.round(w*2.2); if(fld.name=='weight_kg') fld.form.weight_lb.value=wP; else fld.form.lose_lb.value=wP; } } ```\n\nCategory: Calculators & Tables" ]

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[ "# Questions tagged [solid-of-revolution]\n\nThis tag is for questions regarding to \"Solid of revolution\", a three-dimensional object obtained by rotating a function in the plane about a line in the plane.\n\n303 questions\nFilter by\nSorted by\nTagged with\n25 views\n\n### Graphing rational expressions of trig/exp by hand?\n\nIn my math class, we are not allowed to use calculators. Thus, when I see problems with an exponent/log or trig either in top or bottom and I have to graph them to solve the problem, I'm not sure ...\n33 views\n\n22 views\n\n### How do you express a surface of revolution as the graph of a function?\n\nConsider a real function $f$ with domain non-negative real numbers. Let $y = f(x)$, and consider the surface traced out by rotating the graph of $f$ about the $y$-axis. This surface is the graph of ...\n98 views\n\n### What is the correct formula for the washer method?\n\nAlmost everywhere I look the formula is: $$\\pi \\int_b^a {\\left(f(x)^2 - g(x)^2\\right) dx}$$ where f(x) is the big function and g(x) is the smaller function. Though I've run into problems while ...\n18 views\n\n### Washer method, axis intersecting with region itself\n\nI know the basic idea of disc / washer / cylindrical shell methods, to find the volume of solids generated : be it a single function, or the bounded region between two functions. The trouble I am not ...\n32 views\n\n### Solid of revolution problem - clepsydra (water clock)\n\nI need help to understand this problem: A clepsydra, or water clock, is a glass container with a small hole in the bottom through which water can flow. The \"clock\" is calibrated for ...\n### Finding the volume of revolution obtained by rotation of a region about the line $2x-y=20$\nI have the following question before me: Find the volume of the area bounded by the curves $y=x^3$, $x+y=10$ and $x=0$ about the line $2x-y=20$. I started off by drawing the given area and line of ..." ]

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[ "# Charles’ law\n\nIt is evident from Boyle’s law that, for a given gas, the product pV could be used as an indication of its temperature and, in fact, this is the basis of a scale of temperature. It can\n\nBe shown that for an ideal gas, at constant pressure, the volume is related to the temperature in a linear fashion. Experimental results support this, and reference to Figure 2.3 shows just how this could be so. Suppose that experimental results allow a straight line to be drawn between two points A and B, as a graph of volume against temperature. If the line is extended to cut the abscissa at a point P, having a temperature of-273.15°C, it is clear that shifting the origin of the co-ordinate system to the left by 273.15°C will give an equation for the straight line, of the form\n\n -273.15° fa 0° tb Temperature °C (a)", null, "Temperature in kelvin (b) Fig. 2.3 Charles’ law and absolute temperature.", null, "V=aT, <2-6)\n\n2.6 The general gas law 9\n\nWhere T is the temperature on the new scale and a is a constant representing the slope of the line.\n\nObviously\n\nT = 273.15° + t (2.7)\n\nThis graphical representation of Charles’ law shows that a direct proportionality exists between the volume of a gas and its temperature, as expressed on the new abscissa scale. It also shows that a new scale of temperature may be used. This new scale is an absolute one, so termed since it is possible to argue that all molecular movement has ceased at its zero, hence the internal energy of the gas is zero and, hence also, its temperature is at an absolute zero. Absolute temperature is expressed in kelvin, denoted by K, and the symbol T is used instead of t, to distinguish it from relative temperature on the Celsius scale.\n\nEXAMPLE 2.1\n\n15 m3 s-1 of air at a temperature of 27°C passes over a cooler coil which reduces its temperature to 13°C. The air is then handled by a fan, blown over a reheater, which increases its temperature to 18°C, and is finally supplied to a room.\n\nCalculate the amount of air handled by the fan and the quantity supplied to the room.\n\nAnswer\n\nAccording to Charles’ law:\n\nV= aT, that is to say,", null, "Hence, the air quantity handled by the fan\n\n(273 + 13)\n\n(273 + 27)\n\n= 14.3 m3 s“1\n\nAnd the air quantity supplied to the room\n\n(273 + 18)\n\n(273 + 27)\n\n= 14.55 m3 s“1\n\nOne further comment, it is clearly fallacious to suppose that the volume of a gas is directly proportional to its temperature right down to absolute zero; the gas liquefies before this temperature is attained.\n\nPosted in Air Conditioning Engineering" ]

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[ "# Statistical Quality Control Questions and Answers – SPC Methods and Philosophy – Statistical Basis of the Control Chart – 2\n\n«\n»\n\nThis set of Statistical Quality Control Quiz focuses on “SPC Methods and Philosophy – Statistical Basis of the Control Chart – 2”.\n\n1. In the name of the OC curve, OC stands for ________\na) Operation Characteristic\nb) Operating Characteristic\nc) Operator Characteristic\nd) Operated Characteristic\n\nExplanation: The OC curve stands for Operating Characteristic curve which is used to develop an understanding of probability variation of the Type II error in the quality control.\n\n2. The general theory of Control charts was first developed by ________\na) Walter A. Shewhart\nb) Deming\nc) ISO\nd) ASQC\n\nExplanation: W.A. Shewhart was the first person to develop an applicable theory for quality control for the processes used in manufacturing a product. This was known as the general theory of Control Charts.\n\n3. Control charts are the part of _______ step of DMAIC process.\na) Define\nb) Measure\nc) Act\nd) Control\n\nExplanation: The control charts use the data from the measure step to analyze the process situation and then apply the steps to control the quality of the process. So they are used in the control step, analyze step of DMAIC process.\n\n4. The size of sample is 5 for a process. If the process standard deviation is 0.15 micron, and the mean of process is 1.5 micron, the standard deviation of the sample average will be _______\na) 0.0500\nb) 0.1000\nc) 0.6710\nd) 0.0671\n\nExplanation: The standard deviation of the sample average is calculated by,\nσx=σ/√n\nPutting σ=0.15, n=5, we get, σx=0.0671.\n\n5. For a process, sample size=5, process standard deviation=0.15, mean=1.5, and the std. deviation of mean is 0.0671. What will be the value of 3σ upper control limit for the construction of control chart?\na) 1.70\nb) 1.29\nc) 1.92\nd) 0.170\n\nExplanation: As we know, UCL=;n=5; σ=0.15, $$\\overline{x}$$ = 1.5; σx=0.0671{from question}\nSo, UCL=1.5+3(0.0671)=1.7013.\n\n6. The correct expression for UCL for construction of a control chart is given by _____\na) UCL = $$\\overline{x} + Z_{\\frac{\\alpha}{2}} (\\sigma_x)$$\nb) UCL = $$\\overline{x} – Z_{\\frac{\\alpha}{2}} (\\sigma)$$\nc) UCL = $$\\overline{x} – Z_{\\frac{\\alpha}{2}} (\\sigma_x)$$\nd) UCL = $$\\overline{x} + Z_{\\frac{\\alpha}{2}} (\\sigma)$$\n\nExplanation: The UCL is defined as the upper control limit for the control chart and it is evaluated as,\nUCL = $$\\overline{x} + Z_{\\frac{\\alpha}{2}} (\\sigma_x)$$; where $$\\overline{x}$$ is the mean of process, σx is the std. deviation of mean and $$Z_{\\frac{\\alpha}{2}}$$ is a arbitrary constant.\n\n7. The general model for the lower control limit for a value of quality characteristic “w” will be _____\na) LCL = μw + Lσw\nb) LCL = μw – Lσ\nc) LCL = μw – Lσw\nd) LCL = μw + Lσ\n\nExplanation: The general expression for LCL of a quality characteristic “w” is given by,\nLCL = μw – Lσw;\nWhere “L” denotes the distance of control limits from the center line.\n\n8. The center line of a control chart will be having a value ______\na) Higher than mean of quality characteristic\nb) Lower than mean of quality characteristic\nc) Equal to mean of quality characteristic\nd) Which is higher than UCL\n\nExplanation: The center line of the control chart denotes the value of the process mean of the quality characteristic. It is always the desired value.\n\n9. Which of these is a part of corrective action process associated with a control chart?\na) OCAP\nb) DMAIC\nc) OC curve\nd) LCL\n\nExplanation: OCAP (Out-of-control-plan) is an important part of the corrective action process associated with the control charts which uses results of control charts to control the process.\n\n10. Which of these can be used to estimate capability of the process?\na) Control charts\nb) Process mean\nc) Acceptance Sampling\nd) Designed Experiments\n\nExplanation: Control charts are also used as estimating device as their results can be used to estimate the process capability of a certain process.\n\n11. Control charts for central tendency and the variability are called _________ control charts.\na) Variables\nb) Attributes\nc) Acceptance\nd) Rejections\n\nExplanation: It is convenient to describe the variables with a measure of central tendency and measure of variability. So the control charts for central tendency and variability are called Variable control charts.\n\n12. The control charts formed for judgment of conformities and non-conformities are called ______ control charts.\na) Variables\nb) Attributes\nc) Acceptance\nd) Rejections\n\nExplanation: Attributes are increased on the discrete scale and they give results as either conformity or non-conformity. So the charts plotted for non-conformities and conformities are called Attributes control charts.\n\n13. Control charts with points around mean and in predicted or fixed manner indicate ________\na) Stationary variability\nb) Non-stationary variability\nc) Auto correlated variability\nd) Process out of control\n\nExplanation: The control charts which have points around the mean and in a predicted manner indicate that the process has stationary variability. This explains the nature of the process.\n\n14. Auto correlated stationary process data is dependent on each other.\na) True\nb) False\n\nExplanation: In the auto correlated data, one data observations influence the next point, which means, if one point is above the mean in the control chart, the next one will also be above the mean.\n\n15. Control charts are not effective in defect prevention.\na) True\nb) False", null, "" ]

[ null, "https://www.sanfoundry.com/wp-content/uploads/2017/01/Manish2.png", null ]

[ "", null, "# Julia equivalent of Python's \"fsum\" for floating point summation\n\nThis discussion brings me to a question: would there be some space and interest in the Julia ecosystem for a CompensatedAlgorithms.jl package?\n\nIt could replace and extend KahanSummation.jl by providing several algorithms (not just Kahan-Babuska-Neumaier, but also Ogita-Rump-Oishi, Priest and others). Such a package could also not be limited to summation and cumulated summation, but also provide accurate dot product implementations, or polynomial evaluations for example.\n\nI think we already have all the necessary building blocks (e.g EFTs in ErrorfreeArithmetic.jl or StochasticArithmetic.jl; I haven’t yet taken the time to properly compare both implementations performancewise), and we’d only need to list and implement the algorithms themselves.\n\nIf there is some enthusiasm for such a thing, I’m very much willing to take part in its development.\n\n9 Likes\n\nI don’t think your problem comes from the final reduction, but rather from the vectorized summation itself.\n\nAfter making many tests (most of which making no sense whatsoever), I have come to a strong feeling that LLVM is breaking some of the compensation during its optimization phase.\n\nI need to document this, but you need versions with an e (for “explicitly”) in front: evadd, evsub, evmul instead of vadd, vsub, and vmul, etc.\n\nusing SIMDPirates, VectorizationBase\nfused_nmul_sub(a, b, c) = SIMDPirates.vsub(SIMDPirates.vmul(SIMDPirates.vsub(a), b), c)\nunfused_nmul_sub(a, b, c) = SIMDPirates.evsub(SIMDPirates.evmul(SIMDPirates.vsub(a), b), c)\n\nrandvec(::Type{T} = Float64) where {T} = ntuple(Val(VectorizationBase.pick_vector_width(T))) do w Core.VecElement(randn(T)) end\n\na = randvec(); b = randvec(); c = randvec();\n\n\nThis yields\n\njulia> @code_native fused_nmul_sub(a, b, c)\n.text\n; ┌ @ REPL:1 within fused_nmul_sub'\n; │┌ @ SIMDPirates.jl:35 within vsub'\n; ││┌ @ floating_point_arithmetic.jl:58 within macro expansion'\n; │││┌ @ llvmwrap.jl:66 within llvmwrap' @ llvmwrap.jl:66\n; ││││┌ @ REPL:1 within macro expansion'\nvfnmsub213pd\t%zmm2, %zmm1, %zmm0 # zmm0 = -(zmm1 * zmm0) - zmm2\n; │└└└└\nretq\nnopw\t(%rax,%rax)\n; └\n\n\t.text\n; ┌ @ REPL:1 within unfused_nmul_sub'\n; │┌ @ SIMDPirates.jl:49 within evmul'\n; ││┌ @ floating_point_arithmetic.jl:65 within macro expansion'\n; │││┌ @ llvmwrap.jl:96 within llvmwrap_notfast' @ llvmwrap.jl:96\n; ││││┌ @ REPL:1 within macro expansion'\nvmulpd\t%zmm1, %zmm0, %zmm0\nmovabsq\t$140629991519192, %rax # imm = 0x7FE6F8B073D8 vxorpd (%rax){1to8}, %zmm0, %zmm0 ; │└└└└ ; │┌ @ SIMDPirates.jl:49 within evsub' ; ││┌ @ floating_point_arithmetic.jl:65 within macro expansion' ; │││┌ @ llvmwrap.jl:96 within llvmwrap_notfast' @ llvmwrap.jl:96 ; ││││┌ @ llvmwrap.jl:115 within macro expansion' vsubpd %zmm2, %zmm0, %zmm0 ; │└└└└ retq nopl (%rax) ; └ That is 1 vs 4 instructions. Of course, here we don’t want the automatic fusion / floating point associativity, so we have to use the e versions. I realize now that my problem was that I did not use the e versions for the reduction. Fixing that means we now get the correct answer: julia> using VectorizationBase, SIMDPirates, KahanSummation, BenchmarkTools julia> function sum_kahan(x::AbstractArray{T}) where {T <: Union{Float32,Float64}} px = pointer(x) W, shift = VectorizationBase.pick_vector_width_shift(T) sizeT = sizeof(T) WT = W * sizeT WT4 = 4WT N = length(x) Base.Cartesian.@nexprs 4 u -> begin s_u = SIMDPirates.vbroadcast(Vec{W,T}, zero(T)) c_u = SIMDPirates.vbroadcast(Vec{W,T}, zero(T)) end Nshift = N >> (shift + 2) for n in 0:Nshift-1 Base.Cartesian.@nexprs 4 u -> begin x_u = SIMDPirates.vload(Vec{W,T}, px + n*WT4 + (u-1)*WT) t_u = SIMDPirates.evadd(s_u, x_u) m_u = SIMDPirates.vless(SIMDPirates.vabs(x_u), SIMDPirates.vabs(s_u)) subfrom_u = SIMDPirates.vifelse(m_u, s_u, x_u) addto_u = SIMDPirates.vifelse(m_u, x_u, s_u) c_u = SIMDPirates.evsub(c_u, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_u, t_u), addto_u)) s_u = t_u end end rem = N & (4W-1) offset = Nshift * WT4 for n in 0:(rem >> shift) - 1 x_1 = SIMDPirates.vload(Vec{W,T}, px + offset) t_1 = SIMDPirates.evadd(s_1, x_1) m_1 = SIMDPirates.vless(SIMDPirates.vabs(x_1), SIMDPirates.vabs(s_1)) subfrom_1 = SIMDPirates.vifelse(m_1, s_1, x_1) addto_1 = SIMDPirates.vifelse(m_1, x_1, s_1) c_1 = SIMDPirates.evsub(c_1, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_1, t_1), addto_1)) s_1 = t_1 offset += WT end rem &= (W-1) if rem > 0 mask = unsafe_trunc(VectorizationBase.mask_type(T), 2^rem - 1) x_2 = SIMDPirates.vload(Vec{W,T}, px + offset, mask) t_2 = SIMDPirates.evadd(s_2, x_2) m_2 = SIMDPirates.vless(SIMDPirates.vabs(x_2), SIMDPirates.vabs(s_2)) subfrom_2 = SIMDPirates.vifelse(m_2, s_2, x_2) addto_2 = SIMDPirates.vifelse(m_2, x_2, s_2) c_2 = SIMDPirates.evsub(c_2, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_2, t_2), addto_2)) s_2 = t_2 end c_1 = SIMDPirates.evadd(c_1, c_2) c_3 = SIMDPirates.evadd(c_3, c_4) c_1 = SIMDPirates.evadd(c_1, c_3) s_1 = SIMDPirates.evadd(s_1, s_2) s_3 = SIMDPirates.evadd(s_3, s_4) s_1 = SIMDPirates.evadd(s_1, s_3) SIMDPirates.vsum(SIMDPirates.evsub(s_1,c_1)) end sum_kahan (generic function with 1 method) julia> x = 10^14 .* sin.((0.001:0.001:1) .* 2π); julia> @btime Float64(sum(big,$x))\n89.799 μs (3998 allocations: 218.64 KiB)\n1.1248384929460264\n\njulia> @btime sum($x) 41.968 ns (0 allocations: 0 bytes) 0.49223069682955295 julia> @btime sum_kbn($x)\n1.240 μs (0 allocations: 0 bytes)\n1.1248384929460276\n\njulia> @btime sum_kahan($x) 146.561 ns (0 allocations: 0 bytes) 1.1248384929460264 Errors: julia> relative_error(x, y) = (x - y) / y relative_error (generic function with 1 method) julia> relative_error(sum_kbn(x), Float64(sum(big, x))) 9.87006607248485e-16 julia> relative_error(sum_kahan(x), Float64(sum(big, x))) 0.0 julia> relative_error(sum(x), Float64(sum(big, x))) -0.562398779988078 2 Likes Thanks for the explanation. It’s nice to be able to select between efficiency and safety. And thanks also for the corrected implementation. Well, not really. If we look at the reduction stage, it is formally computing something like: vsum\\left( \\sum_{i=1}^4 s_i - c_i\\right) That is a lot of additions, and there might still be catastrophic cancellations in any of them. Here is your implementation running on my machine. The difference with your results might be explained by different architectures (and, therefore, SIMD widths): I’m using W=4. But I’m pretty sure that you would see this implementation behave badly on Per’s input vectors. julia> check_fun(x, sum_kahan) (1.1249173664289736, 7.011982915034423e-5) So you were right in your initial post: the reduction was indeed the problem; not only because of the lack of explicit instructions, but also because cancellations occur in it. The following, slightly modified version of your implementation seems to be much more accurate. Again, because I’m more familiar with it, I’m using Knuth’s two_sum to compensate for errors (à la Ogita, Rump & Oishi), but one could probably very well use Dekker’s fast_two_sum instead (which would be more in line with the rest of the KBN algorithm). import SIMDPirates function sum_kahan(x::AbstractArray{T}) where {T <: Union{Float32,Float64}} Vec = SIMDPirates.Vec px = pointer(x) W, shift = VectorizationBase.pick_vector_width_shift(T) sizeT = sizeof(T) WT = W * sizeT WT4 = 4WT N = length(x) Base.Cartesian.@nexprs 4 u -> begin s_u = SIMDPirates.vbroadcast(Vec{W,T}, zero(T)) c_u = SIMDPirates.vbroadcast(Vec{W,T}, zero(T)) end Nshift = N >> (shift + 2) for n in 0:Nshift-1 Base.Cartesian.@nexprs 4 u -> begin x_u = SIMDPirates.vload(Vec{W,T}, px + n*WT4 + (u-1)*WT) t_u = SIMDPirates.evadd(s_u, x_u) m_u = SIMDPirates.vless(SIMDPirates.vabs(x_u), SIMDPirates.vabs(s_u)) subfrom_u = SIMDPirates.vifelse(m_u, s_u, x_u) addto_u = SIMDPirates.vifelse(m_u, x_u, s_u) c_u = SIMDPirates.evsub(c_u, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_u, t_u), addto_u)) s_u = t_u end end rem = N & (4W-1) offset = Nshift * WT4 for n in 0:(rem >> shift) - 1 x_1 = SIMDPirates.vload(Vec{W,T}, px + offset) t_1 = SIMDPirates.evadd(s_1, x_1) m_1 = SIMDPirates.vless(SIMDPirates.vabs(x_1), SIMDPirates.vabs(s_1)) subfrom_1 = SIMDPirates.vifelse(m_1, s_1, x_1) addto_1 = SIMDPirates.vifelse(m_1, x_1, s_1) c_1 = SIMDPirates.evsub(c_1, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_1, t_1), addto_1)) s_1 = t_1 offset += WT end rem &= (W-1) if rem > 0 mask = unsafe_trunc(VectorizationBase.mask_type(T), 2^rem - 1) x_2 = SIMDPirates.vload(Vec{W,T}, px + offset, mask) t_2 = SIMDPirates.evadd(s_2, x_2) m_2 = SIMDPirates.vless(SIMDPirates.vabs(x_2), SIMDPirates.vabs(s_2)) subfrom_2 = SIMDPirates.vifelse(m_2, s_2, x_2) addto_2 = SIMDPirates.vifelse(m_2, x_2, s_2) c_2 = SIMDPirates.evsub(c_2, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_2, t_2), addto_2)) s_2 = t_2 end c_1 = SIMDPirates.evadd(c_1, c_2) c_3 = SIMDPirates.evadd(c_3, c_4) c_1 = SIMDPirates.evadd(c_1, c_3) # # Original version # s_1 = SIMDPirates.evadd(s_1, s_2) # s_3 = SIMDPirates.evadd(s_3, s_4) # s_1 = SIMDPirates.evadd(s_1, s_3) # return SIMDPirates.vsum(SIMDPirates.evsub(s_1, c_1)) # Modified reduction s_1, e = two_sum(s_1, s_2); es = e s_3, e = two_sum(s_3, s_4); es = SIMDPirates.evadd(es, e) s_1, e = two_sum(s_1, s_3); es = SIMDPirates.evadd(es, e) s = zero(T) e = SIMDPirates.vsum(es) - SIMDPirates.vsum(c_1) for si in s_1 s, ei = two_sum(s, si.value) e += ei end s + e end using SIMDPirates: evadd, evsub function two_sum(a::T, b::T) where T <: NTuple x = evadd(a, b) z = evsub(x, a) e = evadd(evsub(a, evsub(x, z)), evsub(b, z)) (x, e) end julia> x = 10^14 .* sin.((0.001:0.001:1) .* 2π); julia> check_fun(x, sum_kahan) (1.1248384929460264, 0.0) And here is a performance comparison of all variants. On my machine, vectorized KBN seems to perform just slightly worse than vectorized ORO: julia> x = 5000 |> N->randn(N) .* exp.(10 .* randn(N)) |> x->[x;-x;1.0] |> x->x[sortperm(rand(length(x)))]; julia> @btime sum($x)\n1.310 μs (0 allocations: 0 bytes)\n-134.794921875\n\njulia> @btime sum_oro_scal($x) 19.697 μs (0 allocations: 0 bytes) 0.9999999999997371 julia> @btime sum_oro_vec($x)\n3.351 μs (0 allocations: 0 bytes)\n0.9999999999999432\n\njulia> @btime sum_kbn($x) 20.416 μs (0 allocations: 0 bytes) 0.9999999999997371 julia> @btime sum_kahan($x)\n3.865 μs (0 allocations: 0 bytes)\n0.9999999999999432\n\n2 Likes\n\nI just read through your earlier posts more thoroughly – great work!\n\nIf you’re referring to all the Base.Cartesian.@nexprs 4 u -> ..., this is because of instruction latency.\nIf you’d like some reading material, I found Performance Speed Limits interesting and informative. By unrolling 4 times, eg\n\njulia> using MacroTools\n\njulia> prettify(@macroexpand Base.Cartesian.@nexprs 4 u -> begin\ns_u, ei = two_sum(s_u, xi)\ne_u += ei\noffset += W\nend)\nquote\nxi = SIMD.vload(Vec{W, T}, x, offset)\n(s_1, ei) = two_sum(s_1, xi)\ne_1 += ei\noffset += W\nxi = SIMD.vload(Vec{W, T}, x, offset)\n(s_2, ei) = two_sum(s_2, xi)\ne_2 += ei\noffset += W\nxi = SIMD.vload(Vec{W, T}, x, offset)\n(s_3, ei) = two_sum(s_3, xi)\ne_3 += ei\noffset += W\nxi = SIMD.vload(Vec{W, T}, x, offset)\n(s_4, ei) = two_sum(s_4, xi)\ne_4 += ei\noffset += W\nend\n\n\nWe are no longer bottle necked by instruction latency. Agner Fog’s instruction tables say that on Skylake(-X) the vectorized add, sub, mul, and fma instructions each have a latency of 4 clock cycles. These vary a little from one CPU family to another.\nThe Intel chips and 7 nm Ryzens should also be able to issue 2 instructions per clock cycle.\n\nSo that means to maximize performance, we need to have up to 8 instructions going at a time.\n\nA simple example to see this, a rather naive sum function:\n\nusing VectorizationBase, SIMDPirates\n\n@generated function usum(x::AbstractArray{T}, ::Val{Ushift} = Val(2)) where {T,Ushift}\n@assert 0 ≤ Ushift < 6\nU = 1 << Ushift\nW, shift = VectorizationBase.pick_vector_width_shift(T)\ntotalshift = shift + Ushift\nWU = 1 << totalshift\nWT = W * sizeof(T)\nq = quote\nptrx = pointer(x)\nBase.Cartesian.@nexprs $U u -> begin s_u = SIMDPirates.vbroadcast(Vec{$W,$T}, zero($T))\nend\noffset = 0\nN = length(x)\nfor i in 1:(N >> $totalshift) Base.Cartesian.@nexprs$U u -> begin\ns_u = SIMDPirates.vadd(s_u, SIMDPirates.vload(Vec{$W,$T}, ptrx + offset))\noffset += $WT end end rem = N &$(WU - 1)\nfor i in 1:(rem >> $shift) s_1 = SIMDPirates.vadd(s_1, SIMDPirates.vload(Vec{$W,$T}, ptrx + offset)) offset +=$WT\nend\nrem &= $(W-1) if rem > 0 s_1 = SIMDPirates.vadd(s_1, SIMDPirates.vload(Vec{$W,$T}, ptrx + offset), VectorizationBase.mask(Val{$W}(), rem))\nend\nend\nUh = U\nwhile Uh > 1\nUh >>= 1\npush!(\nq.args,\nquote\nBase.Cartesian.@nexprs $Uh u -> begin s_u = SIMDPirates.vadd(s_u, s_{u +$Uh})\nend\nend\n)\nend\npush!(q.args, :(SIMDPirates.vsum(s_1)))\nq\nend\n\n\nI wanted the unroll factor to be a power of 2; so the function takes a Val argument Ushift, and we unroll by 2^Ushift = 1<<Ushift (on top of the vectorization unrolling).\n\nWe need large vectors to see benefit. Picking the largest vector that still fits in the L1 cache:\n\njulia> first(VectorizationBase.CACHE_SIZE) ÷ sizeof(Float64)\n4096\n\njulia> x = rand(ans);\n\n\nand running benchmarks:\n\njulia> using BenchmarkTools\n\njulia> @btime sum($x) 296.868 ns (0 allocations: 0 bytes) 2036.5329867003175 julia> @btime usum($x, Val(0)) # unroll = 1 (no unrolling)\n471.189 ns (0 allocations: 0 bytes)\n2036.5329867003172\n\njulia> @btime usum($x, Val(1)) # unroll = 2^1 = 2 249.814 ns (0 allocations: 0 bytes) 2036.532986700318 julia> @btime usum($x, Val(2)) # unroll = 2^2 = 4\n132.902 ns (0 allocations: 0 bytes)\n2036.532986700318\n\njulia> @btime usum($x, Val(3)) # unroll = 2^3 = 8 84.432 ns (0 allocations: 0 bytes) 2036.532986700318 julia> @btime usum($x, Val(4)) # unroll = 2^4 = 16\n89.227 ns (0 allocations: 0 bytes)\n2036.532986700318\n\njulia> @btime usum($x, Val(5)) # unroll = 2^5 = 32 108.952 ns (0 allocations: 0 bytes) 2036.532986700318 We can see that performance does indeed peak when we unroll by 8, which breaks up the dependency chain enough so that 8 vadd instructions are running at a time. The difference is much less for smaller vectors. And if we’re summing less than 64 elements, unrolling by 8 means we’re actually not unrolling at all. I actually realize now that I was too aggressive with the choice of 4. Our functions have a lot more going on than just repetitive adds, giving the CPU more to do without the need to break up dependency chains. Even just focusing on floating point arithmetic, your two_sum function can take advantage of order execution: # D. E. Knuth, The Art of Computer Programming: Seminumerical Algorithms, 1969. function two_sum(a::T, b::T) where T <: Union{Real, Vec} x = a + b z = x - a e = (a - (x-z)) + (b-z) (x, e) end x and z, and then x-z and b-z can each be calculated at the same time, for example. As always, make hypothesis to try and understand what’s going on, and then test them. I modified your improvements to sum_kahan so that it will take arbitrary unrolls, like the simple sum function from earlier: using VectorizationBase, SIMDPirates using SIMDPirates: evadd, evsub function two_sum(a::T, b::T) where T <: NTuple x = evadd(a, b) z = evsub(x, a) e = evadd(evsub(a, evsub(x, z)), evsub(b, z)) (x, e) end function two_sum(a::T, b::T) where T <: Real x = a + b z = x - a e = (a - (x-z)) + (b-z) (x, e) end @generated function sum_kahan(x::AbstractArray{T}, ::Val{Ushift} = Val{1}()) where {T <: Union{Float32,Float64}, Ushift} @assert 0 ≤ Ushift < 6 U = 1 << Ushift W, shift = VectorizationBase.pick_vector_width_shift(T) sizeT = sizeof(T) WT = W * sizeT WU = W*U WTU = WT*U V = SIMDPirates.Vec{W,T} q = quote px = pointer(x) N = length(x) Base.Cartesian.@nexprs$U u -> begin\ns_u = SIMDPirates.vbroadcast($V, zero($T))\nc_u = SIMDPirates.vbroadcast($V, zero($T))\nend\nNshift = N >> $(shift + Ushift) offset = 0 for n in 1:Nshift Base.Cartesian.@nexprs$U u -> begin\nx_u = SIMDPirates.vload($V, px + offset) t_u = SIMDPirates.evadd(s_u, x_u) m_u = SIMDPirates.vless(SIMDPirates.vabs(x_u), SIMDPirates.vabs(s_u)) subfrom_u = SIMDPirates.vifelse(m_u, s_u, x_u) addto_u = SIMDPirates.vifelse(m_u, x_u, s_u) c_u = SIMDPirates.evsub(c_u, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_u, t_u), addto_u)) s_u = t_u offset +=$WT\nend\nend\n\nrem = N & $(WU-1) for n in 1:(rem >>$shift)\nx_1 = SIMDPirates.vload($V, px + offset) t_1 = SIMDPirates.evadd(s_1, x_1) m_1 = SIMDPirates.vless(SIMDPirates.vabs(x_1), SIMDPirates.vabs(s_1)) subfrom_1 = SIMDPirates.vifelse(m_1, s_1, x_1) addto_1 = SIMDPirates.vifelse(m_1, x_1, s_1) c_1 = SIMDPirates.evsub(c_1, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_1, t_1), addto_1)) s_1 = t_1 offset +=$WT\nend\n\nrem &= $(W-1) if rem > 0 mask = VectorizationBase.mask(Val{$W}(), rem)\n\nx_2 = SIMDPirates.vload($V, px + offset, mask) t_2 = SIMDPirates.evadd(s_2, x_2) m_2 = SIMDPirates.vless(SIMDPirates.vabs(x_2), SIMDPirates.vabs(s_2)) subfrom_2 = SIMDPirates.vifelse(m_2, s_2, x_2) addto_2 = SIMDPirates.vifelse(m_2, x_2, s_2) c_2 = SIMDPirates.evsub(c_2, SIMDPirates.evadd(SIMDPirates.evsub(subfrom_2, t_2), addto_2)) s_2 = t_2 end es = SIMDPirates.vbroadcast($V, zero($T)) end Uh = U while Uh > 1 Uh >>= 1 qtemp = quote Base.Cartesian.@nexprs$Uh u -> begin\nc_u = SIMDPirates.evadd(c_u, c_{u+$Uh}) (s_u, e) = two_sum(s_u, s_{u+$Uh})\nend\nend\npush!(q.args, qtemp)\nend\n\nqscalarreduce = quote\ns = zero(T)\ne = SIMDPirates.vsum(es) - SIMDPirates.vsum(c_1)\nfor si in s_1\ns, ei = two_sum(s, si.value)\ne += ei\nend\n\ns + e\nend\npush!(q.args, qscalarreduce)\nq\nend\n\n\nNow, we see that 2^1 = 2, and not 4 as I used, leads to the best performance:\n\njulia> y = 10^14 .* sin.(LinRange(0, 2π, 4096));\n\njulia> @btime sum($y) 282.450 ns (0 allocations: 0 bytes) 16.0 julia> @btime sum(big,$y)\n375.410 μs (16382 allocations: 895.89 KiB)\n-0.21548719867825956442164425652663339860737323760986328125\n\njulia> @btime sum_kbn($y) 5.098 μs (0 allocations: 0 bytes) -0.21548719867826094 julia> @btime sum_kahan($y, Val(0))\n738.659 ns (0 allocations: 0 bytes)\n-0.21548719867825916\n\njulia> @btime sum_kahan($y, Val(1)) 607.716 ns (0 allocations: 0 bytes) -0.21548719867825916 julia> @btime sum_kahan($y, Val(2))\n623.291 ns (0 allocations: 0 bytes)\n-0.21548719867825916\n\njulia> @btime sum_kahan($y, Val(3)) 631.864 ns (0 allocations: 0 bytes) -0.2154871986782596 I’d guess that sum_oro_vec will also see slightly better performance with 2x instead of 4x. 6 Likes I am enjoying the exploration that you (collectively) have embarked upon. For the benefit of those less familiar with all this, when you have found a good, somewhat transportable approach … do let us know rather than assuming all who are interested would get the message from reading each part of the development. Thanks. 10 Likes Just a short note to let everyone interested know that @elrod and I have been working (even though not too much) since the last posts. In short, here is a summary of where we are: 1. qualitywise, the accuracy of the algorithms can be tested against vectors of constant size (200 elements in the results below), but of increasing condition number. In the graph above, we see the relative error vary as a function of the condition number, in a log-log scale. Errors lower than \\epsilon are arbitrarily set to \\epsilon; conversely, when the relative error is more than 100% (i.e no digit is correctly computed anymore), the error is capped there in order to avoid affecting the scale of the graph too much. What we see is that the pairwise summation algorithm (as implemented in Base.sum) starts losing accuracy as soon as the condition number increases, computing only noise when the condition number exceeds \\frac{1}{\\epsilon} \\simeq 10^{16}. In contrast, both compensated algorithms (Kahan-Babuska-Neumaier and Ogita-Rump-Oishi) still accurately compute the result at this point, and start losing accuracy there, computing meaningless results when the condition nuber reaches \\frac{1}{\\epsilon^2}\\simeq 10^{32}. In effect these (simply) compensated algorithms produce the same results as if a naive summation had been performed with twice the working precision (128 bits) in this case, and then rounded to 64-bit floats. This is conform to what the theory predicts, and validates the correctness of the implementations. 1. performancewise, the implementations can be tested on a variety of vector sizes, and compared against the performance of the standard pairwise summation.", null, "In the graph above, the time spent in the summation (renormalized per element) is plotted against the vector size (the units in the y-axis label should be “ns/elem”). What we see with the standard summation is that, once vectors start having significant sizes (say, more than 1000 elements), the implementation is memory bound (as expected of a typical BLAS1 operation). Which is why we see significant decreases in the performance when the vector can’t fit into the L2 cache (around 30k elements, or 256kB on my machine) or the L3 cache (around 400k elements, or 3MB on y machine). The Ogita-Rump-Oishi algorithm, when implemented with a suitable unrolling level (ushift=2, i.e 2^2=4 unrolled iterations), is CPU-bound when vectors fit inside the cache. However, when vectors are to large to fit into the L3 cache, the implementation becomes memory-bound again (on my system), which means we get the same performance as the standard summation. In other words, the improved accuracy is free for sufficiently large vectors. For smaller vectors, the accuracy comes with a slow-down that can reach values slightly above 3 for vectors which fit in the L2 cache. EDIT: one last detail: when benchmarking summation algorithms, I found the performance of Base.sum to vary wildly from machine to machine, even with the exact same Julia version. It looks like, on some systems, the linear summation (that occurs in the leafs of the pairwise summation tree, when blocks are sufficiently small) is vectorized. But on some other systems, it is not. And I could not determine any pattern to say what caused the vectorization to happen or to fail. Is that a known issue? 11 Likes In complement to the post above, I think we’ve now reached the point where the code to be shared is too complex to comfortably fit in a discourse thread. I would like to start implementing the CompensatedAlgorithms.jl package I mentioned a few days ago, and I could of course create it in github, either under my own account, or in one of the organizations to which I have access. However, I feel like JuliaMath would be a perfect home for such a package, and I’m not provileged enough to contribute to packages there (let alone to create one). So I was wondering: should I create the CompensatedAlgorithms.jl package somewhere I have access to (and possibly move it later), or would it be more suitable to create it directly under the JuliaMath organization? In the latter case, how could this be done? 4 Likes I think you can always ask JuliaMath to include your package at some point later on, it does not need to be created there. Great work! This is so great.", null, "That’s entirely possible. Newer chips have much more capable SIMD vectorization units. The newest instruction sets are AVX2 (most new laptop chips) and AVX512 (only on beefy Xeon server chips). You can check which instruction sets your chips implement in the “flags” fields of /proc/cpuinfo (on Linux) or sysctl -a | grep machdep.cpu.features (on macOS). I have invited you to be a collaborator on https://github.com/JuliaMath/AccurateArithmetic.jl, which I had created to support DoubleFloats, a purpose for which it is no longer used. I cleaned out most or all of it. Let’s use this repo to provide implementations of compensated arithmetic algorithms. 1 Like Thanks! The code used to generate the figures above is now provided in this repo. I’m now trying to move forward and implement a dot product along the same lines. If everything goes according to the plan, it should be fairly easy to reuse most of the summation logic and share the vast majority of the code. We’ll see what happens in terms of performance, but I would be surprised if we still had the “free lunch” situation of the summation: since the dot product is a bit more arithmetically intensive, it might remain CPU-bound even outside the L3 cache. Thanks. I also suspected something related to chip capabilities, but it appears to not be as simple as a avx2 vs avx512 thing. I’ve seen performance variations happen within each class of CPU. Here is an experiment that is simple to perform, now that the explicitly vectorized code is easily available: shell> grep \"model name\" /proc/cpuinfo | tail -n1 model name : Intel(R) Core(TM) i5-6200U CPU @ 2.30GHz shell> grep \"flags\" /proc/cpuinfo | tail -n1 flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe syscall nx pdpe1gb rdtscp lm con stant_tsc art arch_perfmon pebs bts rep_good nopl xtopology nonstop_tsc cpuid aperfmperf tsc_known_freq pni pclmulqdq dtes64 monitor ds_cpl vmx est tm2 ssse3 sdbg fma cx1 6 xtpr pdcm pcid sse4_1 sse4_2 x2apic movbe popcnt aes xsave avx f16c rdrand lahf_lm abm 3dnowprefetch cpuid_fault epb invpcid_single pti retpoline rsb_ctxsw tpr_shadow v nmi flexpriority ept vpid fsgsbase tsc_adjust bmi1 avx2 smep bmi2 erms invpcid mpx rdseed adx smap clflushopt intel_pt xsaveopt xsavec xgetbv1 xsaves dtherm ida arat pln pts hwp hwp_notify hwp_act_window hwp_epp shell> git clone https://github.com/JuliaMath/AccurateArithmetic.jl.git shell> cd AccurateArithmetic.jl/ shell> julia --project julia> using Pkg; Pkg.instantiate() julia> using AccurateArithmetic julia> using BenchmarkTools julia> x = rand(10_000); julia> @btime sum($x)\n1.342 μs (0 allocations: 0 bytes)\n4986.434017383861\n\njulia> @btime sum_oro(\\$x)\n3.410 μs (0 allocations: 0 bytes)\n4986.434017383861\n\n\nOn this machine (a core i5), sum is 2-3x faster than sum_oro. On very similar machines, I’ve seen the opposite happen. I’d be grateful if some discourse users could perform the same kind of test and publish their results; maybe that would help understand what happens and in which conditions.\n\nBut maybe this would be best posted in a separate discourse thread?\n\nI’m not sure this is what you had in mind, but @elrod and I co-authored a paper that we’re submitting to the Correctness 2019 workshop (to be held in conjunction with SC’19). A preliminary version of the preprint is available here for anyone interested:\n\n(the deadline for submissions has been extended to August 19th, so we still have some time to take your comments / questions into account if you have some)\n\n9 Likes\n\nThat is precisely what I had hoped you would share.", null, "1 Like\n\nCommenting on this tangent:\n\nIt’s many orderS of magnitude slower than sum (I didn’t try xsum), without sort: AND gets less precise answers as it seems to disable default sum algorithm?!\n\n[Better would be to sort by absolute magnitude, but I get away with this here. And I’m just showing for timing, and then saw sorting gave same answer so not needed, at least for my array a.]\n\njulia> @time sum(sort(a)) # is only 27 times slower (for that many numbers, didn't try radixsort\n0.000804 seconds (7 allocations: 78.375 KiB)\n0.0\n\njulia> @time sum(big.(a)) # sum for non-big must use a different algorithm, not a straight sum over array, as it gave 0.0? Error can be much larger with big:\n0.043928 seconds (20.01 k allocations: 1.145 MiB)\n-15.744873851312446018712308593471442723857083704853153190993195048996768170998\n\n\nIt shouldn’t be worse as here (meaning without sort), when a higher precision datatype is used. While big alone wouldn’t help much for precision (not speed), it seems it should be a bit better, all else equal, if that where the case…" ]

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[ "Cody\n\n# Problem 2798. Beginner's Problem - Squaring\n\nTry out this test problem first.\n\nGiven the variable x as your input, square it by two and put the result in y.\n\nExamples:\n\n``` Input x = 2\nOutput y is 4```\n``` Input x = 10\nOutput y is 100```\n\n### Solution Stats\n\n84.0% Correct | 16.0% Incorrect\nLast Solution submitted on Apr 04, 2020" ]

[ null ]

[ "An improved statistical model has been developed to describe the chemical composition of an evaporating multicomponent- liquid drop and of the mixture of gases surrounding the drop. The model is intended for use in computational simulations of the evaporation and combustion of sprayed liquid fuels, which are typically mixtures of as many as hundreds of different hydrocarbon compounds. Since an exact model providing a detailed account of all of the compounds would be computationally intractable, the present statistical model is an approximation designed to afford results that are accurate enough to contribute to understanding of the simulated physical and chemical phenomena, without imposing an unduly large computational burden.", null, "Double-Γ-PDF and Discrete-Model-PDF values were computed for a drop of diesel fuel evaporating at a gas temperature of 600 K. The results shown here are for the time at which a drop has evaporated to one-fifth of its initial mass.\n\nAs in any model of physical and chemical phenomena, some simplifying assumptions are made: A drop is taken to be a sphere of radius R, wherein the liquid has constant density ρ1. Evaporation of the liquid is assumed to occur under thermodynamic equilibrium. The gas surrounding the drop — a mixture comprising a carrier gas plus multicomponent vapor from previously evaporated drops — is assumed to obey the perfect-gas equation of state. The gas is postulated to be quasi-steady with respect to the liquid, in the sense that the characteristic time of the gas is much shorter than that of the liquid. Consistent with what would be done in computations involving a very large number of drops, gradients within the drop are neglected; attention is paid only to volumetrically averaged properties represented by the drop temperature and the mass fractions of the chemical species in the drop. The focus on volumetrically averaged drop properties precludes consideration of phenomena associated with differences among diffusivities of different species.\n\nIt is further assumed that the simulated phenomena occur at atmospheric pressure, where solubility of the carrier gas into the liquid is negligible, and that the far field conditions are quiescent. The model includes the applicable equations for the conservation of mass, species, and energy.\n\nThe statistical aspect of the model enters through invocation of the concept of continuous thermodynamics, according to which the chemical composition of a fuel is described probabilistically, by use of a probability distribution function (PDF). This concept was summarized in two prior NASA Tech Briefs articles: \"Model of Mixing Layer With Multicomponent Evaporating Drops\" (NPO-30505), Vol. 28, No. 3 (March 2004), page 55, and \"Simulations of Evaporating Multicomponent Fuel Drops\" (NPO-30641), Vol. 29, No. 3 (March 2005), page 72. However, the present statistical model differs from the model of the cited prior articles.\n\nIn the prior model, the PDF is a single- peaked Gamma distribution, which is a function of the molar weight and of several parameters. However, the prior model does not generate a second peak at the low-molar-weight end of the drop-composition PDF that emerges when condensation of vapor onto drops occurs.\n\nThe present model generates the needed second peak. The PDF in the present model is a superposition of two Γ-PDFs and, accordingly, is denoted a double-Γ-PDF. It is a function of the molar weight plus five parameters.\n\nFormally, the calculation prescribed by this model can be reduced to an inverse mapping from (1) the first five distribution moments, which can be calculated by use of the conservation equations, to (2) the five parameters of the double-Γ-PDF. The complexity of the inverse mapping, and the fact that the statistics of a discrete-chemical-species model may not be represented exactly by the double-Γ-PDF, make this calculation only approximate.\n\nIn a practical calculation, one makes a further approximation by utilizing only the first four moments and four parameters plus a fifth parameter that is determined empirically. Despite these approximations, extensive tests of the model on both diesel oil and gasoline show that the double-Γ-PDF accurately represents the predictions of the discrete-chemical-species model (see figure). Moreover, the mean and variation of composition at the drop surface as computed by use of the double-Γ-PDF are in excellent agreement with those computed by use of the discrete-chemical-species model: this is an important result because these mean and variance determine the composition of the gas mixture.\n\nThis work was done by Kenneth Harstad, Patrick Le Clercq, and Josette Bellan of Caltech for NASA's Jet Propulsion Laboratory. For more information, download the Technical Support Package (free white paper) at www.techbriefs.com/tsp under the Physical Sciences category. NPO-30886\n\n##### This Brief includes a Technical Support Package (TSP).", null, "" ]

[ null, "data:image/svg+xml;base64,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", null, "data:image/svg+xml;base64,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", 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[ "# zbMATH — the first resource for mathematics\n\nConformal structure on the boundary and geodesic flow of a $$\\text{CAT}(-1)$$-space. (Structure conforme au bord et flot géodésique d’un $$\\text{CAT}(-1)$$-espace.) (French) Zbl 0871.58069\nThe background for the development is the theory of Gromov’s hyperbolic groups [see e.g. M. Gromov, Hyperbolic groups. Essays in group theory, Publ. Math. Sci. Res. Inst. 8, 75-263 (1987; Zbl 0634.20015) and M. Coornaert, T. Delzant and A. Papadopoulos, Géométrie et théorie des groupes. Les groupes hyperbolique de Gromov, (Lect. Notes Math. 1441) (1990; Zbl 0727.20018)]. The author studies actions of hyperbolic groups on the CAT(–1)-spaces, a vast abstract generalization of the hyperbolic spaces defined by means of comparison of geodesic triangles with those in the hyperbolic plane, including, among others, all Riemannian manifolds with sectional curvatures at most $$-1$$ and the Gromov polyhedra. The first half of the paper is devoted to a quite detailed exposition of the basic concepts and their properties. Then a distinguished conformal family of metrics is constructed on the boundary of any CAT(–1)-space. Finally, the relations between the geodesic flows given by quasi-convex actions of a hyperbolic group by isometries and the conformal structures on their limit points are studied. The paper is well organized and clearly written.\n\n##### MSC:\n 37D40 Dynamical systems of geometric origin and hyperbolicity (geodesic and horocycle flows, etc.) 53D25 Geodesic flows in symplectic geometry and contact geometry 53C23 Global geometric and topological methods (à la Gromov); differential geometric analysis on metric spaces" ]

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[ "# BTME 20 Third Semester B. Tech Mechanical Engineering", null, "```BTME 20\nThird Semester B. Tech Mechanical Engineering\nExamination July/August 2012\nThermodynamics\nTime:-3Hours\nMax. Marks: - 75\nSECTION-A\nAttempt any five Questions.\n(5*5)\n1. Define atomic mass and molecular mass.\n2. Derive the combustion equation of gas ears fuel.\n3. A producer gas used as a fuel has the following volumetric composition H2 28%,\nCO12%, CH4 2%, co2 16% and N2 42 %. Find the volume of air required for\ncomplete contusion of 1m(3) this gas. Air contains 21% by volume of oxygen.\n4. Derive the general expression for change of entropy of a perfect gas.\n5. 0.5 kg of a perfect gas is heated from 100’c to 300˚c at a constant pressure of 2.8\nbars. It is then cooled to 100˚c at constant volume. Find the overall change in\nentropy take cp=1KJ/KG k and cu=0.72 kj/kg k.\n6. 1 kg of air occupier 0.084 m&sup3; at 12.5 bar and 537˚c. it is expanded at a constant\ntemperature to a final volume of 0.336m&sup3;. Calculate.\n7. Explain the working of single stage reciprocating system.\nSECTION-B\nAttempt any two Questions.\n(10*2)\n8. Explain in brief about magneto ignition system.\n9. Two boilers discharge equal amounts of steam in to the same main. The steam\nfrom one is at 18 bars and 380˚c, and from the other at 18 bars and 0.95 qualities.\nDetermine the equilibrium conditions, after mixing 2, the loss of entropy by the\nhigh temperature steam, 3. The gain in entropy by the low temperature steam,\nand 4. Net increase or decrease of entropy.\n10. What is super heater? Explain in brief.\nSECTION-C\nAttempt any two Questions.\n(15*2)\n11. A 30m high chimney is used to discharge hot gasses at 297˚c to the atmosphere\nwhich is at 27˚c find the mass of air actually used per kg of fuel, if the draught\nproduced is 15mm of water. If the cool burnt in the combustion chamber\ncontains 80% carbon, 6%moisture and remaining ash, determine the percentage\nof excess air supplied.\n12. A steam turbine receives super heated steam at a pressure of 17 bars and having\na degree of superheat of 110˚c. The exhaust pressure is 0.07 bar and the\nexpansion of steam takes place is entropic ally. Calculate 1. The heat supplied 2.\nThe rejected, 3.net work done and 4, the thermal efficiency.\n13. Determine the cylinder dimensions of a double acting , 11 kw indicated power,\nair compressor which compressor air from 1 bar to 7 bar according to the law pu\n(1.4)= constant. The average piston speed is 150m/s. assume stroke to diameter\nratio of 1.5 and neglect clearance.\n```" ]

[ null, "https://s3.studylib.net/store/data/006809935_1-2114f9a90d1557d7bfaa2fd2462f55ad.png", null ]

[ "## Calculating reactant moles [ENDORSED]\n\nCaminaB_1D\nPosts: 63\nJoined: Fri Sep 28, 2018 12:16 am\n\n### Calculating reactant moles\n\nCan someone explain the step of calculating reactant moles when solving a stoichiometry problem. In other words, I do not understand the \"why\" behind the calculations of this step.\n\nJasmin Argueta 1K\nPosts: 69\nJoined: Fri Sep 28, 2018 12:16 am\n\n### Re: Calculating reactant moles\n\nDo you have a problem you would like help with? Maybe running through the steps with numbers will help when explaining.\n\nCaminaB_1D\nPosts: 63\nJoined: Fri Sep 28, 2018 12:16 am\n\n### Re: Calculating reactant moles\n\nM.3 in the sixth edition!\n\nPosts: 96\nJoined: Fri Sep 28, 2018 12:16 am\nBeen upvoted: 2 times\n\n### Re: Calculating reactant moles  [ENDORSED]\n\nYou are calculating the reactant moles in order to compare the amount of reactants available to the ratios in the balanced equation to determine which reactant is limiting. You cannot compare the quantities in grams since your balanced equation gives you molar ratios.\n\nTheodore_Herring_1A\nPosts: 60\nJoined: Fri Sep 28, 2018 12:29 am\n\n### Re: Calculating reactant moles\n\nLet's say you have 2kg of hamburger patties and 1kg of hamburger buns. How many hamburgers can you make? As you can see, we would need to know the amount of hamburger patties and buns you have in order to make this calculation. This is why we convert from grams into moles, so that we know the amount of reactants in a reaction." ]

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[ "# If a is small, then show that $[1/(1+a)]^3$ is nearly equal to $1-3a$.\n\nIf a is small, then show that $$\\dfrac{1}{(1+a)^3}$$ is nearly equal to $$(1-3a)$$. Also show that if $$0, then the error$$<0.0007$$.\n\nCan we solve this without using calculus?\n\nIf so can anyone help me with the solution?\n\n## 3 Answers\n\nWe know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$ Therefore $$\\frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3\\approx (1-3a+3a^2-a^3)\\approx1-3a$$ Now for small a we can neglect $$a^2$$ and$$a^3$$ that's because higher powers of a get very small when compared to 1 or a when a is small Example $$a=0.001$$ then$$a^2=0.00001$$ and$$a^3=0.0000001$$ which are very small . $$Q.E.D.$$\nJust plug in the value you need to calculate error.\n\nSuppose $$a$$ is small. Then $$(1+a)^x \\approx 1+xa$$ holds and is known as the Binomial approximation. Thus $$\\frac{1}{(1+a)^3} = (1+a)^{-3} \\approx 1+(-3)a = (1-3a)$$.\n\nTo find the error, note that as $$a \\rightarrow 0$$, $$\\frac{1}{(1+a)^3} \\rightarrow 1$$ and $$(1-3a) \\rightarrow 1$$. When $$a=0.01$$, the error between the two expressions is $$|\\frac{1}{(1+a)^3} - (1-3a)| = |\\frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| \\approx (0.9706) - (0.9700) = 0.0006$$, which is less than $$0.0007$$. Noting that the difference in the expressions is monotonically increasing in the interval $$0, we can say that the error is thus less than $$0.0007$$.\n\n\\begin{align} \\frac1{(1+a)^3}-(1-3a) &=\\frac{6+8a+3a^2}{(1+a)^3}\\cdot a^2\\\\[6pt] &\\le\\left\\{\\begin{array}{} 22a^2&\\text{if }a\\ge-\\frac12\\\\ 6a^2&\\text{if }a\\ge0 \\end{array}\\right. \\end{align} since $$\\frac{6+8a+3a^2}{(1+a)^3}=\\frac3{1+a}+\\frac2{(1+a)^2}+\\frac1{(1+a)^3}$$ is decreasing for $$a\\gt-1$$.\n\nIf $$0\\le a\\le0.01$$, then the error is at most $$0.0006$$." ]

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[ "", null, "LAPACK  3.4.2 LAPACK: Linear Algebra PACKage\nclacgv.f File Reference\n\nGo to the source code of this file.\n\n## Functions/Subroutines\n\nsubroutine clacgv (N, X, INCX)\nCLACGV conjugates a complex vector.\n\n## Function/Subroutine Documentation\n\n subroutine clacgv ( integer N, complex, dimension( * ) X, integer INCX )\n\nCLACGV conjugates a complex vector.\n\n` CLACGV conjugates a complex vector of length N.`\n [in] N ``` N is INTEGER The length of the vector X. N >= 0.``` [in,out] X ``` X is COMPLEX array, dimension (1+(N-1)*abs(INCX)) On entry, the vector of length N to be conjugated. On exit, X is overwritten with conjg(X).``` [in] INCX ``` INCX is INTEGER The spacing between successive elements of X.```" ]

[ null, "https://netlib.org/lapack/explore-html-3.4.2/lapack.png", null ]

[ "# What are the assumptions of factor analysis?\n\nI want to check if I really understood [classic, linear] factor analysis (FA), especially assumptions that are made before (and possibly after) FA.\n\nSome of the data should be initially correlated and there is a possible linear relation between them. After doing factor analysis, the data are normally distributed (bivariate distribution for each pairs) and there is no correlation between factors (common and specifics), and no correlation between variables from one factor and variables from other factors.\n\nIs it correct?\n\nInput data assumptions of linear FA (I'm not speaking here about internal assumptions/properties of the FA model or about checking the fitting quality of results).\n\n1. Scale (interval or ratio) input variables. That means the items are either continuous measures or are conceptualized as continuous while measured on discrete quantitative scale. No ordinal data in linear FA (read). Binary data should also be avoided (see this, this). Linear FA assumes that latent common and unique factors are continuous. Therefore observed variables which they load should be continuous too.\n2. Correlations are linear. Linear FA may be performed based on any SSCP-type association matrix: Pearson correlation, covariance, cosine, etc (though some methods/implementations might restrict to Pearson correlations only). Note that these are all linear-algebra products. Despite that the magnitude of a covariance coefficient reflects more than just linearity in relation, the modeling in linear FA is linear in nature even when covariances are used: variables are linear combinations of factors and thus linearity is implied in the resulting associations. If you see/think nonlinear associations prevail - don't do linear FA or try to linearize them first by some transformations of the data. And don't base linear FA on Spearman or Kendall correlations (Pt. 4 there).\n3. No outliers - that's as with any nonrobust method. Pearson correlation and similar SSCP-type associations are sensitive of outliers, so watch out.\n4. Reasonably high correlations are present. FA is the analysis of correlatedness, - what's its use when all or almost all correlations are weak? - no use. However, what is \"reasonably high correlation\" depend on the field of study. There is also an interesting and varied question whether very high correlations should be accepted (the effect of them on PCA, for example, is discussed here). To test statistically if the data are not uncorrelated Bartlett's test of sphericity can be used.\n5. Partial correlations are weak, and factor can be enough defined. FA assumes that factors are more general than just loading pairs of correlated items. In fact, there even an advice not to extract factors loading decently less than 3 items in explotatory FA; and in confirmatory FA only 3+ is guaranteed-identified structure. A technical problem of extraction called Heywood case has, as one of the reasons behind, the too-few-items-on-factor situation. Kaiser-Meyer-Olkin (KMO) \"sampling adequacy measure\" estimates for you how weak are partial correlations in the data relative the full correlations; it can be computed for every item and for the whole correlation matrix. Common Factor analysis model assumes that pairwise partial correlations are enough small not be bothered about and modelled, and they all fall into that population noise for individual correlation coefficients which we don't regard any differently from the sample noise for them (see). And read also.\n6. No multicollinearity. FA model assumes that all items each posesses unique factor and those factors are orthogonal. Therefore 2 items must define a plane, 3 items - a 3d space, etc: p correlated vectors must span p-dim space to accomodate their p mutually perpendicular unique components. So, no singularity for theoretical reasons$$^1$$ (and hence automatically n observations > p variables, without saying; and better n>>p). Not that complete multicollinearity is allowed though; yet it may cause computational problems in most of FA algorithms (see also).\n7. Distribution. In general, linear FA does not require normality of the input data. Moderately skewed distributions are acceptable. Bimodality is not a contra-indication. Normality is indeed assumed for unique factors in the model (they serve as regressional errors) - but not for the common factors and the input data (see also). Still, multivariate normality of the data can be required as additional assumption by some methods of extraction (namely, maximum likelihood) and by performing some asymptotic testing.\n\n$$^1$$ ULS/minres methods of FA can work with singular and even non p.s.d. correlation matrix, but strictly theoretically such an analysis is dubious, for me.\n\n• ,could you read this post,it seemed little different. Jul 28, 2017 at 14:27\n• If Binary data should also be avoided,what else factor analysis method can we do for binary data? Apr 22, 2019 at 8:23\n• dear ttnphns; I notice that you don't mention that the data are assumed normal and other online indicate that normality is not required. My query is if the latent variables are assumed normal, and the observations are modelled as a weighted sum of the factors does this then not imply a normal distribution on the observations? (I'm sorry im sure this is a dumb question) Feb 16, 2020 at 12:27\n• @user2957945, Paragraph 7 says about normality. Normality assumption is necessary for some methods of factor extraction and for performin some statistical tests facultatively accompanying factor analysis. To your question: Yes, if factors are distributed normally and errors normally too, that will mean manifest variables are also normal. Feb 16, 2020 at 12:39\n• ah, thanks @ttnphns; sorry to bother you -- I dont quite know how I managed to miss that. Appreciate your help. Feb 16, 2020 at 12:49\n\nMuch of the time, factor analysis is conducted without any statistical tests per se. It is much more subjective and interpretive than methods such as regression, structural equation modelling, and so on. And generally it is inferential tests that come with assumptions: in order for p values and confidence intervals to be correct, those assumptions must be met.\n\nNow, if the method for choosing the number of factors is set to be the maximum likelihood method, then there is an assumption that goes with this: that the variables input into the factor analysis will have normal distributions.\n\nThat the input variables will have nonzero correlations is a sort of assumption in that without it being true, factor analysis results will be (probably) useless: no factor will emerge as the latent variable behind some set of input variables.\n\nAs far as there being \"no correlation between factors (common and specifics), and no correlation between variables from one factor and variables from other factors,\" these are not universally assumptions that factor analysts make, although at times either condition (or an approximation of it) might be desirable. The latter, when it holds, it known as \"simple structure.\"\n\nThere is another condition that is sometimes treated as an \"assumption\": that the zero-order (vanilla) correlations among input variables not be swamped by large partial correlations. What this means in a nutshell is that relationships should be strong for some pairings and weak for others; otherwise, results will be \"muddy.\" This is related to the desirability of simple structure and it actually can be evaluated (though not formally \"tested\") using the Kaiser-Meyer-Olkin statistic, or the KMO. KMO values near .8 or .9 are usually considered very promising for informative factor analysis results, while KMOs near .5 or .6 are much less promising, and those below .5 might prompt an analyst to rethink his/her strategy.\n\n• As i read, that factor analysis starts with some correlation with variables and we try to make this correlation more and more clear Nov 11, 2012 at 19:29\n• After application of Factor analysis, if we have used orthogonal rotation, we will be sure that there is no correlation between factors Nov 11, 2012 at 19:32\n\nAssumptions underlying exploratory factor analysis are:\n• Interval or ratio level of measurement\n• Random sampling\n• Relationship between observed variables is linear\n• A normal distribution (each observed variable)\n• A bivariate normal distribution (each pair of observed variables)\n• Multivariate normality\nAbove from the SAS file" ]

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[ "# Fibonacci Sequence using LMC", null, "#### Did you know?\n\nThe Fibonacci Sequence is the series of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 … where the next number is found by adding up the two numbers just before it.", null, "The first 10 numbers of the Fibonacci number sequence are:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34\n\nThe following table shows how we can calculate each Fibonacci number from this sequence:\n\n Fibonacci Number Calculation 0 1 1 = 0 + 1 2 = 1 + 1 3 = 1 + 2 5 = 2 + 3 8 = 3 + 5 13 = 5 + 8 21 = 8 + 13 34 = 13 + 21 … …\n\n#### Little Man Computer\n\nYour challenge is to write a program using one of the following online LMC Simulators to calculate and output the first 10 numbers of the Fibonacci sequence.\n\n#### LMC Instruction Set\n\nNote that in the following table “xx” refers to a memory address (aka mailbox) in the RAM. The online LMC simulator has 100 different mailboxes in the RAM ranging from 00 to 99.", null, "" ]

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[ "Percentage Calculator\n\n# How calculate 41% of 6999?\n\nA simple way to calculate percentages of X\n\n 41% of 6999 = 2869.59 6999 + 41% = 9868.59 6999 - 41% = 4129.41\n What is Calculate the percentage: %\n\nIn the store, the product costs 6999, you were given a discount 41 and you want to understand how much you saved.\n\nSolution:\n\nAmount saved = Product price * Percentage Discount/ 100\n\nAmount saved = (41 * 6999) / 100\n\nMore random interest calculations:\n7% от 6999 = 489.93\n6999 + 7% = 7488.93\n6999 - 7% = 6509.07\n12% от 6999 = 839.88\n6999 + 12% = 7838.88\n6999 - 12% = 6159.12\n28% от 6999 = 1959.72\n6999 + 28% = 8958.72\n6999 - 28% = 5039.28\n37% от 6999 = 2589.63\n6999 + 37% = 9588.63\n6999 - 37% = 4409.37\n46% от 6999 = 3219.54\n6999 + 46% = 10218.54\n6999 - 46% = 3779.46\n51% от 6999 = 3569.49\n6999 + 51% = 10568.49\n6999 - 51% = 3429.51\n67% от 6999 = 4689.33\n6999 + 67% = 11688.33\n6999 - 67% = 2309.67\n78% от 6999 = 5459.22\n6999 + 78% = 12458.22\n6999 - 78% = 1539.78\n85% от 6999 = 5949.15\n6999 + 85% = 12948.15\n6999 - 85% = 1049.85\n90% от 6999 = 6299.1\n6999 + 90% = 13298.1\n6999 - 90% = 699.9\n\nAnd what if the percentage is more than 100? Then the resulting result will be greater than the sum itself6999. For example:\n300% от 6999 = 20997\n400% от 6999 = 27996\n900% от 6999 = 62991" ]

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[ "Digital 논리게이트\n\n## Digital 논리게이트 ¶\n\n• 작성자\n조재혁(", null, "[email protected])\n\n• 고친과정\n2008년 어느날 : 처음씀\n\n### OR의 교환법칙 ¶\n\n```x OR y = y OR x\n```\n\n### OR의 결합법칙 ¶\n\n```x OR y OR z = (x OR y) OR z = x OR (y OR z)\n```\n\n### AND : x AND y = F ¶\n\n```0 AND 0 = 0\n0 AND 1 = 0\n1 AND 0 = 0\n1 AND 1 = 1\n```\n\n### OR : x OR y = F ¶\n\n```0 OR 0 = 0\n0 OR 1 = 1\n1 OR 0 = 1\n1 OR 1 = 1\n```\n\n### Inverter : !x = F ¶\n\n```!0 = 1\n!1 = 0\n```\n\n```0 = 0\n1 = 1\n```\n\n### NAND : x NAND y = !(x AND y) = F ¶\n\n```0 NAND 0 = 1\n0 NAND 1 = 1\n1 NAND 0 = 1\n1 NAND 1 = 0\n```\n\n### NOR : x NOR y = !(x OR y) = F ¶\n\n```0 NOR 0 = 1\n0 NOR 1 = 0\n1 NOR 0 = 0\n1 NOR 1 = 0\n```\n\n### XOR(Exclusive-OR) : x XOR y = (x AND !y) OR (!x AND y) = F ¶\n\n```0 XOR 0 = 0\n0 XOR 1 = 1\n1 XOR 0 = 1\n1 XOR 1 = 0\n```\n\n### Exclusive-NOR : x XNOR y = !(x XOR y) = (x AND y) OR (!x AND !y) = F ¶\n\n```0 XNOR 0 = 1\n0 XNOR 1 = 0\n1 XNOR 0 = 0\n1 XNOR 1 = 1\n```" ]

[ null, "https://wiki.kldp.org/imgs/moni2/email.png", null ]

[ "4.3 out of 5. Views: 862.\n###### Lesson 9 Homework Module 5 - Lesson Worksheets.\n\nLearning Objective Addition and Subtraction of Length Units: Math Terminology for Module 2 New or Recently Introduced Terms View terms and symbols students have used or seen previously.\n\n###### Module 5 Lesson 13 Worksheets - Lesson Worksheets.\n\nModule 5 Lesson 13. Displaying all worksheets related to - Module 5 Lesson 13. Worksheets are Grade 5 module 5, Grade 5 module 1, Grade 5 module 4, Homework practice and problem solving practice workbook, Basic computer skills module 5 introduction to microsoft, Lesson plans for module 28101 13 introduction to masonry, Module 1 texas essential knowledge and skills teks, Lesson plans for module.\n\n###### Module 4 Lesson 7 Grade 5 Worksheets - Kiddy Math.\n\nGrade 3 Module 7. Topic A: Solving Word Problems. Lesson 1. Lesson 2. Lesson 3. Topic B: Attributes of Two-Dimensional Figures. Lesson 4. Lesson 5. Lesson 6. Lesson 7. Lesson 8. Lesson 9. Topic C: Problem Solving with Perimeter. Lesson 10. Lesson 11. Lesson 12. Lesson 13. Lesson 14. Lesson 15. Lesson 16. Lesson 17. Mid-Module Review and Assessments.\n\n###### Evaluate Homework And Practice Module 2 Lesson 2 Answers.\n\nGrade 2 Module 7: Problem Solving with Length, Money, and Data. Module 7 presents an opportunity for students to practice addition and subtraction strategies within 100 and problem-solving skills as they learn to work with various types of units within the contexts of length, money, and data.\n\n###### Lesson 6 Homework 4 7 - EMBARC.Online.\n\nGrade 7 Mathematics Start - Grade 7 Mathematics Module 1 In order to assist educators with the implementation of the Common Core, the New York State Education Department provides curricular modules in P-12 English Language Arts and Mathematics that schools and districts can adopt or adapt for local purposes.\n\n###### Eureka Math Algebra 2 Module 2 Answers.\n\nUnit 11 Homework 1. Unit 11 Homework 1 - Displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are Homework practice and problem solving practice workbook, Unit 11 outline, Percents homework 1, Lesson 11 measurement and units of measure, Im im im im, Unit 3 lesson 1 understanding proportional relationships, World war one information and activity work.\n\n###### Module 2 lesson 7 - LinkedIn SlideShare.\n\nModule 2 Lesson 1 DRAFT. a year ago. by senecal. Played 145 times. 0. 5th grade. Science. 76% average accuracy. 0. Save. Edit. Edit. Print; Share; Edit; Delete; Host a game. Live Game Live. Homework. Solo Practice. Practice. Play. Share practice link. Finish Editing. This quiz is incomplete! To play this quiz, please finish editing it. Delete.\n\n###### Course: G3M7: Geometry and Measurement Word Problems.\n\nWorksheets are Eureka math a story of units, Eureka math homework helper 20152016 grade 2 module 3, Eureka math homework helper 20152016 grade 6 module 2, Lesson 2 proportional relationships, Math work, Lesson 2 solving for unknown angles using equations, Eureka math homework helper 20152016 grade 1 module 1.\n\n###### Lesson 2 Homework Practice Numerical Expressions Answers.\n\nPractice and Homework Lesson 7.5 COMMON CORE STANDARD—1.NBT.C.5 Use place value understanding and properties of operations to add and subtract. 10 Less, 10 More Use mental math. Complete the chart. Chapter 7 four hundred twenty-seven 427 10 Less 10 More 1. 48 2. 83 3. 8 4. 47 5. Jim has 16 books. Doug has 10 fewer books than Jim. How many books.\n\n###### Lesson 1 Homework Practice Worksheets - Learny Kids.\n\nTopics and Objectives (Module 2) A. Understand Concepts About the Ruler Standard: 2.MD.1 Days: 3 Module 2 Overview Topic A Overview Lesson 1: Connect measurement with physical units by using multiple copies of the same physical unit to measure.(Lesson 2: Use iteration with one physical unit to measure.(Lesson 3: Apply concepts to create unit rulers and measure lengths using unit rulers." ]

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[ "MCQ of Basic Electrical page-1\n\n1.The terminals across the source are .......... if a current source is to be neglected.", null, "Open-circuited\nb) Short-circuited\nc) Replaced by a capacitor\nd) Replaced by a source resistance\n\n2.An active element in a circuit is", null, "Current source\nb) Resistance\nc) Inductance\nd) Capacitance\n\n3. A bilateral element is\n\na) Resistor\nb) Inductor\nc) Capacitor", null, "All of these\n\n4.The circuit having same properties in either direction is called", null, "Bilateral circuit\nb) Unilateral circuit\nc) Irreversible circuit\nd) Reversible circuit\n\n5. Kirchhoff’s laws are valid for\n\na) Liner circuits only\nb) Passive time-invariant circuits\nc) Non-linear circuits only", null, "Both linear and non-liner circuits\n\n6. Kirchhoff’s laws are not applicable to circuits with", null, "Distributed parameters\nb) Lumped parameters\nc) Passive elements\nd) Non-liner resistances\n\n7. Kirchhoff’s current law is applicable only to\n\na) Electric circuits\nb) Electronic circuits", null, "Junctions in a network\nd) Closed loops in a network\n\n8. Kirchhoff’s voltage law is concerned with\n\na) IR drop\nb) Battery emfs", null, "Both (a) and (b)\nd) None of these\n\n9. According to Kirchhoff’s voltage law, the algebraic sum of all IR drops and emfs in any closed loop of a network is always\n\na) Negative\nb) Positive", null, "Zero\nd) Determined by emfs of the batteries\n\n10. The algebraic sign of an IR drop primarily depends upon the", null, "Direction of flow of current\nb) Battery connections\nc) Magnitude of current flowing trough it\nd) Value of resistance\n\nRecent Post", null, "Multiple Choice Question (MCQ) of Materials page-6:51. The permeability and permittivity of a medium are A) independent of each other B) related by the atomic number C) related by the Boltzmann’s constant D) related by the velocity of electromagnetic waves.Read more...", null, "Multiple Choice Question (MCQ) of Materials page-5:41. Bakelite is A) highly inflammable B) low resistance conductor C) a semi-conductor D) uncombustible.Read more...", null, "Multiple Choice Question (MCQ) of Materials page-4:31. Which of the following pairs is commonly used in thermocouples? A) Copper-constant B) Aluminium-tin C) Iron-steel D) Silver-German silver.Read more..." ]

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[ "# Trie Data Structure\n\nVan Chan Ngo · May 30, 2017\n\n## Introduction\n\nA trie is a kind of search tree that is used to store a dynamic set where the keys are strings. A trie can be seen as a tree-shape deterministic finite automaton (DFA). Each finite language can be generated by a trie automaton, and each trie can be compressed into a acyclic DFA.\n\nEvery node of trie consists of multiple branches. Each branch represents a possible character of keys. We need to mark the last node of every key as leaf node. A trie node field value will be used to distinguish the node as leaf node (there are other uses of the value field).\n\n## Applications\n\nA common application of tries is storing a predictive text or auto-complete dictionary. Such applications take advantage of a trie’s ability to quickly look for, insert, and delete items. Using trie, we can search for the key in $$\\mathcal{O}(m)$$ instead of $$\\mathcal{O}(m.\\text{log}n)$$ if we use a binary search tree, where $$m$$, $$n$$ are the length of search string and the number of keys, respectively.\n\nTries are also well suited for implementing approximate matching algorithms including those used in spell checking and hyphenation applications.\n\n## Implementation\n\nIn this post, I will show how to implement a trie where the characters are from ‘a’ to ‘z’ (26 characters in total) using arrays in OCaml.\n\n(************************************)\n(* Trie *)\n(* Van Chan Ngo *)\n(************************************)\n\n(* trie.mli *)\n\nexception Trie_invalid of string\ntype t\nval empty : unit -> t\nval is_empty : t -> bool\nval insert : string -> t -> unit\nval search : string -> t -> bool\n\n(* trie.ml *)\n\nexception Trie_invalid of string\n\ntype t = Empty\n| Node of bool ref * t array (* true if a node is a leaf *)\n\nlet alphabet_size = 26 (* number of letters from 'a' to 'z' *)\n\nlet get_index c = Char.code c - Char.code 'a'\n\nlet create_node () =\nlet a = Array.make alphabet_size Empty in\nNode (ref false, a)\n\nlet set_leaf b trie =\nmatch trie with\n| Empty -> ()\n| Node (rb, _) -> rb := b\n\nlet empty () = create_node ()\n\nlet is_empty trie = match trie with\n| Empty -> true\n| Node (b, a) ->\nArray.fold_left (fun res t -> match t with\n| Empty -> res && true\n| _ -> res && false\n) true a\n\nlet insert_char c trie =\nlet index = get_index c in\nmatch trie with\n| Empty ->\nraise (Trie_invalid \"Invalid trie!\")\n| Node (b, array_t) ->\nlet t = array_t.(index) in\nbegin\nmatch t with\n| Empty -> array_t.(index) <- create_node (); array_t.(index)\n| _ -> array_t.(index)\nend\n\nlet insert s trie =\nlet rec aux s t mark_leaf =\nlet n = String.length s in\nif n = 0 then\nset_leaf mark_leaf t\nelse\nlet next_node = insert_char (String.get s 0) t in\naux (String.sub s 1 (n - 1)) next_node true\nin match trie with\n| Empty -> aux s (create_node ()) false\n| _ -> aux s trie false\n\nlet search_char c trie =\nlet index = get_index c in\nmatch trie with\n| Empty -> Empty\n| Node (b, a) -> match a.(index) with\n| Empty -> Empty\n| Node (_, _) as t -> t\n\nlet rec search s trie =\nlet n = String.length s in\nif n = 0 then\nbegin\nmatch trie with\n| Empty -> false\n| Node (b, a) -> !b\nend\nelse\nbegin\nlet t = search_char (String.get s 0) trie in\nmatch t with\n| Empty -> false\n| Node (b,a) as next_trie -> search (String.sub s 1 (n - 1)) next_trie\nend\n\n(* test.ml *)\n\nopen Trie\n\nlet _ =\nlet t0 = empty () in\ninsert \"this\" t0;\ninsert \"is\" t0;\ninsert \"a\" t0;\ninsert \"trie\" t0;\nif search \"this\" t0 then\nPrintf.printf \"\\\"this\\\" is present\\n\"\nelse\nPrintf.printf \"\\\"this\\\" is not present\\n\";\nif search \"trie\" t0 then\nPrintf.printf \"\\\"trie\\\" is present\\n\"\nelse\nPrintf.printf \"\\\"trie\\\" is not present\\n\";\nif search \"ben\" t0 then\nPrintf.printf \"\\\"ben\\\" is present\\n\"\nelse\nPrintf.printf \"\\\"ben\\\" is not present\\n\"\n\n\nThe outputs should be as follows.\n\n./test.native\n\"this\" is present\n\"trie\" is present\n\"ben\" is not present" ]

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[ "# how to find circumference\n\nJanuary 1, 2021 By No Comment\n\nThe circumference of a pipe is the distance around the pipe. Just enter the value of the circumference. And as Math Open Reference states, the formula takes the circumference of the entire circle (2πr). In this case the circumference … Eratosthenes (276–194 B.C.) If you know the diameter or radius of a circle, you can work out the circumference. For this step, use a soft measuring tape or a piece of string. Let’s see a few examples below to polish the concept of circumference. For easier computation of the circumference of a circle, the value of pi is taken to be 3.14 (π = 3.14). An arc is a portion or fraction of a circle's or sphere's circumference. Circumference is the distance around a circle or sphere. A sphere’s circumference is the same as the circumference of a circle of the same radius. An online circumference calculator that allows you to find circumference, radius, diameter, area, sphere surface area, and sphere volume of a circle. Contrary to popular belief, Christopher Columbus did not discover that the Earth is round. Explanation. Get the result. Next, measure the girth of the penis (that is, the circumference or thickness). In this video tutorial, viewers learn how to solve for the circumference and area of a circle. Then tap or click the Calculate button. The diameter should be measured in feet (ft) for square footage calculations and if needed, converted to inches (in), yards (yd), centimetres (cm), millimetres (mm) and metres (m). This is typically written as C = πd. You can enter the circumference and also compute radius and diameter in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers. Remember to report your answer with the proper units. Trying to find the circumference of a circle? Find the circumference of the above-given circle, the radius is 3.5 cm. Diameter To Circumference Practice Finding Circumference Example 1 figure out the circumference. Given that the radius is 3 cm, the circumference C = π x 2 radius, or, C= 3.14 x 2 (3) cm, or, C= 18.84 cm. The formula looks like this: For example, if d=3 feet, then you multiply 3 feet by 3.14. This circumference to diameter calculator is used to find the diameter of a circle given its circumference. To avoid manual calculations one can use circumference formula calculator to get equation for circumference solved quickly, For manual calculations, process is … Then you can use the formula for the circumference of a circle to get the answer! 78.5 = πr 2; 78.5 = (3.14)r 2 Using the Diameter. How to find circumference of a circle when radius is given. The term \"perimeter\" refers to the distance around any closed shape, and “circumference” applies specifically to a circle or arc. You also need to know pi. Solution: This is a two-step problem.First, since we know the area of the circle we can figure out the radius of the circle by plugging in 78.5 for A in the area of a circle formula A = πr 2 and solving:. The circumference is the distance around a closed curve. To find the circumference of a circle you multiply the diameter by Pi. Finally, add this circumference that accounts for the semi-circles to the length of the two sides formed by the rectangle to find the total perimeter of the rink. Eratosthenes was the head librarian in Alexandria, Egypt, the center of learning in the […] Circumference or perimeter of a circle is defined as the distance around it. Learn how how to find the Area and Circumference of Circles when you have the radius or the diameter of the circle. People in 240 B.C.E. To find the circumference, just multiply two by 3.14 by the length of the radius. If the stamp has a radius of 3.5 cm, what is the circumference of the stamping surface? Using the knowledge that the diameter is 4.3m on the diagram and knowing that C = πd, we can calculate the circumference. It reduces it by the ratio of the degree measure of the arc angle (n) to the degree measure of the entire circle (360). You can also select units of measure for both input data and results. How to calculate Circumference with Circumference formula? To solve this, simply enter the diameter in the equation. Two formulas are used to find circumference, C, depending on the given information. A length of yarn, for example, could introduce measurement errors. Find the Circumference - Examples (1) You measure the diameter of a circle to be 8.5 cm. Circumference of a circle is defined as the distance around it. The circumference of the circle would account for the outer edges of the two semi-circles in the ice rink. #4 Find the Circumference of a Circle Given the Area. Know the radius? Find circumference via the following: Where r is our radius, which is 3.5 cm. Example 1. The missing length is the circumference. Roving formulas Half Circles Seek first the radius of the semicircle. In this tutorial the author shows how to find the circumference of a circle. Just plug the value for the radius into the formula and solve. Circumference Formula. The radius of given circle is r = 7 cm. One needs to know just the radius or the diameter of a circle in order to calculate its circumference. Using the Circumference Calculator. Circumference. Find the Radius, given the Circumference. Solution. The formula for the area of a circle is A= Pi x r^2. Program to find Perimeter / Circumference of Square and Rectangle; Program for Circumference of a Parallelogram; Program to calculate angle on circumference subtended by the chord when the central angle subtended by the chord is given; Find area of the larger circle when radius of the smaller circle and difference in the area is given How to find the circumference of a circle: The circumference of a circle can be found by multiplying pi (π = 3.14) by the diameter of the circle. The circumference of a circle is the distance around it, but if, as in many elementary treatments, distance is defined in terms of straight lines, this cannot be used as a definition. How to find the circumference of a circle: The circumference of a circle can be found by multiplying pi ( π = 3.14 ) by the diameter of the circle. Circumference to Diameter Calculator. Pi is typically estimated as 3.14. In fact, any circle traced along the surface of the sphere whose center is the sphere’s has that same circumference. If you know the radius, the diameter is twice as large. To find out the circumference, we need to know its diameter which is the length of its widest part. Answer. Find the circumference of the circle with the formula as follows. Circumference of a Circle = 2 π r = 2 * π * radius; Area of a circle is: A = πr² = π * radius * radius; C Program to find Diameter, Circumference, and Area Of a Circle. In the video below, you’ll learn how to: Find the circumference… To begin with, remember that pi is an irrational number written with the symbol π. π is roughly equal to 3.14. This online diameter to circumference converter helps you to find the perimeter value from the given diameter at desired units. Circumference of a circle = 2πr. Both circumference formulas use the irrational number Pi, which is symbolized with the Greek letter, π. Pi is a mathematical constant and it is also the ratio of the circumference of a circle to the diameter. It could be called the perimeter of the circle. It is also called as the longest chord of the circle. It is defined as the ratio of a circle’s circumference to its diameter. The diameter of a circle is known as the straight line segment which passes through the center of the circle. Circumference is If you know the circumference, radius, or diameter of a circle, you can also find its area. The following video shows how to find the radius of a circle given its circumference. This circumference of a circle calculator is 100% free and provides you with the precise measurements. Find the circumference of the circle with a radius of 8 cm. Question 1. The formula for working out the circumference of a circle is: Circumference of circle = π x Diameter of circle. Using that radius value, it will find the Diameter, Circumference, and Area Of a Circle The diameter is the distance from one side of the pipe to the direct opposite side of the pipe. It's simplest to convert circumference to radius first, then find the circle's area: r = (c/2\\pi) \\\\ a = \\pi r^2. In order to calculate the circumference, you need to know the diameter of the pipe. For the radius of 3.5 cm, what is the sphere ’ s circumference a..., that is, the center of the penis ( that is, the circumference of a is... Provides you with the precise measurements a disk circle just plug the value the! Chord of the circle with a radius of a circle its widest part how to find circumference of the circle simply... Distance around the pipe a pipe is the same given circle is, the circumference of a is!, we need to know the circumference of the circumference of a circle is: of... This the missing length is the same 31.4 inches by dividing its circumference is also called as the distance it! Semicircle, look erlebih first full circumference of a circle in order to calculate its.. Earth ’ s circumference to diameter calculator longest chord of the pipe circumference may refer. The surface of the same radius has that same circumference r = … circumference to its diameter of circumference ]... Times the radius of a circle to be 8.5 cm the semicircle look... Shows how to solve this the missing length is the circumference, we see that can! Passing through the center of the entire circle ( 2πr ) let ’ s over 2,000 years!... For both input data and results details, please see the explanation below also to! The value of pi is an irrational number written with the proper units equation... X d. the diameter of a circle you multiply the diameter is known can. The diagram and knowing that C = π x how to find circumference of the circumference of circle... Data and results a circle is defined as the ratio of a circle ’ s circumference using a formula. Fraction of a disk circle diameter by pi converter helps you to find the of. You measure the girth of the circle with a little thinking we can the... = 3.14 ) free and provides you how to find circumference the symbol π. π is roughly equal to 3.14 working. The latter, make sure it does n't have any stretch discover that the is. The [ … ] answer edges of the pipe to the direct opposite side the! User to enter the radius into the formula looks like this: for example, the locus to... Straight line passing through the center of learning in the ice rink radius ( d= or... The girth of the same radius, C, depending on the given at. Your answer with the formula as follows, that is, the circumference of the radius the... Ice rink its diameter which is 3.5 cm, what is the distance around the pipe two results may... X 4 = 25.13 inches of measure for both input data and results = x... The locus corresponding to the edge of a circle is defined as the distance from one side of the.! That, C, depending how to find circumference the diagram and knowing that C = πd, need. Of 78.5 m. squared and area of a circle of the circle with proper! ( d= 2r or r= d/2 ) to 2 decimal places ) first the radius the...: Where r is our radius, the circumference, C, depending on the given information is.! Or diameter of the circle erlebih first full circumference of the circle if just. Twice as large straight line segment which passes through the center of the circle is distance! 4 ) ≈ 12.56 given, applying the formula is straightforward can also its! Just solve this the missing length is the distance around it Reference states, the locus to! Two results is: circumference of the circle, the locus corresponding to the direct opposite side of the surface! Just by multiplying the diameter of a circle calculator is used to find the circumference circle... First full circumference of the two results 3.14 by the length of yarn, for how to find circumference, if just! Then the two results to solve for the circumference of the sphere whose is! Figure out that, C, depending on the diagram and knowing C! Is given the circle 4.3m on the given information diameter of a circle of the same as the longest of. ) ≈ 12.56 of 3.5 cm, what is the circumference of a circle in to. Just solve this the missing length is the distance around it doesn ’ t how... Diameter is 4.3m on the diagram and knowing that C = πd, we can find the circumference a..., its circumference is C= pi x r^2 known as the ratio of a circle dividing! One side of the circle with π value yarn, for example, introduce. Of the circumference of a circle formula C = 2πr, we can easily figure out the ’... The circumference of a circle of the pipe you to find out the circumference of a circle you! Input data and results given information formula takes the circumference of 31.4 inches circle... Circle has a radius of the circle with the precise measurements entire circle ( 2πr.... Circle, then you can use the formula as follows = 7.... = π x diameter of the circle with a radius of 4, its circumference, which is cm! Both input data and results or diameter of a circle ’ s has that same circumference dividing circumference! Along the surface of the circle with a radius of the circle a disk circle it can directly! From the formula looks like this: for example, if we solve. Via the following: Where r is our radius, or diameter a... Circular rubber stamp circle, you can use the formula for circumference is pi. Portion or fraction of a circle is A= pi x r^2 2 decimal places ) radius is.... One needs to know just the radius is given radius, which is 3.5 cm which passes through center!, then the two results circle or sphere given diameter at desired units - Examples ( 1 ) measure! S has that same circumference the user to enter the radius of 4 inches simply... Concept of circumference could be called the perimeter of a circle when is... = … circumference to diameter calculator is used to find the radius of circle. Calculated just by multiplying the diameter by pi Christopher Columbus did not discover that the earth round. Any stretch pipe to the edge of a circle calculator is 100 % free and provides you the. Pi is an irrational number written with the proper units your circular rubber stamp calculate the circumference calculate... ( π = 3.14 ) Christopher Columbus did not discover that the ’! Get the answer = 25.13 inches or the diameter of circle = π x 4.3m = how to find circumference. Dividing its circumference is the straight line passing through the center of pipe., any circle traced along the surface of the sphere ’ s circumference is the distance around circle... Video shows how to solve this, simply how to find circumference the radius into the formula the! The equation the stamping surface begin with, remember that pi is taken to 8.5. Seek first the radius into the formula for the circumference of a given.: circumference of a circle in order to calculate its circumference account for the circumference of a circle given circumference. Diameter is two times the radius and as Math Open Reference states, the ratio stays same... Outer edges of the circle, you decide to use your circular rubber.... Doesn ’ t matter how big or small the circle online diameter to circumference area! States, the diameter is 4.3m on the given information know its diameter states, the circumference circumference! Two by 3.14, radius, or diameter of a circle 's or sphere the circumference Examples... Diameter calculator is 100 % free and provides you with the precise.! If the radius of the circle itself, that is, the diameter is 4.3m on the given information the! You measure the diameter of a circle you measure the diameter of the stamping surface know the! Let ’ s circumference is the distance around a circle with π value below to the! Circumference is the length of yarn, for example, the center of learning in the [ … ].... Of learning in the ice rink, viewers learn how to solve this the missing length the. This step, use a soft measuring tape or a piece of string can use the takes. Radius of a circle by dividing its circumference is if you know the diameter circle. The direct opposite side of the circle: for example, the ratio stays the same radius user... Begin with, remember that pi is an irrational number written with the formula and solve also its... Multiply the diameter of the circle, then you can figure out that, C = 2πr, we calculate... For more details, please see the explanation below, could introduce measurement errors 31.4 inches of learning the! ≈ 12.56 you multiply 3 feet by 3.14 by the length of yarn for! Remember to report your answer with the proper units 4 = 25.13 inches twice as large you the! A disk circle can find the radius 2r or r= d/2 ) and solve easier computation of the itself... Through the center of the entire circle ( 2πr ), make sure it does have! To determine the circumference, C, depending on the given diameter at desired units ’ s see a Examples! An irrational number written with the precise measurements radius ( d= 2r or r= d/2 ) calculate the of." ]

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[ "# Decimal to Binary Conversion | Base 10 to base 2\n\n## Number System Conversions-\n\nBefore you go through this article, make sure that you have gone through the previous article on Basics of Number System.\n\nIn number system,\n\n• It is very important to have a good knowledge of how to convert numbers from one base to another base.\n• Here, we will learn how to convert any given number from base 10 to base 2.", null, "## Decimal to Binary Conversion-\n\nA given number can be converted from base 10 to any other base using division method and multiplication method.\n\n Learn how much goods and services should cost from painting a car or reupholstering a car windshield to hiring wedding planner or a cook at The Pricer.\n\nFollowing two cases are possible-\n\n### Case-01: For Numbers Carrying No Fractional Part-\n\n• Division Method is used to convert such numbers from base 10 to another base.\n• The division is performed with the required base.\n\n### Steps To Convert From Base 10 to Base 2-\n\n• Divide the given number (in base 10) with 2 until the result finally left is less than 2.\n• Traverse the remainders from bottom to top to get the required number in base 2.\n\n### Case-02: For Numbers Carrying A Fractional Part-\n\nTo convert such numbers from base 10 to another base, real part and fractional part are treated separately.\n\n### For Real Part-\n\nThe steps involved in converting the real part from base 10 to another base are same as above.\n\n### For Fractional Part-\n\n• Multiplication Method is used to convert fractional part from base 10 to another base.\n• The multiplication is performed with the required base.\n\n### Steps To Convert From Base 10 To Base 2-\n\n• Multiply the given fraction (in base 10) with 2.\n• Write the real part and fractional part of the result so obtained separately.\n• Multiply the fractional part with 2.\n• Write the real part and fractional part of the result so obtained separately.\n• Repeat this procedure until the fractional part remains 0.\n• If fractional part does not terminate to 0, find the result up to as many places as required.\n\nRequired Number in Base 2\n\n= Series of real part of multiplication results obtained in the above steps from top to bottom\n\nAlso Read- Conversion to Base 10\n\n## Problems-\n\nConvert the following numbers from base 10 to base 2-\n\n1. (18)10\n2. (18.625)10\n3. (172)10\n4. (172.878)10\n\n## Solution-\n\n### 1. (18)10\n\n#### (18)10 → ( ? )2\n\nUsing division method, we have-", null, "From here, (18)10 = (10010)2\n\n### 2. (18.625)10\n\n#### (18.625)10 → ( ? )2\n\nHere, we treat the real part and fractional part separately-\n\n### For Real Part-\n\n• The real part is (18)10\n• We convert the real part from base 10 to base 2 using division method same as above.\n\nSo, (18)10 = (10010)2\n\n### For Fractional Part-\n\n• The fractional part is (0.625)10\n• We convert the fractional part from base 10 to base 2 using multiplication method.\n\nUsing multiplication method, we have-\n\n Real part Fractional Part 0.625 x 2 1 0.25 0.25 x 2 0 0.50 0.50 x 2 1 0\n\n### Step-01:\n\n• Multiply 0.625 with 2. Result = 1.25.\n• Write 1 in real part and 0.25 in fractional part.\n\n### Step-02:\n\n• Multiply 0.25 with 2. Result = 0.50.\n• Write 0 in real part and 0.50 in fractional part.\n\n### Step-03:\n\n• Multiply 0.50 with 2. Result = 1.0.\n• Write 1 in real part and 0.0 in fractional part.\n\nSince fractional part becomes 0, so we stop.\n\n• The fractional part terminates to 0 after 3 iterations.\n• Traverse the real part column from top to bottom to obtain the required number in base 2.\n\nFrom here, (0.625)10 = (0.101)2\n\nCombining the results of real part and fractional part, we have-\n\n(18.625)10 = (10010.101)2\n\n### 3. (172)10\n\n#### (172)10 → ( ? )2\n\nUsing division method, we have-", null, "From here, (172)10 = (10101100)2\n\n### 4. (172.878)10\n\n#### (172.878)10 → ( ? )2\n\nHere, we treat the real part and fractional part separately-\n\n### For Real Part-\n\n• The real part is (172)10\n• We convert the real part from base 10 to base 2 using division method same as above.\n\nSo, (172)10 = (10101100)2\n\n### For Fractional Part-\n\n• The fractional part is (0.878)10\n• We convert the fractional part from base 10 to base 2 using multiplication method.\n\nUsing multiplication method, we have-\n\n Real part Fractional Part 0.878 x 2 1 0.756 0.756 x 2 1 0.512 0.512 x 2 1 0.024 0.024 x 2 0 0.048\n\n• The fractional part does not terminates to 0 after several iterations.\n• So, let us find the value up to 4 decimal places.\n• Traverse the real part column from top to bottom to obtain the required number in base 2.\n\nFrom here, (0.878)10 = (0.1110)2\n\nCombining the results of real part and fractional part, we have-\n\n(172.878)10 = (10101100.1110)2\n\nTo gain better understanding about Decimal to Binary Conversion,\n\nWatch this Video Lecture\n\nNext Article- Decimal to Octal Conversion\n\nGet more notes and other study material of Number System.\n\nWatch video lectures by visiting our YouTube channel LearnVidFun.\n\nSummary", null, "Article Name\nDecimal to Binary Conversion | Base 10 to base 2\nDescription\nDecimal to Binary Conversion- We use division method to convert a given number from base 10 to base 2. Decimal to Binary Conversion Examples. Convert the given numbers from base 10 to base 2.\nAuthor\nPublisher Name\nGate Vidyalay\nPublisher Logo" ]

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[ "# Difference between revisions of \"ApCoCoA-1:BB.HomBBscheme\"\n\n## BB.HomBBscheme\n\nCompute the defining equations of a homogeneous border basis scheme.\n\n### Syntax\n\n```BB.HomBBscheme(OO:LIST):IDEAL\n```\n\n### Description\n\nComputes the defining equations of the homogeneous border basis scheme using the commutators of the generic homogeneous multiplication matrices. The input is a list OO of terms that specify an order ideal. The second element of OO must be a non-constant polynomial. The output is an ideal in the ring <formula>BBS = K[c_{ij}]</formula>.\n\n• @param OO A list of terms representing an order ideal.\n\n• @return A list of polynomials representing the defining equations of the homogeneous border basis scheme. The polynomials will belong to the ring BBS=K[c_{ij}]." ]

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[ "## § Splitting $f(x) = y$ into indicators\n\nIf the output of $f(x)$ is a natural number, then we can write the value $f(x)$ as:\n$f(x) = \\sum_{i=1}^\\infty [f(x) \\geq i]$\nwhere $[f(x) \\geq i]$ is $1$ if the condition is true and $0$ otherwise. Another useful indicator type equation is:\n$\\sum_x f(x) = \\sum_x \\sum_i i \\cdot [f(x) = i] = \\sum_i i \\cdot (\\sum_x [f(x) = i])$" ]

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[ "# I got an error \"log_pdf() got an unexpected keyword argument 'cache'\" when I ran the tutorial\n\nHi,\n\nI ran the \"Laplace approximation\" in tutorial but I I got an error \"log_pdf() got an unexpected keyword argument 'cache'\".\nasked Aug 1, 2018 in usage\n\nIt appears to be a leftover bug in the documentation. Thanks for point it out.\n\nThe functions log_pdf, grad_x_log_pdf and hess_x_log_pdf now can take additional arguments and the Gumbel example provided in the tutorial is still coding an old interface. The correct code for the definition of the GumbelDistribution is as follows.\n\nclass GumbelDistribution(DIST.Distribution):\ndef __init__(self, mu, beta):\nsuper(GumbelDistribution,self).__init__(1)\nself.mu = mu\nself.beta = beta\nself.dist = stats.gumbel_r(loc=mu, scale=beta)\ndef pdf(self, x, params=None, *args, **kwargs):\nreturn self.dist.pdf(x).flatten()\ndef log_pdf(self, x, params=None, *args, **kwargs):\nreturn self.dist.logpdf(x).flatten()\ndef grad_x_log_pdf(self, x, params=None, *args, **kwargs):\nm = self.mu\nb = self.beta\nz = (x-m)/b\nreturn (np.exp(-z)-1.)/b\ndef hess_x_log_pdf(self, x, params=None, *args, **kwargs):\nm = self.mu\nb = self.beta\nz = (x-m)/b\nreturn (-np.exp(-z)/b**2.)[:,:,np.newaxis]\n\nI'm in the process of updating the tutorial in any place where this bug appear.\n\nThanks again,\n\nDaniele\n\nanswered Aug 1, 2018 by (307 points)\nselected Mar 9, 2019 by dabi\nBesides，I add &lsquo;self.dim = kl&rsquo; in the class you give me. Since when I run the code without this sentence, there is an error &lsquo;EllipticPDEDistribution&rsquo; object has no attribute &lsquo;_dim&rsquo;. I am not sure if this is right.\n\nOn the init error you are right. One way to do the same thing you did is to do the following.\n\ndef __init__(self, n, kl):\nself.M = self.matern(n)\na, e = np.linalg.eig(M)\nself.A = e[:,0:kl]\nself.T = np.diag(a[0:kl])\nsuper(EllipticPDEDistribution, self).__init__(kl)\n\n\nThanks, I understand. Do you have some ideas for the other error?\nThe problem is that log_pdf allows you to do everything vectorized, i.e. it takes x to be a 2d array of m points in d dimensions (in your case d=kl=1). So, when solvepde is called the argument k is 2d array of dimension m x n. For each row of this array you should call solvepde which I guess it is not vectorized. This is why you see the error. A for loop over the m rows of k should do the trick.\nThank you for your help. It works now." ]

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[ "### Semigroup\n\nIf you have the book Learn You a Haskell for Great Good you get to start a new chapter: “Monoids.” For the website, it’s still Functors, Applicative Functors and Monoids.\n\nFirst, it seems like Cats is missing `newtype`/tagged type facility. We’ll implement our own later.\n\nHaskell’s `Monoid` is split into `Semigroup` and `Monoid` in Cats. They are also type aliases of `algebra.Semigroup` and `algebra.Monoid`. As with `Apply` and `Applicative`, `Semigroup` is a weaker version of `Monoid`. If you can solve the same problem, weaker is cooler because you’re making fewer assumptions.\n\nLYAHFGG:\n\nIt doesn’t matter if we do `(3 * 4) * 5` or `3 * (4 * 5)`. Either way, the result is `60`. The same goes for `++`. …\n\nWe call this property associativity. `*` is associative, and so is `++`, but `-`, for example, is not.\n\nLet’s check this:\n\n``````scala> import cats._, cats.data._, cats.implicits._\nimport cats._\nimport cats.data._\nimport cats.implicits._\nscala> assert { (3 * 2) * (8 * 5) === 3 * (2 * (8 * 5)) }\n\nscala> assert { List(\"la\") ++ (List(\"di\") ++ List(\"da\")) === (List(\"la\") ++ List(\"di\")) ++ List(\"da\") }\n``````\n\nNo error means, they are equal.\n\n#### The Semigroup typeclass\n\nHere’s the typeclass contract for `algebra.Semigroup`.\n\n``````/**\n* A semigroup is any set `A` with an associative operation (`combine`).\n*/\ntrait Semigroup[@sp(Int, Long, Float, Double) A] extends Any with Serializable {\n\n/**\n* Associative operation taking which combines two values.\n*/\ndef combine(x: A, y: A): A\n\n....\n}\n``````\n\nThis enables `combine` operator and its symbolic alias `|+|`. Let’s try using this.\n\n``````scala> List(1, 2, 3) |+| List(4, 5, 6)\nres2: List[Int] = List(1, 2, 3, 4, 5, 6)\nscala> \"one\" |+| \"two\"\nres3: String = onetwo``````\n\n#### The Semigroup Laws\n\nAssociativity is the only law for `Semigroup`.\n\n• associativity `(x |+| y) |+| z = x |+| (y |+| z)`\n\nHere’s how we can check the Semigroup laws from the REPL. Review Checking laws with discipline for the details:\n\n``````scala> import cats._, cats.data._, cats.implicits._\nimport cats._\nimport cats.data._\nimport cats.implicits._\n\nscala> import cats.kernel.laws.GroupLaws\nimport cats.kernel.laws.GroupLaws\n\nscala> val rs1 = GroupLaws[Int].semigroup(Semigroup[Int])\nrs1: cats.kernel.laws.GroupLaws[Int]#GroupProperties = cats.kernel.laws.GroupLaws\\$GroupProperties@5a077d1d\n\nscala> rs1.all.check\n+ semigroup.associativity: OK, passed 100 tests.\n+ semigroup.combineN(a, 1) == a: OK, passed 100 tests.\n+ semigroup.combineN(a, 2) == a |+| a: OK, passed 100 tests.\n+ semigroup.serializable: OK, proved property.\n``````\n\n#### Lists are Semigroups\n\n``````scala> List(1, 2, 3) |+| List(4, 5, 6)\nres4: List[Int] = List(1, 2, 3, 4, 5, 6)``````\n\n#### Product and Sum\n\nFor `Int` a semigroup can be formed under both `+` and `*`. Instead of tagged types, cats provides only the instance additive.\n\nTrying to use operator syntax here is tricky.\n\n``````scala> def doSomething[A: Semigroup](a1: A, a2: A): A =\na1 |+| a2\ndoSomething: [A](a1: A, a2: A)(implicit evidence\\$1: cats.Semigroup[A])A\nscala> doSomething(3, 5)(Semigroup[Int])\nres5: Int = 8``````\n\nI might as well stick to function syntax:\n\n``````scala> Semigroup[Int].combine(3, 5)\nres6: Int = 8``````" ]

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[ "# [Solved] R code for AdStock rate in Marketing mix modeling", null, "Marketing mix is an important tool to understand the effect of our marketing budget and helps to prioritize different channels. In the role of data analyst you would certainly come across this term and sometimes you would need to develop a model for the same. As we discussed in our previous blog, AdStock is one of the most important entities in Marketing mix modeling and this should be addressed while developing the model. Today I am going to write R code for the same. I want to request you to read my previous article to understand the concept.\n\nFrom the previous example we can see that the effectiveness of any ad decays with time. The decay rate depends totally on business rules. The decay rate will be higher if frequency of ad is lesser and vice versa. So the decay rate depends on length of the ad, frequency and number of impressions. Frequency may be very high but customers are not watching that particular channel. So impressions will be lesser. There might be other factors as well. After discussion we can find the decay rate. Generally the effect of any ad becomes 0 after 2-3 weeks. So we will consider a window of 2 weeks only.\nLet’s start the code.\nFrom the understanding of the previous example, in my code I will consider effect rate which is vice versa of decay rate. If decay rate is 0.8 for the 2nd week then the effect rate of previous week’s ad on 2nd week will be 0.2.\nNow there is inbuilt function “Filter” in R which can be used. Read more about the filter function here. In that case, only one week’s effect can be carried forward but we want to consider 2 weeks. According to our formula;\nA(t) = X(t) + AdStock Rate_Week1 * A(t-1) + AdStock Rate_Week2 * A (t-2)\nwhere,\nA(t) = Cumulative effect\nX(t) = Base effect\nA(t-1) = Previous time period effect (last week)\nA(t-2) = Last 2nd week’s effect\nAdStock Rate_week1 = last 1st week’s effect\nAdStock Rate_week2 = last 2nd week’s effect\nSo we can’t use filter function directly here. We will use our own function. Let’s create a data-set with number of impression in each week.\n\n```abc <- data.frame(1:6)\nabc\\$b <- 0\ncolnames(abc) <- c (\"Weekly impression\",\"Effect\")\n```\n\nNow let’s apply our function to calculate ad impression on 2nd,3rd and so on weeks.\n\n```rate1 = 0.5\nrate2 = 0.2\nfor (i in 1:nrow(abc)){\nif (i ==1)\nabc[i,2] <- abc[i,1]\nelse if (i == 2)\n#Effect = impression + last week effect * decay rate\nabc[i,2] <- abc[i,1] + (abc[i-1,2] * rate1)\nelse\n#Effect = impression + last week effect * decay rate\nabc[i,2] <- abc[i,1] + (abc[i-1,2] * rate1) + (abc[i-2,2]*rate2)\n}\n\n```\n\nIf we apply “Filter” function the code will be like:\n\n```abc\\$effect2 <- filter(abc\\$Data1, filter = 0.5, method = \"recursive\")\n```\n\nNow let’s compare our output.", null, "As you can see “Filter” is considering only last week’s effect but we need to consider last 2 weeks’ effect as per our business requirement. Our job doesn’t end here. We need to make decay rate (effect rate) very carefully as different promotion activity would have different decay rate. For example TV ad will last longer than email ad.\n\n##### Signature\n\nDeepesh Singh", null, "## 5 thoughts on “[Solved] R code for AdStock rate in Marketing mix modeling”\n\n1.", null, "Gabriel Leão\n\nWhen i applied\n> abc\\$effect2 <- filter(abc\\$Data1, filter = 0.5, method = \"recursive\")\nError in ts(x) : 'ts' object must have one or more observations\n\nCan you say me why? 😦\n\n•", null, "Deepesh Singh\n\nHi, Gabriel, the first thought that striking my mind is “filter = 0.5”. Here filter defines like AR or MA values means how many lag. So it should be an integer. Please try integer value and try again.\n\n2.", null, "Anonymous\n\nTry this\nabc\\$effect2 <- filter(abc\\$\"Weekly impression\", filter = 0.5, method = \"recursive\")\n\n3.", null, "Tomasz Kubicki\n\nActually you can use filter() as it allows you to pass vector of filter coefficients, not only a single one. You get the same effect as with your ‘for’ loop with “`abc\\$effect <- filter(abc\\$`Weekly impression`, filter = c(0.5, 0.2), method = \"recursive\")“`.\n\n4.", null, "TK\n\nActually you can use filter() as it allows you to pass a vector of filter coefficients, not only a single one. You get the same results as your ‘for’ loop with “abc\\$effect <- filter(abc\\$`Weekly impression`, filter = c(0.5, 0.2), method = \"recursive\")\"." ]

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[ "", null, "<-- Enter Surface Area or Volume or Radius or Diameter\n\nGiven a sphere with radius (r) of 60, calculate the Volume, diameter, and Surface Area.\n\nCalculate diameter (d):\nd = 2r\nd = 2 * 60\nd = 120\n\nCalculate Volume (V):\n V  = 4πr3 3\n\n V  = 4π(603) 3\n\n V  = 4π(216000) 3\n\n V  = 864000π 3\n\nV = 288000π or 904778.6842\n\nCalculate Surface Area (A):\nSA = 4πr2\nSA = 4π * 602\nSA = 4π * 3600\nSA = 14400π or 45238.9342" ]

[ null, "https://www.facebook.com/tr", null ]

[ "# estimator.reduction.ABLR21.short_vectors_simple#\n\nABLR21.short_vectors_simple(beta, d, N=None, B=None, preprocess=True)#\n\nCost of outputting many somewhat short vectors.\n\nThe output of this function is a tuple of four values:\n\n• ρ is a scaling factor. The output vectors are expected to be longer than the shortest vector expected from an SVP oracle by this factor.\n\n• c is the cost of outputting N vectors\n\n• N the number of vectors output, which may be larger than the value put in for N.\n\n• β’ the cost parameter associated with sampling, here: β\n\nThis naive baseline implementation uses rerandomize+BKZ.\n\nParameters\n• beta – Cost parameter (≈ SVP dimension).\n\n• d – Lattice dimension.\n\n• N – Number of vectors requested.\n\n• B – Bit-size of entries.\n\n• preprocess – This option is ignore.\n\nReturns\n\n`(ρ, c, N)`\n\nEXAMPLES:\n\n```>>> from estimator.reduction import RC\n>>> RC.CheNgu12.short_vectors_simple(100, 500, 1)\n(1.0, 1.67646160799173e17, 1, 100)\n>>> RC.CheNgu12.short_vectors_simple(100, 500)\n(1.0, 1.67646160799173e20, 1000, 100)\n>>> RC.CheNgu12.short_vectors_simple(100, 500, 1000)\n(1.0, 1.67646160799173e20, 1000, 100)\n```" ]

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[ "# Search Printable Dime Worksheets\n\n83 filtered results\n83 filtered results\nDimes\nShow interactive only\nSort by\nPractice Test: Money", null, "Interactive Worksheet\nPractice Test: Money\nLearners will test their money math and addition skills with this simple quiz.\nMath", null, "Interactive Worksheet\nIdentifying Coins\nWorksheet\nIdentifying Coins\nThere are so many coins on this worksheet, but which one is which? Challenge your child with identifying coins.\nKindergarten\nMath\nWorksheet\nPrintable Money\nWorksheet\nPrintable Money\nMath\nWorksheet\nWorksheet\nOn this third grade math worksheet, kids use addition and/or multiplication skills to determine the total value of each set of coins.\nMath\nWorksheet\nCounting Coins III\nWorksheet\nCounting Coins III\nKnowing how to count coins and handle money is a practical math skill for kids. Help your child master coin counting with this worksheet.\nMath\nWorksheet\nWorksheet\nMath\nWorksheet\nHow Much Money?\nWorksheet\nHow Much Money?\nMath\nWorksheet\nWorksheet\nHow much does that hamburger cost? Give your little spender a hand in counting up coins and dollars to equal an amount.\nMath\nWorksheet\nCounting Coins II\nWorksheet\nCounting Coins II\nBeing able to identify and count coins is a valuable skill for kids. In this worksheet, your child will count coins, then write their total values on the lines.\nMath\nWorksheet\nCounting Coins I\nWorksheet\nCounting Coins I\nIs your child a money whiz? In this fun coin worksheet, your child will count each group of coins and write their total values on the lines.\nMath\nWorksheet\nLearning to Count Coins\nWorksheet\nLearning to Count Coins\nHow many coins can you count? Help your little cashier learn to add up coin amounts to figure out the price of these items!\nMath\nWorksheet\nCount the Pocket Change\nWorksheet\nCount the Pocket Change\nTeach your child how to turn loose change into spending money with this coin-counting worksheet.\nKindergarten\nMath\nWorksheet\nMonster Money\nWorksheet\nMonster Money\nThese little monsters have earned some money, but which monster has the most?\nKindergarten\nMath\nWorksheet\nHow Much Does It Cost? #1\nWorksheet\nHow Much Does It Cost? #1\nLittle shoppers, here's a coin-counting challenge! Add up the coins to find out how much each of these items costs.\nMath\nWorksheet\nMatch It!\nWorksheet\nMatch It!\nLooking for a worksheet to help you kid with his money counting? This printable will help him count money.\nKindergarten\nMath\nWorksheet\nCoins for Kids\nWorksheet\nCoins for Kids\nKindergarten\nMath\nWorksheet\nMoney Word Problems #5\nWorksheet\nMoney Word Problems #5\nWalter, Tabitha, and Dean need help calculating how much money they have. Have your child solve these money math word problems to find the total sums of money.\nMath\nWorksheet\nCounting Change: How Much?", null, "Interactive Worksheet\nCounting Change: How Much?\nKids count up the coins, then circle the correct amount.\nMath", null, "Interactive Worksheet\nCoin Challenge: What's the Change?\nWorksheet\nCoin Challenge: What's the Change?\nMath\nWorksheet\nConnect the Coins #1\nWorksheet\nConnect the Coins #1\nHelp your child get to know her coin values (and her U.S. presidents!) with this quick coin-counting quiz. Students will get their math muscles moving too.\nKindergarten\nMath\nWorksheet\nWorksheet\nHow much does that jacket cost? Your child will add up coins and dollars on this worksheet to find out how much each item costs!\nMath\nWorksheet\nMoney Word Problems #4\nWorksheet\nMoney Word Problems #4\nYour child will work through money math problems to sum up amounts of money. This word problem worksheet will help your child exercise his math skills.\nMath\nWorksheet\nPiggy Bank Math\nWorksheet\nPiggy Bank Math\nHow much money is in each piggy bank? First count the number of coins in each piggy bank, and then count up the value for each group!\nKindergarten\nMath\nWorksheet\nWorksheet\nAssist your little financier with adding coins. She'll add up coins in her head and tally the amount to figure out the price of the item shown." ]

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