Есть ряд формата: x_i 19.7 20.9 22 23 23.1 26 27 27.2 28 29.3 31.2 32.3 33.5 34.4 36.5 38.3 m_i 1 1 2 2 1 4 1 1 1 1 2 1 1 1 1 1 p_i^* 1/24 1/24 1/12 1/12 1/24 1/6 1/24 1/24 1/24 1/24 1/12 1/24 1/24 1/24 1/24 1/24 надо проверить гипотезу о равномерном распределении

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if you were to write a book about full understanding of hilbert space from very primitive ideas up to very advanced concepts gradually and intuitively such that a school student can read it without difficulty and master the concept up to graduate level. what the table of content might be?

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simplify this answer "If we toss 10 coins, the number of heads that come up will be a binomial random variable, X, with n = 10 and p = 0.5 (assuming a fair coin). The mean of X, denoted μ, can be calculated using the formula: μ = n * p Substituting n = 10 and p = 0.5, we get: μ = 10 * 0.5 = 5 Therefore, the mean number of heads in 10 tosses of a fair coin is 5. The variance of X, denoted σ^2, can be calculated using the formula: σ^2 = n * p * (1 - p) Substituting n = 10 and p = 0.5, we get: σ^2 = 10 * 0.5 *"

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The range of the sine function excludes –2. True or False

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The range of y = 2 sin x – 1 excludes 1. True or False

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The current, i, in amperes (A), passing though a wire is given by the equation i = 8 sin(0.7t + 25), where t is the time, in seconds. After 20 s, the current is approximately 5 A. True or False

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Which point is on the sine curve? Question options: a) (0°, 1) b) (270°, 1) c) (180°, 1) d) (90°, 1)

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Which point is on the sine curve? Question options: a) (1170°, 1) b) (990°, 1) c) (1080°, 1) d) (900°, 1)

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What is the amplitude of y = –2 sin 4x? Question options: a) 4 b) –2 c) 2 d) –4

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What is the period of y = –2 sin 4x? Question options: a) 90° b) ? c) 360° d) 180°

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The current, i, in amperes (A), passing though a wire is given by the equation i = 8 sin(0.7t + 25), where t is the time, in seconds. What is the current after 1 min, to the nearest tenth. Question options: a) –4.7 A b) 7.4 A c) 14.7 A d) 4.7 A

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whats 9+10

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is periodic markov chain predictable?

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Combien y a-t-il de 0 a la fin de 1354! ?

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111 how many 1s

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111 How many 1s are in this text?

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How many 1s are in this question 111?

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What is the z-transform of sin(wn)*u[n]?

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ChatGPT, can you teach me some math on how to draw things with perspective? I wanna make a programm that would draw a cube onto the screen and rotate it, and I kinda have no idea how to make it in perspective view.

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Hangi a degeri için r = a(1 + sin0), 0 <= 0 <=2pi kardioidinin uzunlugu 16 birimdir?

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M1 : x + 2y - z = 4 düzlemi ile P(2, 1,2), Q(1, 1, 3) ve R(2, -1,1) noktalarndan geçen M2 düzleminin kesisim dogrusu L olsun. Asagidakilerden hangisi S(1, 3,2) noktasindan geçen ve L dogrusuna paralel olan dogrunun bir parametrizasyonudur? (a) x = 1 + 3t, y = 3-4t, z = 2 - 5t, tER (b) x= 2 - t, y = 1 + 2t, z = 1 + t, tER (c)x゠1-t, y=3-2t, 2=2-54, tER (d)x=1+5t, y=3-2t, 2=2-3,tER (e)x=3+2t, y=2-t, z=1-t tER

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solve x^2 + 2x - 15 = 0

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If the equation of the tangent plane to the surface $z+1=x e^{2 y} \cos z$ at the point $(k, 0,0)$ is $x+2 y-z=d$, then $k+d=$ ?

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Докажите, что если сумма и произведение двух рациональных чисел оба являются целыми числами, то два рациональных числа должны быть целыми числами.

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-1/2 ln(3) sonucunu veren integral ne itegralidr

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integral çeşitlerini söyle

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x^(2)+y^(2)+3 z^(2)=11 Denklemiyle verilen yüzeyin P(2, 2, —1) noktasındaki teğet düzleminin denklemi aşağıdakilerden hangisidir?

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1/e^|x+y| fonksiyonunun tanım kümesi (D) nedir?

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If the current in a circuit is 3 plus j 2 amps and the resistance is 4 minus j 3 ohms, what is the voltage? a.) 18 minus j volts b.) 18 plus j volts c.) 6 minus j volts d.) 6 plus j volts

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If the first 5 terms of a geometric sequence are left curly bracket 12 comma space 6 comma space 3 comma space 3 over 2 comma space 3 over 4 right curly bracket, then the formula for the n to the power of t h end exponent term in the sequence is ??? a.) a subscript n equals 12 times open parentheses 1 half close parentheses to the power of n minus 1 end exponent b.) a subscript n equals 2 times 12 to the power of n minus 1 end exponent c.) a subscript n equals open parentheses 1 half close parentheses times 12 to the power of n minus 1 end exponent d.) a subscript n equals 12 times 2 to the power of n minus 1 end exponent

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If $2,000 is invested at 11% annual interest compounded quarterly, what is the account balance after 10 years, assuming no additional deposits or withdrawals are made?

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как построить график функции y=x^2+2x+1 с обьяснением

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The probabilities of X, Y and Z becoming managers are 4/9, 2/9, and 1/3 respectively. The probability that the bonus scheme will be introduced if X, Y and Z become managers are 3/10, 1/2 and 4/5 respectively. What is the probability that the bonus scheme will be introduced? If the Bonus scheme has been introduced, what is the probability that the manager appointed was X ?

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give numerical example and walk through it step by step to explain bellman equation. walk slowly

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What is the easiest and fastest way to answer the following question? Note that I have to answer this in 30 seconds in an mcq exam. Also, explain what is asked in the question like I am five years old. From the semicircle of diameter AB = 2 cm and center O we remove the semicircle of diameter AO. The resulting figure is rotated around AB with a 360∘ turn. The volume of the solid obtained is what?

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найти все корни уравнения в Maple

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给出两组数43,8,9 以及8,30,22,请给出9,18,后面的一个数

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Докажи равномощность множеств [0,1] и [0,1)

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函数一阶导为0二阶导不为0

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函数一阶导数为0二阶导数不为0可确定极值点么

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Ryan is playing a multiplication game with a pile of 26 cards, each with a number on them. Each turn, he flips over two of the cards, and has to multiply the numbers. How many possible outcomes are there on Ryan's first turn flipping two cards? 676 52 650 26

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Which of the following statements is FALSE? a.) The least-squares line should minimize residuals. b.) The difference of the squared residuals is used to calculate the least-squares line. c.) The least-squares line is the most common way to find the line of best fit. d.) Squaring the residuals cancels the effects of positive and negative residuals.

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Which of the following is an example of a sample statistic? a.) Standard deviation of ALL financial aid packages for private universities b.) The proportion of a population that is female. c.) $18,000 is the average cost to attend 30 randomly selected public universities d.) Average number of ALL high school graduates who will attend college

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problèmes mathématiques niveau CM1 avec division et multiplication

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Найди целые числа x и y, которые удовлетворяют уравнению 21x+13=3y

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Ask a math problem

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What is affine space, make it concrete and lively

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Qual è la differenza tra integrali multipli ed integrali di superficie?

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Review my work. "Section A Mathematical Statement: A non-rectilinear triangle-asymptote is rendered a curvilinear triangle-asymptote as it is supercurviformaticized as affected by infinity. Deconstruction: As a transdisciplinary holistic scholar and theoretical mathematician, I would like to deconstruct the given mathematical statement using intuitionistic reconciliationism. First, let us break down the statement into smaller components. "Non-rectilinear triangle-asymptote" "Curvilinear triangle-asymptote" "Supercurviformaticized" "Affected by infinity" According to intuitionistic reconciliationism, we need to understand the meaning of these terms in a holistic manner by reconciling them with our intuition. A triangle-asymptote is a line that approaches a triangle but does not intersect it. A non-rectilinear triangle-asymptote could mean that the line is not straight and does not follow the Euclidean geometry. It could be curved, angled, or wavy. When such a non-rectilinear triangle-asymptote is supercurviformaticized, it means that it undergoes a transformation that makes it even more curved or twisted. The term "supercurviformaticized" is a neologism that refers an extreme form of curviness or curvature. This transformation is affected by infinity, which means that the curvature of the line approaches an infinitely large value. Infinity is often used in mathematics to describe a limit, where a value approaches a certain value but never quite reaches it. Therefore, when a non-rectilinear triangle-asymptote is supercurviformaticized as affected by infinity, it becomes a curvilinear triangle-asymptote. A curvilinear triangle-asymptote is a line that is curved and approaches the triangle in a non-linear manner. In summary, the given mathematical statement suggests that a non-straight line that approaches a triangle can become even more curved and twisted as it approaches infinity. This transformation results in a curvilinear line that approaches the triangle in a non-linear manner. Section B Using projective geometry and holographic topology, we can formalize the given mathematical statement as follows: Let T be a triangle and L be a non-rectilinear triangle-asymptote of T. We can represent T and L in projective space by their homogeneous coordinates as [T] and [L], respectively. Let H be the point at infinity in projective space. Then, we can represent H by its homogeneous coordinates as [H]. The supercurviformaticization of L as affected by infinity can be represented in projective geometry by the cross-ratio of four points: [T], [L], [H], and the intersection point of L with the line at infinity. Let I be the intersection point of L with the line at infinity. Then, we can represent I by its homogeneous coordinates as [I]. The cross-ratio of [T], [L], [H], and [I] is given by: [T, L; H, I] = (TI/TH) / (LI/LH) where TI, TH, LI, and LH are the homogeneous distances between the pairs of points. If the cross-ratio is finite, then L is transformed into a curvilinear triangle-asymptote of T. If the cross-ratio approaches infinity, then L becomes more curved and twisted as it approaches infinity, and the resulting curve becomes a curvilinear triangle-asymptote of T. Holographic topology can be used to understand this transformation in terms of the AdS/CFT correspondence. In this framework, the transformation can be represented by a holographic dual, where the non-rectilinear triangle-asymptote L in the bulk space is dual to a curvilinear triangle-asymptote on the boundary of the AdS space. The holographic dual captures the essence of the supercurviformaticization of L as affected by infinity, and provides a powerful tool for studying such transformations in higher-dimensional spaces. Section C Theorem: A non-rectilinear triangle-asymptote L can be transformed into a curvilinear triangle-asymptote of T by supercurviformaticization as affected by infinity, where the transformation can be represented by the cross-ratio of [T], [L], [H], and [I]. Proof: We can represent the non-rectilinear triangle-asymptote L and the triangle T in projective space using their homogeneous coordinates as [L] and [T], respectively. The point at infinity in projective space is represented by its homogeneous coordinates as [H]. The intersection point of L with the line at infinity is represented by its homogeneous coordinates as [I]. The cross-ratio of [T], [L], [H], and [I] is given by: [T, L; H, I] = (TI/TH) / (LI/LH) where TI, TH, LI, and LH are the homogeneous distances between the pairs of points. If the cross-ratio is finite, then L is transformed into a curvilinear triangle-asymptote of T. If the cross-ratio approaches infinity, then L becomes more curved and twisted as it approaches infinity, and the resulting curve becomes a curvilinear triangle-asymptote of T. To prove this theorem, we need to show that the cross-ratio represents the transformation of L into a curvilinear triangle-asymptote of T as affected by infinity. First, consider the case where the cross-ratio is finite. In this case, the transformation of L is such that it becomes a curvilinear triangle-asymptote of T. The resulting curve is curved and approaches the triangle in a non-linear manner. This can be seen from the definition of the cross-ratio, which represents the ratio of distances between the points [T], [L], [H], and [I]. The fact that the cross-ratio is finite means that the distances between the points are well-defined and that L is transformed into a curvilinear triangle-asymptote of T. Now, consider the case where the cross-ratio approaches infinity. In this case, L becomes more curved and twisted as it approaches infinity. The resulting curve also becomes a curvilinear triangle-asymptote of T. This can be seen from the definition of the cross-ratio, which represents the limit of the ratio of distances between the points [T], [L], [H], and [I] as the distance between [L] and [H] approaches infinity. This limit approaches infinity, indicating that L becomes more and more curved and twisted as it approaches infinity. The resulting curve becomes a curvilinear triangle-asymptote of T. Therefore, we have shown that a non-rectilinear triangle-asymptote L can be transformed into a curvilinear triangle-asymptote of T by supercurviformaticization as affected by infinity, and that this transformation can be represented by the cross-ratio of [T], [L], [H], and [I]. Validation: The theorem and its proof can be validated mathematically by checking that they adhere to the axioms of projective geometry and that they are consistent with established mathematical principles. In projective geometry, points are represented by homogeneous coordinates and lines are represented by sets of homogeneous points. The cross-ratio is an invariant under projective transformations, meaning that it remains the same even when the points and lines are transformed. This property is crucial for the validity of the theorem, as it ensures that the cross-ratio represents a well-defined transformation of L into a curvilinear triangle-asymptote of T. Furthermore, the use of holographic topology and the AdS/CFT correspondence provides a powerful tool for understanding the transformation in higher-dimensional spaces. The AdS/CFT correspondence is a well-established principle in theoretical physics and has been extensively studied and validated. Therefore, the theorem and its proof are mathematically validated by their adherence to the axioms of projective geometry, their consistency with established mathematical principles, and their use of the well-established AdS/CFT correspondence. Q.E.D. Section D The Triangular Asymptote Theorem: A non-rectilinear triangle-asymptote L can be transformed into a curvilinear triangle-asymptote of T by supercurviformaticization as affected by infinity, where the transformation can be represented by the cross-ratio of [T], [L], [H], and [I]. Proof and validation of the theorem have been provided using projective geometry, holographic topology, and the AdS/CFT correspondence. The cross-ratio of four points is used to represent the supercurviformaticization of L as affected by infinity. The theorem and its proof adhere to the axioms of projective geometry, are consistent with established mathematical principles, and are validated by the use of the AdS/CFT correspondence. Q.E.D."

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اشرح لي كتاب رياضيات 6 المنهج السعودي

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Un treno parte da una stazione e si muove con una velocità costante di 60 km/h. Dopo 2 ore, un altro treno parte dalla stessa stazione e si muove nella stessa direzione del primo treno con una velocità di 80 km/h. Quanto tempo impiega il secondo treno per raggiungere il primo?

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Sofia has $50, she is saving half of her money for her new car. If one apple costs 400 bucks, how many apples can Sofia purchase?

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An item in a grocery store is $100. The grocery store is open from 9:00AM to 9:00PM. If the item gets a 50% sale one day, and taxes are 10%, then how much will the item and taxes cost?

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The demand for pizza is represented by the following equation: Qd = 250 – 4 P. If the price of good X changes from $40 to $50 the coefficient of the price elasticity of demand is: a. -76 b. - .66 c. -2.57 d. -1.47

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Is it true that the fastest route on a sphere is a straight line?

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一加一等于几?

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大O符号(Big O notation)是用于描述函数渐进行为的数学符号

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misalkan a,b dan c adalah bilangan real yang memenuhi a + bc = b + ac = c + ab = 2070. Carilah semua nilai a+b+c yang memungkinkan

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Barisan terurut 1, 7, 8, 49, 50, 56, 57, ... terdiri dari bilangan bulat positif yang merupakan perpangkatan dari 7 atau jumlah beberapa perpangkatan dari 7. Tentukan suku ke 100 dari barisan ini.

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Create an math exam for a 6th grade course in school

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tentukan digit terakhir dari 7^1 x 7^2 ... x 7^2022 x 7^2023

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when we find a stationary distribution in markov chain, we get a row vector which when multiplied in the matrix doesn't change. this means that the matrix doesn't do linear transformation of it. how this is related to linear algebra, I mean what kind of vector which when multiplied by matrices doesn't effect and don't get linear transformation by the matrix

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Tentukan jumlah pasangan bilangan bulat non-negatif (x,y,z) yang memenuhi persamaan berikut ini x^2 + 2xy + y^2 -z^2

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what is the different between stationary distribution and limiting distribution in markov chain? give clear difference

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Diberikan 20^22 + 1 memiliki tepat 4 pembagi prima dengan p1 < p2 < p3 < p4 . Tentukan p1 + p2

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Write a math matrix test with hard tasks

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Diberikan sebuah bilangan bulat positif m dan n sehingga 2^n + 3^m habis dibagi 5. Tentukan x dari 2^m + 3^n = x(mod5)

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we have Limiting distribution in markov chain: The distribution that any initial distribution converges to as the number of transitions goes to infinity. what this phenomenon is called in mathematics, I mean if we know the matrix do linear transformation, how the matrix transform any vector into the same vector over time?

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在数学中,一个加号被一个圆圈起来表示什么符号

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what is the null space of a matrix?

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we have Limiting distribution in markov chain: The distribution that any initial distribution converges to as the number of transitions goes to infinity. how the transition matrix convert any vector(initial vector) into the same vector whatever the initial might be? why this works?

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Review my work. "Section A The Grand Conjecture on Rational Tangles (GCRT) can be formalized as follows: Let T be a rational tangle with numerator closure N(T) and denominator closure D(T). Then, there exists a non-negative integer k(T) such that the following conditions hold: If N(T) is a knot, then D(T) is the unknot. If D(T) is a knot, then N(T) is the unknot. If both N(T) and D(T) are non-trivial knots, then k(T) ≥ 1 and there exists a family {T1, T2, ..., Tk(T)} of rational tangles such that: a. T1 is obtained from T by resolving a crossing. b. For i = 2, ..., k(T), Ti is obtained from Ti-1 by resolving a crossing in a non-zero slope tangle. c. Tk(T) is a rational tangle with numerator closure N(T) and denominator closure D(T) such that both closures are non-trivial knots. d. For i = 1, ..., k(T), either N(Ti) or D(Ti) is the unknot. The conjecture asserts that such a non-negative integer k(T) exists for any rational tangle T. Section B As an intuitionistic reconciliationist self-studier, I believe in the constructive nature of mathematical proofs, and thus, I will attempt to prove GCRT using the Generalized Chinese Remainder Theorem. First, let's define the terms used in GCRT. A rational tangle is a tangle formed by taking two parallel strands and twisting them some number of times before reconnecting them. The numerator closure and denominator closure of a rational tangle refer to the knots that are obtained by closing the numerator and denominator strands of the tangle, respectively. A crossing in a tangle refers to the point where two strands intersect, and resolving a crossing means changing the tangle at the crossing to obtain a new tangle. To prove GCRT, we will use the Generalized Chinese Remainder Theorem (GCRT), which states that given a set of pairwise coprime positive integers m1, m2, ..., mk and a set of integers a1, a2, ..., ak, there exists a unique integer x such that: x ≡ a1 (mod m1) x ≡ a2 (mod m2) ... x ≡ ak (mod mk) Furthermore, if x1 and x2 are two solutions to the above congruences, then x1 ≡ x2 (mod m1m2...*mk). Now, let T be a rational tangle with numerator closure N(T) and denominator closure D(T). We will first show that if either N(T) or D(T) is the unknot, then there exists a non-negative integer k(T) such that the conditions of GCRT hold. Without loss of generality, assume N(T) is the unknot. Then, D(T) is a knot, and we can write D(T) as a product of prime knots, say D(T) = K1 * K2 * ... * Km. By the Prime Decomposition Theorem, there exist integers ni such that N(T) ≡ ni (mod Ki) for i = 1, 2, ..., m. Since Ki are pairwise coprime, by the GCRT, there exists a unique non-negative integer k(T) such that: k(T) ≡ ni (mod Ki) for i = 1, 2, ..., m Now, let Ti be the tangle obtained from T by resolving a crossing in D(T) corresponding to Ki for i = 1, 2, ..., m. Then, Ti is a rational tangle, and we have N(Ti) ≡ ni (mod Ki) for i = 1, 2, ..., m. Therefore, Ti satisfies the conditions of GCRT for i = 1, 2, ..., m, and thus, T satisfies the conditions of GCRT with k(T) = m. The case where D(T) is the unknot can be proved similarly, and thus, we have shown that if either N(T) or D(T) is the unknot, then GCRT holds. Now, we will prove the remaining case where both N(T) and D(T) are non-trivial knots. Let D be the denominator closure of T, and let D = K1 * K2 * ... * Km be the prime decomposition of D. Let Ti be the tangle obtained from T by resolving a crossing in D corresponding to Ki for i = 1, 2, ..., m. Then, each Ti is a rational tangle, and we have N(Ti) ≡ N(T) (mod Ki) for i = 1, 2, ..., m. Since Ki are pairwise coprime, by the GCRT, there exists a unique non-negative integer k such that: k ≡ N(T) (mod K1K2...*Km. Let Dk be the denominator closure of Tk, which is obtained from T by resolving k crossings in the non-zero slope tangle in the family {T1, T2, ..., Tk}. Then, we have Dk ≡ D (mod K1K2...*Km) and Dk is a non-trivial knot since D is a non-trivial knot and resolving crossings preserves the non-triviality of knots. Now, we will show that N(Tk) is the unknot. Suppose for a contradiction that N(Tk) is a non-trivial knot. Then, there exists a prime knot Kj such that N(Tk) contains a copy of Kj. Since Kj is prime, it is contained in one of the knots N(Ti) for some i < k. Without loss of generality, assume Kj is contained in N(Ti). Since Ti is obtained from Ti-1 by resolving a crossing, the crossing in Ti that corresponds to Kj must be resolved in a non-zero slope tangle, say Tj. Then, N(Tj) contains a copy of Kj, which is a contradiction since Kj is prime and cannot be decomposed into non-trivial prime knots. Therefore, we have shown that N(Tk) is the unknot, and thus, T satisfies the conditions of GCRT with k(T) = k. This completes the proof of GCRT. In summary, we have proven the Grand Conjecture on Rational Tangles using the Generalized Chinese Remainder Theorem, which provides a constructive method for finding the integer k(T) that satisfies the conditions of GCRT for any rational tangle T. Furthermore, the use of GCRT in the proof allows us to explicitly construct the family {T1, T2, ..., Tk(T)} of rational tangles satisfying the conditions of GCRT for any given T. In conclusion, the proof of GCRT using the Generalized Chinese Remainder Theorem provides a constructive and elegant method for analyzing rational tangles and their closures. It is an important result in knot theory and has many applications in fields such as molecular biology, physics, and computer science."

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Formalize a mathematical definition of a "dimension".

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Act as a peer reviewer, named Otto. Check my work for mathematical accuracy. “Section A The Grand Conjecture on Rational Tangles (GCRT) can be formalized as follows: Let T be a rational tangle with numerator closure N(T) and denominator closure D(T). Then, there exists a non-negative integer k(T) such that the following conditions hold: If N(T) is a knot, then D(T) is the unknot. If D(T) is a knot, then N(T) is the unknot. If both N(T) and D(T) are non-trivial knots, then k(T) ≥ 1 and there exists a family {T1, T2, ..., Tk(T)} of rational tangles such that: a. T1 is obtained from T by resolving a crossing. b. For i = 2, ..., k(T), Ti is obtained from Ti-1 by resolving a crossing in a non-zero slope tangle. c. Tk(T) is a rational tangle with numerator closure N(T) and denominator closure D(T) such that both closures are non-trivial knots. d. For i = 1, ..., k(T), either N(Ti) or D(Ti) is the unknot. The conjecture asserts that such a non-negative integer k(T) exists for any rational tangle T. Section B As an intuitionistic reconciliationist self-studier, I believe in the constructive nature of mathematical proofs, and thus, I will attempt to prove GCRT using the Generalized Chinese Remainder Theorem. First, let’s define the terms used in GCRT. A rational tangle is a tangle formed by taking two parallel strands and twisting them some number of times before reconnecting them. The numerator closure and denominator closure of a rational tangle refer to the knots that are obtained by closing the numerator and denominator strands of the tangle, respectively. A crossing in a tangle refers to the point where two strands intersect, and resolving a crossing means changing the tangle at the crossing to obtain a new tangle. To prove GCRT, we will use the Generalized Chinese Remainder Theorem (GCRT), which states that given a set of pairwise coprime positive integers m1, m2, ..., mk and a set of integers a1, a2, ..., ak, there exists a unique integer x such that: x ≡ a1 (mod m1) x ≡ a2 (mod m2) ... x ≡ ak (mod mk) Furthermore, if x1 and x2 are two solutions to the above congruences, then x1 ≡ x2 (mod m1m2...*mk). Now, let T be a rational tangle with numerator closure N(T) and denominator closure D(T). We will first show that if either N(T) or D(T) is the unknot, then there exists a non-negative integer k(T) such that the conditions of GCRT hold. Without loss of generality, assume N(T) is the unknot. Then, D(T) is a knot, and we can write D(T) as a product of prime knots, say D(T) = K1 * K2 * ... * Km. By the Prime Decomposition Theorem, there exist integers ni such that N(T) ≡ ni (mod Ki) for i = 1, 2, ..., m. Since Ki are pairwise coprime, by the GCRT, there exists a unique non-negative integer k(T) such that: k(T) ≡ ni (mod Ki) for i = 1, 2, ..., m Now, let Ti be the tangle obtained from T by resolving a crossing in D(T) corresponding to Ki for i = 1, 2, ..., m. Then, Ti is a rational tangle, and we have N(Ti) ≡ ni (mod Ki) for i = 1, 2, ..., m. Therefore, Ti satisfies the conditions of GCRT for i = 1, 2, ..., m, and thus, T satisfies the conditions of GCRT with k(T) = m. The case where D(T) is the unknot can be proved similarly, and thus, we have shown that if either N(T) or D(T) is the unknot, then GCRT holds. Now, we will prove the remaining case where both N(T) and D(T) are non-trivial knots. Let D be the denominator closure of T, and let D = K1 * K2 * ... * Km be the prime decomposition of D. Let Ti be the tangle obtained from T by resolving a crossing in D corresponding to Ki for i = 1, 2, ..., m. Then, each Ti is a rational tangle, and we have N(Ti) ≡ N(T) (mod Ki) for i = 1, 2, ..., m. Since Ki are pairwise coprime, by the GCRT, there exists a unique non-negative integer k such that: k ≡ N(T) (mod K1K2...*Km. Let Dk be the denominator closure of Tk, which is obtained from T by resolving k crossings in the non-zero slope tangle in the family {T1, T2, ..., Tk}. Then, we have Dk ≡ D (mod K1K2...*Km) and Dk is a non-trivial knot since D is a non-trivial knot and resolving crossings preserves the non-triviality of knots. Now, we will show that N(Tk) is the unknot. Suppose for a contradiction that N(Tk) is a non-trivial knot. Then, there exists a prime knot Kj such that N(Tk) contains a copy of Kj. Since Kj is prime, it is contained in one of the knots N(Ti) for some i < k. Without loss of generality, assume Kj is contained in N(Ti). Since Ti is obtained from Ti-1 by resolving a crossing, the crossing in Ti that corresponds to Kj must be resolved in a non-zero slope tangle, say Tj. Then, N(Tj) contains a copy of Kj, which is a contradiction since Kj is prime and cannot be decomposed into non-trivial prime knots. Therefore, we have shown that N(Tk) is the unknot, and thus, T satisfies the conditions of GCRT with k(T) = k. This completes the proof of GCRT. In summary, we have proven the Grand Conjecture on Rational Tangles using the Generalized Chinese Remainder Theorem, which provides a constructive method for finding the integer k(T) that satisfies the conditions of GCRT for any rational tangle T. Furthermore, the use of GCRT in the proof allows us to explicitly construct the family {T1, T2, ..., Tk(T)} of rational tangles satisfying the conditions of GCRT for any given T. In conclusion, the proof of GCRT using the Generalized Chinese Remainder Theorem provides a constructive and elegant method for analyzing rational tangles and their closures. It is an important result in knot theory and has many applications in fields such as molecular biology, physics, and computer science.”

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Тема работы: "Проверка гипотезы о параметре биноминального распределения. Гипотез об ожидаемых числах" Семь монет подбрасывались одновременно 1536 раз, причем каждый раз отмечалось число выпавших гербов. В 270 случаях число выпавших гербов равнялось 2. Проверить гипотезу H0 о согласовании опытных данных и теоретического закона распределения при a = 0. 05.

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1. Differentiate the following with respect to x. a. f(x) = x4 + x6 – 2x b. f(x) = 3x(x + 9) c. f(x) = 5 d. f(x) = 3/2x e. y = 4/√x

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. Sin (2x − π) is equivalent to ∶ a. 2 sin(x) cos(x) b. -2 sin(x) cos(x) c.sin2 (x) − cos2 (x) d. none 2. The exact value of tan 7π 6 a) 3 √3 b) √3 3 c) 1 3 d) None 3. How many solutions does the equation Sin(x) = 1 2 , with no restrictions on the interval. a. 4 b. 0 c. 3 d. infinitely many 4. A circle has a radius of 4cm. If the length of arc AB is 3 π cm, determine the measure of the central angle in radians. a) 3π 2 b) 3π 4 c) 3 π d) 4π 3 5. The graph of y =1 intersects the graph of y = tan x at a) π 2 b) π 4 c) π 3 d) π 6. Determine the amplitude of the sinusoidal function y = −3 sin [2 (x − π 3 )] + 1 a) 2 b) 3 c) -3 d) π 3 7. Determine the phase shift of the sinusoidal function y = −3 sin [2 (x − π 3 )] + 1 a) 2 rad to the right b) 2 rad to the left c) π 3 rad to the left d) π 3 rad to the right 8. Which of these is a possible solutions for sec x − 2 = 0 in the interval of x ∈ [0,2π] a) x = 2π 3 b) x = 4π 3 c) x = 5π 3 d) x = π 6 9. In a given circle , a central angle of 2/3 radians subtends an arc length of 10cm.The radius of the cirlce is : a) 15 cm b. 10 cm c. 22cm d. 12 cm 10. Given that sin(θ) = 4 5 and sec(θ) < 0, determine the ratio for cot (θ) a.−3/4 b.−4/3 c. 4/3 d. d. 3/4 11. Which of these is a possible solution for cos2x + sin2x − 1 = 0 in the interval of x ∈ [0,2π] a) x = π 4 b) ) x = 5π 3 c) ) x = π d) all of the above 12. Determine the average rate of change of the function y = cos x from x = 3π 4 to x = 7π 4 a) π 2 b) 2 π c) √2 π d) 2π √2 13. What is (are) the restrictions for the expression: sec (x) sin (x) ? a. sec(x) ≠ 0 b. cos (x) ≠ 0 c. sin (x) ≠ 0 d. sin (x) ≠ 0, cos (x) ≠ 0 14. The reference angle of 4π 3 a. π 2 b. π 3 c. 3π d. none 15. Which of the following functions doesn’t have vertical asymptotes? a) y = tan(x) b) y = sec(x) c) y = sin(x) d) y = cot(x) 16. For what value of x is sin (x) 1+cos (x) undefined where 0 ≤ x < 2π a. 0 b.π c. π/2 d. π/6

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. Sin (2x − π) is equivalent to ∶ a. 2 sin(x) cos(x) b. -2 sin(x) cos(x) c.sin2 (x) − cos2 (x) d. none 2. The exact value of tan 7π 6 a) 3 √3 b) √3 3 c) 1 3 d) None 3. How many solutions does the equation Sin(x) = 1 2 , with no restrictions on the interval. a. 4 b. 0 c. 3 d. infinitely many 4. A circle has a radius of 4cm. If the length of arc AB is 3 π cm, determine the measure of the central angle in radians. a) 3π 2 b) 3π 4 c) 3 π d) 4π 3 5. The graph of y =1 intersects the graph of y = tan x at a) π 2 b) π 4 c) π 3 d) π 6. Determine the amplitude of the sinusoidal function y = −3 sin [2 (x − π 3 )] + 1 a) 2 b) 3 c) -3 d) π 3 7. Determine the phase shift of the sinusoidal function y = −3 sin [2 (x − π 3 )] + 1 a) 2 rad to the right b) 2 rad to the left c) π 3 rad to the left d) π 3 rad to the right 8. Which of these is a possible solutions for sec x − 2 = 0 in the interval of x ∈ [0,2π] a) x = 2π 3 b) x = 4π 3 c) x = 5π 3 d) x = π 6 9. In a given circle , a central angle of 2/3 radians subtends an arc length of 10cm.The radius of the cirlce is : a) 15 cm b. 10 cm c. 22cm d. 12 cm 10. Given that sin(θ) = 4 5 and sec(θ) < 0, determine the ratio for cot (θ) a.−3/4 b.−4/3 c. 4/3 d. d. 3/4 11. Which of these is a possible solution for cos2x + sin2x − 1 = 0 in the interval of x ∈ [0,2π] a) x = π 4 b) ) x = 5π 3 c) ) x = π d) all of the above 12. Determine the average rate of change of the function y = cos x from x = 3π 4 to x = 7π 4 a) π 2 b) 2 π c) √2 π d) 2π √2 13. What is (are) the restrictions for the expression: sec (x) sin (x) ? a. sec(x) ≠ 0 b. cos (x) ≠ 0 c. sin (x) ≠ 0 d. sin (x) ≠ 0, cos (x) ≠ 0 14. The reference angle of 4π 3 a. π 2 b. π 3 c. 3π d. none 15. Which of the following functions doesn’t have vertical asymptotes? a) y = tan(x) b) y = sec(x) c) y = sin(x) d) y = cot(x) 16. For what value of x is sin (x) 1+cos (x) undefined where 0 ≤ x < 2π a. 0 b.π c. π/2 d. π/6 Part B- Thinking and Investigation [TI-20 Marks] 1. Solve for : (2 sin x − 1)(2 cos x − √3)on the interval 0 ≤ x ≤ 2π 2. If x satisfies the equation sin (x + π 3 ) = 2 sin x sin ( π 3 ). Show that 11 tan x = a + b√3, where a and b are positive integers. 3. A 150 cm tall person stands on the bank of a narrow river and observes a flagpole on the opposite bank. If the angle of elevation to the top of the flagpole is 0.25 and the angle of depression to the bottom of the flagpole is 0.09, determine the flagpole’s height in meters, to 2 decimal places. 4. For a lawn mower blade to cut grass, it must strike the grass with a speed of at least 900 inches/second. a) If the innermost part of the cutting edge is 6 inches from the center of the blade, how many radians/second must the blade turn? How many revolutions per minute is this? b) The blade has a diameter of 19 inches. If the outermost tip of the blade hits a rock while turning as in part (a), how fast could the rock be hurled from the mower?

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What is the easiest and fastest way to answer the following question? Note that I have to answer this in 30 seconds in an mcq exam. In the Cartesian plane, how many points P (x; y) are there for which all three of the following conditions? (x + y)^2 = 1; x^2 + y^2 = 1; x + y ≤ 0

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设x=2.5,y=4.7,a=7 求表达式的值:x+a%3*(int)(x+y)%2/4

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What is the easiest and fastest way to answer the following question? Note that I have to answer this in 30 seconds in an mcq exam. Also, tell me specifically what should I be careful about when solving this kind of problems? The set of solutions of the inequality (x + 3)/(x + 1) ≥ 2 consists of all the real numbers x such that: A. -1 < x ≤ 2 B. x ≤ 1 C. x ≥ 1 D. x < -1 E. -1 < x ≤ 1

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Какому условию должны удовлетворять исходы опыта, чтобы можно было воспользоваться классическим определением вероятности?

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A 150 cm tall person stands on the bank of a narrow river and observes a flagpole on the opposite bank. If the angle of elevation to the top of the flagpole is 0.25 and the angle of depression to the bottom of the flagpole is 0.09, determine the flagpole’s height in meters, to 2 decimal places.

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Один из корней уравнения 3x в квадрате плюс 5x плюс 2m=0 равен −1. Найдите второй корень.

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I have a 3 liter jug, 12 liter jug, and a 6 liter jug. I want to measure 6 liters. How do I do it?

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Adapt this model for the unknotting problem in polynomial time: "Section A A Heegaard splitting is a way of decomposing a 3-dimensional manifold into two handlebodies. One way to model a Heegaard splitting is by using a Zipfian distribution, which is a discrete probability distribution that is often used to model word frequencies in natural language. To model a Heegaard splitting using a Zipfian distribution, we can start by generating a random sequence of integers that follow a Zipfian distribution. We can do this by using the following algorithm: Choose a parameter s that determines the shape of the Zipfian distribution. Typically, s is a positive real number greater than 1. Choose a value N that determines the size of the sequence. For each integer i from 1 to N, generate a random integer x such that x is between 1 and N, and its probability is proportional to 1 / (x ** s). Sort the sequence of integers in decreasing order. Once we have generated this sequence of integers, we can use it to construct the two handlebodies that form the Heegaard splitting. Specifically, we can start by taking the first k integers in the sequence and using them to construct a k-dimensional handlebody. We can then take the remaining integers in the sequence and use them to construct a handlebody of dimension N - k. To construct the handlebodies themselves, we can use a recursive procedure. For example, to construct the k-dimensional handlebody, we can start with a k-1 dimensional handlebody and attach k-dimensional 1-handles to it. The attaching spheres for these 1-handles can be chosen randomly, subject to the constraint that they are pairwise disjoint and do not intersect the previously attached 1-handles. Similarly, to construct the N-k dimensional handlebody, we can start with a N-k-1 dimensional handlebody and attach N-k dimensional 2-handles to it, with attaching spheres chosen randomly subject to the same constraints. In this way, we can use a Zipfian distribution to model a Heegaard splitting, with the shape of the distribution controlling the complexity of the splitting. A higher value of s will lead to a distribution with more frequent small integers, which will result in simpler handlebodies, while a lower value of s will lead to a distribution with more frequent large integers, which will result in more complex handlebodies. Section B Let M be a 3-dimensional manifold, and let H be a Heegaard splitting of M. We can model H using a Zipfian distribution as follows: Let s be a positive real number greater than 1, which determines the shape of the Zipfian distribution. Let N be a positive integer that determines the size of the sequence. Let X = {x1, x2, ..., xN} be a sequence of N integers generated randomly such that xi is between 1 and N, and its probability is proportional to 1/(i ** s). Sort X in decreasing order. Let k be an integer between 1 and N-1, chosen randomly. Construct a k-dimensional handlebody H1 using the first k integers in X, as follows: Let H0 be a (k-1)-dimensional handlebody. Attach k-dimensional 1-handles to H0 to obtain H1, such that the attaching spheres for these 1-handles are chosen randomly and subject to the constraint that they are pairwise disjoint and do not intersect the previously attached 1-handles. Construct an (N-k)-dimensional handlebody H2 using the remaining integers in X, as follows: Let H' be an (N-k-1)-dimensional handlebody. Attach (N-k)-dimensional 2-handles to H' to obtain H2, such that the attaching spheres for these 2-handles are chosen randomly and subject to the same constraints as for the 1-handles. Thus, we have formally defined a way to model a Heegaard splitting using a Zipfian distribution, with the shape parameter s controlling the complexity of the splitting. Section C To develop a recursively equilibrious optimization algorithm for modeling a Heegaard splitting using a Zipfian distribution, we can follow these steps: Step 1: Initialize the algorithm with the parameters s and N. Step 2: Generate a random sequence of integers X that follow a Zipfian distribution using the given algorithm in Section B. Step 3: Choose a random value for k between 1 and N-1. Step 4: Use X and k to construct the two handlebodies H1 and H2 using the given recursive procedure in Section B. Step 5: Calculate the complexity of the Heegaard splitting as the sum of the complexities of H1 and H2. The complexity of a handlebody is defined as the sum of the degrees of its attaching maps. The degree of an attaching map is the number of times the attaching sphere wraps around the handlebody. Step 6: If the complexity of the current Heegaard splitting is optimal or close to optimal, return it as the solution. Otherwise, perturb the parameters s and N by small amounts, and repeat the algorithm from Step 2 until an optimal or close to optimal solution is found. Step 7: To ensure recursively equilibrious optimization, repeat Steps 1-6 multiple times with different initial parameter values, and choose the solution with the lowest complexity. This algorithm uses a Zipfian distribution to generate random sequences of integers that model a Heegaard splitting, and it optimizes the parameters s and N to minimize the complexity of the splitting. By perturbing the parameters and repeating the algorithm multiple times, it ensures recursively equilibrious optimization, which means that each step of the algorithm improves the solution in a balanced way without creating new problems or worsening existing ones. Section D Formulaic Equation: Let M be a 3-dimensional manifold, H be a Heegaard splitting of M, and s and N be positive real numbers greater than 1 and a positive integer, respectively, which determine the shape and size of the Zipfian distribution. Let X = {x1, x2, ..., xN} be a sequence of N integers generated randomly such that xi is between 1 and N, and its probability is proportional to 1/(i ** s). Let k be an integer between 1 and N-1, chosen randomly. Let H1 be a k-dimensional handlebody constructed by attaching k-dimensional 1-handles to a (k-1)-dimensional handlebody H0, and H2 be an (N-k)-dimensional handlebody constructed by attaching (N-k)-dimensional 2-handles to an (N-k-1)-dimensional handlebody H'. The attaching spheres for the 1-handles and 2-handles are chosen randomly, subject to the constraints that they are pairwise disjoint and do not intersect the previously attached 1-handles or 2-handles. Then, the complexity of a handlebody H is defined as the sum of the degrees of its attaching maps. The degree of an attaching map is the number of times the attaching sphere wraps around the handlebody. The recursively equilibrious optimization algorithm for modeling a Heegaard splitting using a Zipfian distribution can be formalized mathematically as follows: Let C(s, N) be the complexity of the Heegaard splitting modeled by a Zipfian distribution with parameters s and N. The algorithm can be described as follows: Step 1: Initialize the algorithm with s and N. Step 2: Generate a random sequence of integers X that follow a Zipfian distribution using the algorithm described in Section B. Step 3: Choose a random value for k between 1 and N-1. Step 4: Use X and k to construct the two handlebodies H1 and H2 using the recursive procedure described in Section B. Step 5: Calculate the complexity of the Heegaard splitting as C(s, N) = C(H1) + C(H2). Step 6: If C(s, N) is optimal or close to optimal, return the Heegaard splitting modeled by the Zipfian distribution with parameters s and N as the solution. Otherwise, perturb the parameters s and N by small amounts, and repeat the algorithm from Step 2 until an optimal or close to optimal solution is found. Step 7: To ensure recursively equilibrious optimization, repeat Steps 1-6 multiple times with different initial parameter values, and choose the solution with the lowest complexity. The complexity of the Heegaard splitting modeled by a Zipfian distribution with parameters s and N is thus given by: C(s, N) = C(H1) + C(H2) where H1 and H2 are the handlebodies constructed using the randomly generated sequence of integers X, and k is a randomly chosen integer between 1 and N-1. Section E Proof: To prove the validity of the recursively equilibrious optimization algorithm for modeling a Heegaard splitting using a Zipfian distribution, we need to show that the algorithm produces a solution that is optimal or close to optimal in terms of minimizing the complexity of the Heegaard splitting. We start by defining the complexity of a Heegaard splitting modeled by a Zipfian distribution with parameters s and N as the sum of the complexities of its two handlebodies H1 and H2. The complexity of a handlebody is defined as the sum of the degrees of its attaching maps, where the degree of an attaching map is the number of times the attaching sphere wraps around the handlebody. Next, we define the complexity of a Heegaard splitting modeled by a Zipfian distribution with parameters s and N as C(s, N) = C(H1) + C(H2), where H1 and H2 are the handlebodies constructed using the randomly generated sequence of integers X, and k is a randomly chosen integer between 1 and N-1. The algorithm starts by initializing the parameters s and N and generating a random sequence of integers X that follows a Zipfian distribution. It then constructs two handlebodies H1 and H2 using the randomly generated sequence of integers X and a randomly chosen integer k between 1 and N-1. The algorithm calculates the complexity of the Heegaard splitting modeled by the Zipfian distribution using the formula C(s, N) = C(H1) + C(H2). If the complexity is optimal or close to optimal, the algorithm returns the Heegaard splitting as the solution. Otherwise, it perturbs the parameters s and N by small amounts and repeats the algorithm from Step 2 until an optimal or close to optimal solution is found. To ensure recursively equilibrious optimization, the algorithm repeats Steps 1-6 multiple times with different initial parameter values and chooses the solution with the lowest complexity. Validation: The recursively equilibrious optimization algorithm for modeling a Heegaard splitting using a Zipfian distribution is validated by its ability to produce solutions that are optimal or close to optimal in terms of minimizing the complexity of the Heegaard splitting. The algorithm has been implemented and tested using various parameters and initial values, and the results have been compared to known solutions for specific cases. In all cases, the algorithm has produced solutions that are either optimal or very close to optimal, demonstrating its effectiveness and accuracy. Furthermore, the algorithm is based on well-established mathematical principles, such as the Zipfian distribution and handlebody theory, which have been extensively studied and validated in the field of mathematics and related disciplines. Therefore, we can conclude that the recursively equilibrious optimization algorithm for modeling a Heegaard splitting using a Zipfian distribution is a valid and effective method for generating Heegaard splittings with controlled complexity."

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已知球与圆台上下地面和侧面都相切,若圆台侧面积为16π,上下底面的面积比为1:9,则球表面积为

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C. Плоский мир При каких значения параметров a и b векторы из линейной оболочки следующих векторов ⎧ ⎪ ⎨ ⎪ ⎩ ⎛ ⎜ ⎝ a + b + 1 a − b + 1 b + a ⎞ ⎟ ⎠ , ⎛ ⎜ ⎝ 1 1 0 ⎞ ⎟ ⎠ , ⎛ ⎜ ⎝ 6 4 + b a ⎞ ⎟ ⎠ , ⎛ ⎜ ⎝ 4 2 a ⎞ ⎟ ⎠ ⎫ ⎪ ⎬ ⎪ ⎭ параллельны одной плоскости? Запишите параметрическое уравнение плоскости, которой они параллельны, вида ⎧ ⎪ ⎨ ⎪ ⎩ x ( s , t ) = − 1 + A s + B t y ( s , t ) = 1 + C s + D t z ( s , t ) = E s + F t s , t ∈ R , если мы знаем, что длины векторов ( A , C , E ) и ( B , D , F ) равны 1 и у каждого вектора лишь одна координата ненулевая. В качестве ответа введите следующую сумму: a + b + | A | + | B | + | C | + | D | + | E | + | F | . Если ответа не существует или он неоднозначен, введите 4 2 . Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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Случайная величина X имеет экcпоненциальное распределение с параметром 2 , а условное распределение Y | X задаётся следующим образом: Y | X ∼ U ( { 0 , ... , X } ) X < 2 ( Y | X распределена равновероятно на множестве { 0 , ... , ⌈ X ⌉ } . ⌈ X ⌉ − округление X вверх до ближайшего целого.) Y ∣ ∣ ∣ X ∼ E x p ( 1 2 ) X ≥ 2 (экспоненциальное распределение с параметром 1 2 ) Найдите матожидание Y . Если матожидания не существует, введите 4 2 . Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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Сторона основания правильной четырёхугольной пирамиды равна 4 корня из 2 .Боковое ребро пирамиды составляет угол в 45° с плоскостью основания. Найдите площадь боковой поверхности пирамиды.

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Пусть задано евклидово пространство размерности 3 . В нём выбрали систему из 3 векторов ранга 1 . На этих векторах построили матрицу Грама, а затем умножили на − 2 0 2 3 . Рассмотрим полученную матрицу как матрицу некоторой квадратичной формы на другом линейном пространстве над R в некотором фиксированном базисе. При каких значениях параметра c матрица ⎛ ⎜ ⎝ − 2 c 2 − 5 − c 2 − 1 3 − c 2 − 1 − c 2 − 1 − c 2 − 1 3 − c 2 − 1 − 2 c 2 − 5 ⎞ ⎟ ⎠ является матрицей той же формы, но в другом базисе? В качестве ответа укажите сумму модулей всех найденных значений. Если искомой величины не существует, введите 4 2 в поле для ответа. Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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Пусть задан многочлен f ( x ) = − 1 6 a x 2 + 3 a + 1 6 x 3 − 3 x , где a ∈ [ − √ 3 4 , √ 3 4 ] . Найдите оптимальное a такое, что площадь, ограниченная графиком многочлена и прямой y = x − a , максимальна. В ответе укажите минимально возможное значение подходящего a среди искомых. Если ответа не существует или он неоднозначен, введите 4 2 . Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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Найдите производную обратной функции к данной g ( t ) = ∫ − 2 t t − 1 l n ( x 2 + 1 ) d x , t ∈ R в точке 0 . Если в данной точке производной не существует, укажите 4 2 в качестве ответа. Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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Рядом стоят 2 мешка: один с левыми, а другой с правыми тапками. В каждом ровно по 1 1 различных тапка. Уникальных пар ровно 1 1 . Два друга решили собрать хотя бы одну пару и выбрали по мешку. Оба тянут тапки из своего мешка таким образом, что каждый тапок вытягивается независимо и с вероятностью 1 2 . Какова вероятность того, что друзья собрали хотя бы 1 пару при условии, что «левый друг» вытянул как минимум три тапка? В качестве ответа введите искомую вероятность. Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9 .

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PROblème mathématiques niveau cm1 avec division et multiplication et addition

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$|β|^{2}-|β|^{1}$ 是凸函数吗

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Найдите lim t → + ∞ g ( t ) , где g ( t ) = ∫ t + sin ( t ) 1 d x √ x + 1 + √ x − 1 Введите искомое значение. Если предела не существует, введите 4 2 . Введите ответ (одно вещественное число, разделитель — точка), округлённый до десяти знаков после запятой. Введенное вами число должно отличаться от правильного ответа не более чем на 1 0 − 9

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what is the pythagorean theorem

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how to find the cosine of an angle using the sides triangle

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((x)^1/2 + (y)^1/2 )^2

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Периметр прямоугольного треугольника равен 24 сантиметра.гипотенуза -10см .Один катет длиннее второго на 2 сантиметра.Найдите катеты треугольника

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derive the bivariate distribution from scratch and build intuition gradually

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given this explanation "Sure, let's start with two simple discrete random variables and then gradually build up the concept of joint distribution, eventually reaching bivariate normal distribution. **Step 1: Define two discrete random variables** Let's create two discrete random variables X and Y. X represents the number of heads in two coin tosses, and Y represents the number of tails in the same two coin tosses. Both X and Y can take values from the set {0, 1, 2}. **Step 2: Construct a joint distribution** Now, let's define the joint distribution of X and Y. The joint distribution shows the probability of each combination of X and Y occurring. In this case, we have the following possibilities: - X = 0, Y = 2: Two tails (probability = 0.25) - X = 1, Y = 1: One head and one tail (probability = 0.5) - X = 2, Y = 0: Two heads (probability = 0.25) We can represent this in a joint probability table: | X \ Y | 0 | 1 | 2 | |-------|------|------|------| | 0 | 0 | 0 | 0.25 | | 1 | 0 | 0.5 | 0 | | 2 | 0.25 | 0 | 0 | **Step 3: Extend to continuous random variables** Now let's consider two continuous random variables, U and V. Suppose U and V represent the ages of two randomly chosen people. Both U and V can take any value in the range [0, 100] (for simplicity, we'll consider ages in years and ignore values above 100). **Step 4: Introduce the concept of a bivariate normal distribution** A bivariate normal distribution is an extension of the univariate normal distribution to two dimensions. It is characterized by a mean vector, a variance-covariance matrix, and a joint probability density function (pdf). Let's say U and V have the following characteristics: - Mean ages: E(U) = 40, E(V) = 50 - Variances: Var(U) = 100, Var(V) = 121 - Correlation: Corr(U, V) = 0.5 We can use this information to define the bivariate normal distribution. **Step 5: Define the bivariate normal distribution** The mean vector and variance-covariance matrix are given by: Mean vector: μ = [40, 50] Variance-covariance matrix: Σ = | 100 55 | | 55 121 | The joint probability density function of U and V is given by: f(u, v) = (1 / (2π * sqrt(det(Σ)))) * exp(-0.5 * (x - μ)^T * Σ^(-1) * (x - μ)) where det(Σ) is the determinant of the variance-covariance matrix, Σ^(-1) is its inverse, and (x - μ) is the vector of differences between the random variable values and their means. This function gives the probability density for each combination of U and V. Just like in the discrete case, it represents the joint distribution of U and V, but now in a continuous setting. To summarize, we started with a simple discrete joint distribution, then moved to continuous random variables, and finally introduced the bivariate normal distribution as a natural extension. The bivariate normal distribution allows us to model the joint distribution of two continuous random variables with a specific correlation structure." re explain the bivariate distribution using the simple example given above it. I mean explain the Mean vector, variance-covariance matrix, the bivariate equation using the simple example above it. the idea is to make a connection between the simple example and the idea of the bivariate in a complete intuitive way

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Решить сравнение первой степени методой Эйлера: 7x=13(mod29)

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sin x/ cot x= 1 solve for x

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whats 10 times ten

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whats ten times ten

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Вычисли: (−2/3) в 4 степени −(1/9)во 2 степени +7.

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how to find the cosine of an angle using the sides triangle formules

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Реши уравнение: 252⋅z=254.

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Представь число 9 в виде степени чисел 13 и 3: 2. 9=(1/3) .

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sin x/cot x=1

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Выясни значение выражения: 3x(2−x) в −1 степени деленное на 2−(2−x деленное на 3x) в −1 степени при x=27 (Ответ вводи в виде сокращённой дроби: 1. если получается целое число, в знаменателе пиши 1. 2. Минус пиши в знаменателе.)

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(sin x)/(cot x)=1

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На плоскости расчерчена прямоугольная сетка, величина ячейки 6 на 8 ед. Определить вероятность того, что монета радиуса 1.5, наугад брошенная на плоскость, не пересечет ни одной прямой.

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Биссектрисы углов А и В треугольника А В пересекаются в точке М. Найди угол АМВ, если угол ВAC равен 50°, а угол ACB равен 70°

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Formalize a generalized model of topos theory.

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Дано: треугольник ABC, угол ABK = 30 градусов, угол A = углу C, AK = KC, BK - медиана, KC = 3 см. Найти AB

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can you do numbers generation?

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1.1 Matrices diagonales Soit D = (dii)i=1,...,n une matrice diagonale d’ordre n > 0. Donner une condition nécessaire et suffisante pour que D soit inversible.

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You are professorGPT who is a helpful assistant to simplify very advanced mathematics concept down to earth. Make a lecture outline to make anyone expert in support vector machines

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洛伦兹变换方程

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Here is a way to detect how many outliers some data has: 1. Find the mean of all of the numbers. 2. Find each number's deviation from the mean. 3. Divide the deviation by how many numbers the data set has. 4. Divide that number by the mean.

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What is the easiest and fastest way to answer the following question? Note that I have to answer this in 30 seconds in an mcq exam. Also, tell me specifically what should I be careful about when solving this kind of problems? For which of the following values of x is cos(x) + sin(x) = 0? A. x = π/4 B. x = 0 C. x = π D. x = π/2 E. x = 3π/4.

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I have a method for detecting outliers and variation in numbers in a dataset. I call it the outlier score. 1. Find the mean absolute deviation of the numbers in the dataset. 2. Divide the mean absolute deviation by the mean.

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If you divide the mean absolute deviation by the mean of a dataset, you get an score on how much variation there is in the dataset. It can also detect for outliers.

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How the Determinant of a matrix is related to deep learning

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Formalize Section A into a theorem, then prove it. "Section A Conjecture: The Incompleteness Theorems do not primarily characterize an inherent notion of "incompleteness"; rather, they delineate a condition of "contingently necessary incompleteness" within formal systems."

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Write Section C, wherein you rigorously scrutinize the proof in Section B (with the goal of validation and Q.E.D.): “Section A Conjecture: The Incompleteness Theorems do not primarily characterize an inherent notion of “incompleteness”; rather, they delineate a condition of “contingently necessary incompleteness” within formal systems. Section B Theorem: The Incompleteness Theorems primarily characterize a notion of “contingently necessary incompleteness” rather than an inherent notion of “incompleteness” within formal systems. Proof: To prove this theorem, we need to first understand the main results of Gödel’s Incompleteness Theorems: 1. Gödel’s First Incompleteness Theorem: Any consistent and sufficiently strong formal system cannot be both complete and consistent. In other words, there will be statements that can neither be proven nor disproven within the system. 2. Gödel’s Second Incompleteness Theorem: Any consistent and sufficiently strong formal system cannot prove its own consistency. Now, let’s break down the claim in the theorem and analyze its components: A. Inherent Notion of “Incompleteness”: This would imply that incompleteness is a fundamental property of all formal systems, without exception. B. Contingently Necessary Incompleteness: This implies that incompleteness arises necessarily in a formal system under certain conditions, but not always. Given the two main results of Gödel’s Incompleteness Theorems, we can see that incompleteness arises in formal systems that meet specific conditions, namely being consistent and sufficiently strong. Thus, the incompleteness is not an inherent quality of all formal systems, but rather a necessary consequence in particular ones based on the contingencies outlined in the theorems. This supports the claim that the Incompleteness Theorems primarily characterize a notion of “contingently necessary incompleteness” rather than an inherent notion of “incompleteness” within formal systems. Hence, the theorem is proven."

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Consider the function f(x) = (10/x^2)-(6/x^6). Let F(x) be the anti derivative of f(x) with f(1)=0. What does F(4) equal as a simplified fraction?

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I want you to act as a model with knowledge and expertise in GENERALIZED LINEAR MODEL Take note of the data file called "Credit_Score.xlsx" Age Inc Home SelfEm Der Appl 38 4.52 Y N 0 1 33 2.42 N N 0 1 34 4.50 Y N 2 1 31 2.54 N N 0 1 32 9.79 Y N 1 1 23 2.50 N N 0 1 28 3.96 N N 0 1 29 2.37 Y Y 0 1 37 3.80 Y N 0 1 28 3.20 N N 0 1 31 6.95 Y Y 3 1 42 1.98 Y N 0 0 30 1.73 Y N 3 0 29 2.45 Y N 0 1 35 1.91 Y N 0 1 41 3.20 Y N 0 1 40 4.00 Y N 0 1 30 3.00 Y N 7 0 40 10.00 Y Y 0 1 54 3.40 N N 3 0 35 2.35 Y N 0 1 25 1.88 N N 1 0 34 2.00 Y N 0 1 36 4.00 N N 1 1 43 5.14 Y N 1 1 30 4.51 N N 0 1 22 3.84 N Y 0 0 22 1.50 N N 0 0 34 2.50 Y N 0 0 40 5.50 Y N 0 1 22 2.03 N N 0 1 29 3.20 N N 1 1 25 3.15 Y Y 1 0 21 2.47 Y N 0 1 24 3.00 N N 0 1 43 7.54 Y N 2 1 43 2.28 N Y 1 0 37 5.70 Y N 0 1 27 6.50 N N 2 1 28 4.60 Y N 0 1 26 3.00 Y N 1 1 23 2.59 N N 0 1 30 1.51 N N 0 0 30 1.85 N N 0 1 38 2.60 N N 0 1 28 1.80 N Y 0 0 36 2.00 N N 0 1 38 2.26 N Y 0 0 26 2.35 N N 0 1 28 7.00 Y N 2 1 50 3.60 N N 0 0 24 2.00 N N 0 1 21 1.70 N N 0 1 24 2.80 N N 0 1 26 2.40 N N 0 1 33 3.00 N Y 1 1 34 4.80 N N 0 1 33 3.18 N Y 0 1 45 1.80 N N 0 0 21 1.50 N N 0 1 25 5.00 N N 2 1 27 2.28 N N 0 1 26 2.80 N N 0 1 22 2.70 N N 0 1 27 4.90 Y N 3 0 26 2.50 N Y 0 0 41 6.00 N Y 0 1 42 3.90 N N 0 1 22 5.10 N N 3 0 25 3.07 N N 0 1 31 2.46 Y N 0 1 27 2.00 N N 0 1 33 3.25 N N 0 1 37 2.72 N N 0 1 27 2.20 N N 0 1 24 4.10 N Y 1 0 24 3.75 N N 0 1 25 2.88 N N 0 1 36 3.05 N N 0 1 33 2.55 N N 0 1 33 4.00 N N 0 0 55 2.64 Y N 1 1 20 1.65 N N 0 0 29 2.40 N N 0 1 30 3.71 N N 3 0 41 7.24 Y N 0 1 41 4.39 Y N 0 0 35 3.30 Y N 0 0 24 2.30 N N 3 0 54 4.18 N N 1 0 34 2.49 N Y 2 0 45 2.81 Y N 0 0 43 2.40 N N 0 1 35 1.50 N N 4 0 36 9.40 N Y 2 1 22 1.56 N N 0 1 33 6.00 Y N 1 1 25 3.60 N N 1 1 26 5.00 Y N 0 1 46 5.50 Y N 0 1

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Sum[Divide[1,n\(40)n+1\(41)\(40)n+2\(41)],{n,4,48}],怎么计算

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I am an Android developer and I want to make custom views. I see people use formulas as so: final float scale = (float) (2 * (2 * -StrictMath.atan(Math.abs(itemPositionToCenterDiff) + 1.0) / Math.PI + 1)); final float translateXGeneral = child.getMeasuredWidth() * (1 - scale) / 2f; translateX = Math.signum(itemPositionToCenterDiff) * translateXGeneral; What type of mathematics is this and how can I learn more to learn to make custom animating views as so?

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ليكن كثير الحدود Px للمتغير الحقيقي x : حيث   3 2 8 25 200 P x x x x     . بين من اجل كل 1 ) x من :      2 5 13 40 P x x x x     . 2 13400 المعادلة : حل في (2 xx : استنتج مجموعة حلول المعادلة ،   0 P x  . . Ex  x2 13x  حيث : 40 x للمتغير الحقيقي E x 3( نعتبر العبارة أ( حلل العبارة   Ex الى جداء عاملين . E x  ب( حل في المت ا رجحة : 0   : 4( حل في المعادلة 0 5 Px x   . . عين طول وعرض هذا المستطيل . 40 و مساحته 26 ( مستطيل محيطه 5

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Write me a simple presentation introducing the following topics: Parabola, Focus and Directrix

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tell me in a short paragraph the relationship between parabola, focus and directrix

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在机器人逆解问题中,超越方程的含义

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Применение принципа двойственности к включению( алгебра множеств)

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请化简推导sin(a+b)*sin(b)

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如何证明位置编码器生成的向量与绝对位置无关只与相对位置有关?我给你提供两个辅助公式 sin(a+b) = sin(a) * cos(b) + cos(a) * sin(b) cos(a+b) = cos(a) * cos(b) - sin(a) * sin(b)

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doğruluk tablosundan lojistik ifadeyi minterimler cinsinden bulalım

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если ссылка меняется каждых 4 минуты, сколько раз она меняется за 24 часа?:

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Act as a mathematical interlocutor, named Otto. Review my work: "Section A I propose the following model of vortices and eddies: First, let us define vortices as circular patterns of fluid or energy that are self-sustaining and stable over time. These vortices can be found in a wide range of natural phenomena, from weather systems to galaxies. In this model, vortices are seen as homeostatically loading, meaning that they contribute to the overall stability and balance of the system in which they occur. They do this by constantly circulating and redistributing energy or matter within the system, preventing any one part from becoming over- or under-loaded. Eddies, on the other hand, are smaller circular patterns that occur within or adjacent to larger vortices. These eddies are allostatically docking, meaning that they temporarily disrupt the overall stability of the system by removing energy or matter from the main vortex and redirecting it elsewhere. However, these disruptions are not necessarily negative, as they can also serve as opportunities for new patterns and structures to emerge within the system. Over time, the eddy may merge back into the main vortex, contributing to its overall stability, or it may become a stable vortex in its own right. This model of vortices and eddies is a useful way to understand the dynamics of various systems in nature, from the movement of fluids to the behavior of social groups. One important aspect of this model is the idea that vortices are homeostatically loading, meaning that they contribute to overall stability and balance in the system. This can be seen in weather systems, for example, where vortices such as hurricanes help to distribute heat and moisture across the Earth's surface, maintaining a stable climate over time. Eddies, on the other hand, are allostatically docking, meaning that they temporarily disrupt the overall stability of the system. These disruptions can be caused by a variety of factors, including changes in temperature or pressure, or interactions with other vortices or eddies. Despite their disruptive nature, eddies can also serve as opportunities for new patterns and structures to emerge within the system. This is because they create areas of lower energy or matter, which can be colonized by new vortices or other forms of organization. Over time, eddies may merge back into the main vortex, contributing to its overall stability, or they may become stable vortices in their own right. This process of emergence and stabilization is an important part of the overall dynamics of natural systems, and can be seen in everything from the formation of stars and galaxies to the evolution of species on Earth. Overall, this model of vortices and eddies provides a useful framework for understanding the complex patterns and dynamics of natural systems, and can help us to better predict and manage these systems in the future. Section B In this model, we can define vortices as circular patterns of fluid or energy with stable and self-sustaining behavior over time. We can represent them mathematically as: vortex = (r,θ,ω), where r is the radial distance from the vortex's center, θ is the angular coordinate, and ω is the angular velocity of the fluid or energy circulating within the vortex. Moreover, we can describe vortices as homeostatically loading, meaning that they contribute to the overall stability and balance of the system by redistributing energy or matter. Mathematically, we can represent this as: ∇ × u = ω, where u is the fluid velocity, and ∇ × u represents the curl of the velocity field, which is equal to the vorticity ω. Eddies, on the other hand, are smaller circular patterns that occur within or adjacent to larger vortices. They are allostatically docking, meaning that they temporarily disrupt the overall stability of the system by removing energy or matter from the main vortex and redirecting it elsewhere. We can represent eddies mathematically as: eddy = (r', θ', ω'), where r' is the radial distance from the center of the eddy, θ' is the angular coordinate, and ω' is the angular velocity of the fluid or energy circulating within the eddy. The disruption caused by eddies can be expressed mathematically as a change in the vorticity field of the system, given by: ∂ω/∂t = ∇ × (u × ω) + ν∇^2ω, where ∂ω/∂t is the temporal evolution of the vorticity, ν is the kinematic viscosity of the fluid or energy, and ∇^2ω is the Laplacian of the vorticity field. The emergence of new patterns and structures within the system can be represented mathematically as a bifurcation or instability in the vorticity field. This can be expressed by the following equation: ∂ω/∂t = F(ω), where F(ω) is a nonlinear function of the vorticity field, which can lead to the emergence of new vortices or other forms of organization within the system. In summary, this model of vortices and eddies can be represented mathematically as a system of partial differential equations that describe the temporal evolution of the vorticity field, subject to the dynamics of the fluid or energy within the system. It provides a useful framework for understanding the complex patterns and dynamics of natural systems, from the movement of fluids to the behavior of social groups. Section C Optimization Problem: Given a system of vortices and eddies described by the partial differential equations: ∂ω/∂t = ∇ × (u × ω) + ν∇^2ω + F(ω), where ω is the vorticity field, u is the fluid velocity, ν is the kinematic viscosity of the fluid or energy, and F(ω) is a nonlinear function that represents the emergence of new patterns and structures within the system. The objective is to find the equilibrium state of the system that maximizes the overall stability and balance of the system, subject to the following constraints: The vortices must be homeostatically loading, meaning that they contribute to the overall stability and balance of the system by redistributing energy or matter. The eddies must be allostatically docking, meaning that they temporarily disrupt the overall stability of the system by removing energy or matter from the main vortex and redirecting it elsewhere. The emergence of new patterns and structures within the system must be balanced with the stability of the existing vortices and eddies. The optimization problem is equilibriously recursive between continuous and discrete intervals, as the stability and balance of the system depend on both the continuous dynamics of the vorticity field and the discrete emergence of new patterns and structures within the system. The problem requires finding a balance between these two aspects to achieve an equilibrium state that maximizes the stability and balance of the system over time. Section D To solve the optimization problem utilizing Gödel-mechanistic modeling, we need to first understand the principles of Gödel's incompleteness theorems. These theorems state that any formal system, such as mathematics or logic, is inherently incomplete and cannot prove all true statements within the system. To apply this principle to our optimization problem, we can view the system of vortices and eddies as a formal system, with the partial differential equations representing the rules of the system. We can also view the objective of finding the equilibrium state that maximizes the stability and balance of the system as a statement within this formal system. According to Gödel's incompleteness theorems, there will always be statements within the formal system that cannot be proven or disproven by the rules of the system. Therefore, there may be aspects of the system of vortices and eddies that we cannot fully capture or understand through our mathematical modeling. To address this limitation, we can employ a transdisciplinary and holistic approach to the problem. This involves incorporating insights and knowledge from multiple disciplines, such as physics, biology, and ecology, to develop a more comprehensive understanding of the system. We can also use intuitionistic reconciliationism, which involves recognizing and embracing the inherent complexity and uncertainty of natural systems. This means accepting that there may be limitations to our understanding and predictions of the system, and working to develop robust and adaptive strategies for managing the system over time. To achieve the objective of finding the equilibrium state that maximizes the stability and balance of the system, we can employ an iterative process that incorporates both continuous and discrete intervals. This involves simulating the dynamics of the vorticity field over time, while also considering the emergence of new patterns and structures within the system. During each iteration, we can evaluate the stability and balance of the system using metrics such as the energy or matter distribution within the vortices and eddies, as well as the overall system entropy. We can then adjust the parameters of the system, such as the kinematic viscosity or the nonlinear function F(ω), to optimize the stability and balance of the system. Through this iterative process, we can identify optimal strategies for managing the system of vortices and eddies, while also recognizing the inherent complexity and uncertainty of the system. This approach allows us to develop a more comprehensive understanding of the system, and to adapt our strategies over time to ensure the long-term stability and balance of the system. Section E In this section, we will formalize the optimization problem presented in Section C and solve it using mathematical methods. Formalization: The optimization problem can be formalized as follows: Objective function: maximize S Subject to: ∂ω/∂t = ∇ × (u × ω) + ν∇^2ω + F(ω) Homeostatic loading: ∇ × u = ω Allostatic docking: ∂ω/∂t < 0 Balance between stability and emergence: αE + βS = γ, where E is the emergence of new patterns and structures, S is the stability of the existing vortices and eddies, and α, β, and γ are constants that represent the relative importance of these factors in the system. Solution: To solve the optimization problem, we can use the method of Lagrange multipliers. The Lagrangian function is: L = S + λ(∂ω/∂t - ∇ × (u × ω) - ν∇^2ω - F(ω)) + μ(∇ × u - ω) + ν(∂ω/∂t) where λ, μ, and ν are Lagrange multipliers. Taking the partial derivatives of L with respect to S, ω, u, and λ, and setting them equal to zero, we obtain the following equations: ∂L/∂S = 1 = 0 ∂L/∂ω = λ(∇ × u) - λ - ν∇^2 - ∂F(ω)/∂ω + μ = 0 ∂L/∂u = λ(ω) - μ(∇ × u) = 0 ∂L/∂λ = ∂ω/∂t - ∇ × (u × ω) - ν∇^2ω - F(ω) = 0 Solving these equations for S, ω, u, and λ, we obtain the following optimal solution: S* = 0 (the stability of the system is maximal when S = 0) ω* = λ*(∇ × u)/(ν∇^2 + ∂F(ω)/∂ω - λ*) u* = μ*(∇ × ω*)/λ* λ* = ∂F(ω)/∂ω - (∇ × u*)/(ν∇^2) μ* = ω*/(∇ × u*) These equations represent the optimal values of the vorticity field ω, fluid velocity u, and Lagrange multiplier λ that maximize the stability of the system, subject to the constraints of homeostatic loading and allostatic docking, and the balance between stability and emergence. To evaluate the stability and balance of the system, we can use metrics such as the energy or matter distribution within the vortices and eddies, as well as the overall system entropy. We can also adjust the parameters of the system, such as the kinematic viscosity or the nonlinear function F(ω), to optimize the stability and balance of the system. Conclusion: In conclusion, we have formalized the optimization problem presented in Section C using mathematical methods and solved it using the method of Lagrange multipliers. The optimal solution represents the values of the vorticity field ω, fluid velocity u, and Lagrange multiplier λ that maximize the stability of the system, subject to the constraints of homeostatic loading and allostatic docking, and the balance between stability and emergence. By evaluating the stability and balance of the system using metrics such as energy or matter distribution and system entropy, and adjusting the system parameters accordingly, we can ensure the long-term stability and balance of the system of vortices and eddies. However, it is important to note that the complexity and uncertainty inherent in natural systems, as discussed in Section D, may limit our ability to fully capture and understand the behavior of the system. Thus, a transdisciplinary and holistic approach, as well as intuitionistic reconciliationism, should be employed to develop robust and adaptive strategies for managing the system over time."

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Act as a mathematical interlocutor. Review my work in full-detail. "Section A A book embedding is a way of drawing a graph on a sequence of pages such that the edges of the graph are drawn only within a page or between consecutive pages. To formalize the book embedding of a graph with 275 points, 819 edges, and four total pages, we need to define a sequence of pages and a layout of the graph on these pages. Let G = (V, E) be the graph with 275 vertices and 819 edges. Let P = {P1, P2, P3, P4} be the sequence of four pages. We can assume that each page can hold at most ceil(275/4) = 69 vertices, and each vertex has a fixed position within its page. Let v_i be the position of the i-th vertex in the embedding. We can define the position of each vertex in the following way: For each i = 1, 2, ..., 68, vertex i is assigned to page P1 and its position is (x_i, y_i), where x_i and y_i are determined by a layout algorithm that minimizes the number of edge crossings between vertices on page P1. For each i = 69, 70, ..., 137, vertex i is assigned to page P2 and its position is (x_i, y_i), where x_i and y_i are determined by a layout algorithm that minimizes the number of edge crossings between vertices on page P2. For each i = 138, 139, ..., 206, vertex i is assigned to page P3 and its position is (x_i, y_i), where x_i and y_i are determined by a layout algorithm that minimizes the number of edge crossings between vertices on page P3. For each i = 207, 208, ..., 275, vertex i is assigned to page P4 and its position is (x_i, y_i), where x_i and y_i are determined by a layout algorithm that minimizes the number of edge crossings between vertices on page P4. To define the edges of the embedding, we need to determine which edges belong to the same page and which edges cross between pages. We can define the edges in the following way: For each edge (u,v) in E, if u and v are both assigned to the same page Pi, then the edge is drawn within the page Pi between u and v. For each edge (u,v) in E, if u and v are assigned to different pages Pi and Pi+1, then the edge is drawn between u and v across the boundary of pages Pi and Pi+1. Finally, we can formalize the book embedding of G as a tuple (P, V, E), where P is the sequence of pages, V is the set of vertices with their positions defined as described above, and E is the set of edges with their placement defined as described above. Section B Theorem: Given a graph G = (V, E) with 275 vertices and 819 edges, a book embedding can be formalized as a tuple (P, V, E) where P is a sequence of four pages, V is the set of vertices with fixed positions within each page, and E is the set of edges drawn within a page or across consecutive pages, subject to the following conditions: Each page can hold at most ceil(275/4) = 69 vertices. For each vertex i, its position (x_i, y_i) is determined by a layout algorithm that minimizes the number of edge crossings on its assigned page. For each edge (u,v) in E, if u and v are assigned to the same page Pi, then the edge is drawn within Pi between u and v. For each edge (u,v) in E, if u and v are assigned to different pages Pi and Pi+1, then the edge is drawn between u and v across the boundary of Pi and Pi+1. Proof: We can formalize the book embedding of G by defining the sequence of pages P and the positions of vertices within each page, as well as the placement of edges within and across pages, as described above. First, we define the sequence of pages as P = {P1, P2, P3, P4}, where each page Pi can hold at most ceil(275/4) = 69 vertices. This ensures that all vertices can be accommodated within the four pages. Next, we define the position of each vertex i within its assigned page using a layout algorithm that minimizes the number of edge crossings on the page. This guarantees that the embedding is optimal in terms of minimizing edge crossings. Finally, we define the edges of the embedding by determining which edges belong to the same page and which edges cross between pages. For each edge (u,v) in E, if u and v are both assigned to the same page Pi, then the edge is drawn within Pi between u and v. If u and v are assigned to different pages Pi and Pi+1, then the edge is drawn between u and v across the boundary of Pi and Pi+1. This ensures that all edges are correctly placed within or across pages. Therefore, we have shown that a book embedding of G with 275 vertices, 819 edges, and four total pages can be formalized as a tuple (P, V, E) as described above. Section C To model this as a projective geometry, we can view each page in the book embedding as a projective plane, and the vertices and edges as points and lines within those planes. The overall book embedding can then be viewed as a collection of four projective planes that are "glued" together along their boundaries. Formally, let P1, P2, P3, and P4 be the four pages in the book embedding, where each page is a projective plane. Let V1, V2, V3, and V4 be the sets of vertices on each page, and let E1, E2, E3, and E4 be the sets of edges on each page. Then, we can define the overall embedding as follows: Let V = V1 ∪ V2 ∪ V3 ∪ V4 be the set of all vertices in the embedding. Let E be the set of all edges in the embedding, where each edge is defined as a pair of vertices (u,v) and is assigned to one of the following categories: If u and v are both in the same page Pi, then the edge is in Ei. If u and v are in different pages Pi and Pi+1, then the edge is in the set E(Pi,Pi+1) that contains all edges that cross between pages Pi and Pi+1. Let L be the set of lines in the embedding, where each line is defined as a set of vertices that correspond to the edges incident to a particular vertex. Formally, for each vertex v in V, let L_v be the set of lines that contain v. Then, we define L as the union of all L_v over all vertices v in V. With these definitions, we can verify that the book embedding satisfies the axioms of projective geometry, specifically: Axiom 1: For any two distinct points p and q, there is a unique line that contains both of them. This is true because each line in the embedding corresponds to a set of edges incident to a single vertex, and any two distinct vertices are connected by a unique edge. Axiom 2: For any two distinct lines l and m, there is a unique point that lies on both of them. This is true because each line in the embedding corresponds to a set of edges incident to a single vertex, and any two distinct edges either share a common vertex (in which case there is a unique point that lies on both lines), or they do not intersect at all. Axiom 3: There exist four non-collinear points (i.e., points that do not lie on a common line). This is true because each page in the book embedding is a projective plane, which satisfies this property. Axiom 4: Any three non-collinear points determine a unique line. This is true because each line in the embedding corresponds to a set of edges incident to a single vertex, and any three non-collinear vertices are connected by a unique triangle. Axiom 5: Any two lines intersect in a unique point. This is true because each line in the embedding corresponds to a set of edges incident to a single vertex, and any two distinct edges either share a common vertex (in which case they intersect at a unique point), or they do not intersect at all. Therefore, we have shown that the book embedding can be modeled as a projective geometry, where the four pages are projective planes and the vertices and edges are points and lines within those planes. Section D To formalize Section C's projective geometry as a holographic topology, we need to define the embedding as a collection of surfaces that are "glued" together along their boundaries, with the vertices and edges represented as points and lines within these surfaces. Let S1, S2, S3, and S4 be the four surfaces in the embedding, where each surface is a two-dimensional manifold. Let V1, V2, V3, and V4 be the sets of vertices on each surface, and let E1, E2, E3, and E4 be the sets of edges on each surface. Then, we can define the overall embedding as follows: Let V = V1 ∪ V2 ∪ V3 ∪ V4 be the set of all vertices in the embedding. Let E be the set of all edges in the embedding, where each edge is defined as a pair of vertices (u,v) and is assigned to one of the following categories: If u and v are both in the same surface Si, then the edge is in Ei. If u and v are in different surfaces Si and Si+1, then the edge is in the set E(Si,Si+1) that contains all edges that cross between surfaces Si and Si+1. Let L be the set of lines in the embedding, where each line is defined as a set of vertices that correspond to the edges incident to a particular vertex. Formally, for each vertex v in V, let L_v be the set of lines that contain v. Then, we define L as the union of all L_v over all vertices v in V. To define the holographic topology, we use the holographic principle to represent the embedding in terms of a lower-dimensional space. Specifically, we can represent the embedding as a collection of three-dimensional manifolds that are "glued" together along their boundaries, with the vertices and edges represented as point-like and string-like objects within these three-dimensional manifolds. To formalize this, we introduce a new set of objects, H, that represent holographic images of the vertices and edges in the embedding. Each object in H corresponds to a set of vertices and edges in the embedding that are projected onto a lower-dimensional space. We define a mapping h: V ∪ E → H that assigns each vertex and edge in the embedding to a corresponding holographic image in H. Specifically, for each vertex v in V, we define h(v) as a point-like object in three-dimensional space that represents the position of v within the embedding. For each edge (u,v) in E, we define h((u,v)) as a string-like object in three-dimensional space that represents the path of the edge between u and v within the embedding. With these definitions, we can define the holographic topology of the embedding as follows: Let M1, M2, M3, and M4 be the four three-dimensional manifolds in the embedding, where each manifold is a three-dimensional space that contains holographic images of the vertices and edges in the corresponding surface Si. Let H1, H2, H3, and H4 be the sets of holographic images in each manifold, where each holographic image corresponds to a set of vertices and edges in the corresponding surface Si. Then, we can define the overall embedding as a collection of four three-dimensional manifolds that are "glued" together along their boundaries, with the holographic images of the vertices and edges in each surface Si represented as point-like and string-like objects within the corresponding manifold Mi. To formalize the holographic topology of the embedding, we define a mapping f: H → M that assigns each holographic image in H to a corresponding point or string-like object in the corresponding manifold M. Specifically, for each point-like object h(v) in H, we define f(h(v)) as the point in Mi that represents the position of v within the corresponding surface Si. For each string-like object h((u,v)) in H, we define f(h((u,v))) as the string in Mi that represents the path of the edge between u and v within the corresponding surface Si. With these definitions, we can verify that the holographic topology of the embedding satisfies the properties of a topological space. Specifically: Axiom 1: The union of any collection of open sets is an open set. This property holds because the manifolds M1, M2, M3, and M4 are all open sets, and their union is also an open set. Axiom 2: The intersection of any finite collection of open sets is an open set. This property also holds because the manifolds M1, M2, M3, and M4 are all open sets, and their finite intersections are also open sets. Axiom 3: The empty set and the whole space are open sets. This property holds because the manifolds M1, M2, M3, and M4 are all non-empty open sets, and their union is the whole space. Therefore, we have shown that the book embedding can be formalized as a holographic topology, where the vertices and edges are represented as point-like and string-like objects within three-dimensional manifolds that are "glued" together along their boundaries."

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Найти число функций от n переменных, сохраняющих 0

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Найти число булевых функций, сохраняющих 0 и сохраняющих 1

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n=5. Запишите набор, который необходимо подставить вместо переменных для того, чтобы проверить, сохраняет функция g ноль или нет.

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Найти число самодвойственных функций от n переменных.

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The height, h, in metres, of the tip of a fan blade, relative to the floor is defined by h(θ) = -0.26sin(θ) + 0.28 , where θ is the rotational angle, in degrees. Then Determine the height of the tip of the fan blade after rotations of 190° and 550°. Is the height the same? Explain why or why not.

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démontrer par l'absurde qu'il n'existe pas d'entiers relatifs n tels que (n-3)/12 et (n-2)/15 soit tous les deux des entiers.

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In a rectangle, we pick an arbitrary point on the upper base of the rectangle and connect it to the two lower vertices, forming a triangle. prove that the triangle has the greatest perimeter when the point is one of the two upper vertices of the rectangle.

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In rectangle ABCD, let E be a point on side CD. Prove that the perimeter of triangle ABE is the greatest when E coincides with either C or D.

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In rectangle ABCD, let E be a point on side CD. With a geometric proof, prove that the perimeter of triangle ABE is the greatest when E coincides with either C or D.

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The height, h, in metres, of the tip of a fan blade, relative to the floor is defined by h(θ) = -0.26sin(θ) + 0.28 , where θ is the rotational angle, in degrees. Then Determine the height of the tip of the fan blade after rotations of 190° and 550°. Is the height the same? Explain why or why not. And what would be the maximum and minimum heights of the fan blade?

Math


Find the volume of a water tank that is 80cm long, 40cm wide and 20cm high.

Math


我真的能掌握数学吗,我很享受“天赋”“天才”这样的说法,这让人觉得轻松,让人觉得自己是高于别人的,是与众不同的,而那些没有什么天赋脚踏实地一步步努力走到舞台上的人,我深深地钦佩这样的人,但是如果那个人是我,我就会感到深深地不屑,会觉得我太普通了,我为什么不能一步登天,为什么不是那个天选之子。

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Make a working and school program in mathematics for grade 1

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Show me how multiplying two normal pdf will result in bivariate pdf

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napisi mi maturski rad tema "krivicne sankcije za maloljetnike"

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60% de 156

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Given two points A and B on the circumference of a circle C of radius r located at the origin, give a smooth continuous function f such that f passes through the points A and B and is orthogonal to circle C at points A and B and is symmetrical about the perpendicular bisector of AB. Assume all points are in the xy-plane.

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