Jawaban tugas kalkulus SF
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The document provides answers to homework problems about minimizing energy loss in a branching pipe system: 1) It expresses the total loss L as a function of the branching angle θ in terms of the system parameters. 2) It finds the critical value of θ that minimizes L by taking the derivative and setting it equal to 0. 3) Given a ratio of pipe radii, it estimates the optimal branching angle to the nearest degree.
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Here are the steps to solve these problems using Euclid's division algorithm:
1) Find the HCF of 135 and 225:
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135 = 90 * 1 + 45
90 = 45 * 2 + 0
Therefore, the HCF of 135 and 225 is 45.
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Therefore, the HCF of 196 and 38220 is 196.
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Here are the steps to solve these problems using Euclid's division algorithm:
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Real numbers - Euclid’s Division Algorithm for class 10th/grade X maths 2014
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Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring.
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Answer True/False and explain for each part. (Short Answer) 1. Say that w...
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Real numbers - Euclid’s Division Algorithm for class 10th/grade X maths 2014
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Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring.
Our Mission-
Our aspiration is to be a renowned unpaid school on Web-World
Maths project
Euclid's Division Lemma states that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. An algorithm is a series of well-defined steps to solve a problem. The word algorithm comes from the name of mathematician al-Khwarizmi. Euclid's algorithm uses the division lemma to find the highest common factor (HCF) of two integers by repeatedly dividing the larger number by the smaller number.
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The document describes the Knuth-Morris-Pratt (KMP) string matching algorithm. KMP finds all occurrences of a pattern string P in a text string T. It improves on the naive algorithm by not re-checking characters when a mismatch occurs. This is done by precomputing a function h that determines how many characters P can skip ahead while still maintaining the matching prefix. With h, KMP ensures each character is checked at most twice, giving it O(m+n) time complexity where m and n are the lengths of P and T.
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2) Hilbert discusses extending Kronecker's theorem that every abelian number field arises from rational numbers by composing fields of roots of unity. He believes extending this theorem to imaginary quadratic number fields can be proved using Weber's theory of complex multiplication.
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This document discusses a method for determining if Goldbach's conjecture is true for a given even number 2n. It presents a formula (formula 3) that checks if two prime numbers p and q exist such that their sum is 2n. The formula uses sine functions and multiplications to test all possible combinations of p and q between 2 and 2n-2. An example application of the formula in a computer program finds a solution, providing evidence for Goldbach's conjecture being true for that even number. The document concludes by stating the remaining step is a formal proof that the formula always produces a valid q.
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The document describes and analyzes two algorithms for finding the convex hull of a set of points: a brute force algorithm and a divide and conquer algorithm.
The brute force algorithm iterates through all points three times, checking all possible line combinations, resulting in O(n3) time complexity.
The divide and conquer algorithm recursively divides the point set into halves at each step by finding the furthest point from the current left-right boundary line. It has O(n log n) time complexity.
An experiment comparing runtimes on sample point sets showed the divide and conquer approach was significantly faster than the brute force approach.
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The Matlab residue command performs partial fraction expansion by decomposing the ratio of two polynomials into partial fractions. It returns the residues, poles, and direct term coefficients. The residues are the coefficients of the terms with denominators of (z-pole), poles are the locations of the singularities, and the direct term contains coefficients of terms without poles. The command works for cases with real or complex poles, repeated poles, and when the numerator degree exceeds the denominator degree.
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Unit 3
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Single Variable Calculus Assignment Help
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- 1. Answer of Calculus Home Work for School of Pharmacy Dadang Amir Hamzah May 7, 2013 a. Express the total loss L as a function of θ, in term of a, b, R, r and θ. Answer: From the figure we know that d1 = a − d2 cos θ and d2 = b sin θ . So the total loss L = k d1 R4 + d2 r4 expressed in term of a, b, R, r and θ is L = k a − b cot θ R4 + b csc θ r4 . b. Find the Critical value of θ such that the energy loss is minimal. Answer: The critical value of θ exceed when L (θ) = 0. From answer of point a L (θ) = k csc2 θ b R4 − b r4 cos θ . The critical value of θ is the value of θ which satisfied k csc2 θ = 0, , cos θ = r4 R4 . So the value of θ is θ = cos−1 r4 R4 . c. If the ratio of the pipe radii is r/R = 5/6. estimate to the nearest degree the optimal branching angle given in part b. Answer: If we substitute the value of r = 5 6 R to the answer of point b we get the value of θ which minimized the loss function L that is θ = cos−1 r4 R4 = cos−1 5 6 4 ≈ 60.49o 1
- 2. d. Find the critical value of θc such that the cost of maintaining loss is minimal, which is not the same as the value θ that minimizes the previous functional L. Answer: First lets express Lc in term of a, b, R, r and θ. Thus we get Lc(θ) = K (a − b cot θ) R2 + b sin θ r2 . By the same argument as previous problem we get Lc(θ) = K b csc2 θ R2 − r2 cos θ . The critical value of θ is the value of θ which satisfied K b csc2 θ = 0 , R2 − r2 cos θ = 0. So the possible value of theta is θ = cos−1 R r 2 . e. We can combine these functionals, L and Lc, into one functional Lt = L + Lc. Find the critical value θ that minimizes Lt. Answer: After we combine the previous answer we can get Lt in term of a, b, R, r and θ. The critical value of Lt exceeds when Lt(θ) = 0. By differenti- ating Lt(θ) we get Lt(θ) = csc2 θ kb R4 + KR2 b − kb r4 + kbr2 cos θ . The value of θ which minimizes Lt is the value of θ which satisfied csc2 θ = 0 , kb R4 + KR2 b − kb r4 + kbr2 cos θ = 0. The possible value of θ is θ = cos−1 r R 4 kb + KbR6 kb + Kbr6 . f(i). By taking lim r→R cos−1 r R 4 kb + KbR6 kb + Kbr6 = cos−1 (1) so we get θ = 0. 2
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