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Calculus

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MasterMathMentor.com!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!"!Q#!"!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Stu Schwartz Function Analysis - Classwork =*!+70!,1'+!,7!&+&(.S)+:!61+5,)7+2!A)&!5&(51(12;!=*!3)3!27!)+!D'*5&(51(12!4.!3*,*'9)+)+:!,-*!S*'72!76!,-*!61+5,)7+ X0-*'*!),!5'722*2!,-*!x"&E)2G!&+3!,-*!2):+!76!,-*!61+5,)7+!4*,0**+!S*'72;!L-&,!:&A*!12!279*!)+67'9&,)7+!&471,!,-* 61+5,)7+!41,!(),,(*!5(1*!&2!,7!),2!&5,1&(!2-&D*;!@&(51(12!0)((!D'7A)3*!,-*!9)22)+:!D)*5*2!,7!,-*!D1SS(*; =*!0)((!3*6)+*!&!61+5,)7+!&2!)+5'*&2)+:!)6/!&2!x!97A*2!,7!,-*!'):-,/!,-*!y"A&(1*!:7*2!1D;!N!61+5,)7+!)2!3*5'*&2)+:!)6/ &2!x!97A*2!,7!,-*!'):-,/!,-*!y"A&(1*!:7*2!370+;!N!61+5,)7+!)2!57+2,&+,!)6/!&2!x!97A*2!,7!,-*!'):-,/!,-*!y"A&(1* 37*2+`,!5-&+:*;

disc. math 2

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rosen 7e solutions

CS 330 : Discrete Computational Structures Fall Semester, 2013 Assignment #1 Due Date: Friday, Sept 6 Suggested Reading: Rosen Sections 1.1 - 1.3; Lehman et al. Sections 1.1, 3.1 - 3.4 These are the problems that you need to hand in for grading. For more practice, you are encouraged to work on the other problems. Always explain your answers and show your reasoning. 1. [6 Pts] Zhenbi Hu Rosen, Section 1.1: Exercise 12 (a) (c) (e) Solution: (a) If you have the flu, then you miss the final examination. (c) If you miss the final examination, then you cannot pass the course. (e) Either if you have the flu, then you cannot pass the course, or if you miss the final examination, then you cannot pass the course. 2. [8 Pts] Zhenbi Hu Rosen, Section 1.1: Exercise 14 (b) (c) (e) (f)

comp sci

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discrete math

CS 330 : Discrete Computational Structures Fall Semester, 2013 Assignment #5 Due Date: Friday, Oct 11 Suggested Reading: Rosen Section 5.1 - 5.2; Lehman et al. Chapter 5.1 - 5.3 These are the problems that you need to turn in. For more practice, you are encouraged to work on the other problems. Always explain your answers and show your reasoning. 1. Rosen, Section 5.1: Exercise 4 2. Rosen, Section 5.1: Exercise 6 3. Rosen, Section 5.1: Exercise 10 4. Rosen, Section 5.1: Exercise 16 5. Rosen, Section 5.1: Exercise 20 6. Rosen, Section 5.1: Exercise 32 7. Rosen, Section 5.1: Exercise 44 8. Let P (n) be the statement that n-cent postage can be formed using just 3-cent and 8-cent stamps. Prove that P (n) is true for all n ? 14, using the steps below.

Ya Mama

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M. C. Carter Ms. Ruggiero Pd. 8 Themes and Motifs Introduction?????????????????. THEMES Keeping Your Religion

Calculus challenge 9 solution

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Calculus Challenge #9 Solution Houdini?s Escape Houdini plans to have his feet shackled on the top of a concrete block which was placed on the bottom of a giant flask. The cross-sectional radius of the flask, measured in feet, is given as a () 10 function of the height y from the ground by the formula ry = , with the bottom of the flask y at y = 1 foot. The flask is to be filled with water at a constant rate of 22pcubic feet per minute. Houdini?s job is to escape the shackles before he drowns! Houdini knows that he can escape the shackles in exactly 10 minutes. For dramatic effect, he wants to escape at the moment the water level reaches the top of his head. Houdini is 6 feet tall.

Ch9 SG

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372 CHAPTER 9 Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y? = 2x2ex 3/3 = x2y and y(0) = 2 by inspection. 2. y? = x3 ? 2 sinx, y(0) = 3 by inspection. 3. (a) ?rst order; dy dx = c; (1 + x) dy dx = (1 + x)c = y (b) second order; y? = c1 cos t? c2 sin t, y?? + y = ?c1 sin t? c2 cos t+ (c1 sin t+ c2 cos t) = 0 4. (a) ?rst order; 2 dy dx + y = 2 ( ? c 2 e?x/2 + 1 ) + ce?x/2 + x? 3 = x? 1 (b) second order; y? = c1et ? c2e?t, y?? ? y = c1et + c2e?t ? ( c1et + c2e?t ) = 0 5. 1 y dy dx = x dy dx + y, dy dx (1? xy) = y2, dy dx = y2 1? xy 6. 2x+ y2 + 2xy dy dx = 0, by inspection. 7. (a) IF: ? = e3 ? dx = e3x, d dx [ ye3x ] = 0, ye3x = C, y = Ce?3x separation of variables: dy y = ?3dx, ln |y| = ?3x+ C1, y = ?e?3xeC1 = Ce?3x

Ch8 SG

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Sum rule in integration

317 CHAPTER 8 Principles of Integral Evaluation EXERCISE SET 8.1 1. u = 3? 2x, du = ?2dx, ? 1 2 ? u3 du = ?1 8 u4 + C = ?1 8 (3? 2x)4 + C 2. u = 4 + 9x, du = 9dx, 1 9 ? u1/2 du = 2 3 ? 9u 3/2 + C = 2 27 (4 + 9x)3/2 + C 3. u = x2, du = 2xdx, 1 2 ? sec2 u du = 1 2 tanu+ C = 1 2 tan(x2) + C 4. u = x2, du = 2xdx, 2 ? tanu du = ?2 ln | cosu |+ C = ?2 ln | cos(x2)|+ C 5. u = 2 + cos 3x, du = ?3 sin 3xdx, ? 1 3 ? du u = ?1 3 ln |u|+ C = ?1 3 ln(2 + cos 3x) + C 6. u = 3x 2 , du = 3 2 dx, 2 3 ? du 4 + 4u2 = 1 6 ? du 1 + u2 = 1 6 tan?1 u+ C = 1 6 tan?1(3x/2) + C 7. u = ex, du = exdx, ? sinhu du = coshu+ C = cosh ex + C 8. u = lnx, du = 1 x dx, ? secu tanu du = secu+ C = sec(lnx) + C 9. u = cotx, du = ? csc2 xdx, ? ? eu du = ?eu + C = ?ecot x + C

precalc test chap 22

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