Unformatted text preview: Math 127, Fall 2010 Assignment 7 solutions Page 1 . Answers to Part A problems: Question 1a 1b 1c 1d 1e 2a 2b 3 4 5 6 1. Answer 120 (ln 2/5)t t/2 120e √ or 120(5) 3000 5 ≈ 6708 60(ln 5)e5(ln 5)/2 ≈ 5398 (2/ ln 5) ln(5000/3) ≈ 9.2 hours − ln 2/ ln(0.945) ≈ 12.25 years − ln 5/ ln(0.945) ≈ 28.45 years (1 + x)−1/2 ≈ 1 − 1 x 2 θ = π/3 + 2nπ , 5π/3 + 2nπ , 4π/3 + 2nπ max = 19, min = 1 max = 2ln 2, min=1 # marks (total 10 + 1 bonus) 1 1 1 1 1 1 1 1 1 1 1 Solutions to Part A problems: Math 127, Fal...
View
Full Document
 Spring '10
 RonaldGreen
 Calculus, Mean Value Theorem, Mean

Click to edit the document details