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hw2_answers

# hw2_answers - Instructor Prof Doyoon Kim TA Diogo Bessam...

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Unformatted text preview: Instructor: Prof. Doyoon Kim TA: Diogo Bessam Please, during (and only during) Fall 2008, report any typo/error you may nd directly to Diogo Bessam ([email protected]) 2.1.36 HW2 answers, math118, 2008 (a) h(40) = 400 feet (b) (h(40) - h(0))/40 = 10 feet/second (c) h (40) = -10 feet/second; negative sign indicates falling (a) Qaver = Q(3,100)-Q(3,025) 3,100-3,025 2.1.40 aprox. 28 (b) Q (3, 025) = 28, 2 2.2.24 2.2.36 2.2.48 5x4 - 18x2 + 14x y - 4 = 2(x - 4) or y = 2x - 4 (a) N (8) = 1652 users per week (b) N (8) - N (7) = 1, 514 riders (a) v(t) = 3t2 - 18t + 15;a(t) = 6t - 18 (b) t = 1 and t = 5 y = -14x ou y = 6x f (t) = t2 -4t-5 (t2 +3t-1)2 2.2.66 2.2.72 2.3.14 2.3.18 2.3.26 2.3.33 2.3.44 2.3.62 f (x) = 2x - 2/x3 y - (-1) = x(x - 0 or y = 2x - 1 book -5 4x3/2 + 18 x4 + 1 4x5/2 (a) a(t) = v (t) = 20/3 - 4/3t (b) a(6) = -4/3 decreasing at a rate of aprox. 1.33 kilometers per hour (c) v(7) - v(6) = -2 km per hour 1/18 3x g (x) = - 6x-3x+1 2 2.4.16 2.4.28 2.4.36 2.4.40 2.4.54 2.4.56 f (x) = -2(25x2 +45x+1) 3(3+2x)4/3 y - 8 = -6(x - 3) or y = -6x + 26 y = 48x(1 - 2x3 )2 (11x3 - 1) f (u) = 12(15u2 +1) (3u2 -1)4 2.4.66 (a) Q(4) = 3 (no units!), p(3) = 239 thousand 2 (b) Q (t) = 2t +2t-4 , Q (4) = 4/9 (2t+1)2 (c) p (4) = p (Q)|Q=3 Q (t)|t=4 = 22(4/9), increasing at aprox. 9.778 per year +6x+18 (a) C (x) = 4/7x, R (x) = 2 x (x+3)2 2 2.5.6 (b) (c) (d) (e) 2.5.12 C (3) = 12/7 C(4) - C(3) = 2 R (3) = 2.50 R(4) - R(3) = 2.43 (a) R (80) = 232 dollars is a good approximation to the additional revenue of increasing one unit when at 80 (b) R(81) - R(80) = 231.95 dollars End. ...
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