MATH 151 Midterm: MATH 151 TAMU Y2015 2015c X3H Solutions
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Wed, 09/dec c(cid:13)2015 art belmonte: with ln(cid:0)x2 2x 2(cid:1) 0, the solve command yields. 1 x < 1 3 or 1 + 3 < x 3. 1 y y = x sec2 x tan x 1 ln(tan x) x2. & y = x cot x sec2 x ln(tan x) x2 tan x1/x: let y be the mass at time t. then y = y0ekt. Given a half-life of 28 days, we have 25 = 50e28k. 30 ln 22 (a) for q0 = 185 and ta = 75, q = 110ekt + 75. So 150 = 110e30k + 75 gives k = 1. 22 + 75 136. 93 f. (b) when the turkey has cooled to 100 f, we have. 100 = 110(cid:0) 15 t = 30 ln(22/5) ln(22/15) 116 minutes. 4 . x = 0 is y = x: with y = tan 1(cid:16)x 1 + x2(cid:17), the tangent line at, recall y = sin(cid:0)2 cos 1 x(cid:1).
