Question

Please help with Calc AB

Answer 1

Answer:

Step-by-step explanation:

36.

p(t)=(2t-3)e^{2-t}

p(1)=(2*1-3)e^{2-1}=-e

p(3)=(2*3-3)e^{2-3}=3e^{-1}

$average~velocity=\frac{3e^{-1}-(-e) }{3-1} \\=\frac{3+e^{2} }{2e} \\\approx 1.9109\\\approx 1.91$

37.

$p=(2t-3)e^{2-t} \\v=\frac{dp}{dt}\\=(2t-3)e^{2-t}(-1) +2e^{2-t} \\=(5-2t)e^{2-t}$

d(u v)=u dv+v du