Answer
$U(5)=18.62$
$U(10)=25.1$
Work Step by Step
$U(t)=T+(U_0-T)e^{kt},$
$T=35, U_0=8$
$U(3)=35+(8-35)e^{3k}=15,$
$-27e^{3k}=-20,$
$e^{3k}=0.741,$
$3k=\ln{0.741},$
$k=-0.1$
$U(5)=35-27e^{-0.5}=18.62$
$U(10)=35-27e^{-1}=25.1,$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Section 6.8 - Exponential Growth and Decay Models; Newton's Law: Logistic Growth and Decay Models - 6.8 Assess Your Understanding - Page 486: 15
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