Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 114: 99


$P \approx 1481.64$

Work Step by Step

To find the present value for an interest rate $r=0.06$ compounded continuously for $t=5$ years $A=Pe^{rt}$ $2000=Pe^{0.06(5)}$ $P = \frac{2000}{e^{0.06(5)}} \approx 1481.64$
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