Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - 5.5 Activities - Page 373: 11

Answer

$T(x)=200 (\frac{0.93^x}{\ln 0.93})+CDVDs $

Work Step by Step

$t(x)=200(0.93^x)DVDs $per week $T(x)=\int200(0.93^x)dx $ $T(x)=200 \int (0.93^x)dx $ Using the formula $\int a^x=\frac{a^x}{\ln a}$ $T(x)=200 (\frac{0.93^x}{\ln 0.93})+CDVDs $
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