College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Review Exercises - Page 380: 89



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ t=\dfrac{k}{s} $ and solve for the value of $k$ with the given $ t $ and $ s $ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $t$ varies inversely as $s,$ then $ t=\dfrac{k}{s} .$ Substituting the given values, $ t=3 $ and $ s=5 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} t=\dfrac{k}{s} \\\\ 3=\dfrac{k}{5} \\\\ 5(3)=\left(\dfrac{k}{5}\right)5 \\\\ 15=k .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} t=\dfrac{k}{s} \\\\ t=\dfrac{15}{s} .\end{array} If $t=20,$ then \begin{array}{l}\require{cancel} t=\dfrac{15}{s} \\\\ 20=\dfrac{15}{s} \\\\ s(20)=\left(\dfrac{15}{s}\right)s \\\\ 20s=15 \\\\ s=\dfrac{15}{20} \\\\ s=\dfrac{\cancel{5}\cdot3}{\cancel{5}\cdot4} \\\\ s=\dfrac{3}{4} .\end{array}
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