College Algebra (11th Edition)

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

Chapter 3 - Review Exercises - Page 380: 88

Answer

$z=\dfrac{1}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ z=\dfrac{k}{w} $ and solve for the value of $k$ with the given $ z $ and $ w $ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $z$ varies inversely as $w,$ then $ z=\dfrac{k}{w} .$ Substituting the given values, $ z=10 $ and $ w=\dfrac{1}{2} ,$ then the value of $k$ is \begin{array}{l}\require{cancel} 10=\dfrac{k}{\dfrac{1}{2}} \\\\ 10=k\div\dfrac{1}{2} \\\\ 10=k\cdot2 \\\\ \dfrac{10}{2}=k \\\\ k=5 .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} z=\dfrac{k}{w} \\\\ z=\dfrac{5}{w} .\end{array} If $w=10,$ then \begin{array}{l}\require{cancel} z=\dfrac{5}{w} \\\\ z=\dfrac{5}{10} \\\\ z=\dfrac{1}{2} .\end{array}
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