College Algebra (11th Edition)

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

Chapter 3 - Section 3.6 - Variation - 3.6 Exercises - Page 365: 12



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ y=kx $ and solve for the value of $k$ with the given $x$ and $y$ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $y$ varies directly as $x$, then $y=kx.$ Substituting the given values, $ y=9 $ and $ x=30 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} 9=k(30) \\\\ \dfrac{9}{30}=k \\\\ k=\dfrac{3}{10} .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} y=kx \\\\ y=\dfrac{3}{10}x .\end{array} If $x=40,$ then \begin{array}{l}\require{cancel} y=\dfrac{3}{10}x \\\\ y=\dfrac{3}{10}\cdot40 \\\\ y=\dfrac{3}{\cancel{10}}\cdot\cancel{10}(4) \\\\ y=12 .\end{array}
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