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 366: 19



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ a=k\cdot\dfrac{mn^2}{y^3} $ and solve for the value of $k$ with the given $ a,m,n $ and $ y $ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $a$ is directly proportional to $m$ and $n^2$ and inversely proportional to $y^3,$ then $ a=k\cdot\dfrac{mn^2}{y^3} .$ Substituting the given values, $ a=9, m=4,n=9 $ and $ y=3 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} a=k\cdot\dfrac{mn^2}{y^3} \\\\ 9=k\cdot\dfrac{(4)(9)^2}{(3)^3} \\\\ 9=k\cdot\dfrac{(4)(81)}{27} \\\\ 9=k\cdot\dfrac{(4)(\cancel{81}(3))}{\cancel{27}} \\\\ 9=k\cdot12 \\\\ \dfrac{9}{12}=k \\\\ k=\dfrac{3}{4} .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} a=k\cdot\dfrac{mn^2}{y^3} \\\\ a=\dfrac{3}{4}\cdot\dfrac{mn^2}{y^3} \\\\ a=\dfrac{3mn^2}{4y^3} .\end{array} If $m=6,n=2,$ and $y=5,$ then \begin{array}{l}\require{cancel} a=\dfrac{3(6)(2)^2}{4(5)^3} \\\\ a=\dfrac{3(6)(4)}{4(125)} \\\\ a=\dfrac{3(6)(\cancel4)}{\cancel4(125)} \\\\ a=\dfrac{18}{125} .\end{array}
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