## College Algebra (11th Edition)

$a=\dfrac{18}{125}$
$\bf{\text{Solution Outline:}}$ 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}