Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 423: 10



Work Step by Step

To find out the solution we need to follow the four-step procedure: Step (1) In the question it is provided that $y$ varies directly as $m$ , ${{n}^{2}}$ and varies inversely as $p$. Thus, we have $y=k\frac{m{{n}^{2}}}{p}$. Here $k$ is a constant. Step (2) Find out the value of $k$. Substitute the values $m=2,\ n=1$ and $p=6$ then $y=15$ in $y=k\frac{m\cdot {{n}^{2}}}{p}$. $\begin{align} & 15=k\frac{2\times 1}{6} \\ & k=\frac{15\times 6}{2} \\ & k=45 \\ \end{align}$ Step (3) Substitute the value of $k$ in $y=k\frac{m\cdot {{n}^{2}}}{p}$. $y=\left( 45 \right)\frac{m\cdot {{n}^{2}}}{p}$ Step (4) In this step, substitute the values $m=3,n=4,p=10$. $\begin{align} & y=45\times \frac{3\times 16}{10} \\ & y=\frac{9\times 3\times 16}{2} \\ & y=216 \\ \end{align}$
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