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: 14

Answer

$m=36$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ m=kzp $ and solve for the value of $k$ with the given $m,x$ and $y$ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $m$ varies jointly as $z$ and $p$, then $m=kzp.$ Substituting the given values, $ m=10,z=2, $ and $ p=7.5 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} m=kzp \\\\ 10=k(2)(7.5) \\\\ 10=k(15) \\\\ \dfrac{10}{15}=k \\\\ k=\dfrac{2}{3} .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} m=kzp \\\\ m=\dfrac{2}{3}zp .\end{array} If $z=6$ and $p=9,$ then \begin{array}{l}\require{cancel} m=\dfrac{2}{3}zp \\\\ m=\dfrac{2}{3}(6)(9) \\\\ m=\dfrac{2}{\cancel3}\cancel3(2)(9) \\\\ m=2(2)(9) \\\\ m=36 .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.