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: 13

Answer

$m=60$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ m=kxy $ 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 $x$ and $y$, then $m=kxy.$ Substituting the given values, $ m=10,x=2, $ and $ y=14 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} m=kxy \\\\ 10=k(2)(14) \\\\ 10=k(28) \\\\ \dfrac{10}{28}=k \\\\ k=\dfrac{5}{14} .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} m=kxy \\\\ m=\dfrac{5}{14}xy .\end{array} If $x=21$ and $y=8,$ then \begin{array}{l}\require{cancel} m=\dfrac{5}{14}xy \\\\ m=\dfrac{5}{14}(21)(8) \\\\ m=\dfrac{5}{\cancel7(2)}\cancel7(3)(8) \\\\ m=\dfrac{5}{2}(3)(8) \\\\ m=\dfrac{5}{\cancel2}(3)\cancel2(4) \\\\ m=5(3)(4) \\\\ m=60 .\end{array}
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