Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 7 - Section 7.3 - Direct Variation - Exercises - Page 285: 52

Answer

$25.98\;gal$.

Work Step by Step

Given values are Distance $d_1=495\;mi$. Amount of gasoline $g_1=14.1\;gal$. Distance $d_2=912\;mi$. Amount of gasoline $g_2$. Let the rate is $r=\frac{d}{g}$ rate is same. $r=\frac{d_1}{g_1}=\frac{d_2}{g_2}$ Substitute all the values. $\Rightarrow \frac{495\;mi}{14.1\;gal}=\frac{912\;mi}{g_2}$ Simplify. $\Rightarrow (912\;mi)(\frac{14.1\;gal}{495\;mi})=g_2$ $\Rightarrow 25.98\;gal=g_2$.
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