Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.5 - Compound Inequalities - Exercise Set - Page 96: 92

Answer

$14^oF \text{ to } \left( \dfrac{322}{5} \right)^oF$

Work Step by Step

Using $C=\dfrac{5}{9}(F-32),$ then the equivalent range of Fahrenheit values for $-10^oC$ to $18^oC$ is \begin{array}{l}\require{cancel} -10\le\dfrac{5}{9}(F-32)\le18 .\end{array} Using the properties of inequality, then \begin{array}{l}\require{cancel} \dfrac{9}{5}\left( -10 \right) \le\dfrac{9}{5}\cdot\dfrac{5}{9}(F-32)\le\dfrac{9}{5}\cdot18 \\\\ -\dfrac{90}{5} \le F-32\le\dfrac{162}{5} \\\\ -\dfrac{90}{5}+32 \le F-32+32\le\dfrac{162}{5}+32 \\\\ -\dfrac{90}{5}+\dfrac{160}{5} \le F-32+32\le\dfrac{162}{5}+\dfrac{160}{5} \\\\ \dfrac{70}{5}\le F\le\dfrac{322}{5} \\\\ 14\le F\le\dfrac{322}{5} .\end{array} Hence the equivalent range of temperature in Fahrenheit is $ 14^oF \text{ to } \left( \dfrac{322}{5} \right)^oF .$
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