Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 1, Equations and Inequalities - 1.4 Rewrite Equations and Formulas - 1.4 Exercises - Problem Solving - Page 32: 35


$C=\dfrac{5}{9}(F-32)$ and $10^{\circ}$

Work Step by Step

As we are given that, $F-32=\dfrac{9}{5} C$ The formula for $C$ can be defined as: $C=\dfrac{5}{9}(F-32)$ Plug $50^{\circ}$ for $F$ Then, we have $C=\dfrac{5}{9}(50-32)=10^{\circ}$ Hence, $C=\dfrac{5}{9}(F-32)$ and $10^{\circ}$.
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