Intermediate Algebra (6th Edition)

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

Chapter 2 - Section 2.3 - Formulas and Problem Solving - Exercise Set: 20

Answer

$F=\frac{9C+160}{5}$

Work Step by Step

We are given that $C=\frac{5}{9}(F-32)$. We can use the distributive property to multiply out the terms on the right side. $C=\frac{5}{9}\times(F-32)=\frac{5F}{9}-\frac{160}{9}$ Multiply all terms by 9 in order to eliminate fractions. $9C=5F-160$ Add 160 to both sides. $9C+160=5F$ Divide both sides by 5. $F=\frac{9C+160}{5}$
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