Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 1 - Section 1.8 - Introduction to Variables, Algebraic Expressions, and Equations - Exercise Set: 44

Answer

$50^oF=10^oC$ $59^oF=15^oC$ $68^oF=20^oC$ $77^oF=25^oC$

Work Step by Step

$\frac{5(F-32)}{9}$ substitute each given value for F and solve $\frac{5(50-32)}{9}=\frac{5(18)}{9}=\frac{(5)(2)(/\!\!9)}{/\!\!9}=10$ $\frac{5(59-32)}{9}=\frac{5(27)}{9}=\frac{(5)(3)(/\!\!9)}{/\!\!9}=15$ $\frac{5(68-32)}{9}=\frac{5(36)}{9}=\frac{(5)(4)(/\!\!9)}{/\!\!9}=20$ $\frac{5(77-32)}{9}=\frac{5(45)}{9}=\frac{(5)(5)(/\!\!9)}{/\!\!9}=25$
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