College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.2 - One-to-One Functions; Inverse Functions - 6.2 Assess Your Understanding - Page 422: 94

Answer

a. $C(F)=\frac{5F-160}{9},$ b. $C(F(C))=C,$ or $F(C(F))=F,$ c. $C(70)=23.33$

Work Step by Step

$F(C)=\frac{9}{5}C+32,$ a. $C(F)=\frac{5F-160}{9},$ b. $C(F(C))=\frac{5(\frac{9}{5}C+32)-160}{9}=\frac{9C+160-160}{9}=C,$ or $F(C(F))=\frac{9}{5}(\frac{5F-160}{9})+32=F-32+32=F,$ c. $C(70)=\frac{350-160}{9}=\frac{70}{3}=23.33$
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