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: 92

Answer

a. $C(H)=\frac{H+10.53}{2.15},$ b. $H(C(H))=H,$ $C(H(C))=C,$ c. $C(26)=16.99$

Work Step by Step

$H(C)=2.15C-10.53,$ a. $C(H)=\frac{H+10.53}{2.15},$ b. $H(C(H))=2.15(\frac{H+10.53}{2.15})-10.53=H+10.53-10.53=H,$ $C(H(C))=\frac{2.15C-10.53+10.53}{2.15}=C,$ c. $C(26)=\frac{26+10.53}{2.15}=16.99$
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