Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-7 - Cumulative Review - Page 500: 29

Answer

99

Work Step by Step

We plug in $-10$ for $x$ in the $g(x)$ equation to find: $$ g(x)=x^2-1 \\ g(-10)=(-10)^2-1 \\ g(-10)=99$$
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