Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Review Exercises - Page 173: 31



Work Step by Step

$(fg)(x)=f(x)g(x)$ $f(-3)=(-3)^2-2(-3)=9+6=15$ $g(-3)=-3-5=-8$ $(fg)(-3)=f(-3)g(-3)=15 \cdot(-8)=-120$
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