Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 3 - Section 3.2 - Introduction to Functions - Exercise Set: 64c

Answer

$-\frac{1}{9}$

Work Step by Step

We are given that $h(x)=-x^{2}$. $h(\frac{1}{3})=-(\frac{1}{3})^{2}=-(\frac{1}{3}\times\frac{1}{3})=-(\frac{1}{9})=-\frac{1}{9}$
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