Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.2 - Rational Exponents - Vocabulary, Readiness & Video Check: 1



Work Step by Step

$9^{-\frac{1}{2}}=\frac{1}{(9)^{\frac{1}{2}}}=\frac{1}{\sqrt 9}=\frac{1}{3}$ $\frac{1}{3}$ is greater than 0, so it is a positive number We know that $\sqrt 9=3$, because $3^{2}=9$
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