Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.10 nth Roots; Rational Exponents - A.10 Assess Your Understanding - Page A88: 80

Answer

$-\dfrac{1}{5}$

Work Step by Step

Use the rule $a^{-m}=\left(\dfrac{1}{a}\right)^m$: $-25^{-\frac{1}{2}}=-\left(\frac{1}{25}\right)^{\frac{1}{2}}$ $=-\left(\frac{1}{5^2}\right)^{\frac{1}{2}}$ Use the rule $\left(\dfrac{a}{b}\right)^n=\dfrac{a^m}{b^m}$, then simplify to obtain: $=-\dfrac{1^{\frac{1}{2}}}{\left(5^2\right)^{ \frac{1}{2}}}$ $=-\dfrac{1}{\left(5^2\right)^{ \frac{1}{2}}}$ Use the rule $\left(a^m\right)^n=a^{mn}$, then simplify to obtain: $=-\dfrac{1}{5^{2\cdot \frac{1}{2}}}$ $=-\dfrac{1}{5^{1}}$ $=-\dfrac{1}{5}$ Hence, the correct answer is $-\frac{1}{5}$.
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