Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 23

Answer

$\dfrac{1}{9}$

Work Step by Step

RECALL: (i) For any non-negative real numbers $a$ and $b$, $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ and $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$. (ii) For any non-negative real number $a$, $\sqrt{a^2}=a$. Use rule (i) above to obtain $\dfrac{\sqrt{1}}{\sqrt{81}} \\=\dfrac{1}{\sqrt{9^2}}.$ Use rule (ii) above to obtain $\dfrac{1}{9}.$
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