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: 13

Answer

$5\sqrt{2}$

Work Step by Step

RECALL: For any non-negative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. Simplify the given expression by using the rule above to obtain $\sqrt{25} \cdot \sqrt{2} \\=\sqrt{5^2} \cdot \sqrt{2} \\=5\sqrt{2}.$
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