Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 6

Answer

$\color{blue}{12}$

Work Step by Step

RECALL: (1) For $a\gt0$ and $b\gt0$, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (2) For $a\ge0$, $\sqrt{a^2} = a$ Use rule (1) above to obtain: $\sqrt{6} \cdot \sqrt{24} \\=\sqrt{6\cdot24} \\=\sqrt{144} \\=\sqrt{12^2}$ Use rule (2) above to obtain: $=\color{blue}{12}$
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