College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 48: 14

Answer

$3\sqrt{3}$

Work Step by Step

$\sqrt{27}$ $=\sqrt{9\cdot3}$ $=\sqrt{9}\cdot \sqrt{3}$ $=3\sqrt{3}$
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