Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set - Page 433: 32

Answer

$3\sqrt 3$

Work Step by Step

$\sqrt 27=\sqrt 9 \times\sqrt 3=3\sqrt 3$ We know that $\sqrt 9=3$, because $3^{2}=9$.
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