Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 459: 3

Answer

Choice B

Work Step by Step

Using the properties of radicals, the expression, $ \sqrt{48} $, is equivalent to \begin{align*} & \sqrt{16\cdot3} \\&= \sqrt{16}\cdot\sqrt{3} &(\text{use }\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}) \\&= 4\sqrt{3} .\end{align*} With $\sqrt{48}$ equivalent to $4\sqrt{3}$, then $\sqrt{48}$ can be simplified. Hence, Choice B.
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