Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - Chapter 5 Test: 15

Answer

$3\sqrt2$

Work Step by Step

Rationalize the denominator by multiplying $\sqrt2$ to both the numerator and the denominator: $=\dfrac{6 \cdot \sqrt2}{\sqrt2 \cdot \sqrt2} \\=\dfrac{6\sqrt2}{2}$ Simplify by canceling common factors: $\require{cancel} \\=\dfrac{\cancel{6}3\sqrt2}{\cancel{2}} \\=3\sqrt2$
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