Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 6 - Polygons and Quadrilaterals - Algebra Review - Page 399: 12


$2 \sqrt {15}$

Work Step by Step

First, we rewrite the problem in the form of a fraction: $\sqrt {300} \div \sqrt {5} = \frac{\sqrt {300}}{\sqrt 5}$ Rewrite both numbers under one radical: $\sqrt {\frac{300}{5}}$ Evaluate what is under the radical sign: $\sqrt {60}$ Rewrite $60$ as the product of a perfect square and another number: $\sqrt {4 • 15}$ Take the square root of $4$ to solve: $2 \sqrt {15}$
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