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: 6


$2 \sqrt {3}$

Work Step by Step

First, we write both numbers under the same radical: $\frac{\sqrt {36}}{\sqrt {3}} = \sqrt \frac{36}{3}$ Evaluate what is under the radical sign: $\sqrt {12}$ Rewrite $12$ as a product of a perfect square and another factor. We can use $4$ as the perfect square factor: $\sqrt {4 • 3}$ Take the square root of $4$ to solve: $2 \sqrt {3}$
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