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: 11

Answer

$7 \sqrt {2}$

Work Step by Step

First, we rewrite the problem in the form of a fraction: $28 \div \sqrt {8} = \frac{28}{\sqrt 8}$ Let's look at the denominator. We can rewrite $8$ as the product of a perfect square and another factor. $8$ can be rewritten as $4 • 2$: $\frac{28}{\sqrt {4 • 2}}$ Take the square root of $4$: $\frac{28}{2 \sqrt {2}}$ Divide both the numerator and the denominator by their greatest common factor, $2$: $\frac{14}{\sqrt {2}}$ We do not like to leave radicals in the denominator. To get rid of the radical in the denominator, we multiply both the numerator and denominator by $\sqrt {2}$: $\frac{14 \sqrt {2}}{2}$ Divide the numerator and denominator by their greatest common factor, $2$: $7 \sqrt {2}$
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