Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises: 8b

Answer

$2 \sqrt {10}$

Work Step by Step

$d = \sqrt {(-2 - 0)^{2} + (6 - 0)^{2}}$ $d = \sqrt {(-2)^{2} + (6)^{2}}$ $d = \sqrt {4 + 36}$ $d = \sqrt {40}$ $d = \sqrt {4} \sqrt {10}$ $d = 2 \sqrt {10}$
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