Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.1 - The Rectangular Coordinate System - Exercises - Page 449: 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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