Answer
$2\sqrt {505}$
Work Step by Step
Apply the distance formula ,ie,
$d = \sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$d = \sqrt {(50-12)^{2}+(42-18)^{2}}=\sqrt {1444+576}=\sqrt 2020=$$2\sqrt {505}$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Prerequisites - P.6 - The Rectangular Coordinate System and Graphs - Exercises - Page 58: 38
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