## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Prerequisites - P.6 - The Rectangular Coordinate System and Graphs - Exercises - Page 58: 38

#### 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}$

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