#### Answer

$d = \sqrt {a^{2} + b^{2}}$

#### Work Step by Step

$d = \sqrt {(0 - a)^{2} + (b - 0)^{2}}$
$d = \sqrt {(-a)^{2} + (b)^{2}}$
$d = \sqrt {a^{2} + b^{2}}$

Chegg costs money, GradeSaver solutions are free!

Published by
Brooks Cole

ISBN 10:
9781285195698

ISBN 13:
978-1-28519-569-8

$d = \sqrt {a^{2} + b^{2}}$

$d = \sqrt {(0 - a)^{2} + (b - 0)^{2}}$
$d = \sqrt {(-a)^{2} + (b)^{2}}$
$d = \sqrt {a^{2} + b^{2}}$

You can help us out by revising, improving and updating this answer.

Update this answerAfter you claim an answer you’ll have **24 hours** to send in a draft. An editor
will review the submission and either publish your submission or provide feedback.