Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 610: 34


x = $\sqrt 37 w$

Work Step by Step

Since the triangle is a right angle triangle we use the pythagorean theorem. $a^{2} + b^{2} = c^{2}$ In this problem a=6w, b=w and c= x $(6w)^{2} + (w)^{2} = x^{2}$ $ 6w \times 6w + (w)^{2} = x^{2}$ $36w^{2} + w^{2} = x^{2}$ $37w^{2} = x^{2}$ We take square root of both sides $\sqrt (37w^{2}) = \sqrt x^{2}$ $\sqrt 37 w$ = x Therefore x = $\sqrt 37 w$
Update this answer!

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

Update this answer

After 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.