College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 119: 101

Answer

Its length was approximately 220.7 feet.

Work Step by Step

Replace the given period into the formula and solve for the length: $16.5=2\pi\sqrt{\frac{l}{32}}$ $16.5\cdot \frac{1}{2\pi}=2\pi\sqrt{\frac{l}{32}}\cdot \frac{1}{2\pi}$ $ \frac{8.25}{\pi}=\sqrt{\frac{l}{32}}$ $( \frac{8.25}{\pi})^2=(\sqrt{\frac{l}{32}})^2$ $ \frac{68.0625}{\pi^2}=\frac{l}{32}$ $ \frac{68.0625}{\pi^2}\cdot32=\frac{l}{32}\cdot32$ $l=\frac{2178}{\pi^2}\approx220.7$
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