Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 637: 27


Amplitude $=5$ ; $ f=\dfrac{1}{3}$ and Period $=3$

Work Step by Step

Amplitude, $|A|=|-5|=5$ Given: $\omega =\dfrac{2 \pi}{3}$ So, $ f=\dfrac{\omega}{2 \pi}=\dfrac{2\pi/3}{2 \pi}=\dfrac{1}{3}$ Now, period, $ P=\dfrac{2 \pi}{\omega}=\dfrac{2 \pi}{2\pi/3}=3$ Hence, Amplitude $=5$ ; $ f=\dfrac{1}{3}$ and Period $=3$
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.