# 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 $=3$

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