## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 4 - Review Exercises - Page 192: 53d

296

#### Work Step by Step

According to the data in the question 'If $x$ is the time in years, with $x = 0$ representing January 1 of the base year, $x=0.5$ representing July 1 of the same year and $x=1.75$ representing October 1 of the following year'. This means that $x=1.5$ represents July 1 of the following year. Therefore, we substitute $x=1.5$ in the equation to find the pollution level y on the aforementioned date: $y=7(1-\cos2\pi x)(x+10)+100e^{0.2x}$ $y=7[1-\cos2\pi (1.5)](1.5+10)+100e^{0.2\times1.5}$ $y=7(1-\cos 3\pi)(11.5)+100e^{0.3}$ $y=7[1-(-1)](11.5)+100(1.349859)$ $y=7(2)(11.5)+134.9859$ $y=161+134.9859$ $y=295.9859\approx296$ Therefore, the pollution level on July 1 of the following year is 296.

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