Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Review Exercises - Page 192: 53d



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.
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.