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: 53c



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 $x=1$ represents January 1 of the following year. Therefore, we substitute $x=1$ in the equation to find the pollution level y on the aforementioned date: $y=7(1-\cos2\pi x)(x+10)+100e^{0.2x}$ $y=7[1-\cos2\pi (1)](1+10)+100e^{0.2\times1}$ $y=7(1-\cos 2\pi)(11)+100e^{0.2}$ $y=7(1-1)(11)+100(1.22140)$ $y=7(0)(11)+122.140$ $y=0+122.14$ $y=122.14\approx122$ Therefore, the pollution level on January 1 of the following year is 122.
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