Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 4 - Graphs of the Circular Functions - Section 4.1 Graphs of the Sine and Cosine Functions - 4.1 Exercises - Page 147: 59d

Answer

$58^{\circ}$F

Work Step by Step

Since June $25$ represents Day $175$, we substitute $x=175$ in the formula: $T(175)=37+21\sin[\frac{2\pi}{365}(x-91)]$ $T(175)=37+21\sin[\frac{2\pi}{365}(175-91)]$ $T(175)=37+21\sin[\frac{2\pi}{365}(84)]$ $T(175)=37+21\sin[1.446]$ $T(175)=37+21(0.992)$ $T(175)=37+20.84$ $T(175)=57.84\approx58^{\circ}$F Therefore, on Day $175$, the temperature is $58^{\circ}$F.
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