Trigonometry (11th Edition) Clone

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

Chapter 4 - Graphs of the Circular Functions - Section 4.1 Graphs of the Sine and Cosine Functions - 4.1 Exercises - Page 152: 57b



Work Step by Step

Since April $5$ corresponds to Day $95$, we substitute $x=95$ in the formula: $T(95)=37+21\sin[\frac{2\pi}{365}(x-91)]$ $T(95)=37+21\sin[\frac{2\pi}{365}(95-91)]$ $T(95)=37+21\sin[\frac{2\pi}{365}(4)]$ $T(95)=37+21\sin[0.0689]$ $T(95)=37+21(0.0688)$ $T(95)=37+1.45$ $T(95)=38.45\approx38^{\circ}$F Therefore, on Day $95$, the temperature is $38^{\circ}$F.
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