## Trigonometry (10th Edition)

The temperature on March 1 is $1.0^{\circ}F$
March 1 is day 60, so we can use $x = 60$ in the equation: $T(x)=37~sin[\frac{2\pi}{365}~(x−101)]+25$ $T(60)=37~sin[\frac{2\pi}{365}~(60−101)]+25$ $T(60)=37~sin(-0.70578)+25$ $T(60)=37~[-sin(0.70578)]+25$ $T(60)=(37)(-0.64863)+25$ $T(60) = 1.0^{\circ}F$ The temperature on March 1 is $1.0^{\circ}F$