Solve the problem.
-The maximum afternoon temperature in Granderson is modeled by $t = 62 - 30 \cos \frac { x \pi } { 6 }$ where $t$ represents the maximum afternoon temperature in degrees Fahrenheit in month $x$, with $x = 0$ representing January, $x = 1$ representing February, and so on. Find the maximum afternoon temperature in June.
A) $88 ^ { \circ } \mathrm { F }$
B) $87 ^ { \circ } \mathrm { F }$
C) $36 ^ { \circ } \mathrm { F }$
D) $92 ^ { \circ } \mathrm { F }$

Question

Quizplus · Accepted Answer

To find the maximum afternoon temperature in June, substitute $x = 5$ (since June is the 6th month and $x = 0$ represents January) into the given equation: $t = 62 - 30 \cos \left( \frac{5\pi}{6} \right)$. The cosine of $\frac{5\pi}{6}$ is $-\sqrt{3}/2$, so the equation becomes $t = 62 - 30 \left(-\sqrt{3}/2\right)$, which simplifies to $t = 62 + 15\sqrt{3}$. Numerically, this is approximately $62 + 15(1.732) = 62 + 25.98 = 87.98$, which rounds to $88^\circ F$.

Solve the Problem t=62−30cos⁡xπ6t = 62 - 30 \cos \frac { x \pi } { 6 }t=62−30cos6xπ​

Solve the Problem $t = 62 - 30 \cos \frac { x \pi } { 6 }$