question 7(multiple choice worth 1 points) (rates of change in polar functions mc) determine the average…

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. For a polar function $r = f(\theta)$ over the interval $[\alpha,\beta]$, the average rate of change is $\frac{f(\beta)-f(\alpha)}{\beta-\alpha}$. Here, $\alpha=\frac{7\pi}{6}$, $\beta = \frac{5\pi}{3}$, and $f(\theta)=5\sin\theta$.

Step2: Calculate $f(\beta)$ and $f(\alpha)$

First, find $f(\frac{5\pi}{3})$: [ \begin{align*} f\left(\frac{5\pi}{3}\right)&=5\sin\left(\frac{5\pi}{3}\right)\ &=5\times\left(-\frac{\sqrt{3}}{2}\right)\ &=-\frac{5\sqrt{3}}{2} \end{align*} ] Next, find $f(\frac{7\pi}{6})$: [ \begin{align*} f\left(\frac{7\pi}{6}\right)&=5\sin\left(\frac{7\pi}{6}\right)\ &=5\times\left(-\frac{1}{2}\right)\ &=-\frac{5}{2} \end{align*} ]

Step3: Calculate the average rate of change

[ \begin{align*} \frac{f\left(\frac{5\pi}{3}\right)-f\left(\frac{7\pi}{6}\right)}{\frac{5\pi}{3}-\frac{7\pi}{6}}&=\frac{-\frac{5\sqrt{3}}{2}-\left(-\frac{5}{2}\right)}{\frac{10\pi - 7\pi}{6}}\ &=\frac{-\frac{5\sqrt{3}}{2}+\frac{5}{2}}{\frac{3\pi}{6}}\ &=\frac{\frac{- 5\sqrt{3}+5}{2}}{\frac{\pi}{2}}\ &=\frac{-5\sqrt{3}+5}{\pi} \end{align*} ]

Answer:

A. $\frac{-5\sqrt{3}+5}{\pi}$