use the polar function to answer the followin...

# Explanation: ## Step1: Identify type of $r = 8\cos(4\theta)$ The general form of a rose - curve is $r = a\cos(n\theta)$ or $r=a\sin(n\theta)$. When $n$ is even, the number of petals is $2n$. Here $n = 4$, so it is a rose - curve. Type: Rose; Petals: $2\times4 = 8$; Cycle: The period of $\cos(4\theta)$ is $\frac{2\pi}{4}=\frac{\pi}{2}$, so cycle is $\frac{\pi}{2}$. When $\theta=\frac{\pi}{6}$, $r = 8\cos(4\times\frac{\pi}{6})=8\cos(\frac{2\pi}{3})=8\times(-\frac{1}{2})=- 4$. When $\theta=\pi$, $r = 8\cos(4\pi)=8\times1 = 8$. ## Step2: Identify type of $r=-7\sin(\theta)$ The general form of a circle in polar coordinates is $r = 2a\sin(\theta)$ or $r = 2a\cos(\theta)$ centered at $(0,a)$ or $(a,0)$ respectively. For $r=-7\sin(\theta)$, it is a circle. The circle $r = - 7\sin(\theta)$ is centered at $(0,-\frac{7}{2})$ and opens downwards. Type: Circle; Opens: Downwards; Petals: N/A; Cycle: The period of $\sin(\theta)$ is $2\pi$, so cycle is $2\pi$. When $\theta=\frac{\pi}{3}$, $r=-7\sin(\frac{\pi}{3})=-7\times\frac{\sqrt{3}}{2}=-\frac{7\sqrt{3}}{2}$. When $\theta=\frac{\pi}{2}$, $r=-7\sin(\frac{\pi}{2})=-7$. # Answer: | | $r = 8\cos(4\theta)$ | $r=-7\sin(\theta)$ | |--|--|--| | Type | Rose | Circle | | Opens | N/A | Downwards | | Petals | 8 | N/A | | Cycle | $\frac{\pi}{2}$ | $2\pi$ | | $\theta=\frac{\pi}{6},r$ | - 4 | N/A | | $\theta=\pi,r$ | 8 | N/A | | $\theta=\frac{\pi}{3},r$ | N/A | $-\frac{7\sqrt{3}}{2}$ | | $\theta=\frac{\pi}{2},r$ | N/A | - 7 |