Answered

On the interval [tex]$[0, 2\pi)$[/tex], which points are intersections of [tex]$r = 5 + 4 \sin (\theta)$[/tex] and [tex]$r = -6 \sin (\theta)$[/tex]? Check all that apply.

A. [tex]$\left(-3, \frac{7\pi}{6}\right)$[/tex]

B. [tex]$\left(-3, \frac{11\pi}{6}\right)$[/tex]

C. [tex]$\left(3, \frac{5\pi}{6}\right)$[/tex]

D. [tex]$\left(3, \frac{7\pi}{6}\right)$[/tex]

E. [tex]$\left(3, \frac{11\pi}{6}\right)$[/tex]



Answer :

GinnyAnswer
To find the intersection points of the curves defined by the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex], we need to check if the given candidate points satisfy both equations simultaneously.

### Candidate Points:
1. [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
2. [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
3. [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
4. [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
5. [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]

We will check each point one by one.

#### Check 1: [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].

Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex] is NOT an intersection point.

#### Check 2: [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].

Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex] is NOT an intersection point.

#### Check 3: [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(\frac{1}{2}\right) = 5 + 2 = 7 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(\frac{1}{2}\right) = -3 \)[/tex].

Since [tex]\(3 \neq 7\)[/tex] and [tex]\(3 \neq -3\)[/tex], [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex] is NOT an intersection point.

#### Check 4: [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].

Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex] is an intersection point.

#### Check 5: [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].

Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex] is an intersection point.

### Conclusion:
The intersection points of the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]

Thus, the correct points are:
[tex]\[ \left(3, \frac{7\pi}{6}\right), \left(3, \frac{11\pi}{6}\right) \][/tex]

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