1. Graph the polar equation

2. Find each quotient and express it in rectangular form.
2(cos150°+isin150°)/ 8(cos210°+isin210°)

-0.22+0.12i
1/4[sin(-60°)+icos(-60°)]
0.12-0.22i
1/4[cos(-60°)+isin(-60°)]

3. Write the rectangular equation (x+5)^2+y^2=25 in polar form.

R= -10sin theta
R= 5
R= 5cos theta
R= -10cos theta

These are two questions and two answers.

Question 1) Which of the following polar equations is equivalent to the parametric equations below? 

x=t²

y=2t

option A.) r = 4cot(theta)csc(theta)

Explanation:

1) Polar coordinates ⇒ x = r cosθ and y = r sinθ

2) replace x and y in the parametric equations:

r cosθ = t²

r sinθ = 2t

3) work r sinθ = 2t

r sinθ/2 = t 

(r sinθ / 2)² = t²

4) equal both expressions for t²

r cos θ = (r sin θ / 2 )²

5) simplify

r cos θ = r² (sin θ)² / 4

4 = r (sinθ)² / cos θ

r = 4 cosθ / (sinθ)²

r = 4 cot θ csc θ ↔ which is the option A.

Question 2) Which polar equation is equivalent to the parametric equations below?

x=sin(theta)cos(theta)+cos(theta)

y=sin^2(theta)+sin(theta)

option B) r = sinθ + 1

Explanation:

1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ

2) replace x and y in the parametric equations:

a) r cosθ = sin(θ)cos(θ)+cos(θ)

b) r sinθ =sin²(θ)+sin(θ)

3) work both equations

a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1]  ⇒ r = sinθ + 1

b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1

Therefore, the answer is r = sinθ + 1 which is the option B.

9

step-by-step explanation:

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled