Question

3 part question

Answer 1

well parts I and II are pretty straightforward, you just gotta apply the formulas they give you to your (x,y) coordinate

Answer 2

r^2=(-4)^2+4sqrt3)^2
64^2=4096

Answer 3

r^2 = 64 not 64^2

r = ?

Answer 4

8

Answer 5

tan(theta)=(4sqrt3)/-4
=sqrt3

Answer 6

good, so thats part I, now find theta for part II

Answer 7

tan(theta) = -sqrt(3) (you missed the negative sign)

now solve for theta

Answer 8

1/3

Answer 9

arctan(-sqrt(3)) = ?

Answer 10

-pi/3

Answer 11

good, typically we would want a positive theta value so we can just add 2pi to that to get 5pi/3

then for part III just combine the r value and the theta value to get a new set of coordinates

Answer 12

like, we got r = 8 and theta = 5pi/3 so it's literally just (8,5pi/3)