Question

change (13,6) to polar points

Answer 1

also (-9,-6)

Answer 2

You have to use \(r=\sqrt{x^2+y^2}\) and \( \theta = \arctan\frac{b}{a}\)

Answer 3

i can find r but all my angles are nonreal

Answer 4

Make that last one \( \theta=\arctan\frac{y}{x}\)

Answer 5

k let me try

Answer 6

My calculator gives 24.8 degrees for (13,6)

Answer 7

i need radians though

Answer 8

0.43

Answer 9

ok

Answer 10

what about the second one

Answer 11

THe calculator says 0.59 radians. That is wrong, because we're in the third quadrant here.
So add pi radians to get there. Or: subtract pi, doesn't matter.

Answer 12

so its .59+pi

Answer 13

Yes, I would turn that into 3.73, because 0.59 isn't exact either.

Answer 14

ok thanks

Answer 15

YW!

Answer 16

i still says i am wrong for (-9,-6) also (-4,6)