Doug C. answered • 10/25/17

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Since the directions indicate you must use theta between -pi/2 and pi/2 when (x,y) is in the 2nd or 3rd quadrant you will have to use the negative square root of X

^{2}+ y^{2}.For part a) r = sqrt(4+9) = sqrt(13).

Theta = tan

^{-1}(y/x) = tan^{-1}(-3/2) = -0.983 radians. (your calculator will give you inverse tangent between -pi/2 and pi/2)Since (2,-3) is in quadrant 4, use the positive square root for r to set the polar coordinates:

sqrt(13), -0.983 (r, theta)

Hussein H.

10/25/17