Question

Can anybody help me with precalculus???

Answer 1

\[(\sqrt{3},-5\pi/3)\] in rectangula coordinates

Answer 2

The rectangular coordinates are :
x = r cos theta = -1 cos (-4π/3) = -1 cos (4π/3) = -1*-0.5 = 0.5
y = r sin theta = -1 sin (-4π/3) = 1 sin(4π/3) = 1* (-√3 /2) = - 0.5 √3
the polar coordinates of (-6√3, 6) are :
r = (x^2+y^2)^(1/2) = 12
tan theta = y/x = -1/√3 in the second quarter for x is negative & y is positive
theta = 5π/3