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Change ((-3sqrt3)/2, (3/2)) to polar coordinates?
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R=x^2+y^2 Q=arc tan (y/x) (X,y)=(p,q)
Sorry meant (r,q)
ok so what do i plug in for all of that?
Your values for x and y
R^2= (-3sqrt3/2)^2+(3/2)^2= (-3*3/4)+(9/4)=-9/4+9/4=0 Arc tan (-3sqrt3/2/(3/2)=arc tan (-sqrt3)
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Q=2pi/3
ok thanx so much!!!
Answer (0,2pi/3)
I was working it through and checked here to ask if it was right but you wrote the same thing i got!
Awesome glad I could help
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