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anonymous
 5 years ago
Convert the rectangular coordinates to polar coordinates with r 0 and 0 θ 2π.
(6√3, 6)
anonymous
 5 years ago
Convert the rectangular coordinates to polar coordinates with r 0 and 0 θ 2π. (6√3, 6)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use those same equations from the other posts, plug in the corresponding values and solve for the polar coordinates \[(r,\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did, but on my calc i get deciamls and when i enter the answers into my hw program it says they are wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the r 0 and 0 theta and 2pi mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would use \[r=\sqrt{x^2+y^2}\] to solve for r and then use those other equations to find theta.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[r >0 and 0\le \theta <2\pi\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you want to find r and then plug it into the \[x=r \cos \theta\] and \[y=r \sin \theta\] along with your corresponding x and y values. You then want to solve for the value of theta that satisfies both equations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You then want to solve for the value of theta that satisfies both equations How do I do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve for all values of theta, and then find the value that when you take the cosine of it, and then take the sine of that theta, it matches up to your equation. You need to solve both equations for theta, in which the same value satisfies both equations. Use the unit circle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am so confused :( I'm sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[6\sqrt{3}=12\sin \theta\] I found r=12 from that r^2 equation above and the condition that r>0 then \[6=12\sin \theta\] both these equations need to be solved for theta. If you know your trig, there is more than one solution between 0 and 2pi for both equations. You need to find the one theta value that works for both equations.
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