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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)

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  1. anonymous
    • 5 years ago
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    Use those same equations from the other posts, plug in the corresponding values and solve for the polar coordinates \[(r,\theta)\]

  2. anonymous
    • 5 years ago
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    I did, but on my calc i get deciamls and when i enter the answers into my hw program it says they are wrong

  3. anonymous
    • 5 years ago
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    i need exact values

  4. anonymous
    • 5 years ago
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    What is the r 0 and 0 theta and 2pi mean?

  5. anonymous
    • 5 years ago
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    I would use \[r=\sqrt{x^2+y^2}\] to solve for r and then use those other equations to find theta.

  6. anonymous
    • 5 years ago
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    \[r >0 and 0\le \theta <2\pi\]

  7. anonymous
    • 5 years ago
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    So 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.

  8. anonymous
    • 5 years ago
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    You then want to solve for the value of theta that satisfies both equations How do I do that?

  9. anonymous
    • 5 years ago
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    Solve 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.

  10. anonymous
    • 5 years ago
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    I am so confused :( I'm sorry.

  11. anonymous
    • 5 years ago
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    So \[6\sqrt{3}=12\sin \theta\] 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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