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anonymous

  • one year ago

Thanks for helping! This question is really weird :( r=-6sin theta I need to multiply both sides of the equation by r and use r^2=x^2+y^2 to rewrite the equation in terms of x and y.

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  1. anonymous
    • one year ago
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    @Ashleyisakitty @jim_thompson5910 @zepdrix @Nnesha @e.mccormick @wio @sammixboo @kropot72 @mathmate

  2. Astrophysics
    • one year ago
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    Ok so you want to convert it to rectangular, it's good you notice we have to multiply r and use \[r^2 = x^2+y^2\] we know the ratio for sin theta is the following \[\sin \theta = \frac{ y }{ r }\] so we have \[r = - 6\left( \frac{ y }{ r } \right)\] can you finish it off?

  3. anonymous
    • one year ago
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    I'm not sure I understand where to go from there

  4. Astrophysics
    • one year ago
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    \[r^2 = - 6y\]

  5. Astrophysics
    • one year ago
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    What's next?

  6. anonymous
    • one year ago
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    square - 6y?

  7. Astrophysics
    • one year ago
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    Why? Look at what we know, and we want to "get rid" of the polar coordinates.

  8. anonymous
    • one year ago
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    x^2?

  9. Astrophysics
    • one year ago
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    I don't know what that means

  10. anonymous
    • one year ago
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    add it in?

  11. Astrophysics
    • one year ago
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    Hint: \[r^2 = x^2+y^2\]

  12. anonymous
    • one year ago
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    -6y=x^2+y^2?

  13. Astrophysics
    • one year ago
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    Yes, that looks good :)

  14. Astrophysics
    • one year ago
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    You can rearrange it and what not if you wish

  15. anonymous
    • one year ago
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    That is it in terms of x and y?

  16. anonymous
    • one year ago
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    I'm asked to complete the square to produce another equation in my worksheet. Should I do it from this form?

  17. Astrophysics
    • one year ago
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    You may complete the square

  18. Astrophysics
    • one year ago
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    Ah, yes we have to complete the square, you know how to do that right?

  19. Astrophysics
    • one year ago
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    When you have it as such \[-6y=x^2+y^2 \] it's always best to complete the square as it will be in terms of x and y.

  20. anonymous
    • one year ago
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    I know how to do it in regular form, but I'm sort of confused about this one

  21. Astrophysics
    • one year ago
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    \[-6y = x^2+y^2 \implies x^2+y^2 + 6y = 0\]

  22. anonymous
    • one year ago
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    Oh, I thought it was when you added (b/2)^2 to both sides or something like that

  23. Astrophysics
    • one year ago
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    I didn't complete the square...I put in a form so you can complete the square.

  24. anonymous
    • one year ago
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    Oh, okay gotcha

  25. Astrophysics
    • one year ago
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    So we don't need to worry about the x^2, now complete the square for y^2+6y

  26. Astrophysics
    • one year ago
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    When you have \[x^2+ax \implies x^2+ 2 \frac{ ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2\] to complete the square.

  27. anonymous
    • one year ago
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    Wouldn't that make it remain the same?

  28. anonymous
    • one year ago
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    Sorry, I suck at pre calc :(

  29. Astrophysics
    • one year ago
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    You should try it, otherwise me doing everything is not going to help you :P

  30. anonymous
    • one year ago
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    very true

  31. anonymous
    • one year ago
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    But when plugging in y^2+6y in, it cancels back to y^2+6y, no?

  32. Astrophysics
    • one year ago
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    Mhm, no you don't do that, we don't plug in anything I just put if you have the form y^2+6y you do the following -> .....etc

  33. Astrophysics
    • one year ago
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    \[x^2+ax \implies x^2+ 2 \frac{ ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2\] \[x^2+ ax \implies y^2+6y\] in your question

  34. anonymous
    • one year ago
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    y ^2+6x+9-9?

  35. anonymous
    • one year ago
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    wait, no

  36. Astrophysics
    • one year ago
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    |dw:1434416994934:dw| now we just factor

  37. Astrophysics
    • one year ago
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    |dw:1434417167196:dw|

  38. Astrophysics
    • one year ago
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    So now we have \[x^2+y^2+6y+9 = -9\] one step away from completing it

  39. Astrophysics
    • one year ago
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    You were right, but you put 6x instead of 6y :P

  40. anonymous
    • one year ago
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    Now I factor?

  41. Astrophysics
    • one year ago
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    That should be = 9 not -9, yes you want it to look as such \[\huge x^2+(y+3)^2 = 9 \] that's your final answer

  42. anonymous
    • one year ago
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    x^2+(y+3)^2-9=0

  43. Astrophysics
    • one year ago
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    Good!

  44. anonymous
    • one year ago
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    oh whoop, same thing ha ha

  45. anonymous
    • one year ago
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    Thanks for the help!

  46. anonymous
    • one year ago
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    That really helped

  47. Astrophysics
    • one year ago
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    No problem, it makes sense now? :)

  48. anonymous
    • one year ago
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    Yea a lot more!

  49. Astrophysics
    • one year ago
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    Awesome, glad to hear it, make sure you go over it again!

  50. anonymous
    • one year ago
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    Of course :), have a good day

  51. Astrophysics
    • one year ago
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    You to, bye :)

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