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music101 Group Title

Convert the polar equation to rectangular form and identify the graph. r= 7/ (2cosΘ+5sinΘ)

  • 7 months ago
  • 7 months ago

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  1. music101 Group Title
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    |dw:1393722700861:dw|

    • 7 months ago
  2. sourwing Group Title
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    y = r sin(theta) x = r cos(theta) x^2 + y^2 = r^2 plug and chug

    • 7 months ago
  3. music101 Group Title
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    yea so how would i plug it in

    • 7 months ago
  4. music101 Group Title
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    cuz i did this last week and i kinda forgot sorry

    • 7 months ago
  5. myininaya Group Title
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    Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)

    • 7 months ago
  6. music101 Group Title
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    okay. i did that

    • 7 months ago
  7. myininaya Group Title
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    So you have: \[2rcos(\theta)+5rsin(\theta)=7 ?\]

    • 7 months ago
  8. music101 Group Title
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    why would you put r w/ each one of the terms isnt there just 1?

    • 7 months ago
  9. myininaya Group Title
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    \[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\] Distributive property!

    • 7 months ago
  10. music101 Group Title
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    oh ok i gotchu

    • 7 months ago
  11. myininaya Group Title
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    you should have it from here. Use the equations sour gave you.

    • 7 months ago
  12. music101 Group Title
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    thannk you(:

    • 7 months ago
  13. music101 Group Title
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    what if the question's like r= -3sinΘ ?

    • 7 months ago
  14. myininaya Group Title
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    Try multiplying r on both sides recall rsin(theta)=y and r^2=x^2+y^2

    • 7 months ago
  15. music101 Group Title
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    hmm ok...

    • 7 months ago
  16. music101 Group Title
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    so r^2= -3rsinΘ?

    • 7 months ago
  17. myininaya Group Title
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    right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2

    • 7 months ago
  18. music101 Group Title
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    okay

    • 7 months ago
  19. music101 Group Title
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    then youd have?

    • 7 months ago
  20. music101 Group Title
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    the square root of that right?

    • 7 months ago
  21. myininaya Group Title
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    just do exactly what I said replace r sin(theta) with y replace r^2 with x^2+y^2

    • 7 months ago
  22. music101 Group Title
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    i did tht

    • 7 months ago
  23. myininaya Group Title
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    you are done what else do you want?

    • 7 months ago
  24. music101 Group Title
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    is tht my final answer

    • 7 months ago
  25. music101 Group Title
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    oh ok

    • 7 months ago
  26. myininaya Group Title
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    putting an equation in terms of just x and y ( or just x or just y) is called a cartesian equation if that was the directions to just right as a cartesian equation then you are done

    • 7 months ago
  27. myininaya Group Title
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    write*

    • 7 months ago
  28. music101 Group Title
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    right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ

    • 7 months ago
  29. myininaya Group Title
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    multiply both sides by r

    • 7 months ago
  30. music101 Group Title
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    do you square both sides 1st

    • 7 months ago
  31. music101 Group Title
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    thts what i emant

    • 7 months ago
  32. myininaya Group Title
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    replace r^2 with x^2+y^2 replace r sin(theta) with y replace r cos(theta) with x

    • 7 months ago
  33. music101 Group Title
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    ok i did that: x^2+y^2= 2y-4x

    • 7 months ago
  34. myininaya Group Title
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    yep. :) that is right you wrote the poloar equation as a cartesian equation if that is all the directions said to do then you are done.

    • 7 months ago
  35. music101 Group Title
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    do you stop there

    • 7 months ago
  36. music101 Group Title
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    ok

    • 7 months ago
  37. music101 Group Title
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    aand i missed r= 3+3cosΘ

    • 7 months ago
  38. music101 Group Title
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    so i multiplied both sides by r 1st

    • 7 months ago
  39. music101 Group Title
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    and got x^2+y^2=3r +3x but dont i need to get rid of the r?

    • 7 months ago
  40. myininaya Group Title
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    you could subtract both sides by 3x giving you \[x^2+y^2-3x=3r \] then square both sides and replace the r^2 with x^2+y^2

    • 7 months ago
  41. music101 Group Title
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    why square both sides?

    • 7 months ago
  42. myininaya Group Title
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    I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals

    • 7 months ago
  43. music101 Group Title
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    so what would be final answer then

    • 7 months ago
  44. myininaya Group Title
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    what if you with what I was saying take my equation and then square both sides and then replace the r^2 with x^2+y^2 you will have your final answer

    • 7 months ago
  45. music101 Group Title
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    yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?

    • 7 months ago
  46. myininaya Group Title
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    \[(x^2+y^2-3x)^2=(3r)^2 \] the us what you get when you square both sides your equation is not equivalent to that you squared each term sorta but not really it is whatever you do to one side of equation you can do to the other and the equation still holds

    • 7 months ago
  47. myininaya Group Title
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    \[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]

    • 7 months ago
  48. music101 Group Title
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    so..

    • 7 months ago
  49. music101 Group Title
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    im not following are you saying keep left side just squared in parentheses?

    • 7 months ago
  50. myininaya Group Title
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    It is easier than multiplying it out, don't you think?

    • 7 months ago
  51. music101 Group Title
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    yes. so only change right side. like substitue

    • 7 months ago
  52. myininaya Group Title
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    You can multiply it out if you want. Just do it correctly, but the if the directions just say write as a rectangular (aka cartesian equation) than you are done. If the left hand side was already in terms of x and y. r^2 was the only thing that needed to be converted.

    • 7 months ago
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