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music101

  • one year ago

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

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  1. music101
    • one year ago
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    |dw:1393722700861:dw|

  2. sourwing
    • one year ago
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    y = r sin(theta) x = r cos(theta) x^2 + y^2 = r^2 plug and chug

  3. music101
    • one year ago
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    yea so how would i plug it in

  4. music101
    • one year ago
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    cuz i did this last week and i kinda forgot sorry

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

  6. music101
    • one year ago
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    okay. i did that

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

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

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

  10. music101
    • one year ago
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    oh ok i gotchu

  11. myininaya
    • one year ago
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    you should have it from here. Use the equations sour gave you.

  12. music101
    • one year ago
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    thannk you(:

  13. music101
    • one year ago
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    what if the question's like r= -3sinΘ ?

  14. myininaya
    • one year ago
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    Try multiplying r on both sides recall rsin(theta)=y and r^2=x^2+y^2

  15. music101
    • one year ago
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    hmm ok...

  16. music101
    • one year ago
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    so r^2= -3rsinΘ?

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

  18. music101
    • one year ago
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    okay

  19. music101
    • one year ago
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    then youd have?

  20. music101
    • one year ago
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    the square root of that right?

  21. myininaya
    • one year ago
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    just do exactly what I said replace r sin(theta) with y replace r^2 with x^2+y^2

  22. music101
    • one year ago
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    i did tht

  23. myininaya
    • one year ago
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    you are done what else do you want?

  24. music101
    • one year ago
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    is tht my final answer

  25. music101
    • one year ago
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    oh ok

  26. myininaya
    • one year ago
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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

  27. myininaya
    • one year ago
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    write*

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

  29. myininaya
    • one year ago
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    multiply both sides by r

  30. music101
    • one year ago
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    do you square both sides 1st

  31. music101
    • one year ago
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    thts what i emant

  32. myininaya
    • one year ago
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    replace r^2 with x^2+y^2 replace r sin(theta) with y replace r cos(theta) with x

  33. music101
    • one year ago
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    ok i did that: x^2+y^2= 2y-4x

  34. myininaya
    • one year ago
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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.

  35. music101
    • one year ago
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    do you stop there

  36. music101
    • one year ago
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    ok

  37. music101
    • one year ago
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    aand i missed r= 3+3cosΘ

  38. music101
    • one year ago
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    so i multiplied both sides by r 1st

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

  40. myininaya
    • one year ago
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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

  41. music101
    • one year ago
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    why square both sides?

  42. myininaya
    • one year ago
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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

  43. music101
    • one year ago
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    so what would be final answer then

  44. myininaya
    • one year ago
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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

  45. music101
    • one year ago
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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?

  46. myininaya
    • one year ago
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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

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

  48. music101
    • one year ago
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    so..

  49. music101
    • one year ago
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    im not following are you saying keep left side just squared in parentheses?

  50. myininaya
    • one year ago
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    It is easier than multiplying it out, don't you think?

  51. music101
    • one year ago
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    yes. so only change right side. like substitue

  52. myininaya
    • one year ago
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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.

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