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music101

  • 9 months ago

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

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  1. music101
    • 9 months ago
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    |dw:1393722700861:dw|

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

  3. music101
    • 9 months ago
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    yea so how would i plug it in

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

  5. myininaya
    • 9 months 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
    • 9 months ago
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    okay. i did that

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

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

  9. myininaya
    • 9 months 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
    • 9 months ago
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    oh ok i gotchu

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

  12. music101
    • 9 months ago
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    thannk you(:

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

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

  15. music101
    • 9 months ago
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    hmm ok...

  16. music101
    • 9 months ago
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    so r^2= -3rsinΘ?

  17. myininaya
    • 9 months 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
    • 9 months ago
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    okay

  19. music101
    • 9 months ago
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    then youd have?

  20. music101
    • 9 months ago
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    the square root of that right?

  21. myininaya
    • 9 months 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
    • 9 months ago
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    i did tht

  23. myininaya
    • 9 months ago
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    you are done what else do you want?

  24. music101
    • 9 months ago
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    is tht my final answer

  25. music101
    • 9 months ago
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    oh ok

  26. myininaya
    • 9 months 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
    • 9 months ago
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    write*

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

  29. myininaya
    • 9 months ago
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    multiply both sides by r

  30. music101
    • 9 months ago
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    do you square both sides 1st

  31. music101
    • 9 months ago
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    thts what i emant

  32. myininaya
    • 9 months 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
    • 9 months ago
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    ok i did that: x^2+y^2= 2y-4x

  34. myininaya
    • 9 months 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
    • 9 months ago
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    do you stop there

  36. music101
    • 9 months ago
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    ok

  37. music101
    • 9 months ago
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    aand i missed r= 3+3cosΘ

  38. music101
    • 9 months ago
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    so i multiplied both sides by r 1st

  39. music101
    • 9 months 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
    • 9 months 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
    • 9 months ago
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    why square both sides?

  42. myininaya
    • 9 months 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
    • 9 months ago
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    so what would be final answer then

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

  48. music101
    • 9 months ago
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    so..

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

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

  51. music101
    • 9 months ago
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    yes. so only change right side. like substitue

  52. myininaya
    • 9 months 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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