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anonymous

  • one year ago

@rvc

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  1. anonymous
    • one year ago
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    I need your help :/

  2. anonymous
    • one year ago
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    i need medals

  3. rvc
    • one year ago
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    yes @nopen

  4. anonymous
    • one year ago
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    it's part c

  5. anonymous
    • one year ago
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    I got the answer for a and b

  6. anonymous
    • one year ago
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    Part a.) is OC = OT - CT = R- r

  7. anonymous
    • one year ago
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    Part b.) is R*sin*theta = r(1+ sin theta)

  8. rvc
    • one year ago
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    im extremely sorry gtg now

  9. Michele_Laino
    • one year ago
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    we can write this: \[\Large \begin{gathered} 21 = 2R + R2\theta = 2R + 2R\arcsin \left( {\frac{3}{4}} \right) = \hfill \\ \hfill \\ = 2R\left\{ {1 + \arcsin \left( {\frac{3}{4}} \right)} \right\} \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    since: \[\Large ATB = R2\theta \]

  11. anonymous
    • one year ago
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    yea I did that but my problem is why am I to taking the 1 with the given angle?

  12. Michele_Laino
    • one year ago
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    by definition of radians, we have: \[\Large L = \alpha R\] |dw:1436029827922:dw|

  13. dan815
    • one year ago
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    http://prntscr.com/7orri3

  14. anonymous
    • one year ago
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    @dan815 your answer is wrong =_=

  15. anonymous
    • one year ago
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    @Michele_Laino I do remember that arc length formula but what I'm asking is that, if we are given that O is sin theta = 3/4 why are we taking the 1?

  16. dan815
    • one year ago
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    what is the answer

  17. anonymous
    • one year ago
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    2.43

  18. anonymous
    • one year ago
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    @Michele_Laino what you've written in your answer is exactly the same thing that is written in the solution bank I simply wanna know why you talking (1 + arcsin(3/4) instead of just arcsin3/4?

  19. anonymous
    • one year ago
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    *are taking

  20. Michele_Laino
    • one year ago
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    since I have factored out the quantity 2R

  21. dan815
    • one year ago
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    hey michele can you tell me what's wrong in my method, i cant find it

  22. Michele_Laino
    • one year ago
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    your link tells me "500 Internal Server Error" @dan815

  23. anonymous
    • one year ago
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    lol

  24. dan815
    • one year ago
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    1 Attachment
  25. dan815
    • one year ago
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    how about that one

  26. anonymous
    • one year ago
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    Ohhhhh Ok I got it!!

  27. dan815
    • one year ago
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    oh oops its the whole perimeter not just the arc perimeter

  28. anonymous
    • one year ago
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    THANK YOU @Michele_Laino! :DDDDD

  29. Michele_Laino
    • one year ago
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    yes! the perimeter is the perimeter of the sector @dan815

  30. Michele_Laino
    • one year ago
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    :) @nopen

  31. dan815
    • one year ago
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    http://prntscr.com/7orzms hehe too clumpsy

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