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dietrich_harmon

  • one year ago

In circle O, CD = 44, OM = 20, ON = 19,CD l OM, EF l ON (The figure is not drawn to scale.)

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  1. dietrich_harmon
    • one year ago
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  2. dietrich_harmon
    • one year ago
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    a.Find the radius. If your answer is not an integer, express it in radical form. b.Find FN. If your answer is not an integer, express it in radical form. c.Find EF. Express it as a decimal rounded to the nearest tenth.

  3. dietrich_harmon
    • one year ago
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    @jim_thompson5910

  4. jim_thompson5910
    • one year ago
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    |dw:1360903051401:dw|

  5. jim_thompson5910
    • one year ago
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    |dw:1360903107022:dw|

  6. dietrich_harmon
    • one year ago
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    i hate geometry..lol

  7. jim_thompson5910
    • one year ago
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    |dw:1360903151843:dw| solve for x to find the radius a^2 + b^2 = c^2 20^2 + 22^2 = x^2

  8. dietrich_harmon
    • one year ago
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    is this the answer

  9. jim_thompson5910
    • one year ago
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    tell me what you get for x

  10. dietrich_harmon
    • one year ago
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    42?

  11. dietrich_harmon
    • one year ago
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    oops 22

  12. jim_thompson5910
    • one year ago
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    20^2 + 22^2 = x^2 400 + 484 = x^2 x^2 = 884 x = ???

  13. dietrich_harmon
    • one year ago
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    52

  14. dietrich_harmon
    • one year ago
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    i don't know

  15. jim_thompson5910
    • one year ago
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    x^2 = 884 x = sqrt(884) x = 2*sqrt(221)

  16. jim_thompson5910
    • one year ago
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    so thats your radius

  17. jim_thompson5910
    • one year ago
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    now you can use this radius to find EF and FN

  18. dietrich_harmon
    • one year ago
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    202 + 222

  19. jim_thompson5910
    • one year ago
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    solve for y y^2 + 19^2 = (2*sqrt(221))^2 y^2 + 381 = 884 keep going

  20. d_loparsaie
    • one year ago
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    What do you do after you solve for y? @jim_thompson5910

  21. jim_thompson5910
    • one year ago
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    I let y = EN so to find EF, you need to find 2y this is because EF = 2*EN EF = 2y

  22. jim_thompson5910
    • one year ago
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    also, FN = EN, so once you get EN, you have found FN

  23. d_loparsaie
    • one year ago
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    y^2 + 361=884?? y^2=523 right?

  24. jim_thompson5910
    • one year ago
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    oh somehow I had 381...hmm lol yeah it's 361

  25. jim_thompson5910
    • one year ago
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    y^2=523 y=sqrt(523)

  26. d_loparsaie
    • one year ago
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    How did you get 884 up there?

  27. jim_thompson5910
    • one year ago
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    (2*sqrt(221))^2 = 4*221 = 884

  28. d_loparsaie
    • one year ago
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    alright so y=29.73?

  29. jim_thompson5910
    • one year ago
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    leave it in radical form

  30. jim_thompson5910
    • one year ago
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    unless it asks about the approximation and rounding it

  31. d_loparsaie
    • one year ago
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    to leave it in radical form do I just leave it as y=sqrt(523)?

  32. jim_thompson5910
    • one year ago
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    yes square roots are considered radicals

  33. d_loparsaie
    • one year ago
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    Solving for y solves for both FN and EF..?

  34. jim_thompson5910
    • one year ago
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    FN yes since FN = EN

  35. jim_thompson5910
    • one year ago
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    but you have to double it to find EF

  36. d_loparsaie
    • one year ago
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    So the one we just solved (y) is FN right? and I add FN and EN together.. but how do I do that since its in radical form?

  37. jim_thompson5910
    • one year ago
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    or you can double FN to get EF

  38. d_loparsaie
    • one year ago
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    1,046? right

  39. d_loparsaie
    • one year ago
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    is EF

  40. jim_thompson5910
    • one year ago
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    no EF = 2*sqrt(523)

  41. d_loparsaie
    • one year ago
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    How do I double Fn to get Ef

  42. jim_thompson5910
    • one year ago
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    I just did it

  43. d_loparsaie
    • one year ago
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    -___- if 2*sqrt(523) is EF then half of that is Fn and En right..

  44. jim_thompson5910
    • one year ago
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    correct

  45. d_loparsaie
    • one year ago
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    Alright lol sorry I was confused on which was which. How do I find out what half of the radical is? just divide 523 by 2?

  46. jim_thompson5910
    • one year ago
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    it's just 2*sqrt(523) and sqrt(523)

  47. d_loparsaie
    • one year ago
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    EF=2*sqrt(523) and EN and FN = sqrt(523) .. wow haha I feel stupid. Thank you so much though!

  48. jim_thompson5910
    • one year ago
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    it's fine, yw

  49. d_loparsaie
    • one year ago
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    to find 2*sqrt(523) as a decimal rounded to the nearest tenth I multiply 523*2 and then take the square root of that number right?

  50. jim_thompson5910
    • one year ago
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    you first take the square root of 523, then you double that result

  51. jim_thompson5910
    • one year ago
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    then you round to whatever they want

  52. d_loparsaie
    • one year ago
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    so the answer would be 45.74?

  53. jim_thompson5910
    • one year ago
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    yep 2*sqrt(523) = 45.74 roughly

  54. d_loparsaie
    • one year ago
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    Thank you again

  55. jim_thompson5910
    • one year ago
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    sure thing

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