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dietrich_harmon

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

  • one year ago
  • one year ago

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  1. dietrich_harmon
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    • one year ago
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  2. dietrich_harmon
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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.

    • one year ago
  3. dietrich_harmon
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    @jim_thompson5910

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

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

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

    • one year ago
  7. jim_thompson5910
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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

    • one year ago
  8. dietrich_harmon
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    is this the answer

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

    • one year ago
  10. dietrich_harmon
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    42?

    • one year ago
  11. dietrich_harmon
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    oops 22

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

    • one year ago
  13. dietrich_harmon
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    52

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

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

    • one year ago
  16. jim_thompson5910
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    so thats your radius

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

    • one year ago
  18. dietrich_harmon
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    202 + 222

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

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

    • 11 months ago
  21. jim_thompson5910
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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

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

    • 11 months ago
  23. d_loparsaie
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    y^2 + 361=884?? y^2=523 right?

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

    • 11 months ago
  25. jim_thompson5910
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    y^2=523 y=sqrt(523)

    • 11 months ago
  26. d_loparsaie
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    How did you get 884 up there?

    • 11 months ago
  27. jim_thompson5910
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    (2*sqrt(221))^2 = 4*221 = 884

    • 11 months ago
  28. d_loparsaie
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    alright so y=29.73?

    • 11 months ago
  29. jim_thompson5910
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    leave it in radical form

    • 11 months ago
  30. jim_thompson5910
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    unless it asks about the approximation and rounding it

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

    • 11 months ago
  32. jim_thompson5910
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    yes square roots are considered radicals

    • 11 months ago
  33. d_loparsaie
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    Solving for y solves for both FN and EF..?

    • 11 months ago
  34. jim_thompson5910
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    FN yes since FN = EN

    • 11 months ago
  35. jim_thompson5910
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    but you have to double it to find EF

    • 11 months ago
  36. d_loparsaie
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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?

    • 11 months ago
  37. jim_thompson5910
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    or you can double FN to get EF

    • 11 months ago
  38. d_loparsaie
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    1,046? right

    • 11 months ago
  39. d_loparsaie
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    is EF

    • 11 months ago
  40. jim_thompson5910
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    no EF = 2*sqrt(523)

    • 11 months ago
  41. d_loparsaie
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    How do I double Fn to get Ef

    • 11 months ago
  42. jim_thompson5910
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    I just did it

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

    • 11 months ago
  44. jim_thompson5910
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    correct

    • 11 months ago
  45. d_loparsaie
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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?

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

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

    • 11 months ago
  48. jim_thompson5910
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    it's fine, yw

    • 11 months ago
  49. d_loparsaie
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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?

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

    • 11 months ago
  51. jim_thompson5910
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    then you round to whatever they want

    • 11 months ago
  52. d_loparsaie
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    so the answer would be 45.74?

    • 11 months ago
  53. jim_thompson5910
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    yep 2*sqrt(523) = 45.74 roughly

    • 11 months ago
  54. d_loparsaie
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    Thank you again

    • 11 months ago
  55. jim_thompson5910
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    sure thing

    • 11 months ago
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