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Emily778

  • 2 years ago

Find the value of x. FG perpendicular to OP, RS perpendicular to OQ, FG=20, RS=24, OP=14

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  1. Emily778
    • 2 years ago
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  2. Emily778
    • 2 years ago
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    @Hunus @jim_thompson5910

  3. jim_thompson5910
    • 2 years ago
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    let r = radius of the circle

  4. jim_thompson5910
    • 2 years ago
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    by the pythagorean theorem, we can say (FP)^2 + (OP)^2 = (OF)^2 10^2 + 14^2 = r^2 solve for r and tell me what you get

  5. Emily778
    • 2 years ago
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    100 + 196 = 296^2. Now what?

  6. jim_thompson5910
    • 2 years ago
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    r^2 = 296 so r = ???

  7. Emily778
    • 2 years ago
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    296

  8. jim_thompson5910
    • 2 years ago
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    no

  9. jim_thompson5910
    • 2 years ago
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    r^2 is 296, not r

  10. Emily778
    • 2 years ago
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    87, 616

  11. jim_thompson5910
    • 2 years ago
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    don't square it though

  12. jim_thompson5910
    • 2 years ago
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    how do you undo squaring something

  13. Emily778
    • 2 years ago
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    sqrt?

  14. jim_thompson5910
    • 2 years ago
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    yep

  15. Emily778
    • 2 years ago
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    16.9?

  16. jim_thompson5910
    • 2 years ago
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    you did sqrt(296) right?

  17. Emily778
    • 2 years ago
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    Yea

  18. jim_thompson5910
    • 2 years ago
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    it's close, but not correct

  19. jim_thompson5910
    • 2 years ago
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    sqrt(296) = 17.2046505340852 round to whatever place you need to

  20. Emily778
    • 2 years ago
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    17.2 isn't one of the answers. Only 16.9

  21. jim_thompson5910
    • 2 years ago
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    but remember, we want x, not r

  22. jim_thompson5910
    • 2 years ago
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    we use r to get x

  23. Emily778
    • 2 years ago
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    the possible answers are: 16.9, 14, 12.3, and 12..

  24. jim_thompson5910
    • 2 years ago
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    \[\large r = \sqrt{296} \rightarrow r^2 = 296\] \[\large (OQ)^2 + (RQ)^2 = (OR)^2\] \[\large (OQ)^2 + (RQ)^2 = r^2\] \[\large x^2 + 12^2 = 296\] \[\large x^2 + 144 = 296\] ... ... ... \[\large x = ???\]

  25. Emily778
    • 2 years ago
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    152

  26. Emily778
    • 2 years ago
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    12.3

  27. jim_thompson5910
    • 2 years ago
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    \[\large x^2 = 152\] so \[\large x = ???\]

  28. jim_thompson5910
    • 2 years ago
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    good, x is roughly 12.3

  29. Emily778
    • 2 years ago
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    What about this one?

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  30. jim_thompson5910
    • 2 years ago
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    what about it? what do you want to find? and what are you given?

  31. Emily778
    • 2 years ago
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    Oh. Lol. Forgot.\ AB is a diameter and AB is perpendicular to CD.

  32. jim_thompson5910
    • 2 years ago
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    alright so that's all the given info?

  33. Emily778
    • 2 years ago
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    I know the angles are all 90 degrees.

  34. jim_thompson5910
    • 2 years ago
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    what do you need to find exactly?

  35. Emily778
    • 2 years ago
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    Find BD for AC= 58 DEGREES

  36. jim_thompson5910
    • 2 years ago
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    AC is a segment, not an angle

  37. Emily778
    • 2 years ago
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    Okay..

  38. jim_thompson5910
    • 2 years ago
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    so is it ACP or ACB that's 58 degrees

  39. Emily778
    • 2 years ago
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    This is exactly what it says. Find m BD for m AC= 58 degrees.

  40. jim_thompson5910
    • 2 years ago
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    oh arc AC, gotcha

  41. jim_thompson5910
    • 2 years ago
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    angle APC = (1/2)*(arc AC + arc BD) 90 = (1/2)*(58 + x) 90*2 = 58 + x 180 = 58 + x solve for x

  42. Emily778
    • 2 years ago
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    122

  43. jim_thompson5910
    • 2 years ago
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    yep

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