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daustin15 Group Title

AC + BC could equal 15 Always Never Sometimes

  • one year ago
  • one year ago

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  1. daustin15 Group Title
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    • one year ago
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  2. UsukiDoll Group Title
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    Well one side is 9. This can get tricky since if A = 1, B = 1, and C = 9 you'll have 18. And you can't have a zero side or a negative side. That's impossible

    • one year ago
  3. ParthKohli Group Title
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    \[\rm AC = 9\]Now apply the Pythagorean Theorem.\[9^2 + x^2 = 15^2\]Any solutions to the above?

    • one year ago
  4. ParthKohli Group Title
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    Turns out they could never, because the only solution is \(12\) but \(9 + 12 = 21\)

    • one year ago
  5. ParthKohli Group Title
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    Also thinking about how AC + BC > 15, we could never get AC + BC = 15.

    • one year ago
  6. UsukiDoll Group Title
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    oh I remember doing that PArth...

    • one year ago
  7. ParthKohli Group Title
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    Erm?

    • one year ago
  8. UsukiDoll Group Title
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    the a^2+b^2 = c^2

    • one year ago
  9. Directrix Group Title
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    Triangle Inequality Theorem: One side of a triangle is less than the sum of the other two. Therefore, 15 < AC + BC which is equivalent to AC + BC > 15. AC + BC could never be equal to 15. There would be no triangle. @daustin15

    • one year ago
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