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anonymous

  • one year ago

Pythagorean Thereom 3-D!!! Picture attached!

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  1. anonymous
    • one year ago
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  2. Jadedry
    • one year ago
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    @angel12310 You have to find DG first. \[DG^2 = DC^2 + CG^2\] \[AG^2 = DG^2 + AD^2\] Can you figure it out now?

  3. anonymous
    • one year ago
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    Nope...

  4. Jadedry
    • one year ago
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    @angel12310 You just have to substitute the names of lines for their lengths, try!

  5. anonymous
    • one year ago
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    :( ok

  6. Jadedry
    • one year ago
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    @angel12310 Just give it another bash, if you really, really have trouble please don't hesitate to ping me. c:

  7. Jadedry
    • one year ago
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    @angel12310 That equation is not possible. \[DG^2 = 5^2 + 12^2\]

  8. anonymous
    • one year ago
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    IDK!!!

  9. anonymous
    • one year ago
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    DG^2= 25+ 144

  10. anonymous
    • one year ago
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    DG^2=169

  11. anonymous
    • one year ago
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    Square root of 169

  12. anonymous
    • one year ago
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    13

  13. anonymous
    • one year ago
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    DG^2=13?

  14. anonymous
    • one year ago
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    13.0 cm?

  15. Jadedry
    • one year ago
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    Okay, \[DG^2 = 5^2 +12^2\] Therefore DG = \[\sqrt 169=13\] \[AG^2 = DG^2 + 4^2\] =\[AG^2 = 13^2 +4^2\] =\[AG^2 = 185\] \[\sqrt 185 = 13.6\] Therefore the answer is 13.6

  16. anonymous
    • one year ago
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    ok thx!

  17. Jadedry
    • one year ago
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    Remember that \[AG^2 = Dg^2 + 4^2\] No problem! c:

  18. anonymous
    • one year ago
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    Can u help me with a few more?

  19. Jadedry
    • one year ago
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    @angel12310 Sure, I'll be on for a bit longer! c:

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