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Nimwhitted

  • 3 years ago

New Math Question! Help!

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  1. Nimwhitted
    • 3 years ago
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  2. mathslover
    • 3 years ago
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    i suppose the given info may be : AF : AC :: AD : AB ?

  3. Ganpat
    • 3 years ago
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    are you aware of similarity of triangles ??

  4. mathslover
    • 3 years ago
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    @Ganpat dude not a case of similarity or congruency but instead a ratio problem.. i do tink

  5. katiebugg
    • 3 years ago
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    lol its double 5 it has to be so x is 10

  6. Nimwhitted
    • 3 years ago
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    But thats not a answer?

  7. mathslover
    • 3 years ago
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    @Nimwhitted ... so can you tell me : \[\large\frac{AF}{AC}=\frac{AD}{AB}\]

  8. katiebugg
    • 3 years ago
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    wait lol x is 15 lol sorry

  9. Nimwhitted
    • 3 years ago
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    I was thinking it was 15

  10. mathslover
    • 3 years ago
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    First of all AF = 5 , AC = 9 AD = x - 5 BD = 8

  11. Ganpat
    • 3 years ago
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    @mathslover : for taking ratios we do consider similarity na ??

  12. mathslover
    • 3 years ago
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    yes u r thinking right :

  13. mathslover
    • 3 years ago
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    yes right @Ganpat but mentioning that r u aware of ratios would have been more nice : just kidding ...

  14. Nimwhitted
    • 3 years ago
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    Everyone should get a medal, so each of you give eachother medals and ill give someone a medal too

  15. Nimwhitted
    • 3 years ago
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    Someone give ganpat and kaitie one

  16. Ganpat
    • 3 years ago
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    :).. h aha ha.. good team work !! :)

  17. Nimwhitted
    • 3 years ago
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    :D Yay everyone got a medal!

  18. mathslover
    • 3 years ago
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    :D i gave ..

  19. katiebugg
    • 3 years ago
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    i didnt get one lol ohh welll

  20. Nimwhitted
    • 3 years ago
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    Yea you did!

  21. katiebugg
    • 3 years ago
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    ohhh ok lol nermind..

  22. Ganpat
    • 3 years ago
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    i gave medal to you @katiebugg : be proud , you r the 1st one to get medal from Troll ... :D

  23. katiebugg
    • 3 years ago
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    hahaha oh ok thatnks!

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