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alwaysdancing18

Classify the quadrilateral. Then, find the value(s) of the variable(s).

  • one year ago
  • one year ago

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  1. alwaysdancing18
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    |dw:1355420237417:dw|

    • one year ago
  2. mayankdevnani
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    it's a KITE

    • one year ago
  3. alwaysdancing18
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    wow... not helpful

    • one year ago
  4. mayankdevnani
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    opposite sides are equal

    • one year ago
  5. alwaysdancing18
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    obviously

    • one year ago
  6. mayankdevnani
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    and find the values

    • one year ago
  7. alwaysdancing18
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    I dont know how to set the equation....

    • one year ago
  8. AravindG
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    opposit sides are not equal, it is the adjacent sides that are equal

    • one year ago
  9. AravindG
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    so 3x+6=6x+1

    • one year ago
  10. alwaysdancing18
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    well thats what I was thinking when he said that

    • one year ago
  11. mayankdevnani
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    opp. sides are equal so that 6x+1=3x+6 and 2y+6=4y-3 and then find values of x and y

    • one year ago
  12. mayankdevnani
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    yaaa correct!!!! @AravindG

    • one year ago
  13. AravindG
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    2y+6=4y-3

    • one year ago
  14. AravindG
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    solve for x and y and be happy :)

    • one year ago
  15. alwaysdancing18
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    I just look at the lines there. if they are the same then whala and this one question will not make me happy.... I have so many more

    • one year ago
  16. mayankdevnani
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    can you find the values? @alwaysdancing18

    • one year ago
  17. alwaysdancing18
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    y is not a whole number? I got 4 R1

    • one year ago
  18. mayankdevnani
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    |dw:1355410031710:dw|

    • one year ago
  19. alwaysdancing18
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    oh you wanted to do x first... ok.. umm 5/3

    • one year ago
  20. mayankdevnani
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    right

    • one year ago
  21. mayankdevnani
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    and same as second comment

    • one year ago
  22. alwaysdancing18
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    so y would be 9/2?

    • one year ago
  23. mayankdevnani
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    correct!!! you got it

    • one year ago
  24. alwaysdancing18
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    Thx... if you dont mind helping with two more?

    • one year ago
  25. mayankdevnani
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    sorry...g2g

    • one year ago
  26. alwaysdancing18
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    oh... thats ok Thanks anyway!!

    • one year ago
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