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imehs18

  • one year ago

Find the number of sides of each of the two polygons if the total numbers of sides of the polygon is 15, and the sum of the number of diagonals is 36.

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  1. Decart
    • one year ago
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    the number of sides is 30

  2. imehs18
    • one year ago
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    i was able to get the number of sides but i don't know how to show the solution

  3. imehs18
    • one year ago
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    the number of side is 9 and 6 i just need a solution how to get this

  4. Decart
    • one year ago
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    then the numbers of sides of the polygon should be polygons

  5. kc_kennylau
    • one year ago
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    |dw:1384944170133:dw|

  6. kc_kennylau
    • one year ago
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    http://www.dummies.com/how-to/content/how-to-find-the-number-of-diagonals-in-a-polygon.html

  7. kc_kennylau
    • one year ago
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    So the number of diagonals of a polygon with \(n\) sides is \(\dfrac{n(n-3)}2\)

  8. kc_kennylau
    • one year ago
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    Why don't you set up equations :)

  9. imehs18
    • one year ago
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    number of diagonals i already given. i need a solution.

  10. imehs18
    • one year ago
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    can you help me?

  11. kc_kennylau
    • one year ago
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    You could try letting the numbers of sides of the two polygons be \(m\) and \(n\) respectively :)

  12. imehs18
    • one year ago
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    can you pls show me the solution. i got this m(m-3)/2 + n(n-3)/2 = 36

  13. kc_kennylau
    • one year ago
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    Ok I'll set up the equations for you but you'll work out the steps yourself :) \[m+n=15\]\[\frac{m(m+3)}2+\frac{n(n+3)}2=36\]

  14. kc_kennylau
    • one year ago
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    ok here's a hint: \(m=15-n\)

  15. imehs18
    • one year ago
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    m^2-3m+n^2-3n = 72 m^2+n^2-3(m+n) = 72 we know that m+n = 15 m^2+n^2-3(15) = 72 m^2+n^2 = 27 after that i dont know

  16. Decart
    • one year ago
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    substitute 15-n for m

  17. imehs18
    • one year ago
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    huh?

  18. kc_kennylau
    • one year ago
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    substitute \(m=15-n\) to \(\dfrac{m(m-3)}2+\dfrac{n(n-3)}2=36\) :)

  19. kc_kennylau
    • one year ago
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    into*

  20. Decart
    • one year ago
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    \[\left( 15-n ^{} \right)^{2}-3\left( 15-n \right)+n ^{2}-3n=72\]

  21. kc_kennylau
    • one year ago
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    yep, and this will be a quadratic :)

  22. imehs18
    • one year ago
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    im stuck

  23. imehs18
    • one year ago
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    i got -108 = -27n+n^2

  24. kc_kennylau
    • one year ago
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    Have you learnt quadratic? :)

  25. kc_kennylau
    • one year ago
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    Use quadratics pl0x

  26. imehs18
    • one year ago
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    can you just pls provide me the whole solution im stuck. sorry if im stupid about this..

  27. imehs18
    • one year ago
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    i stop studying for 5years and i need to recall everything about geometry

  28. Decart
    • one year ago
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    2n^2-30n+108=0

  29. Decart
    • one year ago
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    (_n+_)(_n-_) factor

  30. Decart
    • one year ago
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    you did it right you just got 27 instead of 30

  31. imehs18
    • one year ago
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    how did you get the 30?

  32. imehs18
    • one year ago
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    got it

  33. Decart
    • one year ago
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    from the first equation 225-30n+n^2-45+3n+n^2-3n=72

  34. Decart
    • one year ago
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    combine like terms

  35. imehs18
    • one year ago
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    after that. what is next?

  36. Decart
    • one year ago
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    (15-n)^2 gave me the thirty

  37. imehs18
    • one year ago
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    how do i get the n now?

  38. Decart
    • one year ago
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    then you factor the 2n^2-30n+108 you can divide both sides by two to make it easier n^2-15n+54

  39. Decart
    • one year ago
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    two factors of 54 that total to -15

  40. Decart
    • one year ago
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    (x-_)*(x-_)=0

  41. imehs18
    • one year ago
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    (n-9)(n-6)

  42. Decart
    • one year ago
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    yes so those are your sides

  43. imehs18
    • one year ago
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    thank you so much for your patience with me. i really appreciate it.

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