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imehs18

  • 2 years 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
    • 2 years ago
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    the number of sides is 30

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

  4. Decart
    • 2 years ago
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    then the numbers of sides of the polygon should be polygons

  5. kc_kennylau
    • 2 years ago
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    |dw:1384944170133:dw|

  6. kc_kennylau
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Why don't you set up equations :)

  9. imehs18
    • 2 years ago
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    number of diagonals i already given. i need a solution.

  10. imehs18
    • 2 years ago
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    can you help me?

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

  12. imehs18
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    ok here's a hint: \(m=15-n\)

  15. imehs18
    • 2 years 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
    • 2 years ago
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    substitute 15-n for m

  17. imehs18
    • 2 years ago
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    huh?

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

  19. kc_kennylau
    • 2 years ago
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    into*

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

  21. kc_kennylau
    • 2 years ago
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    yep, and this will be a quadratic :)

  22. imehs18
    • 2 years ago
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    im stuck

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

  24. kc_kennylau
    • 2 years ago
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    Have you learnt quadratic? :)

  25. kc_kennylau
    • 2 years ago
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    Use quadratics pl0x

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

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

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

  29. Decart
    • 2 years ago
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    (_n+_)(_n-_) factor

  30. Decart
    • 2 years ago
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    you did it right you just got 27 instead of 30

  31. imehs18
    • 2 years ago
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    how did you get the 30?

  32. imehs18
    • 2 years ago
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    got it

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

  34. Decart
    • 2 years ago
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    combine like terms

  35. imehs18
    • 2 years ago
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    after that. what is next?

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

  37. imehs18
    • 2 years ago
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    how do i get the n now?

  38. Decart
    • 2 years 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
    • 2 years ago
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    two factors of 54 that total to -15

  40. Decart
    • 2 years ago
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    (x-_)*(x-_)=0

  41. imehs18
    • 2 years ago
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    (n-9)(n-6)

  42. Decart
    • 2 years ago
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    yes so those are your sides

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

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