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imehs18

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.

  • 5 months ago
  • 5 months ago

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  1. Decart
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    the number of sides is 30

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

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

    • 5 months ago
  4. Decart
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    then the numbers of sides of the polygon should be polygons

    • 5 months ago
  5. kc_kennylau
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    |dw:1384944170133:dw|

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

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

    • 5 months ago
  8. kc_kennylau
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    Why don't you set up equations :)

    • 5 months ago
  9. imehs18
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    number of diagonals i already given. i need a solution.

    • 5 months ago
  10. imehs18
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    can you help me?

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

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

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

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

    • 5 months ago
  15. imehs18
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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

    • 5 months ago
  16. Decart
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    substitute 15-n for m

    • 5 months ago
  17. imehs18
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    huh?

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

    • 5 months ago
  19. kc_kennylau
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    into*

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

    • 5 months ago
  21. kc_kennylau
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    yep, and this will be a quadratic :)

    • 5 months ago
  22. imehs18
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    im stuck

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

    • 5 months ago
  24. kc_kennylau
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    Have you learnt quadratic? :)

    • 5 months ago
  25. kc_kennylau
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    Use quadratics pl0x

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

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

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

    • 5 months ago
  29. Decart
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    (_n+_)(_n-_) factor

    • 5 months ago
  30. Decart
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    you did it right you just got 27 instead of 30

    • 5 months ago
  31. imehs18
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    how did you get the 30?

    • 5 months ago
  32. imehs18
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    got it

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

    • 5 months ago
  34. Decart
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    combine like terms

    • 5 months ago
  35. imehs18
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    after that. what is next?

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

    • 5 months ago
  37. imehs18
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    how do i get the n now?

    • 5 months ago
  38. Decart
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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

    • 5 months ago
  39. Decart
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    two factors of 54 that total to -15

    • 5 months ago
  40. Decart
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    (x-_)*(x-_)=0

    • 5 months ago
  41. imehs18
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    (n-9)(n-6)

    • 5 months ago
  42. Decart
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    yes so those are your sides

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

    • 5 months ago
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