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anonymous

  • 4 years ago

George is thinking about a regular polygon. If he takes the number of diagonals in the polygon and adds the number of degrees in an exterior angle, the result is 84. What kind of polygon is George thinking about?

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  1. anonymous
    • 4 years ago
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    I've gotten to n(n-3)/2 + 360/n = 84. Then I was unsure what to do but I multiplied both sides by 2n to get rid of denominators and got n^2(n-3) + 720 = 168n which is the same as n^3 - 3n^2 - 168n + 720 = 0 but I can't factor that.

  2. Directrix
    • 4 years ago
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    Dodecagon (12 sides). Exterior angle of regular 12-gon is 30 and number of diagonals is 54. See the following link for solutions to Mr. Id's (above) equation. http://www.wolframalpha.com/input/?i=n%5E3+-+3n%5E2+-+168n+%2B+720+%3D+0

  3. anonymous
    • 4 years ago
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    How did you do that algebraically though. I mean sure you can guess and check...

  4. anonymous
    • 4 years ago
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    Ah, so my cubic equation was right. How would I factor that though?

  5. Directrix
    • 4 years ago
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    Using the rational root theorem, find that 12 is actually a root.

  6. Directrix
    • 4 years ago
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    @ Mr. ID --> the other two roots are irrational. No guessing and checking this time. :)

  7. anonymous
    • 4 years ago
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    Yes, I was just wondering how you got the 12. Did you factor my equation? If so, how?

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