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07242007  #1 
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geometry. i think?
the number of diagonals of a regular polygon is subtracted from the number of sides of the polygon, and the result is zero. what is the number of sides of this polygon?

07242007  #2 
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too drunk to answer this! sorry

07242007  #3 
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wont pentagons, hexagons, octagons etc have the same number of sides and diagnols?

07242007  #4 
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If there are n sides, each vertex has (n3) diagonals leading from it (there are n1 other vertices, two of which are its neighbours; a line to any other is a diagonal). So the number of diagonals is n (n3) / 2 (each diagonal can be started from either end). So we have to solven  n (n3) / 2 = 0 n (n3)  2n = 0 n (n  5) = 0 n = 0 or 5, obviously 0 doesn't count. So the answer is 5.

07242007  #5 
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the number of diagonls=1/2n(n3)0=1/2n(n3)n3=0n=3Remarkn is the number of sides

07242007  #6 
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the number of diagonals is determined by the equation n(n3)/2 where n equals the number of sides.. then in your problem, we will be having an equation;n(n3)/2  n = 0n(n3)  2n=0n^2  5n = 0n=5there are 5 sides.

07242007  #7 
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no of diagnols is n(n1)/2n if the number of sides is 'n'so the equation becomes n(n1)/22*n=0solving this one gets n=5

07242007  #8 
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n(n3)/2  n = 0n(n3)  2n=0n^2  5n = 0n=0 n=5there are 5 sides.

07242007  #9 
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i think in every polygon the maximum diagonals you can have is 1/2 of the number of its sides. for example a square has 4 sides and its has a max. of 2 diagonals. so i think a zero answer is impossible.

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