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?

too drunk to answer this! sorry

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

If there are n sides, each vertex has (n-3) diagonals leading from it (there are n-1 other vertices, two of which are its neighbours; a line to any other is a diagonal). So the number of diagonals is n (n-3) / 2 (each diagonal can be started from either end). So we have to solven - n (n-3) / 2 = 0 n (n-3) - 2n = 0 n (n - 5) = 0 n = 0 or 5, obviously 0 doesn't count. So the answer is 5.

the number of diagonls=1/2n(n-3)0=1/2n(n-3)n-3=0n=3Remarkn is the number of sides

the number of diagonals is determined by the equation n(n-3)/2 where n equals the number of sides.. then in your problem, we will be having an equation;n(n-3)/2 - n = 0n(n-3) - 2n=0n^2 - 5n = 0n=5there are 5 sides.

no of diagnols is n(n-1)/2-n if the number of sides is 'n'so the equation becomes n(n-1)/2-2*n=0solving this one gets n=5

n(n-3)/2 - n = 0n(n-3) - 2n=0n^2 - 5n = 0n=0 n=5there are 5 sides.

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.