Question

Me ajudem com a 3º questão

Answer 1

I can't read that language but I'll guess it says write a second degree equation then solve for n when d=10.

$10 = \dfrac{n(n-3)}{2}$

$20 = n^2 - 3n$

$n^2 - 3n - 20 = 0$

Answer: n² - 3n - 20 = 0

That doesn't factor so there is no integer n solution.

That means there are no polygons with 10 diagonals.

$n = \frac 1 2(3 \pm \sqrt{89})$