Question

Number 7. Please asap.

Answer 1

7a)

$\frac{n(n - 3)}{2} = 665\\n(n - 3) = 665 \cdot 2\\n(n - 3) = 1330\\n^2 - 3n = 1330\\n^2 - 3n - 1330 = 0\\n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1330)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 5320}}{2} = \frac{3 \pm \sqrt{5329}}{2} = \frac{3 \pm 73}{2}\\n_1 = \frac{3 + 73}{2} = \frac{76}{2} = 38\\n_2 = \frac{3 - 73}{2} = \frac{-70}{2} = -35$

If n is the amount of sides then there's 38 sides in a polygon with 665 diagonals

7b)

We do the same as 7a) but with 406. We get that $n_1 = 30.04$ and $n_2 = -27.04$. A polygon cant have 30.04 sides

Answer 2

Answer: A. 38 sides

B. When solved, you get -27.04 sides or 30.04 sides. Neither of these are possible because the number of sides cannot be negative or a decimal.

Step-by-step explanation: Part A:

If the polygon has 665 diagonals, set your original n-sided polygon equal to it.

n(n - 3)/2 = 665

Multiply both sides by 2 so it’s not in the denominator.

n(n - 3) = 1330

Distribute n and subtract 1330 from both sides so it is equal to zero.

n² - 3n - 1330 = 0

Factor it:

(n - 38)(n * 35) = 0

Solve for n by setting it equal to zero:

n = 38 & n = -35

The number of sides cannot be negative, therefore, the answer to Part A is 38 sides.

Part B:

Set your original n-sided polygon equal to 406 and solve to prove that a polygon cannot have 406 diagonals.

n(n - 3)/2 = 406

Multiply both sides by 2 so it’s not in the denominator.

n(n - 3) = 812

Distribute n and subtract 812 from both sides so that it is equal to zero.

n² - 3n - 812 = 0

This cannot be factored, so plug it into the Quadratic Formula. Here are the values you need for the Quadratic Formula based on the polynomial above:

a = 1

b = -3

c = -812

When you solve this, you will get:

n = -27.04 and n = 30.04

A polygon cannot have negative sides and it cannot have decimal sides. It has to be a positive integer (positive whole number).