Help me solve this algebra 2 question.

A

Step-by-step explanation:

Alright, this may come out weird on computer, but here we go.

First, you want to change this into a multiplication problem, with Keep change flip. So now it's (c^2-c-20/c^2-6c+5) * (3c-3/c^2-16). You then want to simplify some number down.

You can simplify the first part using the X game (not sure if you know this), but you put the c value at the top, and the B value at the bottom, and the two sides blank. These two values must add to equal the C value, but multiply to equal the B value, so you get -5 and 4 for the numerator of the top. Following this formula, and using difference of perfect squares, you'll end up with the following.

(c-5)(c+4)/(c-5)(c-1) * 3(c-1)/(c+4)(c-4)

You then need to cancel out your different groups, so you'll cancel out (c-5), (c+4), (c-1).

After all that, you'll end up with 3/c-4, so the answer is A.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2