Can anyone me with these three questions i don’t get this and i’m struggling plz

This is not an answer but a hint. a^2+b^2=c^2, where a and b are the shorter lengths of the right triangle and c is the longest. So for problem 18 and 19, it's pretty straightforward.

Answer for #20

For problem 20, the answer is 6sqrt2+6, because x and y are both 3sqrt2

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!