25 pts to who ever can with this question: a. find the perimeter of the square b. find the perimeter of the rectangle c. find the sum of the perimeters of the two figures if s=5 d. the perimeter of the rectangle is greater than the perimeter of the square. find the difference between the perimeters of the two figures if s=7

A. 5s+2

B. 8s+8

C. 75

D. 27

Step-by-step explanation:

Is difficult to read the exact expressions. If a mistake is present then fix the number and use the same method.

A. The perimeter of the square is all 4 equal sides added together. The length of a side is . Since there are four sides then:

Adding 1/4+1/4+1/4+1/4=1. This applies here leaving 5s+2 as the answer.

B. The perimeter of the rectangle is all four sides added together. A rectangle has two pairs of equal side so P = 2l+2w.

C. If s=5 then the sum of the two perimeters is:

5s+2+8s+8

5(5)+2+8(5)+8

25+2+40+8

75

D. Subtract the two perimeters by computing s=7:

8s+8 -(5s+2)

8s+8-5s-2

8(7)+8-5(7)-2

56+8-35-2

64-35-2

29-2

27

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)