4.63 clark's country pet resort is fencing a new play area for dogs. the manager has purchased 130 yd of fence to enclose a rectangular pen. the area of the pen must be 1050 ydsquared. what are the dimensions of the pen?

Let W and L be width and length of the rectangular pen respectively.

Therefore,
Circumference, C = 2W+2L= 130 yd
Area, A = LW = 1050 yd^2=> L = 1050/W

Using the circumference expression and substituting for L;
130 = 2W + 2(1050/W) = 2W+2100/W
130*W = 2W*W + 2100
130W = 2W^2 +2100
2W^2-130W+2100 = 0

Solving for W;
W= [-(-130)+/- Sqrt ((-130)^2-4(2)(2100)]/2*2 = 32.5+/- 2.5
W = 30 or 35 yd

When W = 30, L = 1050/30 = 35
When W = 35, L = 1050/35 = 30

Therefore, W = 30 yd and L = 35 yd.

J= r + 9 because since jessica rode 9 more miles than roger, you add 9 miles to rogers mileage.