Arectangular athletic field is twice as long as its wide. if the perimeter of the athletic field is 180 yards, what are the dimensions?

The dimension of the rectangle is 30x60.

The opposite sides are parallel and equal.

Given

Length (L) = 2 x Width (W)

Perimeter (P) = 180 yards

Then

Length = 2W = 2x30 = 60

Thus, the dimension of rectangle is 30x60.

More about the rectangle link is given below.

link

l = 60 yd; w = 30 yd

Step-by-step explanation:

We have two bits of information

(a) The perimeter of the rectangle

The formula for the perimeter of a rectangle is

             P = 2l + 2w

             P = 180 yd     Insert into the formula

(1) 2l + 2w = 180

(b) The length-to-width ratio

(2) l = 2w

Substitute the value of l from Equation (2) into Equation (1)

2×(2w) + 2w = 180

Solve for w

4w + 2w = 180     Combine like term s

        6w = 180     Divide each side by 6

          w = 30 yd

Solve for l

Substitute the value for w into Equation (1).

2l +2×30 = 180

   2l +60 = 180     Subtract 60 from each side

          2l = 120      Divide each side by 2

            l = 60 yd

The field is 60 yd long and 30 yd wide.

Check:

P = 2×60 + 2×30

P = 120 + 60

P = 180 yd

Also

60 = 2 × 30

60 = 60

choice b is correct

step-by-step explanation:

we have been given the following system of linear inequalities;

we are required to graphically determine the solution to these system of inequalities. using a graphing tool, the solution is as shown in the attachment below;