Anew community center is being built in oak valley. the perimeter of the rectangular playing field is 382 yards. the length of the field is 9 yards less than triple the width. what are the dimensions of the playing field?

The width is 50 yards and the length is 141 yards.

Step-by-step explanation:

Let's call: L the length of the field and W the width of the field.

From the sentence, the perimeter of the rectangular playing field is 382 yards we can formulate the following equation:

2L + 2W = 382

Because the perimeter of a rectangle is the sum of two times the length with two times the width.

Then, from the sentence, the length of the field is 9 yards less than triple the width, we can formulate the following equation:

L = 3W - 9

So, replacing this last equation on the first one and solving for W, we get:

2L + 2W = 382

2(3W - 9) + 2W = 382

6W -18 +2W = 382

8W - 18 = 382

8W = 382 + 18

8W = 400

W = 400/8

W = 50

Replacing W by 50 on the following equation, we get:

L = 3W - 9

L = 3(50) - 9

L = 141

So, the width of the rectangular field is 50 yards and the length is 141 yards.

x/3-10=8

x/3=18

x=48

900 degrees

step-by-step explanation: