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A new youth center is being built in erie . The perimeter of the rectangular playing field is 296 yards. The length of the field is 2 yards less than double the width. What are the dimensions of the playing​ field?

Respuesta :

snog

Answer:

The dimensions are 50 yards by 98 yards

Step-by-step explanation:

If the width is w, then the length is 2w - 2. Perimeter can be expressed by 2 * (length + width), therefore:

2 * (2w - 2 + w) = 296

2 * (3w - 2) = 296

3w - 2 = 148

3w = 150

w = 50 so 2w - 2 = 2 * 50 - 2 = 98

09pqr4sT

Answer:

[tex]\Large \boxed{\mathrm{98 \ yards \cdot 50 \ yards}}[/tex]

Step-by-step explanation:

Let the length be l.

Let the width be w.

[tex]l = 2w - 2[/tex]

The perimeter of the rectangle is 296 yards.

[tex]P = 2l + 2w[/tex]

Plug in the value for l and P.

[tex]296 = 2(2w-2) + 2w[/tex]

Evaluate and solve for w.

[tex]296 = 4w - 4 + 2w \\\\296 = 6w - 4 \\\\300 = 6w \\\\50 = w[/tex]

The width is 50 yards.

Let w = 50 for l.

Evaluate and solve for l.

[tex]l = 2(50) - 2 \\ \\ l = 100 - 2 \\\\l = 98[/tex]

The length is 98 yards.

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