The perimeter of an NBA-sized basketball court is 288 ft.  The length (l) is 44 ft longer than its width (w). Find the width of the court.  Write only the numerical value to the nearest ft, you may omit the units of ft.

The perimeter of a rectangle (which is the shape of a basketball court) is given by the formula 2l + 2w, where l is the length and w is the width.

From the problem, we know that the perimeter is 288 ft and that the length is 44 ft longer than the width. This gives us two equations:

2l + 2w = 288 l = w + 44

We can substitute the second equation into the first to get:

2(w + 44) + 2w = 288 2w + 88 + 2w = 288 4w + 88 = 288 4w = 200 w = 50

So, the width of the court is 50 ft.

The perimeter of a rectangle (which is the shape of a basketball court) is given by the formula 2l + 2w, where l is the length and w is the width.

From the problem, we know that the perimeter is 288 ft and that the length is 44 ft longer than the width. This gives us two equations:

2l + 2w = 288 l = w + 44

We can substitute the second equation into the first to get:

2(w + 44) + 2w = 288 2w + 88 + 2w = 288 4w + 88 = 288 4w = 200 w = 50

So, the width of the court is 50 ft.

The perimeter of a rectangle is given by the formula 2*(length + width). In this case, we know that the perimeter is 288 ft and the length is 44 ft longer than the width. We can set up the following equation to solve for the width:

288 = 2*(width + (width + 44))

Solving this equation step by step:

1. Simplify the equation inside the parentheses:

288 = 2*(2*width + 44)

2. Distribute the 2:

288 = 4*width + 88

3. Subtract 88 from both sides:

200 = 4*width

4. Divide both sides by 4 to solve for width:

50 = width

So, the width of the court is 50 ft.