The length of a new rectangular playing field is 55 yards longer than tripletriple the width. If the perimeter of the rectangular playing field is 426426 yards, what are its dimensions?

The length of the rectangle is 159923.5 and the width of 53289.5

Step-by-step explanation:

We will make 2 equations one that is the perimeter and another that is the length with respect to the width

b = w*3+55

2b + 2 w = 426426

we will replace b with (w * 3 + 55) in the second equation

2b + 2 w = 426426

2(w*3+55) + 2w = 426426

we clear w to find its value

6w + 110 + 2w = 426426

8w = 426426 - 110

8w = 426316

w = 426316/8

w = 53289.5

Now that we have the value of w we replace it in the first equation and solve to find b

b = w*3 + 55

b = 53289.5*3 + 55

b = 159868.5 + 55

b = 159923.5

length = 173.5 yards

Width = 39.5 yards

Step-by-step explanation:

Let L represent the length of the rectangular garden.

Let W represent the width of the rectangular garden.

The formula for determining the perimeter of a rectangle is expressed as

Perimeter = 2(L + W)

If the perimeter of the rectangular playing field is 426 yards. This means that

2(L + W) = 426

Dividing through by 2, it becomes

L + W = 213 - - - - - - - - - - - -1

The length of a new rectangular playing field is 55 yards longer than triple the width. This means that

L = 55 + 3W

Substituting L = 55 + 3W into equation 1, it becomes

55 + 3W + W = 213

4W + 55 = 213

4W = 213 - 55

4W = 158

W = 158/4

W = 39.5

L = 55 + 3W = 55 + 3 × 39.5

L = 55 + 118.5

L = 173.5

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation: