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Serhud [2]
3 years ago
11

The length of a new rectangular playing field is 55 yards longer than tripletriple the width. If the perimeter of the rectangula

r playing field is 426426 ​yards, what are its​ dimensions?
Mathematics
2 answers:
asambeis [7]3 years ago
6 0

Answer:

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

Lyrx [107]3 years ago
4 0

Answer: 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

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At 8:00 a.m., there were t inches of snow on the ground. At 5:00 p.m., there were 3t inches of snow on the ground. How many more
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<h3>How many more inches of snow were on the ground at 5:00 p.m. than at 8:00 a.m.?</h3>

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