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sweet [91]
3 years ago
15

A rectangular athletic field is twice as long as its wide. If the perimeter of the athletic field is 180 yards, what are the dim

ensions?
Mathematics
1 answer:
andrezito [222]3 years ago
6 0

Answer:

l = 60 yd; w = 30 yd

Step-by-step explanation:

We have two bits of information

(a) The perimeter of the rectangle

The formula for the perimeter of a rectangle is

             P = 2l + 2w

             P = 180 yd     Insert into the formula

(1) 2l + 2w = 180

===============

(b) The length-to-width ratio

(2) l = 2w

Substitute the value of l from Equation (2) into Equation (1)

2×(2w) + 2w = 180

===============

Solve for w

4w + 2w = 180     Combine like term s

        6w = 180     Divide each side by 6

          w = 30 yd

===============

Solve for l

Substitute the value for w into Equation (1).

2l +2×30 = 180

   2l +60 = 180     Subtract 60 from each side

          2l = 120      Divide each side by 2

            l = 60 yd

The field is 60 yd long and 30 yd wide.

===============

<em>Check</em>:

P = 2×60 + 2×30

P = 120 + 60

P = 180 yd

Also

60 = 2 × 30

60 = 60

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Answer:

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Step-by-step explanation:

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\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

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20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

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Here’s the diagram:

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