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The local high school is having tryouts for the cross country team. The school does not have a track, so the runners run around the school’s football field. The cross country coach determines that the width of the field is seventy yards shorter than the length.

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Complete Question

The local high school is having tryouts for the cross country team. The school does not have a track, so the runners run around the school's football field. The cross country coach determines that the width of the field is seventy yards shorter than the length.

Part A

Write an expression that represents the perimeter of the football field. Let x Represent the length of the football field. Include parentheses in your expression. Next, write an equivalent expression that does not include parentheses.

Part B

The cross country coach later determines that the length of the football field is 120 yards. All students must run five laps. Using your answer from Part A, determine the actual number of yards that each athlete must run in the tryouts. In order to make the team, students must complete the laps in 6 minutes. How quickly must they run each lap?

Answer:

 [tex]p=4A-140[/tex]

[tex]T =or 62sec[/tex]

Step-by-step explanation:

From the question we are told that

The width of the field is seventy yards shorter than the length.

a) Generally the equation for width and length is given as

         [tex]B=x-70[/tex]

Mathematically  in finding the perimeter [tex]p[/tex] around the field

         [tex]p=2(A+B)[/tex]

         [tex]p=2(2A-70)[/tex]

         [tex]p=4A-140[/tex]

b)Generally[tex]A=120[/tex]

Mathematically

         [tex]p= 4(120) -140[/tex]

         [tex]p=480-140[/tex]

         [tex]p= 340[/tex]

Therefore

Yards ran by individual athlete[tex]Y[/tex]

          [tex]340*5=1700\\Y=1700[/tex]

Time elapsed in a lap T

    [tex]T=\frac{340}{1700} *6\\T=1.2min[/tex]

    [tex]T =or \hspace 662sec[/tex]

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