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mamaluj [8]
3 years ago
12

A farmer has 165 feet of fencing material in which to enclose a rectangular grazing area. He wants the

Mathematics
1 answer:
Vikentia [17]3 years ago
8 0

Answer:

2*x + 2*y ≤ 165 ft

x > 50ft

0ft < y ≤ 20ft

Step-by-step explanation:

For a rectangle with length x and width y, the perimeter of the triangle is:

P = 2*x + 2*y

The perimeter needs to be equal or smaller than the length of fencing that the farmer has, then:

2*x + 2*y ≤ 165 ft

We also know that:

(x strictly larger than 50ft)

x > 50ft

(y can be, at most, 20 ft)

0ft < y ≤ 20ft

(where we added the restriction that y needs to be larger than zero, because we can not have a width equal or smaller than zero)

Then the system of inequalities that represent this situation is:

2*x + 2*y ≤ 165 ft

x > 50ft

0ft < y ≤ 20ft

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lidiya [134]

Answer:

C = (2,2)

Step-by-step explanation:

B = (10 ; 2)

M = (6 ; 2)

C = (x ; y )

|___________|___________|

B (10;2)            M (6;2)             C ( x; y)

So:

dBM = dMC

√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]

(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2

0 + (-4)^2 = (y-2)^2 + (x - 6)^2

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Also:

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64 = 16 - (x - 6)^2 + (x - 10)^2

48 = (x - 10)^2 - (x - 6)^2

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Thus:

16 - (x - 6)^2 = (y-2)^2

16 - (2 - 6)^2 = (y-2)^2

16 - (-4)^2 = (y-2)^2

16 - 16 = (y-2)^2

0 =  (y-2)^2

0 = y - 2

2 = y

⇒ C = (2,2)

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