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Travka [436]
2 years ago
13

(-1,8) and (8,5). this is a coordinate distance wuestion that i rly need help on

Mathematics
1 answer:
Inga [223]2 years ago
4 0

Answer:

3\sqrt{10} ≈ 9.45

Step-by-step explanation:

calculate the distance d between the points using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (- 1, 8 ) and (x₂, y₂ ) = (8, 5 )

d = \sqrt{(8-(-1))^2+(5-8)^2}

   = \sqrt{(8+1)^2+(-3)^2}

   = \sqrt{9^2+9}

   = \sqrt{81+9}

   = \sqrt{90}

   = \sqrt{9(10)}

   = \sqrt{9} × \sqrt{10}

   = 3\sqrt{10} ← exact value

   ≈ 9.49 ( to 2 dec. places )

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The complete question in the attached figure

we know that
area each pasture=15000 ft²
area each pasture=x*y
so
15000=x*y-----> equation 1

perimeter two rectangular pastures=1050 ft
perimeter two rectangular pastures=2*[2x+y]+y----> 4x+2y+y---> 4x+3y
so
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clear variable y
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the answer part a) 
y=350-(4/3)x------> <span>it was proved
</span>
Part b)Find the possible lengths and widths of each pasture.
substitute equation 2 in equation 1
15000=x*y-----> 15000=x*[350-(4/3)x]
15000=350x-(4/3)x²----> multiply by 3----> 45000=1050x-4x²
4x²-1050x+45000=0

using a graph tool----> to resolve the second order equation
see the attached figure

the solution is 
x=53.942 ft
x=208.558 ft

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for x=53.94 ft
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for x=208.56 ft
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the possible lengths and widths of each pasture are
case 1
x=53.94 ft
y=278.09 ft

case 2
x=208.56 ft
y=71.92 ft

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