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3 years ago

1. Carter will install fencing all around the flat area of his

Mathematics

1 answer:

MariettaO [177]3 years ago

8 0

Answer:

Amount of fencing required = 75 yards

Step-by-step explanation:

Distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between A(3, 6) and B(3, -2) = $\sqrt{(3-3)^2+(6+2)^2}$

= 8 yards

Distance between B(3, -2) and C(-7, 4) = $\sqrt{(3+7)^2+(-2-4)^2}$

= $\sqrt{136}$

= 11.66 ≈ 12 yards

Distance between C(-7, 4) and D(-7, -2) = $\sqrt{(-7+7)^2+(4+2)^2}$

= 6 yards

Distance between D(-7, -2) and E(-3, -2) = $\sqrt{(-7+3)^2+(-2+2)^2}$

= 4 yards

Distance between E(-3, -2) and F(-3, -8) = $\sqrt{(-3+3)^2+(-2+8)^2}$

= 6 yards

Distance between F(-3, -8) and G(3, -8) = $\sqrt{(-3-3)^2+(-8+8)^2}$

= 6 yards

Distance between G(3, -8) and H(10, -12) = $\sqrt{(3-10)^2+(-8+12)^2}$

= $\sqrt{49+16}$

= $\sqrt{65}$

= 8.06 ≈ 8 yards

Distance between H(10, -12) and J(10, 6) = $\sqrt{(10-10)^2+(-12-6)^2}$

= 18 yards

Distance between A(3, 6) and J(10, 6) = $\sqrt{(10-3)^2+(6-6)^2}$

= 7 yards

Since length of fence required = perimeter of the flat area

Perimeter of the given area = 8 + 12 + 6 + 4 + 6 + 6 + 8 + 18 + 7

= 75 yards

Therefore, amount of fencing required = 75 yards

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Answer:

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Step-by-step explanation:

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Take the derivative again:

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Take the derivative one more time:

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so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

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