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laila [671]
3 years ago
8

Houses cover 51% of the area of a desert community. Use the grid to model the community and complete the statements. The entire

area of the community is represented by squares. The squares that have houses on them represent the area of the community covered with houses.
Mathematics
2 answers:
Mekhanik [1.2K]3 years ago
8 0

Answer:

Too lazy To type put here is an image

pentagon [3]3 years ago
3 0

Answer:

The entire area of the community is represented by  

   49

   51

✔ 100

   151

squares.

The  

   49

✔ 51

   100

   151

squares that have houses on them represent the area of the community covered with houses.

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Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel
timama [110]

Answer:

\displaystyle A=\frac{1}{192}

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

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The amount of fence needed to enclose the external and the internal divisions is

P=4x+3y

We know the total fencing is 1/2 miles long, thus

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Solving for x

\displaystyle x=\frac{\frac{1}{2}-3y}{4}

The total area of the pasture is

A=x.y

Substituting x

\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y

\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}

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\displaystyle A'=\frac{\frac{1}{2}-6y}{4}

Equate to 0

\displaystyle \frac{\frac{1}{2}-6y}{4}=0

Solving for y

\displaystyle y=\frac{1}{12}

And also

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Compute the second derivative

\displaystyle A''=-\frac{3}{2}.

Since it's always negative, the point is a maximum

Thus, the maximum area is

\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}

6 0
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liubo4ka [24]
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