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lbvjy [14]
3 years ago
10

1-x-7/3=x+9/4-3x-8/12

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
8 0

Answer:

x = 2 11/12

Step-by-step explanation:

Multiply the equation by 12. This eliminates fractions.

... 12 -12x -28 = 12x +27 -36x -8

Collect terms.

... -16 -12x = -24x +19

Add 24x+16

... 12x = 35

Divide by the coefficient of x

... x = 35/12 = 2 11/12

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A garden measuring 10 meters by 7 meters is going to have a walkway constructed all around the perimeter, increasing the total a
SashulF [63]

The width of the pathway is 3m

Step-by-step explanation:

Let the width of the pathway be 'w'

The area will be 130 square meters

The dimensions of the garden is 10 x 7

Area = (10+w) (7+w)

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3 years ago
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.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

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

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

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

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}

Differentiating with respect to y

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

Equate to 0

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

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

And also

\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}

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}

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