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Mariulka [41]
3 years ago
5

SCALCET8 4.7.011. Consider the following problem: A farmer with 950 ft of fencing wants to enclose a rectangular area and then d

ivide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens
Mathematics
1 answer:
kaheart [24]3 years ago
6 0

Answer:

For any rectangle, the one with the largest area will be the one whose dimensions are as close to a square as possible.

However, the dividers change the process to find this maximum somewhat.

Letting x represent two sides of the rectangle and the 3 parallel dividers, we have 2x+3x = 5x.

Letting y represent the other two sides of the rectangle, we have 2y.

We know that 2y + 5x = 750.

Solving for y, we first subtract 5x from each side:

2y + 5x - 5x = 750 - 5x

2y = - 5x + 750

Next we divide both sides by 2:

2y/2 = - 5x/2 + 750/2

y = - 2.5x + 375

We know that the area of a rectangle is given by

A = lw, where l is the length and w is the width. In this rectangle, one dimension is x and the other is y, making the area

A = xy

Substituting the expression for y we just found above, we have

A = x (-2.5x+375)

A = - 2.5x² + 375x

This is a quadratic equation, with values a = - 2.5, b = 375 and c = 0.

To find the maximum, we will find the vertex. First we find the axis of symmetry, using the equation

x = - b/2a

x = - 375/2 (-2.5) = - 375/-5 = 75

Substituting this back in place of every x in our area equation, we have

A = - 2.5x² + 375x

A = - 2.5 (75) ² + 375 (75) = - 2.5 (5625) + 28125 = - 14062.5 + 28125 = 14062.5

Step-by-step explanation:

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Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo
kolezko [41]

Let <em>x</em> and <em>y</em> be the unit rates at which one large pump and one small pump works, respectively.

Two large/one small operate at a unit rate of

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2<em>x</em> + <em>y</em> = 0.25

One large/three small operate at the same rate,

(1 pool)/(4 hours) = 0.25 pool/hour

<em>x</em> + 3<em>y</em> = 0.25

Solve for <em>x</em> and <em>y</em>. We have

<em>y</em> = 0.25 - 2<em>x</em>   ==>   <em>x</em> + 3 (0.25 - 2<em>x</em>) = 0.25

==>   <em>x</em> + 0.75 - 6<em>x</em> = 0.25

==>   5<em>x</em> = 0.5

==>   <em>x</em> = 0.1

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In other words, one large pump alone can fill a 1/10 of a pool in one hour, while one small pump alone can fill 1/20 of a pool in one hour.

Now, if you have four each of the large and small pumps, they will work at a rate of

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meaning they can fill 3/5 of a pool in one hour. If it takes time <em>t</em> to fill one pool, we have

(3/5 pool/hour) (<em>t</em> hours) = 1 pool

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3 years ago
Answer with the explanation step by step
andrey2020 [161]

Answer:

The answer is A. \frac{3(x-21)}{(x+7)(x-7)}.

Step-by-step explanation:

To find the difference of this problem, start by simplifying the denominator, which will look like \frac{3}{x+7}-\frac{42}{(x+7)(x-7)}. Next, multiply \frac{3}{x+7} by \frac{x-7}{x-7}  to create a fraction with a common denominator in order to subtract from \frac{42}{(x+7)(x-7)}. The problem will now look like \frac{3}{x+7}*\frac{x-7}{x-7}-\frac{42}{(x+7)(x-7)}.  

Then, simplify the terms in the problem by first multiplying \frac{3}{x+7} and \frac{x-7}{x-7}, which will look like \frac{3(x-7)}{(x+7)(x-7)}-\frac{42}{(x+7)(x-7)}. The next step is to combine the numerators over the common denominator, which will look like \frac{3(x-7)-42}{(x+7)(x-7)}.

Next, simplify the numerator, and to simplify the numerator start by factoring 3 out of 3(x-7)-42, which will look like \frac{3(x-7-14)}{(x+7)(x-7)}. Then, subtract 14 from -7, which will look like \frac{3(x-21)}{(x+7)(x-7)}. The final answer will be \frac{3(x-21)}{(x+7)(x-7)}.    

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