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gogolik [260]
3 years ago
13

The length of a rectangle is 3 more than twice its width. the perimeter is 48 feet. find the width.

Mathematics
1 answer:
zepelin [54]3 years ago
8 0
48 = 2l + 2w
l = 3w

48 = 2(3w) + 2w
48 = 6w + 2w
48 = 8w
6 = w

The width is 6 feet. The length is 18 feet. Hope this helps! :)

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Answer:

Step-by-step explanation:

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Trevor bought 3 shirts that all cost the same amount. He also bought a jacket for $47.15. With the given information in the torn
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Answer: See explanation

Step-by-step explanation:

Your question isn't complete but let me help out. The value of each shirt will be gotten by subtracting $47.15 from the total amount paid for the shirt and jacket. Let's say the total amount paid is $90.02

Since the jacket cost $47.15, the cost of the shirts will be:

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Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an
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Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

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y is the number of investors that contributed 8,000.

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I am going to simplify by 4000. So:

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Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

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I am going to multiply 1) by -1. So we have

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-x + x - y + 2y = -60 + 87

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x = 60-y = 60-27 = 33

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2) Solve the system by row-reducing the corresponding augmented matrix.

2x + y = 17

x + y = 13

This system has the following augmented matrix:

\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]

To help the row reducing, i am going to swap the first with the second line:

L1  L2

So we have:

\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]

Now, reducing the first column.

L2 = L2 - 2L1

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\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]

Now we do:

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And the matrix is:

\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]

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Answer:

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I just looked it up

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