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Tomtit [17]
2 years ago
12

What is the perimeter of this rectangle?

Mathematics
1 answer:
kow [346]2 years ago
5 0

Answer:

I think I would be option C. for your question

Thank you......

Have a good day......

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Find the mean of the following data set<br> 73, 81, 51, 76, 89, 95, 88
kifflom [539]

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mean =  \frac{(73 + 81 + 51 + 76 + 89 + 95 + 88)}{7}  \\  = 79

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2 years ago
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the length of a rectangle is 2 in more than it's w i d t h the area of a rectangle is equal to 1 inch less than three times a pe
Marizza181 [45]
If we let x and y represent length and width, respectively, then we can write equations according to the problem statement.
.. x = y +2
.. xy = 3(2(x +y)) -1

This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).

The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.

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Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.

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2 years ago
Rearrange the following equation to solve for the variable indicated. -4x+2Y=8 solve for y
Alecsey [184]

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y = 8 + 4x / 2

Step-by-step explanation:

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Solve for X. Please help!
Pachacha [2.7K]

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its 2(x-4)

Step-by-step explanation:

4 0
3 years ago
Use the following matrices, A, B, C and D to perform each operation.
Vinvika [58]

Step-by-step explanation:

A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]

B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]

C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]

D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]

1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]

2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]

5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}

6 0
3 years ago
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