Answer:
Relative Extrema: \left(\frac{2}{\sqrt{e}},-\frac{6}{e}\right)
Step-by-step explanation:
There is only one extreme minimum point that is \left(\frac{2}{\sqrt{e}},-\frac{6}{e}\right) and there is no any point of reflection for this function.
You can find it in attached pictures of graphs.
The smaller square has length of 4. So its perimeter is 4*4 = 16
The larger square has length of 8 and its perimeter is 8*4 = 32
The scale factor is 2. Since the larger square is twice the size of the smaller square (Or you can say that the scale factor is 1/2 if you are looking at the perspective of the larger square to the smaller square).
Step-by-step explanation:
135 cu.ft. = <u>5</u>cu.yd.
....
In matrix form, the system is given by

I'll use G-J elimination. Consider the augmented matrix
![\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%20-1%20%26%201%20%26%20-1%20%26%20-20%20%5C%5C%202%20%26%20-1%20%26%201%20%26%2029%20%5C%5C%203%20%26%202%20%26%201%20%26%2029%20%5Cend%7Barray%7D%20%5Cright%5D)
• Multiply through row 1 by -1.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%202%20%26%20-1%20%26%201%20%26%2029%20%5C%5C%203%20%26%202%20%26%201%20%26%2029%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%205%20%26%20-2%20%26%20-31%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%200%20%26%203%20%26%2024%20%5Cend%7Barray%7D%20%5Cright%5D)
• Multiply row 3 by 1/3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%200%20%26%20-3%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).
![\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%200%20%26%200%20%26%209%20%5C%5C%200%20%26%201%20%26%200%20%26%20-3%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
Then the solution to the system is

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
The third row tells us that
. Then in the second row,

and in the first row,

A = 210 + 20(h - 7) <== ur equation
if she earns 410 tutoring...
410 = 210 + 20(h - 7)
410 = 210 + 20h - 140
410 = 20h + 70
410 - 70 = 20h
340 = 20h
340/20 = h
17 = h <=== she spent 17 hrs tutoring