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,

Answer:
The total area for all 16 hamburgers is <u>113.04 in²</u>.
Step-by-step explanation:
Given:
Chris made hamburgers for his family and they each had a diameter of 3 in.
If he made 16 hamburgers.
Now, to find the total area for all 16 hamburgers.
So, the diameter 3 in is given and we need first to find the radius:

Now, to get the area of a hamburger we put formula:
<u><em>(Taking the value of π = 3.14.)</em></u>

Now. to get the total area of 16 hamburgers by multiplying 16 by the area of a hamburger:

Therefore, the total area for all 16 hamburgers is 113.04 in².
Should be <span>10.75, 11.5, 12.25, 13, 13.75, 14.5</span>
Based on the diagram and the ratio of width and height in it, the width that Mr. Howell should make his table is 48 inches .
<h3>How wide should Mr. Howell's table be?</h3>
The width of the table in the diagram is 12 cm and the height is 6cm.
This means that the ratio of width to height is:
12 : 6
2 : 1
As Mr. Howell wants his table to be 24 inches high, the width of the table would be:
2 : 1
x : 24
Cross-multiply to get:
x = 48 inches
Find out more on ratios at brainly.com/question/20594266
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Answer:
h = 60 ; x = 25
Step-by-step explanation:
You are going to want to use Pythagorean Theorem to solve these.
Pythagorean Theorem: a² + b² = c²
h² + 144² = 156²
h² + 20736 = 24336
h² = 3600
h = 60
x² + 60² = 65²
x² + 3600 = 4225
x² = 625
x = 25