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,

From the list of colors: green, red, blue, and white, there are four colors. The concept that should be used in order to determine the number of 4-color code stripes that can be drawn in the sports car is the concept of BASIC PRINCIPLE OF COUNTING.
The equations that would best help us in answering the question is,
n = 4! (that is, four factorial)
For the first pick, there are four choices. When one is already used then, there are only 3 choices for the second color, and so on until the last color.
n = 4 x 3 x 2 x 1 = 24
Thus, there are 24 types of 4-color codes.
So firstly, we have to find the radius of the circular garden before finding the circumference (the amount of fencing needed to surround the garden). To find the radius, use the area formula (
), plug in the area of the garden (36 ft^2) and solve for r as such:

So that we know the radius, plug that into the circumference equation (
) to solve:

Your answer is A. 12√π.
8 divided by 2 is 4, so multiply 3/4 by 2, 6/8, than you have your answer, 6.
Answer:
x = -8 and x = 4
Step-by-step explanation:
given
f(x) = (x+8) (x - 4)
recall that at any point on the x-axis, y = 0 [i.e f(x) = 0]
hence to find where the graph crosses the x-axis, we simply substitue f(x) = 0 into the equation and solve for x
f(x) = (x+8) (x - 4)
0 = (x+8) (x - 4)
Hence
either,
(x+8) = 0 ----> x = -8 (first crossing point)
or
(x-4) = 0 ------> x = 4 (second crossing point)
Hence the graph crosses the x-axis at x = -8 and x = 4