Answer:
Step-by-step explanation:
Step 1: Rearrange -3x + y = -6 into y = mx + b
y = 3x - 6
Step 2: Find the y-intercept
The y-intercept is "b" so in this case the y-intercept is -6
Step 3: Use
to graph this equation
We start at (0,3) the y-intercept and go up by 3 and left by 1 (
)
Answer:
for Part A
Answer is C.
Step-by-step explanation:

Here the angle, a is 60°, Opposite is 20 and adjacent is x.
Therefore we have,

Making x the subject gives us

Part B
Value of x to the nearest tenth

Answer:
Let
be the time for the
th task.
We know these times have a certain structure:
- Any 3 adjacent tasks will take half as long as the next two tasks.
In the form of an equations we have

- The second task takes 1 second

- The fourth task takes 10 seconds

So, we have the following system of equations:

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
Here is the augmented matrix for this system.
![\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcccccc%7Cc%7D%201%20%26%201%20%26%201%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%200%20%26%200%20%5C%5C%5C%5C%200%20%26%201%20%26%201%20%26%201%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%200%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5Cend%7Barray%7D%20%5Cright%5D)
b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.
- Subtract row 2 from row 1
. - Subtract row 2 from row 3
. - Add row 3 to row 2
. - Multiply row 3 by −1
. - Add row 4 multiplied by
to row 1
. - Subtract row 4 from row 3
.
Here is the reduced echelon form for the augmented matrix.
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%2015%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-11%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5Cend%7Barray%7D%20%5Cright%5D)
c) The additional rows are

and the augmented matrix is
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%2015%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-11%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%2020%20%5C%5C%5C%5C%201%20%26%201%20%26%201%20%26%200%20%26%200%20%26%200%20%26%2050%20%5Cend%7Barray%7D%20%5Cright%5D)
d) To solve the system you must use these row operations.
- Subtract row 1 from row 6
. - Subtract row 2 from row 6
. - Subtract row 3 from row 6
. - Swap rows 5 and 6.
- Add row 5 to row 3
. - Multiply row 5 by 2
. - Subtract row 6 multiplied by 1/2 from row 1
. - Add row 6 multiplied by 1/2 to row 3
.
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%205%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%2044%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%2090%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%2020%20%5Cend%7Barray%7D%20%5Cright%5D)
The solutions are:
.
Answer:
8.5
Step-by-step explanation:
To be parallel, this new line has to have the same slope as the other. To find the slope, we look at the number multiplying x.
-x means x is being multiplied by -1, so the slope, or m, is -1.
y = -x + b
Now, input the coordinate point and solve for b.
-2 = -1(2) + b
-2 = -2 + b
Add 2 to both sides.
0 = b
y = -x