Answer:
1) Triangles
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2)

3) Triangles
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Step-by-step explanation:
Hope that helps :D
<u><em>-Jazz</em></u>
Answer:
option D. 126 cm
Step-by-step explanation:
step 1
Find the scale factor
we know that
If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor
In this problem
Triangles PQR and XYZ are similar by AA Similarity Theorem
so

Let
z ---> the scale factor

substitute the given values

step 2
Find the perimeter of triangle XYZ
we know that
If two figures are similar, then the ratio of its perimeters is equal to the scale factor
Let
z ----> the scale factor
p_1 ----> the perimeter of triangle XYZ
p_2 ---> the perimeter of triangle PQR
so

The perimeter of triangle PQR is

we have

substitute


therefore
The perimeter of triangle XYZ is 126 cm
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:
y=-2(x+8)²-7
Step-by-step explanation:
Given: y=-2x²+5
After transformations: y=-2(x+8)²-7
Vertex would be (h,k) -> (-8,-7) instead of (0,5)