Answer:
D
Step-by-step explanation:

Answer:

Step-by-step explanation:
To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:
![A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%261%261%261%5C%5C3%267%26-1%261%5Cend%7Barray%7D%5Cright%5D)
And the vector B is formed with the solution of each equation of the system:![b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=b%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-3%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called
.
![A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A_%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%266%261%261%5C%5C3%261%26-1%261%5Cend%7Barray%7D%5Cright%5D)
The value of y using Cramer's rule is:

Find the value of the determinant of each matrix, and divide:


Answer:
I believe its 57.
Step-by-step explanation:
180-66=57.
57+57+66=180
Answer:
a little late
205 1/5 is your answer
Step-by-step explanation:
break down the 3-D figure
Find area of triangle A=1/2bh
=1/2(8 inch)(6 9/10inch)
=1/2(8 inch)(69/10 inch)
=138/5 square inch
then multiply two because you have 2 triangles
=138/5 (2)
= 276/5 square inch
= 55 1/5 sq.in
Find area of rectangle
A = lw
=6 1/4 in (8 in)
= 25/4 (8 in.)
= 50 sq. in.
then multiply by 3 because of the three rectangles
= 50 sq. inch (3)
= 150 sq. in.
Add all together
55 1/5 sq. inch + 150 sq. inch = 205 1/5 sq. inch