Answer:
The solution of the system of linear equations is 
Step-by-step explanation:
We have the system of linear equations:

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.
The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).
From the system of linear equations that we have, the coefficient matrix is
![\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%26-6%5C%5C1%26-2%263%5C%5C3%261%260%5Cend%7Barray%7D%5Cright%5D)
the variable matrix is
![\left[\begin{array}{c}x&y&z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D)
and the constant matrix is
![\left[\begin{array}{c}12&-2&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%26-2%2613%5Cend%7Barray%7D%5Cright%5D)
We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so
![\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%263%26-6%2612%5C%5C1%26-2%263%26-2%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:
- multiply the 1st row by 1/2
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C1%26-2%263%26-2%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
- add -1 times the 1st row to the 2nd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%26-7%2F2%266%26-8%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
- add -3 times the 1st row to the 3rd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%26-7%2F2%266%26-8%5C%5C0%26-7%2F2%269%26-5%5Cend%7Barray%7D%5Cright%5D)
- multiply the 2nd row by -2/7
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%26-7%2F2%269%26-5%5Cend%7Barray%7D%5Cright%5D)
- add 7/2 times the 2nd row to the 3rd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%260%263%263%5Cend%7Barray%7D%5Cright%5D)
- multiply the 3rd row by 1/3
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add 12/7 times the 3rd row to the 2nd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add 3 times the 3rd row to the 1st row
![\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%260%269%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add -3/2 times the 2nd row to the 1st row
![\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%263%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
From the reduced row echelon form we have that

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.
Answer:
B) Figure 2
Explanation:
In all parallelograms, both pairs of opposite sides are parallel to each other. Thus the name parallelogram. Figure 1, figure 2 , and figure 4 all share that property. Figure 2 does not, as it only has one pair of opposite sides parallel to each other. It is a trapezoid.
Answer:
hhsnghh sihdg hhurihhspig
Step-by-step explanation:
shhgnshhistdh shdj
Just remember the syllables PEMDAS which stand for Parentheses, Exponents, Multiplication, Division, Addition and Subtraction. If you can’t remember it, just write it on the side of your paper. But make sure to use it in that order.
Answer: The initial volume is 593.76mL
Step-by-step explanation:
As you do not say anithing about the pressure, i guess that the pressure remains constant.
If the gas is an ideal gas, we have:
P*V = n*R*T
where P is pressure, n is number of moles and R is a constant.
Now, initially we have:
P*Vi = n*R*315°C
finally we have:
P*825mL = n*R*452°C
Now we can take the quiotient of those two equations and get:
(P*Vi)/(P*852mL) = (n*R*315°C)/( n*R*452°C)
Now we have:
Vi/852mL = 315/452
Vi = (315/452)*852mL = 593.76mL
So when we expand the gas at constant pressure, we increase the temperature.