The matrix is not properly formatted.
However, I'm able to rearrange the question as:
![\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C-2%263%265%7C3%5C%5C3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
Operations:
Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.
Answer:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The first operation:
This means that the new second row (R2) is derived by:
Multiplying the first row (R1) by 2; add this to the second row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by 2
![2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]](https://tex.z-dn.net/?f=2%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&5&7|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%265%267%7C1%5Cend%7Barray%7D%5Cright%5D)
The second operation:
This means that the new third row (R3) is derived by:
Multiplying the first row (R1) by -3; add this to the third row
The row 1 elements are:
Multiply by -3
![-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]](https://tex.z-dn.net/?f=-3%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Hence, the new matrix is:
One of the other angles is 72. Vertically opposite angles are equal.
Two angles that are side by side on this arrangement are supplementary.
That means they are to 180
x + 72 = 180
x = 180 - 72
x = 108
Answer
the for angles are
72 + 72 + 108 + 108
Answer:
Altitude
Step-by-step explanation:
An altitude of the triangle is the line segment from a vertex to the opposite side and perpendicular to the opposite side. The orthocenter will not always be inside the triangle. This is why the orthocenter is the point of concurrency of the lines containing the altitudes rather than just the altitudes.
The number of adenine sim a strand of DNA is equal to the number of thymines.
tan t is negative in quadrant IV and positive in quadrants 1 / 3
