Answer:
Using PEMDAS, you should get the answer of 5.
Answer:
This shows 3 pivot position matrixes.
Step-by-step explanation:
The given matrix is:
![\left[\begin{array}{ccc}1&-2&-5\\0&4&3\\-3&3&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-5%5C%5C0%264%263%5C%5C-3%263%260%5Cend%7Barray%7D%5Cright%5D)
The option D is correct for this matrix.
The matrix is invertible and the given matrix has 3 pivot positions.
The matrix is invertible if its determinant is nonzero.
Multiply the 3rd row by 1/3.we get:
![\left[\begin{array}{ccc}1&-2&-5\\0&4&3\\-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-5%5C%5C0%264%263%5C%5C-1%261%260%5Cend%7Barray%7D%5Cright%5D)
Now, add the first row with third row:
![\left[\begin{array}{ccc}0&-1&-5\\0&4&3\\-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%26-5%5C%5C0%264%263%5C%5C-1%261%260%5Cend%7Barray%7D%5Cright%5D)
Replace third row by first row:
![\left[\begin{array}{ccc}-1&1&0\\0&4&3\\0&-1&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%261%260%5C%5C0%264%263%5C%5C0%26-1%26-5%5Cend%7Barray%7D%5Cright%5D)
This shows 3 pivot position matrixes.
Hence, a matrix is invertible and has 3 pivot positions.
Answer:
y = -1/2x + 4
Step-by-step explanation:
I got -1/2x because for every 2 spaces, the line goes down 1 (rise/run)
I got +4 because 4 is where x = 0, or the y intercept
This breaks into two inequalities:
2x - 3 > 6
and
-2x + 3 > 6
------------------------
2x - 3 > 6
Add 3 to both sides:
2x > 9
Divide 2 to both sides:
x > 4.5
-----------------------
-2x + 3 > 6
Subtract 3 to both sides:
-2x > 3
Divide -2 to both sides:
x < -1.5