0.625 or <span>5/8 left of the apple</span>
Answer:
![\left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
In order to find out the resulting matrix, we will have to multiply the identity matric and the scalar 3:
The 3x3 identity matrix is:
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
Multiplying with scalar 3:
![3\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=3%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
The scalar will be multiplied by each element of the matrix.
Multiplying zeros with scalar 3 will give us zero. So the resulting matrix will be:
![\left[\begin{array}{ccc}3*1&0&0\\0&3*1&0\\0&0&3*1\end{array}\right] = \left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%2A1%260%260%5C%5C0%263%2A1%260%5C%5C0%260%263%2A1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D)
So the resultant matrix will be a scalar matrix with 3 at diagonal positions..
Answer:
F. 
Step-by-step explanation:
if you replace x with 2 and plug these into your calculator, you can see that
has the least value
Answer:
see below
Step-by-step explanation:
1.the equations have different slopes? They will intersect at one point so one solution
2.the equations have the same slope and different y-intercepts. They are parallel lines with a different y intercept so they will never intersect - no solutions
3.the equations have the same slope and same y-intercepts. they are the same line so they have infinite solutions