If you drive 1000 miles, then you would multiply $0.05 * 1000 miles =$50.
Answer:
Choice A: stretched horoxontally by a factor of 2 and translated up by 3 units
Explanation:
The original graph is:
y = log(x)
The modified one is:
y = log(2x) + 3
We can note that 2 changes occured:
1- A 3 has beed added to the whole gaph. This means that the curve has been moved (translated) upwards by 3 units
2- The x inside the log is miltiplied by 2. This means that the graph is stretched horizontally by the factor added (2).
Combining these two, we will find that the correct choice is A
Hope this helps :)
Answer:
One of the obvious non-trivial solutions is 
.
Step-by-step explanation:
Suppose the matrix A is as follows:
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&3_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%263_%7B23%7D%5C%5Ca_%7B31%7D%26a_%7B32%7D%26a_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
The observed system 
after multiplying looks like this
![Ax=0 \iff \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \cdot \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] =0 \iff \\ \\a_{11}x_1+a_{12}x_2+a_{13}x_3=0\\a_{21}x_1+a_{22}x_2+a_{23}x_3=0\\a_{31}x_1+a_{32}x_2+a_{33}x_3=0\\\\](https://tex.z-dn.net/?f=Ax%3D0%20%5Ciff%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%26a_%7B23%7D%5C%5Ca_%7B31%7D%26a_%7B32%7D%26a_%7B33%7D%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5Cx_3%5Cend%7Barray%7D%5Cright%5D%20%3D0%20%5Ciff%20%5C%5C%20%5C%5Ca_%7B11%7Dx_1%2Ba_%7B12%7Dx_2%2Ba_%7B13%7Dx_3%3D0%5C%5Ca_%7B21%7Dx_1%2Ba_%7B22%7Dx_2%2Ba_%7B23%7Dx_3%3D0%5C%5Ca_%7B31%7Dx_1%2Ba_%7B32%7Dx_2%2Ba_%7B33%7Dx_3%3D0%5C%5C%5C%5C)
Since we now that 
, where 
are the columns of the matrix A, we actually know this:
![-2\cdot \left[\begin{array}{ccc}a_{11}\\a_{21}\\a_{31}\end{array}\right] +3\cdot \left[\begin{array}{ccc}a_{12}\\a_{22}\\a_{32}\end{array}\right] -5\cdot \left[\begin{array}{ccc}a_{13}\\a_{23}\\a_{33}\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=-2%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%5C%5Ca_%7B21%7D%5C%5Ca_%7B31%7D%5Cend%7Barray%7D%5Cright%5D%20%2B3%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B12%7D%5C%5Ca_%7B22%7D%5C%5Ca_%7B32%7D%5Cend%7Barray%7D%5Cright%5D%20-5%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B13%7D%5C%5Ca_%7B23%7D%5C%5Ca_%7B33%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Once we multiply and sum up these 3 by 1 matrices, we get that these equations hold:
This actually means that the solution to the previously observed system of equations (or equivalently, our system ) has a non-trivial solution .
