<span>between 35k and 45k
so answer a.</span>
Answer:
C. T is not one-to-one because the standard matrix A has a free variable.
Step-by-step explanation:
Given

Required
Determine if it is linear or onto
Represent the above as a matrix.
![T(x_1,x_2,x_3) = \left[\begin{array}{ccc}1&-5&4\\0&1&-6\\0&0&0\end{array}\right] \left[\begin{array}{c}x_1&x_2&x_3\end{array}\right]](https://tex.z-dn.net/?f=T%28x_1%2Cx_2%2Cx_3%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-5%264%5C%5C0%261%26-6%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5Cend%7Barray%7D%5Cright%5D)
From the above matrix, we observe that the matrix does not have a pivot in every column.
This means that the column are not linearly independent, & it has a free variable and as such T is not one-on-one
If you multiply length times width to get area, it's the same with division. You do area divided by length or width. The equation would be 225 divided by 13 3/4. 225/13 3/4= 16.36 inches. 16.36 > 16 inches, so the answer would be no. (The mirrors length and width are 13 3/4 inches by 16 9/25)