Answer:
Is there a picture you can take
Step-by-step explanation:
Answer: D
<u>Step-by-step explanation:</u>
The first matrix contains the coefficients of the x- and y- values for both equations (top row is the top equation and the bottom row is the bottom equation. The second matrix contains what each equation is equal to.
![\begin{array}{c}2x-y\\x-6y\end{array}\qquad \rightarrow \qquad \left[\begin{array}{cc}2&-1\\1&-6\end{array}\right] \\\\\\\begin{array}{c}-6\\13\end{array}\qquad \rightarrow \qquad \left[\begin{array}{c}-6\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7D2x-y%5C%5Cx-6y%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-1%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
The product will result in the solution for the x- and y-values of the system.
Answer: if Gregory draws the segment with endpoints A and A’, then the midpoint will lie on the line of reflection.
Explanation:
Given that a triangle ABC is reflected in triangle A'B'C'
Here reflection is done on a line
If you imagine the line as a mirror then ABC will have image on the mirror line as A'B'C'
Recall that in a mirror the object and image would be equidistant from the mirror and also the line joining the image and object would be perpendicular to the mirror
But note that corresponding images will only be perpendicular bisector to the line
So A and A' only will be corresponding so AA' will have mid point on line
Option 1 is right
