1. -6(x + 2) - 2 2. -3 - 7(-3 - 6v) *

3. -5(a - 6) + 2a *

4.-5(1 + 2k) - 8(-4 + 5k) *

5. 5(-3p + 7) + 5(p - 1) *

6. -5(x + 2) + 5(x - 5) *

7. 4(3x + 7) - 5x *

8. __3x - 2(-4x + 5) *

9. 8 + (2x – 1)7 *

10. -4(3n – 5) – 7n *

He looks like he answered the question

Step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.