The line AB has midpoint 2,5 A has

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

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'.

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry