Question marks are what i need filled in, will give brainliest. given: a b c d is a square. prove: the diagonals of a b c d are perpendicular. the coordinates of square abcd are a(0, 0) , b(a, 0) , , a), and d(0, the slope of ac¯¯¯¯¯ , when simplified, is equal to . the slope of bd¯¯¯¯¯, when simplified, is equal to −1 . the product of the slopes is equal to . therefore, ac¯¯¯¯¯ is perpendicular to bd¯¯¯¯¯.

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

i ran through it by the other answer is wrong

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."