The function h(t) = −16t2 + 96t + 80 shows the height h(t), in feet, of a projectile after t seconds. a second object moves in the air along a path represented by g(t) = 31 + 32.2t, where g(t) is the height, in feet, of the object from the ground at time t seconds. part a: create a table using integers 2 through 5 for the 2 functions. between what 2 seconds is the solution to h(t) = g(t) located? how do you know? part b: explain what the solution from part a means in the context of the problem.

- "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."

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )