A long jumper leaves the ground at an angle of 22 above the horizontal, at a speed of 10 m/s. The height of the jumper can be modeled by (x) = -0.048x²+0.365x, where his the jumper's height in meters and X is the horizontal distance from the point of

launch

Part: 0/3

Part 1 of 3

(a) At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.

The long Jumper reaches a maximum helght when the horizontal distance from the point of

launch is approximately meters

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0/5 is 0, 2/0 is undefined, and 0/0 is indeterminate.

step-by-step explanation:

hope my answer has you!

idk i would have to be there to under stand it

step-by-step explanation:

idk i would have to be there to under stand it

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

- "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."