A boy throws a ball into the air. The equation h=−16t^2+23t+4 models the path of the ball, where h is the height (in feet) of the ball t seconds after it is thrown. How long is the ball in the air?

The total time for which the ball is in the air is 0.72 seconds and this can be determined by differentiating the given function.

Given :

A boy throws a ball into the air. The equation models the path of the ball, where h is the height (in feet) of the ball t seconds after it is thrown.

Differentiate the given function of height with respect to 't' in order to determine the value of 't'.

Now, equate the above differential equation to zero.

-32t + 23 = 0

23 = 32t

t = 0.72 seconds

So, the total time for which the ball is in the air is 0.72 seconds.

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0.72secs

Step-by-step explanation:

Given the height of the ball in air modeled by the equation:

h=−16t²+23t+4

Required

Total time it spent in the air

To get this we need to calculate its time at the maximum height

At the maximum height,

v = dh/dt = 0

-32t + 23 = 0

-32t = -23

t = 23/32

t = 0.72secs

Hence the total time it spend in the air will be 0.72secs

do i do it on a graph?

step-by-step explanation:

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x