Arocket is launched into the air. the projectile motion of the rocket can be modeled using h(t) = 112t –16t2, where t is the time since launch in seconds and h(t) is the height of the rocket at time t. when will the rocket be 196 feet in the air?

After 3.5 sec rocket will be at 196 feet in the air.

Step-by-step explanation:

The projectile motion of a rocket can be modeled using h(t) = 112t - 16t²

where t is the time since launch in seconds.

h(t) is the height of the rocket at time t.

we have to calculate when height h is 196 feet.

Now from the given function

196 = 112t - 16t²

-16t² + 112t - 196 = 0

-4t² + 28t - 49 = 0

Or 4t² - 28t + 49 = 0

(2t - 7)² = 0

2t - 7 = 0

t = 7/2 = 3.5 sec

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value