A model rocket launched directly upward at a speed of 16 meters per second from a helght of 2 meters . The f(t) = - 4.9t ^ 2 + 16t + 2 models the relationship between the height of the rocket and the time after launch, t. In seconds. Which is the maximum height meters the rocket will reach Round to two decimal places Provide your answer below: meters

f(t) = 15.06 m

Step-by-step explanation:

The relationship between the height of the rocket and the time after launch, t. In seconds is given by :

We need to find the maximum height the rocket will reach the ground.

When it reaches the ground,

f(t) = 0

So,

Maximum height,

f'(t) = 0

Put t = 1.63 in equation (1).

So, the maximum height of the rocket is 15.06 m.

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step-by-step explanation: just trying to get points sorry : (

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!