It is in the air 5.625 seconds.
Using the quadratic formula,

0 is when the ball is first launched, 5.625 is when it hits the ground again.
Answer:
Sorry it's acually (h+3)(h-9)
It would be In between 1/2 and 2
Here is how I would show the work for this problem
5- (a-4) = 3 (a+2)
5-a+4 = 3*a+3*2
5-a+4 = 3a+6
-a+9 = 3a+6
-a+9+a = 3a+6+a
9 = 4a+6
9-6 = 4a+6-6
3 = 4a
4a = 3
4a/4 = 3/4
a = 3/4
So the solution is the fraction a = 3/4 (in decimal form 0.75)
Let me know if you have any questions or issues with any of the steps shown above. It's a lot to take in so feel free to ask about anything really.
Answer:

Step-by-step explanation:
We have the function:
![h(x)=f[f(x)]](https://tex.z-dn.net/?f=h%28x%29%3Df%5Bf%28x%29%5D)
And we want to find:

So, we will differentiate function <em>h</em>. By the chain rule, this yields:
![h^\prime(x)=f^\prime[f(x)]\cdot f^\prime(x)](https://tex.z-dn.net/?f=h%5E%5Cprime%28x%29%3Df%5E%5Cprime%5Bf%28x%29%5D%5Ccdot%20f%5E%5Cprime%28x%29)
Then it follows that:
![h^\prime(1)=f^\prime[f(1)]\cdot f^\prime(1)](https://tex.z-dn.net/?f=h%5E%5Cprime%281%29%3Df%5E%5Cprime%5Bf%281%29%5D%5Ccdot%20f%5E%5Cprime%281%29)
Using the table, we acquire:

And using the table again, we acquire:

Evaluate. Hence:
