For this case we have the following function:

By the time the stone is thrown, x = 0.
We must evaluate this value of x in the function.
We have then:
Answer:
The height of the stone at the time is thrown is given by:
h (0) = 15 meters
Using derivatives, it is found that:
i) 
ii) 9 m/s.
iii) 
iv) 6 m/s².
v) 1 second.
<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
- The velocity is the derivative of the position.
- The acceleration is the derivative of the velocity.
In this problem, the position is:

item i:
Velocity is the <u>derivative of the position</u>, hence:

Item ii:

The speed is of 9 m/s.
Item iii:
Derivative of the velocity, hence:

Item iv:

The acceleration is of 6 m/s².
Item v:
t for which a(t) = 0, hence:




Hence 1 second.
You can learn more about derivatives at brainly.com/question/14800626
Answer:y = k x + 1
(1,3)
3 = k(1) + 1
2 = 1k
k = 2
Step-by-step explanation:
Answer:
He must get 33 hits in his next 46 times at bat to finish the year with a .400 batting average
Step-by-step explanation:
The player has already batted 134 times and will still bat 46 times. So in the end of the year, he is going to have 134 + 46 = 180 at bats.
How many hits does he need to have to hit .400?
This is 40% of 180, which is 0.4*180 = 72.
He has already 39 hits, so in his next 46 at bats, he will need 72 - 39 = 33 hits.