Given, the equation that represents the height of an object:

First, we will find the velocity of the object which is the first derivative of the height using the method of the limits

We will find the value of the function y(t) when t = 3, and when t = 3+h

Substitute y(3+h) and y(3) into the expression of the limit

Where a = 0
d) compute the instantaneous velocity at t = 3

So, the answer will be:
Elimination method
when you make both y equal to each other
Answer:
4.587
Step-by-step explanation:
The answer to your question is 12. 1/.25=4, 4x3=12