Arocket was launched into the air from a podium 6 feet off the ground. the rocket path is represented by the equation h(t)=-16t^2+120t+6, where h(t) represents the height, in feet, and t is the time, in seconds. find the average rate of change from the initial launch to the maximum height.

60

Step-by-step explanation:

The given function is:

The average rate of change of h(t) from t=a to t=b is given by:

We can rewrite this function as:

The maximum height of the rocket is 231 and it occurs at t=3.75

The initial launch occurs at: t=0

and

The average rate of change from the initial launch to the maximum height is

the answer is 13/48

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places