Answer:

or

Considering ![\theta \in (0, 2\pi]](https://tex.z-dn.net/?f=%5Ctheta%20%5Cin%20%280%2C%202%5Cpi%5D)

or

Step-by-step explanation:

We have:

Therefore,

or

---------------------------------

or

The value is -8 of the expression
The height of the rocket is found in terms of the angle as
.. h/(3 mi) = tan(θ)
.. h = (3 mi)*tan(θ)
Then the rate of change of height (vertical velocity) is
.. h' = (3 mi)*sec(θ)^2*θ'
.. h' = (3 mi)*4*(1.5 rad/min)
.. h' = 18 mi/min
The rocket's velocity is 18 miles per minute at that moment.
Answer:
38.480
Hope this helps!!!:)
Step-by-step explanation:
Answer:
the slope of the line is 2/3x
Step-by-step explanation:
i hope this helps :)