Let the angle of elevation is x and the height of the rocket from the ground is y
tanx = y/15
by differentiating both sides with respect to T
sec²x·dx/dt = (dy/dt)/15
at y = 30 , the hypotenuse of the triangle = 15√5
sec²x=(15√5/15)²=5
5 dx/dt = 11/15
dx/dt = 11/75 rad/sec
Answer:
40.1
Step-by-step explanation:
Alright, so the first thing I would do is find the LCM of the dividends, which is 20. So we have 10 15/20 and 6 16/20. You can either leave it there and just carry the one (20 in this case) for 3 19/20.
If you multiply 10 by 20 and add 15, or 6 by 20 and add 16 (which I think is easier) you get 215/20 - 136/20 = 79/20. It can be simplified from here if you teacher wants it in the future.
Answer:
x = 14
Step-by-step explanation:
You want to solve for x, right?
(x-5)^(1/2)+5=2
You have it right so far.
(x - 5)^(1/2) = -3
(x - 5) = (-3)^2
x - 5 = 9
x = 9 + 5 = 14
There are a lot of ways you can do this, depending on what numbers you use, you can divide, multiply, add, or subtract to get 4,384, did you mean a specific term of math?<span />
