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:
5 x 0.6 = 5 x 6tenths, as 6/10 = 0.6
= 30 tenths
7 x 0.8 = 7 x 8tenths, as 8/10 is 0.8
= 56 tenths
Step-by-step explanation:
If you really want a more detailed explanation, just reply to this message.
Answer:
Step-by-step explanation:
Depending on whether or not the g(x) is x^2 or 2x, I have both ways.
If g(x) is x^2 - 9...
--> (3x^2 + 2) - (x^2 - 9) = 2x^2 + 11
If g(x) is 2x...
--> (3x^2 + 2) - (2x - 9) = 3x^2 - 2x + 11
Hope this helps!
The equation would be y = -(3/4)x+1
