Answer:
its teaspoons
Step-by-step explanation:
Answer:
Is there any answer choes or something?
<h3>
Answer: D) 0</h3>
Work Shown:
f(x) = 4x^(-2) + (1/4)x^2 + 4
f ' (x) = 4*(-2)x^(-3) + (1/4)*2x .... power rule
f ' (x) = -8x^(-3) + (1/2)x
f ' (2) = -8(2)^(-3) + (1/2)*(2)
f ' (2) = 0
Answer:
The horizontal distance from the plane to the person on the runway is 20408.16 ft.
Step-by-step explanation:
Consider the figure below,
Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°
BC is the horizontal distance from the plane to the person on the runway.
We have to find distance BC,
Using trigonometric ratio,
![\tan\theta=\frac{Perpendicular}{base}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cfrac%7BPerpendicular%7D%7Bbase%7D)
Here,
,Perpendicular AB = 4000
![\tan\theta=\frac{AB}{BC}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cfrac%7BAB%7D%7BBC%7D)
![\tan 11.1^{\circ} =\frac{4000}{BC}](https://tex.z-dn.net/?f=%5Ctan%2011.1%5E%7B%5Ccirc%7D%20%3D%5Cfrac%7B4000%7D%7BBC%7D)
Solving for BC, we get,
![BC=\frac{4000}{\tan 11.1^{\circ} }](https://tex.z-dn.net/?f=BC%3D%5Cfrac%7B4000%7D%7B%5Ctan%2011.1%5E%7B%5Ccirc%7D%20%7D)
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft