9514 1404 393
Answer:
9.5°, yes
Step-by-step explanation:
The relevant trig relation is ...
Tan = Opposite/Adjacent
The distance opposite the angle of elevation is the plane's height, 500 m. The distance adjacent to the angle of elevation is the horizontal distance to the plane, 3 km = 3000 m. Then the angle is found from ...
tan(α) = 500/3000 = 1/6
α = arctan(1/6) ≈ 9.46°
The plane is approaching at an angle of 9.46°. It is safe to land, since that angle is less than 15°.
_____
<em>Additional comment</em>
The usual descent angle for most commercial air traffic is 3°. Some airport geography demands it be different (steeper). A higher descent angle can put undue stress on the landing gear.
The chord length is the same as the radius, so R^2 = 9.
Area = (1/2)(9)(pi/3 - sqrt(3)/2) = (3/4) (2 pi - 3 sqrt(3))
<h3>
I'll teach you how to solve f(x)=0x63 - 52x – 96</h3>
----------------------------------------------------------
f(x)=0x^3 - 52x – 96
Apply rule:
0 * a= 0
0-52x-96
Group like terms:
-52x+0-96
Add/subtract the numbers:
0-96= -96
-52x-96
Your Answer Is -52x-96
plz mark me as brianliest if this helped :)
Answer:
Step-by-step explanation:
12.80+3.42= 16.22
16.22-9.70= 6.52 that's your answer