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°.
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<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 answer is C because you can have a triangle with 5,9,n 5,10,n 9,10,n and 5,10,9
The reason why is because the sum of two sides is always larger than the value of the third side.
I hope that this helps.
A goes in stock price decreased
B goes in stock price increased
C goes in stock price remained constant
D goes in stock price decreased
E goes is stock price increased
Hope this helps!
I think it is d not sure tho
Yeah I know why ?????,????
