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.
Answer:
20
Step-by-step explanation:
Evaluate for y= −5
(−5)2+(2)(−5)+5
(−5)2+(2)(−5)+5
=20
You first separate the triangle from the rectangle. Then multiply one side be one side.
First you plug in for x,
2(y-3)+y=12
Then you solve for y
2y-5+y=12
-10y+y=12
-9y=12
-9y/9=12/-9
y=-1.333333333333333333333333333333333333333333333333333333333
