Answer:
The direct distance from the plane to the start of the runway is approximately 5.0623 km
Step-by-step explanation:
The locations of the beacons on the runway, x = 5 km from the start of each runway
The angle of elevation of the airplane from the runway, θ = 9°
Let 'R' represent the direct distance from the plane to the start of the runway
The horizontal distance from the start of the runway to the beacon, 'x', the distance from the plane to the start of the runway, 'R', and the vertical height of the plane, 'h' form a right triangle with rides x, R, and h
By trigonometric ratios, we have;

Therefore, we have;

From which we have;

The direct distance from the plane to the start of the runway, R ≈ 5.0623 km
Answer:
f(-2) = 3.
Step-by-step explanation:
Reading off the graph, f(-2) = 3.
Answer- 7n-6/2
Step by step explanation-
solution,
here
Given,
g(n)=2n+6/7
Let g(n) be y
So,
Interchanging the position of n and y we get the reverse So,
n = 2y+6/7
or, 7n = 2y+6
or, 7n-6 = 2y
or, 7n-6/2 = y
Therefore the inverse function of g(n) is 7n-6/2
Thank you.
Answer:
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