David spots a plane at an angle of elevation of 35 degrees. If he is 2650 feet from the plane, what is the altitude of the plane?

1,519.98 feet

Step-by-step explanation:

Given the following

Angle of elevation = 35 degrees

Distance of david from the plane = 2650feet

Altitude = x

Using SOH CAH TOA

sin theta = opp/hyp

sin 35 = x/2650

x = 2650sin35

x = 2650(0.5736)

x = 1,519.98

Hence the altitude of the plane is 1,519.98 feet

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.

the answer is z7+8/5