An airplane takes off at a 50° angle from the ground. It ascends a distance of 12 km. How high above the ground is the plane at this distance? (please show work)

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°

its 50 miles away from the starting point

we have the parent function given by,

1. the function is translated 5 units to the left.

so, we get,   becomes

2. the function is translated 5 units to the right.

so, we get,   becomes

3. the function is translated 5 units upwards.

so, we get,   becomes

4. the function is translated 5 units downwards.

so, we get,   becomes

5. the function is reflected across x-axis.

so, we get,   becomes

6. the function is reflected across y-axis.

so, we get,   becomes .

step-by-step explanation:

mark brainliest.