Making a round trip from Fairview to Cartersville, a distance of 20 miles, a pilot faces 30 mph head wind one way and 30 mph tail wind on the return trip. The return trip takes 45 minutes less than the outbound journey. Find the speed of the plane in still air

50mph

Step-by-step explanation:

Given the following :

Distance (d) of journey = 20 miles

Wind speed = 30mph head wind in one way, 30mph tail wind in the other direction

Return trip = 45 minutes (45/60 = 0.75 hour) less Than the outbound journey

Speed of plane in still air

Outbound trip :

Velocity = Distance / time

Time = distance / velocity

Velocity = (v - 30) due to head wind

Return Velocity (V +30) due to tail wind

Outbound time = return distance

20 / (v - 30) = 20 / (v +30) + 0.75

20v + 600 = 20v + 0.75v^2 + 22.5v - 600-22.5v-675

600 = 0.75v^2 - 1275

0.75v^2 = 1875

v^2 = 1875/ 0.75

v^2 = 2500

v = sqrt(2500)

v = 50mph

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation:

do you have an easier question cause this shh is hard like theheck

step-by-step explanation: