An airplane flies to san francisco from los angeles in 4 hours. it flies back in 3 hours. if the wind is blowing from the north at a velocity of 20 mph during both flights, what was the airspeed of the plane (its speed in still air)?

Airspeed in still air:  140 mph

Step-by-step explanation:

Let s represent the speed in still air.  Recognize that the distance traveled is the same in either direction.

Also note that SF is north of LA.

Distance to SF from LA     =      Distance to LA from SF

(s-20)(mph)(4 hr)                 =    (s+20)(mph)(3 hr)

Then:

4s - 80 = 3s + 60 => s = 60 + 80 + 140 (mph)

The airspeed of the plane was 140 mph in still air.  With a tail wind, the plane  is faster; with a headwind, slower.

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value

idk

step-by-step explanation: