An airplane flew 4 hours with a 25 mph tail wind. the return trip against the same wind took 5 hours. find the speed of the airplane in still air. this similar to the current problem as you have to consider the 25 mph tailwind and headwind. plane on outbound trip of 4 hours with 25 mph tailwind and return trip of 5 hours with 25 mph headwind let r = the rate or speed of the airplane in still air. let d = the distance

a. write a system of equations for the airplane. one equation will be for the outbound trip with tailwind of 25 mph. the second equation will be for the return trip with headwind of 25 mph.
b. solve the system of equations for the speed of the airplane in still air.

r  = 225 Mil/h     speed of the airplane in still air

Step-by-step explanation:

Then:

d  is traveled distance   and r  the speed of the airplane in still air

so the first equation is for a 4 hours trip

as  d = v*t

d  =  4 *  ( r  + 25)    (1)          the speed of tail wind  (25 mil/h)

Second equation the trip back in 5 hours

d  =  5  * ( r  - 25 )    (2)

So we got a system of two equation and two unknown variables  d  and

r

We solve it by subtitution

from equation (1)     d  =  4r  + 100

plugging in equation 2

4r  + 100  = 5r -  125    ⇒   -r  =  -225     ⇒    r  = 225 Mil/h

And distance is :

d  =  4*r  +  100         ⇒ d  =  4 * ( 225)  +  100

d  =  900 +   100

d  = 1000 miles

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: