Two planes leave an airport at the same time and travel opposite directions. One plane travels 98 km/h faster than the other one. If the two planes are 3668 Km apart after 2 hours, what is the rate of each plane. Any help with this would be grateful . In desperate need of help here .any takers ?

Step-by-step explanation:

               Let x = the speed of plane A

  Then 98 + x  = the speed of Plane B

   and 98 + 2x = the speed at which the planes are separating

and 2(98 + 2x) = their separation after 2 h

Check:

OK

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

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s