Plane a leaves tulsa at 2: 00 p. m., averaging 300 mph and flying in a northerly direction. plane b leaves tulsa at 2: 30 p. m., averaging 225 mph and flying due east. at 5: 00 p. m., how far apart will the planes be

1061.32 miles apart

Step-by-step explanation:

Given that both planes A and B leave the same place and travel in perpendicular directions one in north and other in east.

The planes position at any time along with starting point can be visualized as a right triangle with sides equal to distances travelled by both the planes, and hypotenuse the distance between the planes, and the place Tulsa the vertex right angled.

Plan A starts at 2.00 with speed 300 mph

Hence at 5 p.m. distance travelled by plane A = 300(3) = 900 miles

Plane B starts at 2.30 with speed 225 mph

At 5 p.m. distance travelled = 225(2.50) = 562.50

Hence distance between the planes = length of hypotenuse

=

The two planes are 1061.32 miles apart

answer: (x+2) (x+3) are the binomials for your question : )