Two airplanes leave the same place and fly in opposite directions. the average rate of one plane is 32 miles per hour faster than the other one. find the rate of each plane if they are 1098 miles apart after 1 1/2 hours

speed of one plane is 50 miles/hour

Speed of other plane is 82 miles/hour

Step-by-step explanation:

Given : Two airplanes leave the same place and fly in opposite directions. The average rate of one plane is 32 miles per hour faster than the other one.

To Find: . Find the rate of each plane if they are 1098 miles apart after 1 1/2 hours

Solution : Let the speed of one plane be x miles per hour

Time =

So, distance traveled by one plane in 1.5 hours = speed * time

                                                        = 1.5 x

Since we are given that speed of other plane is 32 more than one

So, speed of other plane is x+32 miles per hour

Time = 1.5 hours

So, distance traveled by other plane in 1.5 hours = speed * time

                                                                                 = 1.5 (x+32)

Since they are 1098 miles apart after 1.5 hours

So,

So, speed of one plane is 50 miles/hour

Speed of other plane is x+32=50+32 =82 miles/hour

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

answer:

125 yards

explanation:

75+50=125. that is my answer