Two airplanes left the same airport and arrived at the same destination at the same time. the first airplane left at 8: 00 a. m. and traveled at an average rate of 496 miles per hour. the second airplane left at 8: 30 a. m. and traveled at an average rate of 558 miles per hour. let x represent the number of hours that the first plane traveled. how many hours did it take the first plane to travel to the destination? enter an equation that can be used to solve this problem in the first box. solve for x and enter the number of hours in the second box. equation: x = h

The formula for distance is equal to:

d = v * t

where d is distance, v is velocity or speed, and t is time

Since the distance travelled by the two airplane is similar, therefore we can create the initial equation:

v1 * t1 = v2 * t2

We know that v1 = 496, and v2 = 558 so:

496 t1 = 558 t2

or

x = 558 t2 / 496

We also know that airplane 1 travelled 30 minutes (0.5 hours) earlier than airplane 2, therefore:

x = t2 + 0.5

Hence,

496 (t2 + 0.5) = 558 t2

496 t2 + 248 = 558 t2

t2 = 4 hours

x = t2 + 0.5 = 4 + 0.5

x = 4.5 hours

So the equation is:

x = 558 t2 / 496

And the first plane travelled:

x = 4.5 hours

y=mx+