A jet departs from an airport flying east. At the same time, a second jet departs from the same airport flying west at a speed o
f 20 miles per hour slower than the first jet. After 1.5 hours, the planes are 1,500 miles apart. What is the speed in miles per hour of the jet traveling east? A. 300
Let x mph and x+20 mph be the speeds of the planes. The rate at which their distance from each other increases will be the sum of their speeds which is 2x+20 mph.
Their distance from each other after three hours is ( 1.5 hours)*(2x+20 mph) = 3x+30 miles. We are given that this is 1500 miles so we can solve for x.
3x+30 = 1500
3x = 1500 - 30
3x = 1470
x = 1470/3
x = 490mph
For the second jet :
x +20 = 490+20
x = 510
Thus the speeds of the two planes are 490 mph and 510mph.
f(9) = 28. I hope this helps! I have a g00gle classroom where I help with algebra including this. If you are interested, please consider joining. I will post the class code in the comments if it would let me. Sorry it wouln't let me type g0ogle properly.
