Math
posted by Please help .
A refrigerated truck leaves a rest stop traveling at a steady rate of 56 mi/h. A car leaves the same rest stop 1/4 h later followinf rhe truck at a steady rate of 64 mi/h. How long after the truck leaves the rest stop wil the car overtake the truck. In algebra form.

When the car overtakes the truck, they both would have traveled the same distance, except the truck took 1/4 hour longer
let the time taken by the car be x hours
then the time taken by the truck is x+1/4 hour
distance gone by car = 64x
distance gone by truck = 56(x+1/4)
but didn't we say those distances would be equal?
let me know what you got 
2 but is that for the truck or car

The time x cannot be negative. You must have made a mistake.
The two equations are,
distance gone by car = 64x
distance gone by truck = 56(x+1/4)
Since the distances are equal,
64x = 56(x + 1/4)
64x = 56x + 56/4
64x = 56x + 14
8x = 14
x = 1.75
From above (Reiny's explanation)
Since x = time taken by the car in hours, the car will overtake the truck in 1.75 hrs. 
A computer costs $799. To add memory it costs $25 for 8 megabytes. How much memory can you add if you have at most $1,000 to spend.

It will be 2 hours u til the car catches up tot the truck.Sorry, I don't have the work. :(

relative speed=6456
=8mi/h
distance travelled by truck for 1/4h
=s*t=56*1/4=14mi
time spent by car in overtaking
t=d/s=14/8
=1.75hrs