# Math

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.

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1. 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

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2. -2 but is that for the truck or car

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3. 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.

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4. 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.

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

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6. relative speed=64-56
=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

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