Two buses leave towns 576 kilometers apart at the same time and travel toward each other. One bus travels 12 h
slower than the other. If they meet in 3 hours, what is the rate of each bus?
km
Rate of the slower bus:

Rate of the faster bus:

Rate of slower bus; 90 km/h

Rate of faster bus; 102 km/b

Step-by-step explanation:

We know that formula do distance is;

Distance = speed/time

We are told that One bus travels 12h slower than the other.

Let speed of slower bus be x.

Thus;

Speed of faster bus = x + 12

Speed of slower bus = x

After 3 hours, distance by faster bus = 3(x + 12)

Speed of slower bus = 3x

Since the towns are 576 km apart, then;

3(x + 12) + 3x = 576

Divide through by 3 to get;

x + 12 + x = 192

2x + 12 = 192

2x = 192 - 12

2x = 180

x = 180/2

x = 90 km/h

Faster bus speed = 90 + 12 = 102 km/h

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

one that is bigger than 3 1/2