Exits on a highway are numbered consecutively from 1 to 50. The distance from exit 41 to exit 50 is 100 km. If each exit is at least 6 km from the next exit, what is the longest possible distance, in kilometers, between exit 47 and exit 48?

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."

it is 28 degrees colder in detroit than philadelphia.

step-by-step explanation:

what number adds to 21

-7 + ? = 21

-7 + 28 = 21

hope it

16 would be your answer