What line is perpendicular to y=2/7x+11/7 travel through (-3,8)

y=−7/2x−5/2

Step-by-step explanation:

Find the negative reciprocal of the slope of the original line and use the point-slope formula  

y−y1=m(x−x1)  to find the line perpendicular to  y=2/7x+11/7.

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."