Find the equation of the line with slope 6 passing through (3,20) y=mx+b

y = 6x + 2

Step-by-step explanation:

Step 1: Write known variables

y = 6x + b

Step 2: Solve for b

20 = 6(3) + b

20 = 18 + b

b = 2

Step 3: Rewrite

y = 6x + 2

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."

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.