Write an equation of the line containing the point (6,-4) and perpendicular to. the line y=-2x-3

Step-by-step explanation:

We want to determine the equation of the line that passes through the point (6, -4) and is perpendicular to the line:

Recall that the slopes of perpendicular lines are negative reciprocals of each other.

The slope of the given line is -2.

Hence, the slope of the perpendicular line is 1/2.

Therefore, the slope of our new line is 1/2. We also know that it passes through the point (6, -4). Since we are given the slope and a point, we can consider using point-slope form:

Where m is the slope and (x₁, y₁) is a point.

Let (6, -4) be (x₁, y₁). Substitute:

Simplify. Distribute:

And subtract 4 from both sides:

In conclusion, the equation of our line is:

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."

i think 1: 2

step-by-step explanation: