The equation of line mn is y=3/4x+5 in slope-intercept form, write the equation of the line that is perpendicular to mn and that passes through the point (–6, –2).

For a line to be perpendicular to a reference line, it must have a slope that is the negative reciprocal of the reference line's slope, mathematically:

m1*m2=-1, where m1 and m2 are the slopes of the perpendicular lines for them to be perpendicular.  In this case, the slope of the reference line is 3/4 so:

3m/4=-1

3m=-4

m=-4/3 

So the perpendicular line will have a slope of -4/3, then the line is currently:

y=-4x/3+b, we can solve for b using the point that we must pass through.

-2=-4(-6)/3+b

-2=8+b

-10=b, so our perpendicular line is:

y=-4x/3-10  or more neatly in my opinion:

y=(-4x-30)/3

C. i believe it is (5×(-20))+(100×(0.7))