What is the equation of the line perpendicular to y=1/4-5 trough the point (4,-3)

The equation is y=-4x+13

Solutions:

y=1/4-5 (4,-3)

y=mx+b

The new slope = -4/1x+b

-3=-4/1(4)+b

-3=-16+b

+16 +16

13=b

y=-4x+13

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180