What is the equation of the line that passes through the point (-6, -5) and has a slope of -3/2?

y=3/2x+3/2

Step-by-step explanation:

point at (-6,-5)

slope = m = 3/2

y = mx+b

(-6)=(3/2)(-5)+b >>> substituting x & y to solve for b

-6 = (-15/2)+b >>> simplify previous equation

-6 + (15/2) = b >>> isolate and solve for m

3/2 = b

therefore, the equation of the line passing through the point (-6,-5) with a slope of 3/2 is y=(3/2)x+(3/2)

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i