Find the distance between each pair of parallel lines with the given equations y=3x, y=3x+10

sqrt10

Step-by-step explanation:

Since the closest distance between two parallel lines is a line that is perpindicular to them, we first find the line of the slope that is perpindicular to both lines. Since slope of line y = 3x+10 is 3, the slope of the line that is perpinduclar is -1/3. Therefore y = -1/3x is perpindicluar to both lines. We then find the intersection point of each line. We find said points by setting the lines equal to each other. 3x+10=-1/3x, x = -3 therefore y = 1. For the second line we get we find the origin is the intersection. Therefore since this forms a right triangle with the x axis, we find that the distance is 3^2+1^2 = sqrt10

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

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step-by-step explanation: