What is the correct solution set for the following graph? graph showing two lines, labeled x and y {point on negative x-axis} {point in Quadrant I} {point in Quadrant II} {point in Quadrant III}

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

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

-3.5 is the distance.

i've had this question so choose the one in the absolute value sign.

x:       y:  

-1         4

0         4

1         6

2         18

3         20