| 2 +r | ≥ 3 (lines are absolute value)

r ≤ -5 and r ≥ 1

Step-by-step explanation:

The solution has two parts:

1) that derived from | 2 +r | ≥ 3 when (2 + r) is already positive.  Then:

                   2 + r ≥ 3, or r ≥ 1

and

that derived from | 2 +r | ≥ 3 when (2 + r) is negative.  If (2 + r) is negative, then

|(2 + r)| = -(2 + r) = -2 -r, which is ≥ 3.  Therefore, -2 -r ≥ 3, or -r ≥ 5.  To solve this for r, divide both sides by -1 and reverse the direction of the inequality sign:  r ≤ -5

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.