1. The problem involves evaluating and solving multiple absolute value expressions and equations. 2. For the expressions like $|4|, |-3|, |2 - 5|, |6 - 6|, |-1.7|, |-3 \frac{5}{11}|$, simply remove the absolute value by making the quantity inside positive: $|4| = 4$ $|-3| = 3$ $|2 - 5| = | -3 | = 3$ $|6 - 6| = |0| = 0$ $|-1.7| = 1.7$ $|-3 \frac{5}{11}| = | - \left(3 + \frac{5}{11}\right) | = 3 \frac{5}{11} = \frac{38}{11}$ 3. For equations involving absolute values, solve by considering the definition: $|2y + 3| = 7$ means $2y + 3 = 7$ or $2y + 3 = -7$ Solving gives $y = 2$ or $y = -5$ $|2x - 3| + 8 = 0$ cannot be true since $|2x - 3| \geq 0$ and $+8$ makes it always positive. $4|2x + 3| = 9$ means $|2x + 3| = \frac{9}{4}$ $2x + 3 = \frac{9}{4}$ or $2x + 3 = -\frac{9}{4}$ Solve for $x$: $x = \frac{9}{8} - \frac{3}{2} = -\frac{3}{8}$ or $x = -\frac{9}{8} - \frac{3}{2} = -\frac{21}{8}$ $|2y - 1| = 0$ means $2y - 1 = 0$ so $y = \frac{1}{2}$ $|6x + 1| - 3 = 0$ means $|6x + 1| = 3$ $6x + 1 = 3$ or $6x + 1 = -3$ Solve for $x$: $x = \frac{2}{6} = \frac{1}{3}$ or $x = \frac{-4}{6} = -\frac{2}{3}$ $11 + |x - 2| = 0$ cannot be true since $|x - 2| \geq 0$ and $+11$ is positive. 4. For functions involving absolute values: $y = |x| + 3$ is the basic absolute value graph shifted 3 units up. $y = -|x + 4| - 1$ is a reflection about x-axis, shifted left by 4 and down by 1. $y = |x - 2|$ is the basic absolute value graph shifted right by 2. $y = |-x - 5| + 3 = |-(x + 5)| + 3 = |x + 5| + 3$ shifted left by 5 and up by 3. 5. The description of the graphs: Graph 19 is a classic $y = |x|$ shape vertex at (0,0) with y-axis from 0 to 4. Graph 18 appears to be $y = |x| + 1$ reflected or shifted, vertex at (0,1).