1. **Identify the plane parallel to plane PQT.** Given the figure, plane PQT is a plane formed by points P, Q, and T. To find a plane parallel to PQT, we look for another plane with the same orientation but different position. From the figure, plane SRV is parallel to plane PQT because both are opposite faces of the cuboid-like shape. 2. **Which segment is skew to RV?** Segment RV is an edge of the shape. A segment skew to RV is one that is not parallel, not intersecting, and not in the same plane. From the figure, segment SP is skew to RV. 3. **Identify the relationship between $\angle 3$ and $\angle 10$.** From the angle diagram, $\angle 3$ and $\angle 10$ are corresponding angles formed by parallel lines $p$ and $q$ and a transversal. Corresponding angles are equal when lines are parallel. 4. **Identify the relationship between $\angle 9$ and $\angle 13$.** From the diagram, $\angle 9$ and $\angle 13$ are alternate interior angles. Alternate interior angles are equal when lines are parallel. 5. **Given $p \parallel q$ and $m\angle 3 = 75^\circ$, find $m\angle 5$.** Since $\angle 3$ and $\angle 5$ are alternate interior angles, they are equal. Therefore, $m\angle 5 = 75^\circ$. 6. **Given $p \parallel q$, $m\angle 10 = 3x - 7$, and $m\angle 13 = 4x - 9$, find $x$.** Since $\angle 10$ and $\angle 13$ are alternate interior angles, they are equal. Set up the equation: $$3x - 7 = 4x - 9$$ Subtract $3x$ from both sides: $$\cancel{3x} - 7 = \cancel{3x} + x - 9 \implies -7 = x - 9$$ Add 9 to both sides: $$-7 + 9 = x - 9 + 9 \implies 2 = x$$ 7. **If $p \parallel q$ by the Consecutive Interior Angles Theorem, which angle pair must be supplementary?** Consecutive interior angles (also called same-side interior angles) are supplementary. One such pair is $\angle 4$ and $\angle 8$. 8. **If $m\angle 4 = 7x - 20$ and $m\angle 8 = 5x + 18$, find $x$ so that $p \parallel q$.** Since $\angle 4$ and $\angle 8$ are consecutive interior angles, they must be supplementary: $$m\angle 4 + m\angle 8 = 180$$ Substitute: $$7x - 20 + 5x + 18 = 180$$ Simplify: $$12x - 2 = 180$$ Add 2 to both sides: $$12x = 182$$ Divide both sides by 12: $$x = \frac{182}{12} = \frac{91}{6} \approx 15.17$$ 9. **Find the slope of the line containing points P(-6, 3) and Q(12, 9).** Slope formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Calculate: $$m = \frac{9 - 3}{12 - (-6)} = \frac{6}{18} = \frac{\cancel{6}}{\cancel{18}} = \frac{1}{3}$$ 10. **Find the slope of the line containing points M(-8, 14) and N(2, -11).** Calculate: $$m = \frac{-11 - 14}{2 - (-8)} = \frac{-25}{10} = -\frac{5}{2}$$ 11. **Given A(-1, 4), B(1, 5), and C(-5, 3), which coordinate will make AB parallel to CD?** Calculate slope of AB: $$m_{AB} = \frac{5 - 4}{1 - (-1)} = \frac{1}{2}$$ Check each D option for slope CD equal to $\frac{1}{2}$: - D(-7, 4): $$m_{CD} = \frac{4 - 3}{-7 - (-5)} = \frac{1}{-2} = -\frac{1}{2} \neq \frac{1}{2}$$ - D(-6, 1): $$m_{CD} = \frac{1 - 3}{-6 - (-5)} = \frac{-2}{-1} = 2 \neq \frac{1}{2}$$ - D(-4, 5): $$m_{CD} = \frac{5 - 3}{-4 - (-5)} = \frac{2}{1} = 2 \neq \frac{1}{2}$$ - D(-3, 4): $$m_{CD} = \frac{4 - 3}{-3 - (-5)} = \frac{1}{2} = m_{AB}$$ Answer: D(-3, 4) 12. **Given A(2, 3), B(8, 7), and C(6, 1), which coordinate will make AB perpendicular to CD?** Calculate slope of AB: $$m_{AB} = \frac{7 - 3}{8 - 2} = \frac{4}{6} = \frac{2}{3}$$ For perpendicular lines, slopes satisfy: $$m_{AB} \times m_{CD} = -1$$ So, $$m_{CD} = -\frac{1}{m_{AB}} = -\frac{3}{2}$$ Check each D option: - D(3, 3): $$m_{CD} = \frac{3 - 1}{3 - 6} = \frac{2}{-3} = -\frac{2}{3} \neq -\frac{3}{2}$$ - D(4, 4): $$m_{CD} = \frac{4 - 1}{4 - 6} = \frac{3}{-2} = -\frac{3}{2} = m_{CD}$$ - D(8, 4): $$m_{CD} = \frac{4 - 1}{8 - 6} = \frac{3}{2} \neq -\frac{3}{2}$$ - D(9, 3): $$m_{CD} = \frac{3 - 1}{9 - 6} = \frac{2}{3} \neq -\frac{3}{2}$$ Answer: D(4, 4)