Question

1. ∠urs is being bisected by rq. if m∠urq is 37° and m∠qrs is (6x - 11)°, solve for x.
2. based on how the hands of a clock look at 4 oclock, and how they look at 10 oclock, are those two angles complementary, supplementary, or neither?
3. solve for the variables. (11x + 16)° (8x + 12)° (3y + 13)°

Explanation:

Step1: Use angle - bisection property

Since RQ bisects ∠URS, then m∠URQ = m∠QRS. So we set up the equation 37=6x - 11.

Step2: Solve the equation for x

Add 11 to both sides of the equation: 37 + 11=6x - 11+11, which simplifies to 48 = 6x.

Step3: Isolate x

Divide both sides by 6: $\frac{48}{6}=x$, so x = 8.

Step4: Analyze clock - hand angles at 4 o'clock and 10 o'clock

The angle between the hands of a clock at 4 o'clock: The hour - hand moves 30 degrees per hour (360÷12 = 30). At 4 o'clock, the hour - hand is at 4×30 = 120 degrees from the 12 - o'clock position and the minute - hand is at 0 degrees, so the angle is 120 degrees. At 10 o'clock, the hour - hand is at 10×30 = 300 degrees from the 12 - o'clock position and the minute - hand is at 0 degrees, so the angle is 60 degrees. The sum of 120 and 60 is 180 degrees, so the two angles are supplementary.

Step5: Solve for x and y in the intersecting - lines problem

Vertical angles are equal. So 11x + 16=8x + 12. Subtract 8x from both sides: 11x-8x + 16=8x-8x + 12, 3x+16 = 12. Subtract 16 from both sides: 3x=12 - 16=-4, x=$-\frac{4}{3}$. Also, 3y + 13=11x + 16. Substitute x = $-\frac{4}{3}$ into it: 3y+13=11×($-\frac{4}{3}$)+16. 3y+13=$-\frac{44}{3}+16$. 3y+13=$-\frac{44}{3}+\frac{48}{3}=\frac{4}{3}$. Subtract 13 from both sides: 3y=$\frac{4}{3}-13=\frac{4 - 39}{3}=-\frac{35}{3}$, y=$-\frac{35}{9}$.

Answer:

1. x = 8
2. Supplementary
3. x=$-\frac{4}{3}$, y=$-\frac{35}{9}$