Question

1. (6y + 8)° (6x - 26)° 4x (7y - 12)° 12. 2(5x - 5)° (5y + 5)° 20y° (3x - 5)° 13. error analysis describe and correct the error in using the diagram to find the value of x. (13x + 45)° (6x + 2)° (12x - 40)° (19x + 3)° (13x + 45)°+(19x + 3)° = 180° 32x + 48 = 180 32x = 132 x = 4.125 14. college prep which statements can you conclude from the diagram? select all that apply. a ∠vzt ≅ ∠xzy c ∠vzx ≅ ∠tzy b ∠vzt ≅ ∠vzw d ∠vzw ≅ ∠xzy 15. making an argument your friend claims that ∠1 ≅ ∠4 because they are vertical angles. is your friend correct? support your answer with definitions or theorems. 16. structure find the measure of each angle in the diagram. 10y° (3y + 11)° (4x - 22)° (7x + 4)° 2.6 proving geometric relationships 107

Explanation:

Step1: Identify vertical - angle property

Vertical angles are equal. In problem 13, the error is that the angles \((13x + 45)^{\circ}\) and \((19x+3)^{\circ}\) are not supplementary (they are not a linear - pair). The correct pair of supplementary angles is \((13x + 45)^{\circ}\) and \((12x-40)^{\circ}\) since they form a linear pair.

Step2: Set up the correct equation

We know that for a linear pair of angles, the sum of the angles is \(180^{\circ}\). So, \((13x + 45)+(12x-40)=180\).

Step3: Simplify the left - hand side of the equation

Combine like terms: \(13x+12x + 45-40=180\), which gives \(25x + 5=180\).

Step4: Solve for \(x\)

Subtract 5 from both sides: \(25x=180 - 5=175\). Then divide both sides by 25: \(x=\frac{175}{25}=7\).

For problem 14:

• \(\angle VZT\) and \(\angle XZY\) are vertical angles, so \(\angle VZT\cong\angle XZY\) (A is correct).
• \(\angle VZX\) and \(\angle TZY\) are vertical angles, so \(\angle VZX\cong\angle TZY\) (C is correct).
• \(\angle VZT\) and \(\angle VZW\) are not equal in general, and \(\angle VZW\) and \(\angle XZY\) are not equal in general.

For problem 15:
The friend is correct. By the definition of vertical angles, vertical angles are formed when two lines intersect. \(\angle1\) and \(\angle4\) are vertical angles, and vertical angles are always congruent.

For problem 16:
Since vertical angles are equal, we have \(10y=3y + 11\).

Step1: Solve for \(y\)

Subtract \(3y\) from both sides: \(10y-3y=11\), so \(7y = 11\), and \(y=\frac{11}{7}\).
Also, \(4x-22=7x + 4\).

Step2: Solve for \(x\)

Subtract \(4x\) from both sides: \(-22=7x-4x + 4\), which is \(-22=3x + 4\). Then subtract 4 from both sides: \(-22 - 4=3x\), so \(-26=3x\), and \(x=-\frac{26}{3}\).

Answer:

1. \(x = 7\)
2. A. \(\angle VZT\cong\angle XZY\), C. \(\angle VZX\cong\angle TZY\)
3. The friend is correct because vertical angles are congruent.
4. \(y=\frac{11}{7}\), \(x =-\frac{26}{3}\)