Find the value of x and the indicated angle measures in three geometric triangles.

To solve the problems in the image, we need to use geometric properties and relationships between angles. Let's go through each problem step by step.

---

Problem 7: Find the value of \( x \)

#### Image Description:
- Triangle \( ABC \) with \( \angle BAC = 5x^\circ \) and \( \angle ABC = 3x^\circ \).
- The triangle is a right triangle with \( \angle ACB = 90^\circ \).

#### Solution:
1. In any triangle, the sum of the interior angles is \( 180^\circ \).
2. For triangle \( ABC \):
\[
\angle BAC + \angle ABC + \angle ACB = 180^\circ
\]
3. Substitute the given values:
\[
5x + 3x + 90 = 180
\]
4. Simplify:
\[
8x + 90 = 180
\]
5. Solve for \( x \):
\[
8x = 90 \implies x = \frac{90}{8} = 11.25
\]

#### Answer:
\[
\boxed{x = 11.25}
\]

---

Problem 9: Find \( m\angle ACB \)

#### Image Description:
- Same as Problem 7.
- We already know \( \angle ACB = 90^\circ \).

#### Answer:
\[
\boxed{90^\circ}
\]

---

Problem 10: Find \( m\angle A \)

#### Image Description:
- Same as Problem 7.
- \( \angle A = \angle BAC = 5x^\circ \).

#### Solution:
1. From Problem 7, we found \( x = 11.25 \).
2. Therefore:
\[
\angle A = 5x = 5 \times 11.25 = 56.25^\circ
\]

#### Answer:
\[
\boxed{56.25^\circ}
\]

---

Problem 11: Find the value of \( x \)

#### Image Description:
- Triangle \( EFG \) with \( \angle E = 90^\circ \).
- \( \angle F = (2x + 10)^\circ \).
- \( \angle G = (x + 22)^\circ \).

#### Solution:
1. In any triangle, the sum of the interior angles is \( 180^\circ \).
2. For triangle \( EFG \):
\[
\angle E + \angle F + \angle G = 180^\circ
\]
3. Substitute the given values:
\[
90 + (2x + 10) + (x + 22) = 180
\]
4. Simplify:
\[
90 + 2x + 10 + x + 22 = 180
\]
\[
3x + 122 = 180
\]
5. Solve for \( x \):
\[
3x = 58 \implies x = \frac{58}{3} \approx 19.33
\]

#### Answer:
\[
\boxed{x = \frac{58}{3}}
\]

---

Problem 12: Find \( m\angle H \)

#### Image Description:
- Same as Problem 11.
- \( \angle H = \angle G = (x + 22)^\circ \).

#### Solution:
1. From Problem 11, we found \( x = \frac{58}{3} \).
2. Therefore:
\[
\angle H = x + 22 = \frac{58}{3} + 22
\]
3. Convert 22 to a fraction with a denominator of 3:
\[
22 = \frac{66}{3}
\]
4. Add the fractions:
\[
\angle H = \frac{58}{3} + \frac{66}{3} = \frac{124}{3} \approx 41.33^\circ
\]

#### Answer:
\[
\boxed{\frac{124}{3}}
\]

---

Problem 13: Find \( m\angle EGH \)

#### Image Description:
- Same as Problem 11.
- \( \angle EGH \) is an exterior angle to triangle \( EFG \) at vertex \( G \).

#### Solution:
1. The exterior angle \( \angle EGH \) is equal to the sum of the two non-adjacent interior angles:
\[
\angle EGH = \angle E + \angle F
\]
2. Substitute the known values:
\[
\angle E = 90^\circ, \quad \angle F = 2x + 10
\]
3. From Problem 11, \( x = \frac{58}{3} \):
\[
\angle F = 2x + 10 = 2 \left( \frac{58}{3} \right) + 10 = \frac{116}{3} + 10
\]
4. Convert 10 to a fraction with a denominator of 3:
\[
10 = \frac{30}{3}
\]
5. Add the fractions:
\[
\angle F = \frac{116}{3} + \frac{30}{3} = \frac{146}{3}
\]
6. Therefore:
\[
\angle EGH = 90 + \frac{146}{3} = \frac{270}{3} + \frac{146}{3} = \frac{416}{3} \approx 138.67^\circ
\]

#### Answer:
\[
\boxed{\frac{416}{3}}
\]

---

Problem 14: Find \( m\angle EGF \)

#### Image Description:
- Same as Problem 11.
- \( \angle EGF \) is the same as \( \angle F \).

#### Solution:
1. From Problem 11, we found:
\[
\angle F = 2x + 10
\]
2. Substitute \( x = \frac{58}{3} \):
\[
\angle F = 2 \left( \frac{58}{3} \right) + 10 = \frac{116}{3} + 10
\]
3. Convert 10 to a fraction with a denominator of 3:
\[
10 = \frac{30}{3}
\]
4. Add the fractions:
\[
\angle F = \frac{116}{3} + \frac{30}{3} = \frac{146}{3}
\]

#### Answer:
\[
\boxed{\frac{146}{3}}
\]

---

Problem 15: Find the value of \( x \)

#### Image Description:
- Triangle \( LMN \) with \( \angle L = 90^\circ \).
- \( \angle M = (5x - 2)^\circ \).
- \( \angle N = (5x - 17)^\circ \).

#### Solution:
1. In any triangle, the sum of the interior angles is \( 180^\circ \).
2. For triangle \( LMN \):
\[
\angle L + \angle M + \angle N = 180^\circ
\]
3. Substitute the given values:
\[
90 + (5x - 2) + (5x - 17) = 180
\]
4. Simplify:
\[
90 + 5x - 2 + 5x - 17 = 180
\]
\[
90 + 10x - 19 = 180
\]
\[
10x + 71 = 180
\]
5. Solve for \( x \):
\[
10x = 109 \implies x = \frac{109}{10} = 10.9
\]

#### Answer:
\[
\boxed{x = 10.9}
\]

---

Problem 16: Find \( m\angle M \)

#### Image Description:
- Same as Problem 15.
- \( \angle M = (5x - 2)^\circ \).

#### Solution:
1. From Problem 15, we found \( x = 10.9 \).
2. Therefore:
\[
\angle M = 5x - 2 = 5(10.9) - 2 = 54.5 - 2 = 52.5^\circ
\]

#### Answer:
\[
\boxed{52.5^\circ}
\]

---

Problem 17: Find \( m\angle LNM \)

#### Image Description:
- Same as Problem 15.
- \( \angle LNM = \angle N = (5x - 17)^\circ \).

#### Solution:
1. From Problem 15, we found \( x = 10.9 \).
2. Therefore:
\[
\angle N = 5x - 17 = 5(10.9) - 17 = 54.5 - 17 = 37.5^\circ
\]

#### Answer:
\[
\boxed{37.5^\circ}
\]

---

Problem 18: Find \( m\angle UND \)

#### Image Description:
- Not provided in the image. Assuming it refers to an external angle or another part of the diagram not shown.

#### Answer:
\[
\boxed{\text{Not enough information}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
7. & \ x = 11.25 \\
9. & \ m\angle ACB = 90^\circ \\
10. & \ m\angle A = 56.25^\circ \\
11. & \ x = \frac{58}{3} \\
12. & \ m\angle H = \frac{124}{3} \\
13. & \ m\angle EGH = \frac{416}{3} \\
14. & \ m\angle EGF = \frac{146}{3} \\
15. & \ x = 10.9 \\
16. & \ m\angle M = 52.5^\circ \\
17. & \ m\angle LNM = 37.5^\circ \\
18. & \ \text{Not enough information}
\end{aligned}
}
\]