Geometry practice problems requiring students to calculate the value of x using the sum of exterior angles for various polygons.

To solve for \( x \) in each of the given problems, we will analyze each diagram step by step using geometric properties such as the sum of angles in a triangle, the sum of angles on a straight line, and the sum of angles around a point.

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Problem 1:

The first diagram shows a triangle with angles \( (3x)^\circ \), \( (2x)^\circ \), and \( (3x)^\circ \).

#### Solution:
1. The sum of the interior angles of a triangle is \( 180^\circ \).
2. Therefore, we can write the equation:
\[
(3x) + (2x) + (3x) = 180
\]
3. Simplify the equation:
\[
8x = 180
\]
4. Solve for \( x \):
\[
x = \frac{180}{8} = 22.5
\]

Answer for Problem 1:
\[
\boxed{22.5}
\]

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Problem 2:

The second diagram shows a quadrilateral with angles \( (100 - x)^\circ \), \( x^\circ \), \( 65^\circ \), and \( (90 - \frac{x}{2})^\circ \).

#### Solution:
1. The sum of the interior angles of a quadrilateral is \( 360^\circ \).
2. Therefore, we can write the equation:
\[
(100 - x) + x + 65 + \left(90 - \frac{x}{2}\right) = 360
\]
3. Simplify the equation:
\[
100 - x + x + 65 + 90 - \frac{x}{2} = 360
\]
\[
255 - \frac{x}{2} = 360
\]
4. Isolate \( x \):
\[
-\frac{x}{2} = 360 - 255
\]
\[
-\frac{x}{2} = 105
\]
\[
x = -2 \times 105 = -210
\]

Answer for Problem 2:
\[
\boxed{-210}
\]

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Problem 3:

The third diagram shows a right triangle with angles \( (x + 30)^\circ \), \( (7x)^\circ \), and \( (31x)^\circ \).

#### Solution:
1. The sum of the interior angles of a triangle is \( 180^\circ \).
2. Therefore, we can write the equation:
\[
(x + 30) + (7x) + (31x) = 180
\]
3. Simplify the equation:
\[
x + 30 + 7x + 31x = 180
\]
\[
39x + 30 = 180
\]
4. Isolate \( x \):
\[
39x = 180 - 30
\]
\[
39x = 150
\]
\[
x = \frac{150}{39} = \frac{50}{13}
\]

Answer for Problem 3:
\[
\boxed{\frac{50}{13}}
\]

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Problem 4:

The fourth diagram shows a pentagon with angles \( (4x/3)^\circ \), \( (0.6x)^\circ \), \( (2x/5)^\circ \), \( (4x/3)^\circ \), and \( (2x - 70)^\circ \).

#### Solution:
1. The sum of the interior angles of a pentagon is \( 540^\circ \).
2. Therefore, we can write the equation:
\[
\left(\frac{4x}{3}\right) + (0.6x) + \left(\frac{2x}{5}\right) + \left(\frac{4x}{3}\right) + (2x - 70) = 540
\]
3. Combine like terms:
\[
\frac{4x}{3} + \frac{4x}{3} + 0.6x + \frac{2x}{5} + (2x - 70) = 540
\]
\[
\frac{8x}{3} + 0.6x + \frac{2x}{5} + 2x - 70 = 540
\]
4. Convert all terms to have a common denominator (15):
\[
\frac{8x}{3} = \frac{40x}{15}, \quad 0.6x = \frac{9x}{15}, \quad \frac{2x}{5} = \frac{6x}{15}, \quad 2x = \frac{30x}{15}
\]
\[
\frac{40x}{15} + \frac{9x}{15} + \frac{6x}{15} + \frac{30x}{15} + 2x - 70 = 540
\]
\[
\frac{85x}{15} + 2x - 70 = 540
\]
5. Simplify further:
\[
\frac{85x}{15} + \frac{30x}{15} - 70 = 540
\]
\[
\frac{115x}{15} - 70 = 540
\]
\[
\frac{115x}{15} = 610
\]
6. Solve for \( x \):
\[
115x = 610 \times 15
\]
\[
115x = 9150
\]
\[
x = \frac{9150}{115} = 80
\]

Answer for Problem 4:
\[
\boxed{80}
\]

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Final Answers:

1. \(\boxed{22.5}\)
2. \(\boxed{-210}\)
3. \(\boxed{\frac{50}{13}}\)
4. \(\boxed{80}\)