https://file.aiquickdraw.com/mathbot/file/172783872515455jn8h2l.jpg

Let's break down each of the three problems in the image and address how to solve them.

### 1. Length of CM
We are given the following details:
- The length of \( RJ \) is 16 units.
- \( M \) is the midpoint of \( RJ \), so \( RM = MJ = 8 \) units.
- \( CM \) is the perpendicular bisector.
- The distances of point \( C \) from points \( R \) and \( J \) are given by \( 2x^2 + 2x + 5 \) and \( x^2 - x + 15 \), respectively.

#### To solve for the length of \( CM \):
- Set the distances equal, since \( CM \) is a perpendicular bisector, and solve for \( x \).

\[
2x^2 + 2x + 5 = x^2 - x + 15
\]

Solve the quadratic equation for \( x \), and then use the midpoint formula to find \( CM \).

---

### 2. Triangle with side lengths
We are given the lengths of the sides of a triangle:
- \( \frac{1}{2}x + \frac{3}{2} \)
- \( \frac{3}{2}x + \frac{3}{2} \)
- \( \frac{7x}{3} - \frac{4}{3} \)

#### To find all integral values of \( x \):
- Apply the triangle inequality, which states that the sum of any two sides of a triangle must be greater than the third side.

Set up inequalities for each combination of sides:
1. \( \left(\frac{1}{2}x + \frac{3}{2}\right) + \left(\frac{3}{2}x + \frac{3}{2}\right) > \left(\frac{7x}{3} - \frac{4}{3}\right) \)
2. \( \left(\frac{1}{2}x + \frac{3}{2}\right) + \left(\frac{7x}{3} - \frac{4}{3}\right) > \left(\frac{3}{2}x + \frac{3}{2}\right) \)
3. \( \left(\frac{3}{2}x + \frac{3}{2}\right) + \left(\frac{7x}{3} - \frac{4}{3}\right) > \left(\frac{1}{2}x + \frac{3}{2}\right) \)

Solve these inequalities to find the integral values of \( x \).

---

### 3. Triangle with angle measures
We are given the measures of the angles in the acute triangle \( \triangle ABC \):
- \( \angle CAB = 7x + y \)
- \( \angle ABC = y + 2z \)
- \( \angle BCA = 5z - x \)
Additionally, the exterior angles are given as:
- \( \angle ACD = 102 - 2x \)
- \( \angle EAC = 11x - y \)

#### To solve this:
- Use the fact that the sum of the angles in a triangle is 180 degrees. Thus:

\[
(7x + y) + (y + 2z) + (5z - x) = 180
\]

- Use the given exterior angles to establish additional relationships between \( x \), \( y \), and \( z \).

Solve for \( x \), \( y \), and \( z \), and use these to find the angle measures.

---

Would you like a detailed step-by-step solution for any of these problems, or do you have any specific questions?

### Related Questions:
1. How do you solve quadratic equations like the one in Problem 1?
2. What is the triangle inequality theorem, and how does it apply to Problem 2?
3. How do exterior angles relate to interior angles in Problem 3?
4. How can you check if a solution satisfies all constraints of a triangle's sides or angles?
5. What are the methods for solving systems of equations in geometry?

### Tip:
When working with geometric problems, always look for symmetry or known properties (e.g., midpoints, bisectors) to simplify your calculations.

Consider a line segment RJ with a length of 16 units. Let M be the midpoint of RJ and CM be a perpendicular bisector. Determine the length of CM if the distance of C from R and J are (2x^2 + 2x + 5) and (x^2 - x + 15) units, respectively. The length of the sides of a triangle are (1/2x + 3/2) units, (3/2x + 3/2) units, and (7x/3 - 4/3) units. Find all integral values of x. The measures of the interior angles of an acute triangle, ΔABC, are m∠CAB = 7x + y, m∠ABC = y + 2z, and m∠BCA = 5z - x. Furthermore, ∠ACD and ∠EAC are exterior angles whose measures are (102 - 2x) and (11x - y), respectively. Solve for x, y, and z and determine the angle measure of each interior angle.

Solve a geometry problem involving the midpoint of a line segment, triangle inequality for side lengths, and the measures of angles in a triangle. This problem applies concepts of algebra, quadratic equations, and geometry, and is suitable for high school students in Grades 9-12.

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