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We are given the inequality:

\[
y + 3 > \frac{5}{2}(x - 6)
\]

We need to determine which of the following points satisfies this inequality.

### Steps:
1. **Rewrite the inequality** by isolating \( y \):
   \[
   y > \frac{5}{2}(x - 6) - 3
   \]
   Simplifying the right side:
   \[
   y > \frac{5}{2}x - 15 - 3
   \]
   \[
   y > \frac{5}{2}x - 18
   \]

Now, we will substitute each point into this inequality to see which one satisfies it.

---

### Option 1: \( (8, 0) \)

Substitute \( x = 8 \) and \( y = 0 \) into the inequality:

\[
0 > \frac{5}{2}(8) - 18
\]
\[
0 > 20 - 18
\]
\[
0 > 2
\]

This is false.

---

### Option 2: \( (10, 4) \)

Substitute \( x = 10 \) and \( y = 4 \) into the inequality:

\[
4 > \frac{5}{2}(10) - 18
\]
\[
4 > 25 - 18
\]
\[
4 > 7
\]

This is false.

---

### Option 3: \( (0, -15) \)

Substitute \( x = 0 \) and \( y = -15 \) into the inequality:

\[
-15 > \frac{5}{2}(0) - 18
\]
\[
-15 > -18
\]

This is true.

---

### Conclusion:

The only point that satisfies the inequality is \( (0, -15) \).

So, the correct answer is **\( (0, -15) \)**.

---

Would you like further clarification or details? Here are 5 follow-up questions related to this problem:

1. How do we determine if an inequality is strict or non-strict?
2. How would the solution change if the inequality was \( y + 3 \geq \frac{5}{2}(x - 6) \)?
3. What is the geometric interpretation of this inequality on a coordinate plane?
4. How do we handle fractional coefficients in inequalities when solving for \( y \)?
5. What happens to the inequality sign when multiplying or dividing by a negative number?

**Tip**: When solving inequalities involving fractions, it's helpful to first isolate the variable of interest (e.g., \( y \)) and then substitute the values to check for correctness.

Which of the following points is a solution to the inequality y + 3 > 5/2(x - 6)?

Find out which points satisfy the inequality y + 3 > 5/2(x - 6) by substituting values for x and y. This problem involves linear inequalities and algebraic substitution, suitable for students in Grades 9-10.

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