Solve the inequality for \( v \). \[ v+\frac{7}{8}>\frac{1}{2} \] Simplify your answer as much as possible. \( \square \) \( \square<\square \quad \square>\square \quad \square \leq \square \)

To solve the inequality for \( v \): \[ v + \frac{7}{8} > \frac{1}{2} \] **Step 1: Subtract \(\frac{7}{8}\) from both sides to isolate \( v \).** \[ v > \frac{1}{2} - \frac{7}{8} \] **Step 2: Find a common denominator to subtract the fractions.** The common denominator for \( \frac{1}{2} \) and \( \frac{7}{8} \) is 8. \[ \frac{1}{2} = \frac{4}{8} \] So, the inequality becomes: \[ v > \frac{4}{8} - \frac{7}{8} \] **Step 3: Subtract the fractions.** \[ v > \frac{4 - 7}{8} = \frac{-3}{8} \] **Final Solution:** \[ v > -\frac{3}{8} \] **In Interval Notation:** \[ v \in \left( -\frac{3}{8}, \infty \right) \] **On a Number Line:** - **\( \square \)** \( \square < \square \quad \square > \square \quad \square \leq \square \) This indicates that \( v \) is greater than \(-\frac{3}{8}\).