\( \underset{(59)}{6} \frac{1}{2} + \frac{2}{3} + \frac{1}{6} \)

To solve the expression \( \underset{(59)}{6} \frac{1}{2} + \frac{2}{3} + \frac{1}{6} \), let's break it down step by step! First, we need to compute the fractions. The least common denominator (LCD) for \( \frac{1}{2} \), \( \frac{2}{3} \), and \( \frac{1}{6} \) is 6. Converting each fraction to have a denominator of 6: - \( \frac{1}{2} = \frac{3}{6} \) - \( \frac{2}{3} = \frac{4}{6} \) - \( \frac{1}{6} = \frac{1}{6} \) (no change needed) Now, we can add these fractions: \[ \frac{3}{6} + \frac{4}{6} + \frac{1}{6} = \frac{3 + 4 + 1}{6} = \frac{8}{6} \] Simplifying \( \frac{8}{6} \) gives us \( \frac{4}{3} \). Now, we can include the \( \underset{(59)}{6} \). But it's a little unclear what \( \underset{(59)}{6} \) refers to—perhaps it indicates something context-specific, like a previous value or context. Assuming it should just lead us to add it directly: If we add \( \frac{4}{3} \) this addition stems from a whole number (if we consider 6 as a whole): \[ 6 + \frac{4}{3} = 6 + 1\frac{1}{3} = 7\frac{1}{3} \text{ or } \frac{22}{3} \text{ as an improper fraction.} \] So the final result, depending on the clarification needed on \( \underset{(59)}{6} \), is roughly \( \frac{22}{3} \) or \( 7\frac{1}{3} \) if we treat it as 6 added directly!