The problem is to evaluate the sum of three numbers: $+(1/2)$, $+(2/3)$, and $-(1\frac{1}{6})$.

1. Problem Description

The problem is to evaluate the sum of three numbers: $+(1/2)$, $+(2/3)$, and $-(1\frac{1}{6})$.

2. Solution Steps

First, we can rewrite the expression as:

$\frac{1}{2} + \frac{2}{3} - 1\frac{1}{6}$

Next, we convert the mixed number to an improper fraction:

$1\frac{1}{6} = \frac{1 \times 6 + 1}{6} = \frac{7}{6}$

So the expression becomes:

$\frac{1}{2} + \frac{2}{3} - \frac{7}{6}$

Now, we find a common denominator for the three fractions, which is

6. We convert each fraction to an equivalent fraction with a denominator of 6:

$\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

$\frac{7}{6}$ remains as is.

So the expression now becomes:

$\frac{3}{6} + \frac{4}{6} - \frac{7}{6}$

Now, we can add and subtract the fractions:

$\frac{3+4-7}{6} = \frac{7-7}{6} = \frac{0}{6} = 0$

3. Final Answer

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