1. The problem is to find the sum of the repeating decimals $0.1\dot{}$ and $0.2\dot{8}$.\n\n2. First, let's express each repeating decimal as a fraction.\n\n3. For $0.1\dot{}$, the repeating digit is 1. Let $x = 0.1111\ldots$. Multiply both sides by 10: $10x = 1.1111\ldots$. Subtract the original equation: $10x - x = 1.1111\ldots - 0.1111\ldots$ which simplifies to $9x = 1$. So, $x = \frac{1}{9}$.\n\n4. For $0.2\dot{8}$, the repeating part is 8 after the 2. Let $y = 0.2888\ldots$. Multiply both sides by 10: $10y = 2.888\ldots$. Subtract the original equation: $10y - y = 2.888\ldots - 0.2888\ldots$ which simplifies to $9y = 2.6$. Convert $2.6$ to fraction: $2.6 = \frac{26}{10} = \frac{13}{5}$. So, $9y = \frac{13}{5}$ and $y = \frac{13}{45}$.\n\n5. Now, add the two fractions: $\frac{1}{9} + \frac{13}{45}$. Find a common denominator, which is 45. Convert $\frac{1}{9} = \frac{5}{45}$. So, $\frac{5}{45} + \frac{13}{45} = \frac{18}{45}$. Simplify $\frac{18}{45}$ by dividing numerator and denominator by 9: $\frac{2}{5}$.\n\n6. Therefore, the sum of $0.1\dot{} + 0.2\dot{8}$ is $\frac{2}{5}$ or 0.4 in decimal form.