1.2 Write the recurring decimal \( 0,5 \dot{1} \) as a common fraction. 1.3 Determine between which two consecutive integers does \( \sqrt{76} \) lie.

**Problem 1.2** Let \[ x = 0.5\overline{1} \] which means that the digit \(1\) repeats indefinitely. 1. Write \(x\) with its non‐repeating and repeating parts: \[ x = 0.51111\ldots \] 2. Identify the number of digits in the non-repeating part and the repeating block. Here, the non-repeating part is "5" (1 digit) and the repeating block is "1" (also 1 digit). 3. Multiply \(x\) by \(10\) (since there is 1 digit before the repeating part) to shift the decimal point: \[ 10x = 5.11111\ldots \] 4. Multiply \(x\) by \(10^2=100\) (since 1 digit non-repeating + 1 digit repeating = 2 digits) to shift the repeating part fully to the left: \[ 100x = 51.11111\ldots \] 5. Subtract the equation from step 3 from the equation in step 4 to eliminate the repeating part: \[ 100x - 10x = 51.11111\ldots - 5.11111\ldots \] This gives: \[ 90x = 46 \] 6. Solve for \(x\): \[ x = \frac{46}{90} = \frac{23}{45} \] **Problem 1.3** We need to determine between which two consecutive integers the value of \(\sqrt{76}\) lies. 1. Compare the square of consecutive integers to \(76\): \[ 8^2 = 64 \quad \text{and} \quad 9^2 = 81 \] 2. Since \[ 64 < 76 < 81, \] it follows that: \[ 8 < \sqrt{76} < 9. \] Thus, \(\sqrt{76}\) lies between \(8\) and \(9\).