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.

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**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 $$9$$.
**Problem 1.2** The recurring decimal $$0.5\overline{1}$$ can be written as the common fraction $$\frac{23}{45}$$. **Problem 1.3** The square root of 76 lies between the consecutive integers 8 and 9.

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.

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