Given the following question:
We are given the repeating decimal of 0.45
We will use the formula:
[tex]\begin{gathered} \frac{(d\times10^r)-n}{10^r-1} \\ \frac{0.45\times10^2)-0}{10^2-1} \\ \text{ Simplify} \\ \frac{0.45\times2\cdot10^2}{10^2-1}=\frac{45}{99} \\ \text{ Simplify once more} \\ \frac{45}{99}\div9=\frac{5}{11} \\ =\frac{5}{11} \end{gathered}[/tex]