the idea being, we multiply the value by some power of 10 just to move the repeating digits to the left of the decimal point. We have 3 digits, we'll use 1000, anyhow, without further adieu,
[tex]\bf \begin{array}{|ccl|ll}
\cline{1-3}
\stackrel{\qquad}{x=0.\overline{482}}&&\\
\cline{1-3}
&&\\
\stackrel{\textit{3 digits, we use 3 zeros}}{1000\cdot x}&=&482.\overline{482}\\
&&482+0.\overline{482}\\
&&482+x\\
&&\\
\cline{1-3}
\end{array}\implies 1000x=482+x
\\\\\\
999x=482\implies x=\cfrac{482}{999}[/tex]
