Respuesta :

we'll start off by making the recurring decimal value to a variable, hmmm say "x", thus x = 9.268686868......

we'll next multiply that value by a power of 10 so that the recurring decimal part moves over to the left of the decimal point, the recurring decimals in this case are "68", but we need to bring the 2 in front along as well, so to move all those three fellows, we'll need 1000.

we'll next multiply the same value by a power of 10 that brings two of the recurring decimals over, so 1000 gives us one "68" over, we'll need "6868" over to the left, so we'll use 100000, let's proceed.

[tex]\bf x = 9.2\overline{68}\qquad \qquad \begin{array}{llll} 1000\cdot x&=&9268.\overline{68}\\\\ 100000\cdot x&=&926868.\overline{68} \end{array}~\hfill \begin{array}{rllll} \stackrel{100000x}{926868.\overline{68}}\\\\ -\stackrel{1000x}{9268.\overline{68}}\\ \cline{1-1}\\ 917600.00 \end{array} \\\\\\ 100000x-1000x = 917600\implies 99000x=917600 \\\\\\ x = \cfrac{917600}{99000}\implies x = \cfrac{4588}{495}[/tex]

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