3\overline{51}=0.3+0.0\overline{51}\] \[=0.3+\FRAC{1}{10}\times \LEFT\{ 0.\overline{51} \right\}\] \[=0.3+\frac{1}{10}\times \frac{51}{99}\] \[\frac{3}{10}+~\frac{51}{990}\] \[=\frac{297+51}{990}=\frac{348}{990}=\frac{116}{330}=\frac{58}{165}\].

"> 3\overline{51}=0.3+0.0\overline{51}\] \[=0.3+\FRAC{1}{10}\times \LEFT\{ 0.\overline{51} \right\}\] \[=0.3+\frac{1}{10}\times \frac{51}{99}\] \[\frac{3}{10}+~\frac{51}{990}\] \[=\frac{297+51}{990}=\frac{348}{990}=\frac{116}{330}=\frac{58}{165}\].

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The rational number for \[\mathbf{0}.\mathbf{3}\overline{\mathbf{51}}\] is

7th Class Mathematics in 7th Class 11 months ago

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(b): \[0.3\overline{51}=0.3+0.0\overline{51}\] \[=0.3+\FRAC{1}{10}\times \LEFT\{ 0.\overline{51} \right\}\] \[=0.3+\frac{1}{10}\times \frac{51}{99}\] \[\frac{3}{10}+~\frac{51}{990}\] \[=\frac{297+51}{990}=\frac{348}{990}=\frac{116}{330}=\frac{58}{165}\].

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