frac{frac{1}{3}+frac{1}{3} times frac{1}{3}}{ | vedclass

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(A) To solve the expression $\frac{\frac{1}{3}+\frac{1}{3} 	imes \frac{1}{3}}{\frac{1}{3}+\frac{1}{3} 	ext{ of } \frac{1}{3}}-\frac{1}{9}$, we follo

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

(A) To solve the expression $\frac{\frac{1}{3}+\frac{1}{3} 	imes \frac{1}{3}}{\frac{1}{3}+\frac{1}{3} 	ext{ of } \frac{1}{3}}-\frac{1}{9}$, we follow the $BODMAS$ rule.Step $1$: Simplify the numerator.$\frac{1}{3} + (\frac{1}{3} 	imes \frac{1}{3}) = \frac{1}{3} + \frac{1}{9} = \frac{3+1}{9} = \frac{4}{9}$.Step $2$: Simplify the denominator.$\frac{1}{3} + (\frac{1}{3} 	ext{ of } \frac{1}{3}) = \frac{1}{3} + (\frac{1}{3} 	imes \frac{1}{3}) = \frac{1}{3} + \frac{1}{9} = \frac{3+1}{9} = \frac{4}{9}$.Step $3$: Divide the numerator by the denominator.$\frac{4/9}{4/9} = 1$.Step $4$: Subtract $\frac{1}{9}$ from the result.$1 - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9}$.

$\frac{\frac{1}{3}+\frac{1}{3} \times \frac{1}{3}}{\frac{1}{3}+\frac{1}{3} \text{ of } \frac{1}{3}}-\frac{1}{9} = ?$

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