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

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Question

(NONE) 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 fol

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

(NONE) 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.
$1 - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9}$.
Note: The provided solution in the input was mathematically inconsistent with the question expression. Based on the standard order of operations,the result is $\frac{8}{9}$.

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

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