Question

$[ \frac{6}{7}\times\frac{41}{2} \div 3] \times [ \frac{3}{4} + \frac{7}{3} - ( \frac{ - 7}{3}) ]$
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Answer 1

Step-by-step explanation:

$( \frac{6}{7} \times \frac{41}{2} \div 3) \times ( \frac{3}{4} + \frac{7}{3} - \frac{ - 7}{3} ) \\ = (\frac{ \cancel{6}_{ \: \cancel{2} _{ \: 1} } }{7} \times \frac{41}{ \cancel{2}_{ \: 1} } \times \frac{1}{ \cancel{3}_{ \: 1} }) \times ( \frac{9 + 28 - ( - 28)}{12} ) \\ = \frac{41}{7} \times ( \frac{9 + 28 + 28}{12} ) \\ = \frac{41}{7} \times \frac{65}{12} \\ \large{ \boxed{= \underline{ \underline{ \bf{ \frac{2665}{84} }}}}}$