Question

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Answer 1

$\displaystyle i). \frac{5}{-3 \frac{1}{3} } = \frac{5}{- \frac{10}{3} } =5: \left(- \frac{10}{3} \right)=5 \cdot \left(- \frac{3}{10} \right )=- \frac{15}{10} ^{(5} =- \frac{3}{2} \\ \\ j). \frac{-3+2 \frac{1}{4} }{0,(3)} = \frac{-3+ \frac{9}{4} }{ \frac{3}{9} } = \frac{- \frac{12}{4} + \frac{9}{4} }{ \frac{3}{9} } = \frac{- \frac{3}{4} }{ \frac{3}{9} } =- \frac{3}{4} : \frac{3}{9} =- \frac{\not3}{4} \cdot \frac{9}{\not3} =- \frac{9}{4}$

$\displaystyle k). \frac{-2 \frac{1}{3} }{3 \frac{1}{2} } = \frac{- \frac{7}{3} }{ \frac{7}{2} } =- \frac{7}{3} : \frac{7}{2} =- \frac{\not7}{3} \cdot \frac{2}{\not 7} =- \frac{2}{3}$

$\displaystyle l). \frac{-5+3:4 \frac{1}{2} }{-13} = \frac{-5+3: \frac{9}{2} }{-13} = \frac{-5+3 \cdot \frac{2}{9} }{-13} = \frac{-5+ \frac{6}{9} }{-13} = \frac{- \frac{45}{9}+ \frac{6}{9} }{-13} = \\ \\ = \frac{- \frac{39}{9} }{-13} =- \frac{39}{9} :(-13)=- \frac{39}{9} \cdot \left (- \frac{1}{13} \right )= \frac{39}{117} ^{(39} = \frac{1}{3}$