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Question

# Fill in the blanks: (i) $-4×\frac{7}{9}=\frac{7}{9}×......$ (ii) $\frac{5}{11}×\frac{-3}{8}=\frac{-3}{8}×......$ (iii) $\frac{1}{2}×\left(\frac{3}{4}+\frac{-5}{12}\right)=\frac{1}{2}×......+......×\frac{-5}{12}$ (iv) $\frac{-4}{5}×\left(\frac{5}{7}+\frac{-8}{9}\right)=\left(\frac{-4}{5}×.....\right)×\frac{-8}{9}$

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Solution

## $\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{x}×\mathrm{y}=\mathrm{y}×\mathrm{x}\left(\mathrm{commutativity}\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\frac{5}{11}\phantom{\rule{0ex}{0ex}}\mathrm{x}×\mathrm{y}=\mathrm{y}×\mathrm{x}\left(\mathrm{commutativity}\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\frac{3}{4};\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}×\left(\mathrm{y}+\mathrm{z}\right)=\mathrm{x}×\mathrm{y}+\mathrm{x}×\mathrm{z}\left(d\mathrm{istributivity}\mathrm{of}\mathrm{multiplication}\mathrm{over}\mathrm{addition}\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\frac{5}{7}\phantom{\rule{0ex}{0ex}}\mathrm{x}×\left(\mathrm{y}×\mathrm{z}\right)=\left(\mathrm{x}×\mathrm{y}\right)×\mathrm{z}\left(a\mathrm{ssociativity}\mathrm{of}\mathrm{multiplication}\right)$

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