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# Q.3 Find the product: (i) $$\frac{9}{2}$$ × $$(-\frac{7}{4} )$$ (ii) $$\frac{9}{2}$$ × (-9 ) (iii) $$\frac{-6}{5}$$× $$\frac{9}{11}$$ (iv) $$\frac{3}{7}$$ × $$(-\frac{2}{5} )$$ (v) $$\frac{3}{11}$$× $$\frac{2}{5}$$ (vi) $$\frac{3}{-5}$$× $$\frac{-5}{3}$$

(i) $$\frac{9}{2}$$× $$\frac{-7}{4}$$ = $$\frac{9×-7}{2×4}$$= $$\frac{-63}{8}$$ = $$-7\frac{7}{8}$$
(ii) $$\frac{9}{2}$$ × (-9 )= $$\frac{-81}{2}$$ =$$-40\frac{1}{2}$$
(iii) $$\frac{-6}{5}$$× $$\frac{9}{11}$$ = $$\frac{-6×9}{5×11}$$= $$\frac{-54}{55}$$
(iv) $$\frac{3}{7}$$ × $$(-\frac{2}{5} )$$= $$\frac{-3×2}{7×5}$$ = $$\frac{-6}{35}$$
(v) $$\frac{3}{11}$$× $$\frac{2}{5}$$ = $$\frac{3×2}{11×5}$$=$$\frac{6}{55}$$
(vi) $$\frac{3}{-5}$$× $$\frac{-5}{3}$$ = $$\frac{3×(-5)}{-5×3}$$ =$$\frac{-15}{-15}$$= 1