Question

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Answer 1

$\displaystyle\bf\\3m^{4} n=-2\\\\\\\boxed{m^{4} n=-\frac{2}{3} }\\\\\\1)\\\\-1,2m^{4}n=-1,2\cdot \Big(-\frac{2}{3} \Big)= 0,8\\\\\\2)\\\\m^{8} n^{2} =\Big(m^{4} n^{2} \Big)^{2} =\Big(-\frac{2}{3} \Big)^{2} =\frac{4}{9} \\\\\\3m^{8} n^{2} =3\cdot\frac{4}{9} =\frac{4}{3}=1\frac{1}{3}$