Question Number 86802 by oustmuchiya@gmail.com last updated on 31/Mar/20
Commented by Rio Michael last updated on 31/Mar/20
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Answered by Rio Michael last updated on 31/Mar/20
$$\:\left(\mathrm{3}\right)\:{x}^{\mathrm{0}} \:+\:{x}^{\mathrm{2}} \:=\:\mathrm{5}\:\Rightarrow\:{x}\:=\:\pm\:\mathrm{2}. \\ $$$$\:\left(\mathrm{4}\right)\:\:\left(\mathrm{A}\:\cup\mathrm{B}\right)^{'} \:=\:\left\{\mathrm{7},\mathrm{8}\right\}. \\ $$$$\left(\mathrm{5}\right)\left(\mathrm{a}\right)\:\mathrm{tras}\:\mathrm{A}\:=\:\begin{pmatrix}{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{3}}\end{pmatrix} \\ $$$$\:\:\left(\mathrm{b}\right)\:\mathrm{2}{x}^{\mathrm{2}} \:=\:\mathrm{8}\:\Rightarrow\:{x}\:=\:\pm\:\mathrm{2} \\ $$$$\:\left(\mathrm{6}\right)\:{d}\:=\:\frac{\mathrm{10}}{−\mathrm{6}}\:=\:−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\frac{{v}}{{t}−{a}}\:=\:{d}\:\Rightarrow\:\:{a}\:={t}−\:\frac{{v}}{{d}} \\ $$$$ \\ $$