Question Number 62047 by hhghg last updated on 14/Jun/19 $$\left(\mathrm{4y}\right)^{\mathrm{2}} \\ $$ Commented by maxmathsup by imad last updated on 15/Jun/19 $$=\left(\mathrm{4}{y}\right)×\left(\mathrm{4}{y}\right)\:=\mathrm{16}{y}^{\mathrm{2}} \\ $$ Terms…
Question Number 62048 by hhghg last updated on 14/Jun/19 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$ Commented by maxmathsup by imad last updated on 15/Jun/19 $$=\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{8}} \\…
Question Number 62046 by hhghg last updated on 14/Jun/19 $$\mathrm{4}×\left(\mathrm{3}+\mathrm{2}−\mathrm{3}\right) \\ $$ Commented by maxmathsup by imad last updated on 15/Jun/19 $$=\mathrm{4}×\left(\mathrm{5}−\mathrm{3}\right)\:=\mathrm{4}×\mathrm{2}\:=\mathrm{8} \\ $$ Terms…
Question Number 62045 by hovea cw last updated on 14/Jun/19 $$\mathrm{help} \\ $$$$ \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{4} \\ $$$$\mathrm{find}\:\mathrm{x} \\ $$ Answered by MJS…
Question Number 62041 by mr W last updated on 14/Jun/19 Commented by mr W last updated on 14/Jun/19 $${i}\:{use}\:{these}\:{formulas}\:{to}\:{get}\:{real}\:{roots} \\ $$$${of}\:{cubic}\:{equations}. \\ $$$${maybe}\:{useful}\:{also}\:{for}\:{you}. \\ $$…
Question Number 62023 by aliesam last updated on 14/Jun/19 Commented by MJS last updated on 14/Jun/19 $$\mathrm{French}\:\mathrm{pgcd}\:\left({x},\:{y}\right)\:\mathrm{is}\:\mathrm{English}\:\mathrm{gcd}\:\left({x},{y}\right) \\ $$ Commented by MJS last updated on…
Question Number 193087 by Abdullahrussell last updated on 03/Jun/23 Answered by Frix last updated on 03/Jun/23 $${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{2} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 127547 by Mathgreat last updated on 30/Dec/20 Answered by mr W last updated on 31/Dec/20 Commented by mr W last updated on 31/Dec/20…
Question Number 127525 by Mathgreat last updated on 30/Dec/20 Commented by Mathgreat last updated on 30/Dec/20 $$\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +…+\boldsymbol{{x}}_{\boldsymbol{{n}}} =\:??? \\ $$ Commented by…
Question Number 127519 by O Predador last updated on 30/Dec/20 Answered by mahdipoor last updated on 30/Dec/20 $$\mathrm{0}.\mathrm{0833}…=\mathrm{0}.\mathrm{08}\overset{−} {\mathrm{3}}=\mathrm{0}.\mathrm{08}+\mathrm{0}.\mathrm{00}\overset{−} {\mathrm{3}}= \\ $$$$\frac{\mathrm{8}}{\mathrm{100}}+\frac{\mathrm{1}}{\mathrm{300}}=\frac{\mathrm{25}}{\mathrm{300}}=\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}}=\frac{\mathrm{2}!}{\mathrm{4}!}=\frac{\left[{log}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)−\mathrm{2}\right]!}{\left[{log}_{\mathrm{2}} \left({x}\right)+\mathrm{2}\right]!}…