Question Number 161931 by mnjuly1970 last updated on 24/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\mathrm{I}{f}\:\:\:\:\frac{\:\mathrm{1}−{sin}\left({x}\right)−{cos}\left({x}\right)}{\mathrm{1}+{sin}\left({x}\right)−{cos}\left({x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:{then}\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{tan}\left({x}\right)\:+\:\frac{\mathrm{1}}{{cos}\left({x}\right)}\:=? \\ $$$$\:\:\:\:\:\: \\ $$$$ \\ $$ Commented by cortano…
Question Number 161930 by Rasheed.Sindhi last updated on 24/Dec/21 $$\begin{cases}{{a}_{\mathrm{0}} =−\mathrm{2}\:}\\{{a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{n}}\end{cases}\:\:\:\:\:;\:{a}_{{n}} =? \\ $$ Answered by mr W last updated on 24/Dec/21 $${a}_{{n}}…
Question Number 96390 by bemath last updated on 01/Jun/20 Answered by john santu last updated on 01/Jun/20 $$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{initial}\:\mathrm{corn}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{weight} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{final}\:\mathrm{mixture}.\:\mathrm{therefore} \\ $$$$\left(\mathrm{20}\:\mathrm{bushels}\right)\frac{\mathrm{56}\:\mathrm{pounds}}{\mathrm{bushels}}\:+\:\left({x}\:\mathrm{bushels}\right)\frac{\mathrm{50}\:\mathrm{pounds}}{\mathrm{bushels}}\: \\…
Question Number 30849 by Penguin last updated on 27/Feb/18 $${x}^{\mathrm{7}} +{x}^{\mathrm{6}} +{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}\left[\Re\left({x}_{{k}} \right)\right]^{\mathrm{2}} \:=\:? \\…
Question Number 161914 by mnjuly1970 last updated on 24/Dec/21 $$\:\mathrm{If}\:, \\ $$$$\:\:\:{x}^{\:\mathrm{2}} \:+\:\mathrm{9}{y}^{\:\mathrm{2}} \:+\:\mathrm{4}{x}\:+\mathrm{18}{y}\:−\mathrm{23}=\mathrm{0} \\ $$$$ \\ $$$$\:{then}\:\:{find}\:{the}\:{value}\:\:{of}\:\:,\:\:\mathrm{M}_{\:} {ax}\:\left(\:\mathrm{3}{x}+\mathrm{2}{y}\:\right)\:. \\ $$$$\:−−−−−−−−− \\ $$$$ \\ $$…
Question Number 161900 by HongKing last updated on 23/Dec/21 $$\mathrm{0}<\mathrm{x};\mathrm{y};\mathrm{z}<\mathrm{1} \\ $$$$\left(\mathrm{1}-\mathrm{x}\right)\left(\mathrm{1}-\mathrm{y}\right)\left(\mathrm{1}-\mathrm{z}\right)=\mathrm{xyz} \\ $$$$\mathrm{Find}: \\ $$$$\Omega\:=\:\mathrm{min}\:\left(\frac{\mathrm{1}-\mathrm{x}}{\mathrm{xy}}\:+\:\frac{\mathrm{1}-\mathrm{y}}{\mathrm{yz}}\:+\:\frac{\mathrm{1}-\mathrm{z}}{\mathrm{zx}}\right) \\ $$ Answered by aleks041103 last updated on 24/Dec/21…
Question Number 161899 by HongKing last updated on 23/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:-\infty} {\overset{\:\infty} {\int}}\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:\:;\:\:\mathrm{n}\in\mathbb{Z} \\ $$ Answered by Ar Brandon last updated on…
Question Number 161884 by HongKing last updated on 23/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{-\mathrm{1}} }\:=\:\mathrm{1}\:+\:\sqrt[{\mathrm{7}}]{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:-\:\mathrm{2}^{-\mathrm{1}} } \\ $$ Commented by mr W last updated on…
Question Number 96340 by Rio Michael last updated on 31/May/20 $$\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{two}\:\mathrm{circles}\:{S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:{S}_{\mathrm{1}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}{y}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:{S}_{\mathrm{2}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\:\mathrm{2}{y}\:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{S}_{\mathrm{1}}…
Question Number 161868 by cortano last updated on 23/Dec/21 $$\:\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} \:+\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} }\:=\:\mathrm{2}^{{x}} \\ $$$$\:\:{x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Dec/21 $${a}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:,\:\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}×\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\…