Question Number 159049 by mnjuly1970 last updated on 12/Nov/21 $$ \\ $$$$\:\:\:\:\:\:{solve}\:{the}\:{following}\:{equation}: \\ $$$$\: \\ $$$${sin}^{\:\mathrm{6}} \left({x}\right)\:+{cos}^{\:\mathrm{6}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)+{cos}^{\:\mathrm{4}} \left({x}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$ Answered…
Question Number 27983 by JI Siam last updated on 18/Jan/18 $$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{two}\:\:\mathrm{factors}\:\mathrm{of}\:\mathrm{1000001}\:\mathrm{other}\:\mathrm{than}\:\mathrm{1}\:\mathrm{and}\:\mathrm{1000001} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\mathrm{1}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions}? \\ $$ Answered by mrW2 last updated on 18/Jan/18…
Question Number 27977 by NECx last updated on 18/Jan/18 $$\mid\mathrm{2}{x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 18/Jan/18 $$\mid\mathrm{2x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$$$\pm\left(\mathrm{2x}+\mathrm{1}\right)\leqslant\mathrm{2} \\ $$$$\mathrm{2x}+\mathrm{1}\leqslant\mathrm{2}\:\wedge\:−\mathrm{2x}−\mathrm{1}\leqslant\mathrm{2} \\…
Question Number 159045 by amin96 last updated on 12/Nov/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }=? \\ $$ Answered by mindispower last updated on 16/Nov/21 $$=\underset{{n}\geqslant\mathrm{0}} {\sum}.\frac{\mathrm{1}}{\mathrm{27}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }…
Question Number 27976 by NECx last updated on 18/Jan/18 $${solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$ Commented by abdo imad last…
Question Number 27975 by NECx last updated on 18/Jan/18 $${solve}\:{the}\:{inequality} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}>\mathrm{0} \\ $$ Answered by mrW2 last updated on 18/Jan/18 $$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{x}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 159046 by mnjuly1970 last updated on 12/Nov/21 Commented by cortano last updated on 13/Nov/21 $${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} +\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}\right)+\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}}…
Question Number 159035 by HongKing last updated on 12/Nov/21 Commented by mr W last updated on 12/Nov/21 $${no}\:{unique}\:{solution}\:{possible}. \\ $$$${we}\:{can}\:{select}\:{a}\:{very}\:{small}\:{value}\:{for} \\ $$$${z}=\delta\:{with}\:\delta\rightarrow\mathrm{0}.\:{then}\:{we}\:{have} \\ $$$${m}={min}\left(×,\:×,\:×\right)\:\rightarrow\mathrm{0} \\…
Question Number 93505 by Ar Brandon last updated on 13/May/20 $$\mathrm{Does} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)+\left(\mathrm{x}^{'} ,\mathrm{y}^{'} \right)=\left(\mathrm{x}+\mathrm{x}^{'} ,\:\mathrm{y}+\mathrm{y}^{'} \right) \\ $$$$\mathrm{form}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{space}? \\ $$$$\lambda\left(\mathrm{x},\mathrm{y}\right)=\left(\lambda\mathrm{x},\lambda\mathrm{y}\right) \\ $$ Terms of…
Question Number 159034 by HongKing last updated on 12/Nov/21 Answered by mindispower last updated on 12/Nov/21 $${a}=\sqrt[{{n}}]{\frac{\mathrm{11}+{x}}{\mathrm{11}−{x}}},{b}=\sqrt{\frac{\mathrm{2021}+{x}}{\mathrm{2021}−{x}}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}={b}+\frac{\mathrm{1}}{{b}} \\ $$$$\left({a}−{b}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{ab}}\right)=\mathrm{0} \\ $$$${a}={b},{ab}=\mathrm{1} \\ $$$${a}={b}\Rightarrow\frac{\mathrm{11}+{x}}{\mathrm{11}−{x}}=\frac{\mathrm{2021}+{x}}{\mathrm{2021}−{x}}\:,{x}=\mathrm{0}…