Question Number 104207 by PengagumRahasiamu last updated on 20/Jul/20 Answered by bemath last updated on 20/Jul/20 $${x}^{\mathrm{2}} −{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}=\:\frac{\mathrm{1}\:\pm\:{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\pm\:\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${have}\:{no}\:{real}\:{roots}.\:{Sum}\:{of} \\ $$$${real}\:{root}\:{is}\:{nothing} \\…
Question Number 169738 by Mastermind last updated on 07/May/22 $$\mathrm{2}^{{x}} +{x}=\mathrm{11} \\ $$$${find}\:{x}? \\ $$$$ \\ $$$${Mastermind} \\ $$ Commented by mr W last updated…
Question Number 104199 by hardylanes last updated on 20/Jul/20 $${solve}\:{for}\:{real}\:{values}\:{of}\:{x}\:{the}\:{equation} \\ $$$$\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)=\mathrm{7}. \\ $$$${if}\:{m}\:{and}\:{n}\:{are}\:{positive}\:{real}\:{numbers}\:{other} \\ $$$${than}\:\mathrm{1},\:{show}\:{that}\:{the}\:\mathrm{log}_{{n}} {m}+\mathrm{log}_{\frac{\mathrm{1}}{{m}}} {n}=\mathrm{0} \\ $$ Answered by bemath…
Question Number 169727 by Mastermind last updated on 07/May/22 Commented by Mastermind last updated on 07/May/22 $${Okay},\:{thanks} \\ $$ Commented by mr W last updated…
Question Number 104176 by Dwaipayan Shikari last updated on 19/Jul/20 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}\right) \\ $$ Answered by OlafThorendsen last updated on 19/Jul/20 $${u}_{{n}} \:=\:\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}}…
Question Number 104174 by Dwaipayan Shikari last updated on 19/Jul/20 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$ Answered by OlafThorendsen last updated on 19/Jul/20 $$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 104173 by Dwaipayan Shikari last updated on 19/Jul/20 $$\:\mathrm{For}\:\mathrm{fun}! \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+… \\ $$$$\mathrm{2S}_{\mathrm{n}} =\:\:\:\:\mathrm{2}\:\:+\:\:\:\mathrm{4}\:\:\:\:\:\:+\:\:\mathrm{6}\:\:\:\:\:\:+.. \\ $$$$−\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+….. \\ $$$$−\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)=\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+… \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+….=\frac{\mathrm{1}}{\mathrm{12}} \\…
Question Number 169658 by Tawa11 last updated on 05/May/22 Commented by mr W last updated on 05/May/22 $${frictional}\:{force}={active}\:{force}\:{applied} \\ $$$${on}\:{carton} \\ $$$$\Rightarrow{frictional}\:{force}\:=\mathrm{150}\:{N} \\ $$$${Answer}\:{A}\:{is}\:{correct}. \\…
Question Number 38562 by shakirhsp@gmail.com last updated on 27/Jun/18 $$\left(\sqrt{\mathrm{2}}\:+{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}{i}}\:\right) \\ $$ Answered by MJS last updated on 27/Jun/18 $$\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2i}}\right)=\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{i}\right)\right)= \\ $$$$=\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(−\mathrm{i}\right)=\mathrm{1}−\mathrm{i}\sqrt{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{i}\right)=\sqrt{\mathrm{2}}−\mathrm{2i}+\mathrm{i}+\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{i} \\…
Question Number 38559 by nishant last updated on 27/Jun/18 $${in}\:{a}\:{geometric}\:{series},\:{the}\:{first}\:{term} \\ $$$$={a},\:{common}\:{ratio}={r}.\:{If}\:{S}_{{n}} \:{denotes} \\ $$$${the}\:{sum}\:{of}\:{the}\:{n}\:{terms}\:{and}\:{U}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{S}_{{n},} \\ $$$${then}\:{rS}_{{n}} +\left(\mathrm{1}−{r}\right){U}_{{n}\:\:} {equals}\:{to} \\ $$$$\left({a}\right)\:\:\mathrm{0}\:\:\:\:\:\:\left({b}\right)\:\:{n}\:\:\:\:\:\left({c}\right)\:\:\:\:{na}\:\:\:\:\left({d}\right){nar} \\…