Question Number 1229 by Rasheed Soomro last updated on 16/Jul/15 $${f}\left(\:{f}\left({x}\right)\:\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({Modification}\:{of}\:{Q}\:\mathrm{1147}\right) \\ $$ Commented by 123456 last updated on 17/Jul/15…
Question Number 132302 by Lordose last updated on 13/Feb/21 $$\boldsymbol{\mathrm{Simplify}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$ Answered by Ar Brandon last updated on 13/Feb/21 $$\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\Gamma\left(\mathrm{2m}\right) \\…
Question Number 66765 by John Kaloki Musau last updated on 19/Aug/19 $${Simba}\:{had}\:\mathrm{57}\:{denomination}\:{notes} \\ $$$${which}\:{he}\:{deposited}\:{in}\:{his}\:{account}. \\ $$$${He}\:{had}\:{six}\:{times}\:{as}\:{many}\:{two}-{hundred} \\ $$$${shilling}\:{notes}\:{as}\:{one}-{thousand}\: \\ $$$${shilling}\:{notes}\:{and}\:{twice}\:{as}\:{many}\: \\ $$$${one}-{hundred}\:{shilling}\:{notes}\:{as}\:{two}- \\ $$$${hundred}\:{shilling}\:{notes}.\:{The}\:{rest}\:{were} \\…
Question Number 66760 by John Kaloki Musau last updated on 19/Aug/19 $${By}\:{writing}\:{your}\:{answer}\:{in}\:{the} \\ $$$${form}\:{a}^{{y}} \:{simplify} \\ $$$$\left(\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} \boldsymbol{\div}\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$ Commented by…
Question Number 132293 by Algoritm last updated on 13/Feb/21 Commented by mr W last updated on 13/Feb/21 $${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${the}\:{proof}\:{you}\:{may}\:{ask}\:{for}\:{is}\:{thinking}. \\ $$ Answered…
Question Number 132292 by Algoritm last updated on 13/Feb/21 Answered by mathmax by abdo last updated on 13/Feb/21 $$\frac{\mathrm{1}}{\left[\mathrm{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2x}\right]}=\left[\mathrm{x}\right]+\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{let}\:\left[\mathrm{x}\right]=\mathrm{n}\:\Rightarrow\mathrm{n}\leqslant\mathrm{x}<\mathrm{n}+\mathrm{1}\:\Rightarrow\mathrm{2n}\leqslant\mathrm{2x}<\mathrm{2n}+\mathrm{2} \\ $$$$\mathrm{if}\:\mathrm{2x}\in\left[\mathrm{2n},\mathrm{2n}+\mathrm{1}\left[\:\Rightarrow\mathrm{x}\in\left[\mathrm{n},\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\left[\:\Rightarrow\left[\mathrm{2x}\right]=\mathrm{2n}\right.\right.\right.\right. \\ $$$$\mathrm{e}\Rightarrow\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{9}}{\mathrm{2n}}=\mathrm{3n}+\mathrm{1}\:\Rightarrow\mathrm{9}=\mathrm{2n}\left(\mathrm{3n}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{6n}^{\mathrm{2}}…
Question Number 66759 by aliesam last updated on 19/Aug/19 Commented by mr W last updated on 19/Aug/19 $${question}\:{is}\:{wrong}. \\ $$$${there}\:{is}\:{no}\:{x}\:{which}\:{fulfills}\:\mid\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\mid>\mathrm{2}, \\ $$$${but}\:{such}\:{that}\:{e}^{{x}} >\mathrm{2016},\:{x}>\mathrm{ln}\:\mathrm{2016}\approx\mathrm{7}.\mathrm{6} \\…
Question Number 132295 by shaker last updated on 13/Feb/21 Commented by mathmax by abdo last updated on 13/Feb/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}−\sqrt{\mathrm{5x}−\mathrm{5}}\:\:\:\mathrm{f}\:\mathrm{defined}\:\mathrm{on}\:\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5x}−\mathrm{5}}}\:=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:=\frac{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}} \\ $$$$=\frac{\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}\right)\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}\:=\frac{\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{5x}^{\mathrm{2}} }{\left(…\right)}=\frac{−\mathrm{5x}^{\mathrm{2}}…
Question Number 66756 by John Kaloki Musau last updated on 19/Aug/19 $${solve}\:{the}\:{equations}, \\ $$$${x}+{y}=\mathrm{17} \\ $$$${xy}−\mathrm{5}{x}=\mathrm{32} \\ $$ Answered by $@ty@m123 last updated on 19/Aug/19…
Question Number 132294 by Algoritm last updated on 13/Feb/21 Commented by mr W last updated on 13/Feb/21 $${x}={y}=\mathrm{1} \\ $$$${the}\:{proof}\:{you}\:{may}\:{ask}\:{for}\:{is}\:{thinking}. \\ $$ Terms of Service…