Question Number 70980 by ~ À ® @ 237 ~ last updated on 10/Oct/19 $$\:{Soit}\:\left({E},\mathcal{A},\mu\right)\:{un}\:\:{espace}\:{mesure}\:\:.\:{On}\:{suppose} \\ $$$${qu}'{il}\:{existe}\:{un}\:{X}\in\mathcal{A}\:\:{tel}\:\:\mu\left({X}\right)=+\infty \\ $$$$\left.\mathrm{1}\right){Montrer}\:{que}\:{si}\:\:\mu\:{est}\:{semi}-{finie}\:\:{alors} \\ $$$$\forall\:{r}>\mathrm{0}\:\:{il}\:{existe}\:\:{B}\subseteq{X}\:{tel}\:{que}\:\:{r}<\mu\left({B}\right)<\:+\infty \\ $$$$ \\ $$…
Question Number 5444 by 3 last updated on 15/May/16 $$\mathrm{8} \\ $$ Answered by FilupSmith last updated on 15/May/16 $$=\mathrm{7}+\mathrm{1} \\ $$ Answered by Rasheed…
Question Number 5441 by Rasheed Soomro last updated on 15/May/16 Commented by Rasheed Soomro last updated on 15/May/16 $$\left(\mathrm{a}\right)\:\mathrm{AB}=\mathrm{a}\:,\:\mathrm{BC}=\mathrm{b}\:,\:\mathrm{Area}\:\mathrm{EFGH}=? \\ $$$$ \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{then}\:\mathrm{EFGH}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parallelogram}.…
Question Number 136514 by mohammad17 last updated on 22/Mar/21 $${if}\:{z}=\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{find}\:{arg}\left(−{z}\right) \\ $$ Answered by mr W last updated on 22/Mar/21 $$−{z}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${arg}\left(−{z}\right)=\frac{\mathrm{2}\pi}{\mathrm{3}}=\mathrm{120}° \\ $$…
Question Number 70974 by Tinku Tara last updated on 10/Oct/19 $$\boldsymbol{\mathrm{Password}}\:\boldsymbol{\mathrm{reset}}\:\boldsymbol{\mathrm{changes}} \\ $$$$\mathrm{In}\:\mathrm{case}\:\mathrm{you}\:\mathrm{forget}\:\mathrm{any}\:\mathrm{password}\:\mathrm{set} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{simple}\:\mathrm{reset}\:\mathrm{by}\:\mathrm{going}\:\mathrm{to} \\ $$$$\mathrm{set}/\mathrm{update}\:\mathrm{password}. \\ $$$$ \\ $$$$\mathrm{Leave}\:\mathrm{old}\:\mathrm{password}\:\mathrm{field}\:\mathrm{blank}. \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{will}\:\mathrm{work}\:\mathrm{only}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{trying}…
Question Number 136505 by rexford last updated on 22/Mar/21 Answered by Dwaipayan Shikari last updated on 22/Mar/21 $${Assuming}\:{soda}\:{was}\:{x}\:{litre}\:{in}\:{quantity} \\ $$$${Your}\:{consumption}=\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}^{\mathrm{3}} }+\frac{{x}}{\mathrm{2}^{\mathrm{5}} }+…={x}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{32}}+…\right) \\ $$$$=\frac{\frac{{x}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{2}{x}}{\mathrm{3}} \\…
Question Number 70969 by TawaTawa last updated on 10/Oct/19 Answered by mind is power last updated on 10/Oct/19 $${first}\:{find}\:{how}\:{many}\:{zero}\:{ends}\:\mathrm{80}! \\ $$$${n}=\left[\frac{\mathrm{80}}{\mathrm{5}}\right]+\left[\frac{\mathrm{80}}{\mathrm{25}}\right]=\mathrm{16}+\mathrm{3}=\mathrm{19} \\ $$$${so}\:{we}\:{have} \\ $$$$\mathrm{80}!={d}\ast\mathrm{10}^{\mathrm{19}}…
Question Number 5431 by Rasheed Soomro last updated on 14/May/16 Commented by Yozzii last updated on 14/May/16 $${What}\:{does}\:\boldsymbol{\mathrm{d}}\:{measure}? \\ $$ Commented by Rasheed Soomro last…
Question Number 136500 by rexford last updated on 22/Mar/21 Answered by Dwaipayan Shikari last updated on 22/Mar/21 $${a}−\mathrm{2}{d}+{a}−{d}+{a}+{a}+{d}+{a}+\mathrm{2}{d}+{a}+\mathrm{3}{d}=\mathrm{78} \\ $$$$\Rightarrow\mathrm{6}{a}+\mathrm{3}{d}=\mathrm{78}\Rightarrow\mathrm{2}{a}+{d}=\mathrm{26} \\ $$$${since}\:{a}−\mathrm{2}{d}>\mathrm{1}\Rightarrow{a}>\mathrm{2}{d}+\mathrm{1}\:\:{Greatest}\:{possible}\:{when}\:{a}=\mathrm{2}{d}+\mathrm{1} \\ $$$$\mathrm{2}{a}+{d}=\mathrm{26}\Rightarrow\mathrm{5}{d}+\mathrm{2}=\mathrm{26}\Rightarrow\Rightarrow{d}=\frac{\mathrm{24}}{\mathrm{5}} \\…
Question Number 136497 by mnjuly1970 last updated on 22/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:\:\:{calculus}….. \\ $$$$\:\:\:\:{prove}::\:\: \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{ln}^{\mathrm{2}} \left({x}\right)}{dx}=\int_{\mathrm{0}\:} ^{\:\infty} \frac{{sin}\left({x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$ \\ $$ Commented by…