Question Number 136750 by mnjuly1970 last updated on 25/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:\:{calculus}….. \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:…\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi}\:\:\:\: \\ $$ Answered by Ñï= last…
Question Number 136745 by PGeeman last updated on 25/Mar/21 Commented by mr W last updated on 25/Mar/21 $${the}\:“{forum}\:{user}\:{id}''\:{from}\:{you}\:{is} \\ $$$${PGeeman}\:{for}\:{example}. \\ $$ Terms of Service…
Question Number 5674 by sanusihammed last updated on 23/May/16 $${Find}\:{the}\:{value}\:{of}\:{x}\: \\ $$$$ \\ $$$${x}^{\left({x}\:+\:\mathrm{2}\right)} \:=\:\left({x}\:+\:\mathrm{2}\right)^{{x}} \\ $$$$ \\ $$$${Please}\:{help}. \\ $$ Terms of Service Privacy…
Question Number 5673 by sanusihammed last updated on 23/May/16 $${Using}\:{inductive}\:{method}.\:{prove}\:{that}\:.. \\ $$$$ \\ $$$$\mathrm{7}^{\mathrm{2}{n}} \:+\:\mathrm{16}{n}\:−\:\mathrm{1}\:{is}\:{divisible}\:{by}\:\mathrm{4} \\ $$$$ \\ $$$${please}\:{help}. \\ $$ Commented by Rasheed Soomro…
Question Number 5672 by sanusihammed last updated on 23/May/16 $${Differentiate}\:\:\:\:\frac{{lnx}}{{e}^{{x}} }\:\:\:\:{fom}\:{the}\:{first}\:{principle}. \\ $$$$ \\ $$$${Please}\:{help}\:{me}. \\ $$ Answered by Yozzii last updated on 23/May/16 $${Let}\:{y}=\frac{{lnx}}{{e}^{{x}}…
Question Number 136741 by JulioCesar last updated on 25/Mar/21 Answered by Dwaipayan Shikari last updated on 25/Mar/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}!−\mathrm{1}}{{x}}=\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{1}}{{x}}=\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{1}}=\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$ Terms of Service Privacy…
Question Number 71206 by naka3546 last updated on 13/Oct/19 $${Let}\:\:{p},{q},{r}\:\:{are}\:\:{positive}\:\:{real}\:\:{numbers}\:. \\ $$$$\mathrm{0}\:<\:{r}\:<\:{min}\left\{{p},{q}\right\}. \\ $$$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\sqrt{{p}−{r}}\:+\:\sqrt{{q}−{r}}\:\:\leqslant\:\:{min}\left\{\sqrt{\frac{{pq}}{{r}}}\:,\:\sqrt{\mathrm{2}\left({p}+{q}\:−\:\mathrm{2}{r}\right)}\:\right\} \\ $$ Answered by mind is power last updated…
Question Number 136740 by bramlexs22 last updated on 25/Mar/21 $$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{2f}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)=\:\frac{\mathrm{x}}{\mathrm{2}−\mathrm{x}}\:. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{x}\right)\:? \\ $$ Answered by Olaf last updated on 25/Mar/21 $$ \\ $$$${f}\left(\mathrm{2}{f}^{−\mathrm{1}}…
Question Number 136739 by Ar Brandon last updated on 25/Mar/21 $$\mathrm{Given}\:\mathrm{0}<\mathrm{a}<\mathrm{b},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8b}}\leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\sqrt{\mathrm{ab}}\leqslant\frac{\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{8a}} \\ $$ Answered by snipers237 last updated on 26/Mar/21 $$\frac{{a}+{b}}{\mathrm{2}}−\sqrt{{ab}\:}=\:\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}}…
Question Number 136738 by mohammad17 last updated on 25/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com