Question Number 108290 by 1549442205PVT last updated on 16/Aug/20 $$\mathrm{Why}\:\mathrm{do}\:\mathrm{all}\:\mathrm{active}\:\mathrm{content}\:\mathrm{of}\:\mathrm{forum}\:\mathrm{of} \\ $$$$\mathrm{dates}\:\mathrm{12},\mathrm{13},\mathrm{14was}\:\mathrm{disappear}\:\mathrm{on}\:\mathrm{my} \\ $$$$\mathrm{phone}? \\ $$ Commented by 1549442205PVT last updated on 16/Aug/20 $$\mathrm{Thank}\:\mathrm{Great}\:\mathrm{Tinkutara}\:\mathrm{all}\:\mathrm{datas}\:\mathrm{is} \\…
Question Number 42725 by arpit last updated on 01/Sep/18 Answered by $@ty@m last updated on 02/Sep/18 $${see}\:{Q}.\:{No}.\:\mathrm{42592} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 108259 by Dwaipayan Shikari last updated on 15/Aug/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108239 by Dwaipayan Shikari last updated on 15/Aug/20 $$\left(\frac{\mathrm{7}}{\mathrm{2}}\right)!=? \\ $$$$ \\ $$ Answered by abdomsup last updated on 15/Aug/20 $$\Gamma\left({x}+{n}\right)\:=\Gamma\left({x}+{n}−\mathrm{1}\:+\mathrm{1}\right) \\ $$$$=\left({x}+{n}−\mathrm{1}\right)\Gamma\left({x}+{n}−\mathrm{2}\:+\mathrm{1}\right)…
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Question Number 173630 by MME last updated on 15/Jul/22 $$\left({x}+\mathrm{4}\right)^{\mathrm{2}} ={x}^{\left({x}+\mathrm{2}\right)} \\ $$$${Please}\:\:{Help}… \\ $$ Commented by mr W last updated on 15/Jul/22 $${no}\:{chance}! \\…
Question Number 42550 by Cheyboy last updated on 27/Aug/18 $$\mathrm{2}^{\frac{\mathrm{1}}{{x}}} =\sqrt{{x}} \\ $$$${Find}\:{x} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18 Commented by tanmay.chaudhury50@gmail.com…
Question Number 173620 by eiverzurita last updated on 14/Jul/22 Answered by FelipeLz last updated on 14/Jul/22 $$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{bx}+\mathrm{2}{a}^{\mathrm{2}} {x}+{abx}+{ab}^{\mathrm{2}} \:=\:\mathrm{2}{abx}+\mathrm{2}{a}^{\mathrm{2}} {b} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{bx}+\mathrm{2}{a}^{\mathrm{2}} {x}+{ab}^{\mathrm{2}}…
Question Number 173599 by Mastermind last updated on 14/Jul/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{y}=\mathrm{2}−\mathrm{x}^{\mathrm{2}} \:\mathrm{for}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}. \\ $$ Commented by kaivan.ahmadi last updated on 14/Jul/22 $${x}^{\mathrm{2}} =\mathrm{2}−{x}^{\mathrm{2}}…