Question Number 218813 by Spillover last updated on 15/Apr/25 Answered by nikif99 last updated on 15/Apr/25 Commented by Spillover last updated on 16/Apr/25 $${great}\:{work}.{thanks}\: \\…
Question Number 218749 by hardmath last updated on 15/Apr/25 Answered by MrGaster last updated on 17/Apr/25 $${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{ln}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}}…
Question Number 218799 by depressiveshrek last updated on 15/Apr/25 $$\mathrm{For}\:\mathrm{those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{cryptography}. \\ $$$$\mathrm{The}\:\mathrm{below}\:\mathrm{text}\:\mathrm{has}\:\mathrm{been}\:\mathrm{encrypted}\:\mathrm{using} \\ $$$$\mathrm{Vigenere}\:\mathrm{cipher},\:\mathrm{such}\:\mathrm{that}\:\mathrm{numbers},\:\mathrm{punctuation} \\ $$$$\mathrm{marks}\:\mathrm{and}\:\mathrm{the}\:\mathrm{letter}\:\overset{..} {\mathrm{E}}\:\mathrm{have}\:\mathrm{remained}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{A}\:\mathrm{keyword}\:\mathrm{of}\:\mathrm{length}\:\mathrm{9}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used},\:\mathrm{which} \\ $$$$\mathrm{starts}\:\mathrm{with}\:\mathrm{the}\:\mathrm{letter}\:\mathrm{K}.\:\mathrm{Decrypt}\:\mathrm{the}\:\mathrm{text}. \\ $$ Commented by…
Question Number 218792 by sonukgindia last updated on 15/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218785 by sonukgindia last updated on 15/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218780 by Spillover last updated on 15/Apr/25 Commented by malwan last updated on 17/Apr/25 $${can}\:{you}\:{solve}\:{this}\:{integral} \\ $$$${please}\:? \\ $$ Answered by Spillover last…
Question Number 218781 by Ghisom last updated on 15/Apr/25 $$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$ Answered by breniam last updated on 20/Apr/25…
Question Number 218778 by Spillover last updated on 15/Apr/25 Answered by mr W last updated on 15/Apr/25 Commented by mr W last updated on 17/Apr/25…
Question Number 218779 by Spillover last updated on 15/Apr/25 Answered by breniam last updated on 15/Apr/25 $$ \\ $$$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosh}\:{x}}\mathrm{d}{x}=\int\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}}\mathrm{d}{x}=\int\frac{\mathrm{2}}{{e}^{{x}} +\mathrm{2}+{e}^{−{x}} }\mathrm{d}{x}= \\ $$$$\int\frac{\mathrm{2}}{\left({e}^{\frac{{x}}{\mathrm{2}}}…
Question Number 218775 by sonukgindia last updated on 15/Apr/25 Answered by SdC355 last updated on 15/Apr/25 $$\sqrt{{x}+\mathrm{22}}\in\mathbb{Z} \\ $$$$\mathrm{1}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{8}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{27}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{64}^{\mathrm{1}/\mathrm{3}} …\mathrm{etc} \\ $$$${x}=−\mathrm{21}\:,\:−\mathrm{14}\:,\:\mathrm{5}\:,\:….…