Question Number 225365 by klipto last updated on 23/Oct/25 $$\mathrm{who}'\mathrm{s}\:\mathrm{trying}\:\mathrm{this}\:\mathrm{scholars}? \\ $$ Commented by klipto last updated on 23/Oct/25 Terms of Service Privacy Policy Contact:…
Question Number 225382 by Osefavour last updated on 23/Oct/25 $$\mathrm{Been}\:\mathrm{a}\:\mathrm{while}\:\mathrm{guys} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{xln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by peace2 last updated on 24/Oct/25 $${A}=\int_{\mathrm{0}}…
Question Number 225356 by Ari last updated on 22/Oct/25 Commented by Ari last updated on 22/Oct/25 $${find}\:{volu}\mathrm{8}{e}\:{of}\:{the}\:{glass} \\ $$ Answered by mr W last updated…
Question Number 225354 by ajfour last updated on 22/Oct/25 Commented by ajfour last updated on 22/Oct/25 $${If}\:{P}\left(\mathrm{0},{q}\right)\:{while}\:{origin}\:{centered} \\ $$$${circle}\:{has}\:{radius}\:{R},\:{then}\:{find} \\ $$$${eq}.\:{of}\:{blue}\:{semicircle}\:{and}\:{hence}\: \\ $$$$\:{r},\:{C},\:{B}. \\ $$$${example}:\:{R}=\mathrm{5},\:{q}=\mathrm{3}…
Question Number 225338 by mr W last updated on 22/Oct/25 Commented by mr W last updated on 24/Oct/25 $${as}\:{Q}\mathrm{225146},\:{but}\:{the}\:{wedge}\:{can} \\ $$$${slide}\:{on}\:{the}\:{ground}\:{frictionlessly}. \\ $$ Answered by…
Question Number 225322 by mr W last updated on 21/Oct/25 $${Tinku}\:{Tara}\:{sir}: \\ $$$${it}\:{could}\:{be}\:{a}\:{new}\:{issue}: \\ $$$${when}\:{i}\:{cancel}\:{a}\:{comment}\:{or}\:{an}\:{answer}, \\ $$$${by}\:{tapping}\:{the}\:{back}\:{button},\:{it}\:{should} \\ $$$${popup}\:{a}\:{window}\:{asking}\:{if}\:{discard}\: \\ $$$${or}\:{save}.\:{now}\:{this}\:{popup}\:{window}\: \\ $$$${doesn}'{t}\:{appear}\:{and}\:{the}\:{work}\:{made}\: \\ $$$${goes}\:{automatically}\:{lost}.…
Question Number 225323 by hardmath last updated on 21/Oct/25 $$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}\:-\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .\:\:\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\:-\:\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\mathrm{1}}\:\:=\:? \\ $$ Answered by Raphael254 last updated on 21/Oct/25 $$ \\ $$$$\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}+\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}−\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}}×\sqrt[{\mathrm{4}}]{\frac{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}} \\ $$$$…
Question Number 225307 by fantastic last updated on 21/Oct/25 Commented by fantastic last updated on 21/Oct/25 $${total}\:{area}? \\ $$ Answered by mehdee7396 last updated on…
Question Number 225330 by mathlove last updated on 21/Oct/25 $${if}\:\:\left({fogoh}\right)\left({x}\right)={cos}^{\mathrm{2}} \left({x}+\mathrm{9}\right) \\ $$$${then}\:\:\:{f}\left({x}\right)=?\:,\:\:{g}\left({x}\right)=?\:\:,\:{h}\left({x}\right)=? \\ $$ Answered by Raphael254 last updated on 21/Oct/25 $$ \\ $$$${f}\left({g}\left({h}\left({x}\right)\right)\right)\:=\:{cos}^{\mathrm{2}}…
Question Number 225303 by fantastic last updated on 20/Oct/25 $${Let} \\ $$$${S}_{{n}} \left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{sin}\:\left(\left(\mathrm{2}{r}−\mathrm{3}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{10}{r}+\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)−\mathrm{sin}\left(\:\left(\mathrm{6}{r}−\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{2}{r}+\mathrm{5}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)}{\mathrm{2}^{{r}−\mathrm{1}} \left(\mathrm{sin}\:\left({r}\mathrm{2}^{\mathrm{3}−{r}} −{x}\right)\mathrm{sin}\:\left(\mathrm{2}^{\mathrm{2}−{r}} −{x}\right)\right)} \\ $$$${then}\:{find}\:{the}\:{value}\:{of} \\…