Question Number 215757 by Tawa11 last updated on 17/Jan/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 215754 by universe last updated on 17/Jan/25 Answered by mr W last updated on 17/Jan/25 $${z}=\mathrm{1}+\mathrm{sin}\:\theta\:{with}\:−\frac{\pi}{\mathrm{2}}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${x}={r}\:\mathrm{cos}\:\phi,\:{y}={r}\:\mathrm{sin}\:\phi\:{with}\:\mathrm{0}\leqslant\phi\leqslant\mathrm{2}\pi \\ $$$${I}=\mathrm{2}\pi\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\theta\int_{\mathrm{0}} ^{\mathrm{cos}\:\theta}…
Question Number 215755 by Davidtim last updated on 17/Jan/25 $${Tow}\:{men}\:{complet}\:{a}\:{work}\:{in}\:\mathrm{3}{hr},\:{three} \\ $$$${women}\:{complet}\:{it}\:{in}\:\mathrm{5}{hr},\:{and}\:{five}\: \\ $$$${children}\:{complet}\:{it}\:{in}\:\mathrm{10}{hr}.\:{if}\:{we}\:{consider} \\ $$$${among}\:{these}\:{one}\:{man},\:{one}\:{woman}\:{and}\:{one} \\ $$$${child}\:{in}\:{how}\:{much}\:{time}\:{they}\:{will} \\ $$$${complet}\:{it}\:{together}? \\ $$ Answered by A5T…
Question Number 215748 by MrGaster last updated on 17/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} −\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}}} \\ $$ Answered by MrGaster last updated on 02/Feb/25…
Question Number 215776 by cherokeesay last updated on 17/Jan/25 Answered by A5T last updated on 18/Jan/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{be}\:\mathrm{r} \\ $$$$\mathrm{AC}\parallel\mathrm{OB}\Rightarrow\mathrm{CB}=\sqrt{\mathrm{r}^{\mathrm{2}} +\left(\mathrm{r}−\mathrm{x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2rx}} \\ $$$$\mathrm{Power}\:\mathrm{point}\:\mathrm{of}\:\mathrm{C}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}…
Question Number 215777 by Vettrituition last updated on 17/Jan/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 215778 by Vettrituition last updated on 17/Jan/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 215740 by universe last updated on 16/Jan/25 $$ \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} ^{\mathrm{1}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}}…
Question Number 215737 by Ismoiljon_008 last updated on 16/Jan/25 Commented by Ismoiljon_008 last updated on 16/Jan/25 $$\:\:\:{compare}\:{these}\:{two}\:{numbers} \\ $$$$\:\:\:{help}\:{me},\:{please} \\ $$ Answered by A5T last…
Question Number 215739 by universe last updated on 16/Jan/25 Answered by MrGaster last updated on 19/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{n}^{{k}} }{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{n}} } \\…