Question Number 208322 by alcohol last updated on 11/Jun/24 Answered by Berbere last updated on 11/Jun/24 $$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)^{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left(\mathrm{1}+{k}\right)^{{k}} }{{k}^{{k}} }=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}}…
Question Number 208303 by Simurdiera last updated on 10/Jun/24 $${Resolver} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }\:−\:{x}^{\mathrm{2}} {u}\:=\:{xe}^{\mathrm{4}{y}} \\ $$ Answered by aleks041103 last updated on 12/Jun/24 $${u}={X}\left({x}\right){Y}\left({y}\right)…
Question Number 208280 by Shrodinger last updated on 10/Jun/24 $${L}=\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\pi}} {ln}\left({cosx}\right){dx} \\ $$ Answered by Berbere last updated on 10/Jun/24 $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}.{sin}\left({x}\right){cos}\left({x}\right)\right){dx}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}}…
Question Number 208282 by hardmath last updated on 10/Jun/24 $$\mathrm{If}\:\:\:\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{cos}\left(\alpha\:+\:\beta\right)\:=\:? \\ $$ Commented by mr W last updated on 12/Jun/24 $${what}\:{you}\:{wrote}\:{means} \\ $$$$\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 208292 by universe last updated on 10/Jun/24 $$\:\mathrm{let}\:\mathrm{T}\:\mathrm{be}\:\mathrm{a}\:{n}×{n}\:\mathrm{matrix}\:\mathrm{with}\:\mathrm{integral}\: \\ $$$$\:\mathrm{entries}\:\mathrm{and}\:\:\mathrm{Q}\:=\:\mathrm{T}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\:\:\:\mathrm{where}\:\mathrm{I}\:\mathrm{denote} \\ $$$$\:\:\mathrm{the}\:\mathrm{n}×\mathrm{n}\:\mathrm{identity}\:\mathrm{matrix}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\mathrm{that}\:\mathrm{matrix}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{invertible} \\ $$ Answered by Berbere last updated on 10/Jun/24…
Question Number 208277 by Mastermind last updated on 10/Jun/24 Answered by Rasheed.Sindhi last updated on 10/Jun/24 $$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{2}} ={R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \\ $$$$=\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\:\:\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{-\mathrm{5}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\:\:\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\:\:\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\:\:\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{4}}…
Question Number 208288 by efronzo1 last updated on 10/Jun/24 Answered by A5T last updated on 10/Jun/24 $$\mathrm{8}\left(\mathrm{8}+{x}\right)=\left(\mathrm{2}{r}\right)\left(\mathrm{2}{r}+\mathrm{2}{R}\right)=\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{rR} \\ $$$$\Rightarrow{x}=\frac{{r}^{\mathrm{2}} +{rR}−\mathrm{16}}{\mathrm{2}}…\left({i}\right) \\ $$$$\left(\mathrm{8}+{x}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{r}+{R}\right)^{\mathrm{2}}…
Question Number 208306 by SANOGO last updated on 10/Jun/24 $${calcul}\:/\:{lim}\:{n}\rightarrow+\infty\:\int_{\mathrm{0}} ^{+\infty} \:{f}_{{n}} \left({x}\right) \\ $$$$\:{f}_{{n}} \left({x}\right)=\:{arctan}\left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx} \\ $$ Commented by SANOGO last updated on…
Question Number 208252 by efronzo1 last updated on 09/Jun/24 $$\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{22}}\right)\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{20}\pi}{\mathrm{21}}\right)=? \\ $$ Answered by som(math1967) last updated on 09/Jun/24 $$\:{let}\:\frac{\pi}{\mathrm{21}}={x} \\ $$$${cos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{16}{xcos}\mathrm{20}{x} \\ $$$${cosxcos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{5}{x} \\…
Question Number 208269 by liuxinnan last updated on 09/Jun/24 Commented by liuxinnan last updated on 09/Jun/24 $${how}\:{many}\:{ways}\:{you}\:{can}\:{find}\:{to} \\ $$$${cover}\:{the}\:{hole} \\ $$ Answered by liuxinnan last…