Question Number 205915 by dodo last updated on 02/Apr/24 Answered by Skabetix last updated on 03/Apr/24 $${Placement}\::\:{U}_{{n}} =\:\mathrm{6}\:\mathrm{000}\:\mathrm{000}\:×\mathrm{1}.\mathrm{03}^{{n}} \\ $$$${Valeur}\:{vehicule}\::\:{V}_{{n}} =\:\mathrm{30}\:\mathrm{000}\:\mathrm{000}\:×\mathrm{0}.\mathrm{95}^{{n}} \\ $$$${En}\:{notant}\:{n}\:{le}\:{nombre}\:{d}\:{annees}\:{on}\:{a}\:: \\ $$$$\mathrm{6}\:\mathrm{000}\:\mathrm{000}\:×\:\mathrm{1}.\mathrm{03}^{{n}}…
Question Number 205892 by cortano12 last updated on 02/Apr/24 Answered by A5T last updated on 02/Apr/24 Commented by A5T last updated on 02/Apr/24 $${AC}={ytan}\left(\mathrm{3}\alpha\right);{x}={ytan}\alpha;{z}=\frac{{y}}{{cos}\alpha};{c}={y}\left({tan}\mathrm{3}\alpha−{tan}\alpha\right) \\…
Question Number 205893 by BaliramKumar last updated on 02/Apr/24 Commented by BaliramKumar last updated on 02/Apr/24 $$−\mathrm{33}\:\mathrm{or}\:−\mathrm{1}\:\:?? \\ $$ Commented by A5T last updated on…
Question Number 205910 by Simurdiera last updated on 02/Apr/24 $${Resuelve}\:{la}\:{siguiente}\:{integral} \\ $$$${I}\:=\:\int\frac{{x}}{\mathrm{sinh}^{\mathrm{2}} \left({x}\right)\centerdot\mathrm{ln}\:\left(\mathrm{sinh}\:\left({x}\right)\right)\:−\:{x}\centerdot\mathrm{sinh}\:\left({x}\right)\centerdot\mathrm{cosh}\:\left({x}\right)}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 205904 by RoseAli last updated on 02/Apr/24 Answered by mr W last updated on 02/Apr/24 $${a}+{b}+{c}=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{abc}}\leqslant\frac{{a}+{b}+{c}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${abc}=\left(\sqrt[{\mathrm{3}}]{{abc}}\right)^{\mathrm{3}} \leqslant\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{27}} \\…
Question Number 205884 by Abdullahrussell last updated on 01/Apr/24 Commented by Frix last updated on 01/Apr/24 $${a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 205885 by hardmath last updated on 01/Apr/24 $$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{}\begin{pmatrix}{\mathrm{2n}}\\{\:\:\mathrm{n}}\end{pmatrix}\:=\:? \\ $$ Answered by MM42 last updated on 03/Apr/24 $$\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$${a}=\sqrt[{{n}}]{}\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$\Rightarrow{lna}=\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{2}}{{n}−\mathrm{2}}\right)+…+{ln}\left(\mathrm{2}+\frac{{n}−\mathrm{3}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{2}}{\mathrm{2}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{1}}{\mathrm{1}}\right)\right.…
Question Number 205880 by cortano12 last updated on 01/Apr/24 Commented by cortano12 last updated on 01/Apr/24 $$\:\underbrace{ } \\ $$ Answered by mr W last…
Question Number 205866 by RoseAli last updated on 01/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 205861 by mr W last updated on 01/Apr/24 Answered by A5T last updated on 01/Apr/24 $$\mathrm{9}×\mathrm{9}=\left(\mathrm{2}{r}−{x}\right){x}\Rightarrow\mathrm{2}{r}−{x}=\frac{\mathrm{81}}{{x}} \\ $$$$\mathrm{21}×\mathrm{21}=\left(\mathrm{2}{r}−\mathrm{16}−{x}\right)\left(\mathrm{16}+{x}\right)=\left(\frac{\mathrm{81}}{{x}}−\mathrm{16}\right)\left(\mathrm{16}+{x}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}\Rightarrow{r}=\mathrm{21}.\mathrm{25} \\ $$ Commented…