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Question Number 166754 by mathlove last updated on 27/Feb/22 Answered by mindispower last updated on 28/Feb/22 $${S}=\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right),\underset{\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\mathrm{2}{S} \\ $$$${put}\:{z}_{{k}}…
Question Number 35626 by abdo mathsup 649 cc last updated on 21/May/18 $$\left.\mathrm{1}\right)\:{study}\:{the}\:{diagonalisstion}\:{of}\:{the}\:{matrice} \\ $$$${A}\:=\begin{pmatrix}{\mathrm{1}+{a}^{\mathrm{2}} \:\:\:\:\:{a}\:\:\:\:\:\:\:\mathrm{0}}\\{{a}\:\:\:\:\:\:\:\:\:\mathrm{1}+{a}^{\mathrm{2}} \:\:\:\:\:{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\mathrm{1}+{a}^{\mathrm{2}} \:\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$ Terms…
Question Number 35622 by abdo mathsup 649 cc last updated on 21/May/18 $${find}\:{all}\:{matrices}\:{M}\:\in{M}_{\mathrm{3}} \left({R}\right)\:\:/\:\:{M}^{\mathrm{2}} \:={M} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 101156 by I want to learn more last updated on 30/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\:\:\:\:\:\mathrm{h}\::\:\mathrm{x}\:\:\:=\:\:\mathrm{2}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{x}\right),\:\:\:\:\:\:\mathrm{x}\:\:\geqslant\:\:\mathrm{0} \\ $$ Answered by 1549442205 last updated on…
Question Number 166641 by pete last updated on 23/Feb/22 $$\mathrm{From}\:\mathrm{the}\:\mathrm{standard}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}, \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{origin}\:\left(\mathrm{0},\mathrm{0}\right),\:\mathrm{we}\:\mathrm{deduced}\:\mathrm{the}\:\mathrm{eqution} \\ $$$$\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} . \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{terms}\:\mathrm{do}\:\mathrm{we}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formular}? \\ $$ Commented…
Question Number 101104 by student work last updated on 30/Jun/20 $$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=? \\ $$ Answered by smridha last updated on 30/Jun/20 $$\boldsymbol{{x}}!=\boldsymbol{\Gamma}\left(\boldsymbol{{x}}+\mathrm{1}\right)=\boldsymbol{{x}}\Gamma\boldsymbol{{x}} \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\boldsymbol{{n}}!}{\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{2}\right)……\left(\boldsymbol{{x}}+\boldsymbol{{n}}\right)}\boldsymbol{{n}}^{\boldsymbol{{x}}} \left[\boldsymbol{{E}}{u}\boldsymbol{{ler}}\:\boldsymbol{{definition}}\right]…
Question Number 166633 by mathlove last updated on 23/Feb/22 Answered by mahdipoor last updated on 23/Feb/22 $${xyz}={log}_{\mathrm{2}{a}} {a}×{log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}} \mathrm{3}{a}= \\ $$$$\frac{{lna}}{{ln}\mathrm{2}{a}}×\frac{{ln}\mathrm{2}{a}}{{ln}\mathrm{3}{a}}×\frac{{ln}\mathrm{3}{a}}{{ln}\mathrm{4}{a}}=\frac{{lna}}{{ln}\mathrm{4}{a}} \\ $$$${yz}={log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}}…
Question Number 101096 by student work last updated on 30/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{oblique}\:\mathrm{asymptote}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\centerdot\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \:\:\:\:\: \\ $$$$\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help} \\ $$ Commented by student work last updated on 30/Jun/20 $$\mathrm{i}\:\mathrm{need}\:\mathrm{soon}…
Question Number 166620 by mathlove last updated on 23/Feb/22 Answered by benhamimed last updated on 23/Feb/22 $${x}+\frac{\mathrm{1}}{{x}}\neq\sqrt{\mathrm{3}}\:\:\:\:\:\forall\:{x}\in\mathbb{R}^{\ast} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)'=\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\left.\left(\left.{x}+\frac{\mathrm{1}}{{x}}\right)\in\right]−\infty;−\mathrm{2}\right]\cup\left[\mathrm{2};+\infty\left[\right.\right. \\…