Question Number 50359 by prof Abdo imad last updated on 16/Dec/18 $$\left.\mathrm{1}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{S}_{{n}} \left({p}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:\:\:\frac{{C}_{{n}} ^{{k}} }{{p}+{k}+\mathrm{1}}…
Question Number 50355 by prof Abdo imad last updated on 16/Dec/18 $${let}\:\left(\alpha_{{k}} \right)\:\:\left({k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{the}\:{n}^{{eme}} \:{roots}\:{of}\:\mathrm{1}\right. \\ $$$${calculate}\:{p}\left({x},{y}\right)=\left({x}+\alpha_{\mathrm{0}} {y}\right)\left({x}+\alpha_{\mathrm{1}} {y}\right)….\left({x}+\alpha_{{n}−\mathrm{1}} {y}\right) \\ $$ Terms of Service Privacy…
Question Number 50356 by prof Abdo imad last updated on 16/Dec/18 $${devompose}\:{inside}\:{C}\left[{x}\right]\:{and}\:{R}\left[{x}\right]\:{the}\:{polynom} \\ $$$$\left.\mathrm{1}\right){x}^{\mathrm{4}} +\mathrm{1}\:\:\:\: \\ $$$$\left.\mathrm{2}\right){x}^{\mathrm{6}} −\mathrm{1} \\ $$$$\left.\mathrm{3}\right){x}^{\mathrm{8}} \:+{x}^{\mathrm{4}} \:+\mathrm{1} \\ $$ Terms…
Question Number 50354 by prof Abdo imad last updated on 16/Dec/18 $$\left(\left({x}\right)={a}_{{k}} \right)\:_{\mathrm{1}\leqslant{k}\leqslant{n}} \:{is}\:{a}\:{sequence}\:{of}\:{reals}\:{let} \\ $$$${p}\left({x}\right)\:=\prod_{{k}=\mathrm{1}} ^{{n}} \left({cos}\left({a}_{{k}} \right)+{xsin}\left({a}_{{k}} \right)\right) \\ $$$${if}\:{p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right){q}\:+{r}\:\:\:{find}\:{q}\:{and}\:{r} \\ $$$$…
Question Number 50353 by prof Abdo imad last updated on 16/Dec/18 $${let}\:{p}\left({x}\right)={x}^{\mathrm{4}{n}} −{x}^{\mathrm{3}{n}} +{x}^{\mathrm{2}{n}} −{x}^{{n}} +\mathrm{1}\:{and} \\ $$$${q}\left({x}\right)={x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\:\:{determine}\:{the}\:{integr}\:{n} \\ $$$${to}\:{have}\:{q}\:{divide}\:{p}. \\ $$…
Question Number 181426 by universe last updated on 25/Nov/22 Commented by universe last updated on 25/Nov/22 $${true}\:{or}\:{false}\:? \\ $$ Answered by mr W last updated…
Question Number 50349 by peter frank last updated on 16/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 181415 by Socracious last updated on 24/Nov/22 $$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{solve}}\:\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} } =\mathrm{36} \\ $$$$ \\ $$ Commented by Frix last updated on 25/Nov/22…
Question Number 115871 by bemath last updated on 29/Sep/20 Commented by bemath last updated on 29/Sep/20 $$\begin{pmatrix}{{m}\:\:\:{n}}\\{{n}\:\:\:{m}}\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{3}{mn}}{\mathrm{2}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{{m}^{\mathrm{2}} −{n}^{\mathrm{2}} }\:\begin{pmatrix}{{m}\:\:\:−{n}}\\{−{n}\:\:\:{m}}\end{pmatrix}\:\begin{pmatrix}{\:\:\:\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{3}{mn}}{\mathrm{2}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }}\end{pmatrix}…
Question Number 50328 by peter frank last updated on 15/Dec/18 Answered by peter frank last updated on 16/Dec/18 $$\left.\mathrm{1}\right) \\ $$$${x}+\mathrm{2}{y}−\mathrm{3}{z}={a}…\left({i}\right) \\ $$$$\mathrm{2}{x}+\mathrm{6}{y}−\mathrm{11}{z}={b}…\left({ii}\right) \\ $$$${x}−\mathrm{2}{y}+\mathrm{7}{z}={c}….\left({iii}\right)…