Question Number 115714 by Eric002 last updated on 27/Sep/20 Commented by Eric002 last updated on 27/Sep/20 $${solution}\:{de}\:{question}\:\mathrm{4} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 115711 by DeepakMahato last updated on 27/Sep/20 $${Find}\:{the}\:{value}\:{of}\:\boldsymbol{\mathrm{x}}:- \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{x}}} =\boldsymbol{\pi} \\ $$ Answered by MJS_new last updated on 27/Sep/20 $${x}\approx\mathrm{1}.\mathrm{85410596792} \\ $$…
Question Number 50171 by ggny last updated on 14/Dec/18 $$\mathrm{Sir}\:\mathrm{l}'\mathrm{m}\:\mathrm{sorry}\:\mathrm{l}\:\mathrm{dont}\:\mathrm{understand}\: \\ $$$$\mathrm{you} \\ $$ Commented by mr W last updated on 14/Dec/18 $${dear}\:{sir}! \\ $$$${please}\:{don}'{t}\:{open}\:{a}\:{new}\:{post}\:{for}\:{each}…
Question Number 181238 by SANOGO last updated on 23/Nov/22 $${calculer} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}{artan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$ Answered by qaz last updated on 23/Nov/22 $$\mathrm{arctan}\:\frac{\mathrm{2}}{{n}^{\mathrm{2}}…
Question Number 50156 by ggny last updated on 14/Dec/18 Commented by Redwan1 last updated on 14/Dec/18 $$\mathrm{if}\:\angle\mathrm{BDA}\:=\mathrm{90}° \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$ Terms of Service Privacy…
Question Number 50155 by ggny last updated on 14/Dec/18 $$\mathrm{plz}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir} \\ $$ Commented by peter frank last updated on 14/Dec/18 Terms of Service Privacy Policy…
Question Number 115683 by mohammad17 last updated on 27/Sep/20 Answered by Dwaipayan Shikari last updated on 27/Sep/20 $$\begin{pmatrix}{\mathrm{n}+\mathrm{1}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}=\frac{\mathrm{n}!}{\mathrm{k}!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\mathrm{n}!}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}−\mathrm{1}\right)!}=\frac{\mathrm{n}!\left(\mathrm{n}−\mathrm{k}\right)}{\left(\mathrm{k}+\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{k}\right)!} \\ $$$$\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}+\mathrm{1}}\end{pmatrix}=\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{k}\right)!}\left(\frac{\mathrm{1}}{\mathrm{k}!}+\frac{\mathrm{n}−\mathrm{k}}{\left(\mathrm{k}+\mathrm{1}\right)!}\right)=\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{k}\right)!\left(\mathrm{k}+\mathrm{1}\right)!}\left(\mathrm{k}+\mathrm{1}+\mathrm{n}−\mathrm{k}\right) \\…
Question Number 50138 by ggny last updated on 14/Dec/18 $$\mathrm{D} \\ $$ Commented by peter frank last updated on 14/Dec/18 $$???? \\ $$ Terms of…
Question Number 115674 by naka3546 last updated on 27/Sep/20 $${Solve}\:\:{for}\:\:{x} \\ $$$$\:\:\:\:\:\:{x}\:\:\equiv\:\:\mathrm{1}\:\:{mod}\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:{x}\:\:\equiv\:\:\mathrm{4}\:\:{mod}\:\:\mathrm{5} \\ $$$$\:\:\:\:\:\:{x}\:\:\equiv\:\:\mathrm{6}\:\:{mod}\:\:\mathrm{7} \\ $$ Answered by floor(10²Eta[1]) last updated on 27/Sep/20…
Question Number 181207 by lapache last updated on 23/Nov/22 $${cacul} \\ $$$$\left.\forall{x}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{nx}^{{n}} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22…