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How-Can-we-prove-h-J-h-z-1-

Question Number 204468 by MathedUp last updated on 18/Feb/24 $$\mathrm{How}\:\mathrm{Can}\:\mathrm{we}\:\mathrm{prove}\:\underset{{h}=−\infty} {\overset{\infty} {\sum}}\:{J}_{{h}} \left({z}\right)=\mathrm{1} \\ $$ Answered by Peace last updated on 19/Feb/24 $${J}_{{n}−\mathrm{1}} \left({x}\right)+{j}_{{n}+\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{n}}{{x}}{j}_{{n}}…

Question-204389

Question Number 204389 by Ngarmadji last updated on 15/Feb/24 Answered by Frix last updated on 15/Feb/24 $$\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{2}{x}}{\mathrm{5}}=\mathrm{1}\:\:\frac{{x}}{\mathrm{3}}−\frac{{x}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{5}{x}−\mathrm{3}{x}=\frac{\mathrm{15}}{\mathrm{2}}\:\:{x}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}{x}}{\mathrm{5}}−\frac{\mathrm{2}{x}}{\mathrm{15}}+\frac{{x}}{\mathrm{3}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:\frac{\mathrm{9}{x}−\mathrm{2}{x}+\mathrm{5}{x}}{\mathrm{15}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:…\:{x}=−\frac{\mathrm{7}}{\mathrm{12}} \\ $$ Terms of Service Privacy…

ctg-6-pi-9-9-ctg-4-pi-9-11-ctg-2-pi-9-

Question Number 204344 by SEKRET last updated on 13/Feb/24 $$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{ctg}}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{9}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{11}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)=?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$…

Question-204337

Question Number 204337 by SANOGO last updated on 13/Feb/24 Answered by witcher3 last updated on 13/Feb/24 $$\left(\mathrm{3}\right)\Rightarrow\left(\mathrm{2}\right) \\ $$$$\mathrm{soitU}\:\mathrm{un}\:\mathrm{ouvert}\:\mathrm{de}\:\mathrm{E}\: \\ $$$$\exists\:\mathrm{existe}\:\mathrm{V}\:\mathrm{un}\:\mathrm{ouvert}\:\mathrm{de}\:\mathrm{F} \\ $$$$\mathrm{t}\:\mathrm{elle}\:\mathrm{Que}\:\mathrm{U}=\mathrm{f}^{−} \left(\mathrm{V}\right);\mathrm{car}\:\mathrm{f}\:\mathrm{et}\:\mathrm{bijective}\:\mathrm{donc}\:\mathrm{f}^{−} \\…