Question Number 91378 by jagoll last updated on 30/Apr/20 Commented by jagoll last updated on 30/Apr/20 $${i}\:{have}\:{no}\:{idea}\:{to}\:{solve}\:{this}\: \\ $$$${question} \\ $$ Commented by MJS last…
Question Number 156909 by MathSh last updated on 16/Oct/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{n}\:-\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\left(\mathrm{e}\:-\:\mathrm{1}\right)\centerdot\mathrm{n}}{\mathrm{n}\:+\:\left(\mathrm{e}\:-\:\mathrm{1}\right)\centerdot\mathrm{k}}\right)\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 91377 by A8;15: last updated on 30/Apr/20 Commented by Tony Lin last updated on 30/Apr/20 $$\int\frac{\mathrm{1}}{\:\sqrt{{sinx}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}−\frac{\pi}{\mathrm{2}}\right)}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\right)}}{dx} \\ $$$${let}\:\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}={u},\:\frac{{du}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 156911 by MathSh last updated on 17/Oct/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\left(\left[\mathrm{nx}\right]\:\centerdot\:\mid\mathrm{x}\:-\:\left[\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mid\right]\right)\mathrm{dx} \\ $$$$\left[\ast\right]\:-\:\mathrm{GIF} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 156905 by JSM last updated on 16/Oct/21 $${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y},\:{x}:\mathrm{3}:\mathrm{5}=\mathrm{8}:{y}:\mathrm{9} \\ $$ Answered by Rasheed.Sindhi last updated on 16/Oct/21 $$\mathrm{9}{x}:\mathrm{27}:\mathrm{45}=\mathrm{40}:\mathrm{5}{y}:\mathrm{45} \\ $$$$\mathrm{9}{x}=\mathrm{40}\Rightarrow{x}=\frac{\mathrm{40}}{\mathrm{9}} \\ $$$$\mathrm{5}{y}=\mathrm{27}\Rightarrow{y}=\frac{\mathrm{27}}{\mathrm{5}} \\…
Question Number 156910 by MathSh last updated on 16/Oct/21 Commented by ghimisi last updated on 17/Oct/21 $$\mathrm{9}\:\:\:????????? \\ $$$$ \\ $$ Commented by MathSh last…
Question Number 156904 by JSM last updated on 16/Oct/21 $${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y}\:,\:{x}:\mathrm{3}:\mathrm{5}=\mathrm{2}:{y}:\mathrm{10} \\ $$ Answered by Rasheed.Sindhi last updated on 16/Oct/21 $${x}:\mathrm{3}:\mathrm{5}=\mathrm{2}:{y}:\mathrm{10} \\ $$$$\Rightarrow\mathrm{2}{x}:\mathrm{6}:\mathrm{10}=\mathrm{2}:{y}:\mathrm{10} \\ $$$$\mathrm{2}{x}=\mathrm{2}\Rightarrow{x}=\mathrm{1} \\…
Question Number 91362 by ar247 last updated on 30/Apr/20 Commented by jagoll last updated on 30/Apr/20 $${false} \\ $$ Commented by Tony Lin last updated…
Question Number 156900 by MathSh last updated on 16/Oct/21 $$\Omega_{\mathrm{1}} \:=\:\mathrm{1}\:-\:\frac{\pi}{\mathrm{2}}\:+\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\left(-\:\frac{\mathrm{1}}{\boldsymbol{\pi}}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\Omega_{\mathrm{2}} \:=\:\mathrm{1}\:-\:\frac{\pi}{\mathrm{2}}\:+\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\left(-\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{e}}}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\left.\mathrm{A}\left.\right)\left.\:\Omega_{\mathrm{1}} \:<\:\Omega_{\mathrm{2}} \:\:\:\mathrm{B}\right)\:\Omega_{\mathrm{1}} \:=\:\Omega_{\mathrm{2}}…
Question Number 25822 by abdo imad last updated on 15/Dec/17 $${answer}\:{to}\:{the}\:{question}\:{of}\:{p}\left({X}\right)=\:\left(\mathrm{1}+{iX}\right)^{{n}} −\left(\mathrm{1}−{iX}\right)^{{n}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}}…