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}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}}…
Question Number 25823 by ajfour last updated on 15/Dec/17 $${z}_{\mathrm{1}} =\mathrm{4}−\mathrm{3}{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{9}+\mathrm{2}{i} \\ $$$${Find}\:\:\left({a}\right)\:{z}_{\mathrm{1}} {oz}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:{z}_{\mathrm{1}} ×{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:{angle}\:{between}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} . \\…
Question Number 156880 by MathSh last updated on 16/Oct/21 $$\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{\pi}^{\boldsymbol{\mathrm{n}}} }\:\centerdot\:\left(\frac{\pi}{\mathrm{e}}\right)^{\boldsymbol{\mathrm{k}}} =\:? \\ $$ Answered by Mathspace last updated on 16/Oct/21…
Question Number 156881 by MathSh last updated on 16/Oct/21 $$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{log}\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{log}\left(\mathrm{1}\:-\:\mathrm{y}\right)}{\mathrm{1}\:-\:\mathrm{xy}}\:\mathrm{dxdy}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156867 by cortano last updated on 16/Oct/21 Answered by qaz last updated on 16/Oct/21 $$\frac{\mathrm{1}}{\mathrm{a}+\mathrm{b}}+\frac{\mathrm{9}}{\mathrm{c}}+\frac{\mathrm{16}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\frac{\left(\mathrm{1}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}=\mathrm{8} \\ $$$$\mathrm{Min}=\mathrm{8}\:\mathrm{holds}\:\mathrm{only}\:\mathrm{when}\:\frac{\mathrm{1}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{3}}{\mathrm{c}}=\frac{\mathrm{4}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:\:\mathrm{ie}.\mathrm{a}+\mathrm{b}=\mathrm{1}\:\:\:\mathrm{c}=\mathrm{3} \\ $$ Commented by cortano…
Question Number 25787 by ibraheem160 last updated on 14/Dec/17 $$\mathrm{2}^{{x}} ={x}^{\mathrm{2}\:} ,{hence}\:{x}=\mathrm{2}\:.\:{prove} \\ $$ Commented by mrW1 last updated on 14/Dec/17 $${It}'{s}\:{not}\:{true},\:{Sir}. \\ $$$${If}\:\mathrm{2}^{{x}} ={x}^{\mathrm{2}}…