Question Number 39798 by NECx last updated on 11/Jul/18 $${Please}\:{someone}\:{should}\:{propose}\:{a} \\ $$$${textbook}\:{that}\:{really}\:{explains} \\ $$$${Gauss}\:{law},{electric}\:{flux}\:{and} \\ $$$${capacitance}.{I}\:{need}\:{to}\:{download}\:\mathrm{1}.. \\ $$$${Thanks}. \\ $$ Commented by ajfour last updated…
Question Number 105327 by Anindita last updated on 27/Jul/20 $$\mathrm{If}\:\mathrm{x}\:=\:\frac{\mathrm{11}}{\mathrm{8}}\:\mathrm{and}\:\mathrm{y}\:=\:\frac{\mathrm{66}}{\mathrm{9}}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}\:?\: \\ $$ Commented by Aziztisffola last updated on 27/Jul/20 $$\mathrm{You}\:\mathrm{can}\:\mathrm{try}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it},\:\mathrm{we}\:\mathrm{are}\:\mathrm{here}\:\mathrm{to} \\ $$$$\mathrm{correct}\:\mathrm{and}\:\mathrm{to}\:\mathrm{help}\:\mathrm{you}. \\…
Question Number 170838 by pablo1234523 last updated on 01/Jun/22 $$\mathrm{if}\:{a}<{b},\:\mathrm{show}\:\mathrm{that}\:{a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$$${a},{b},{m},{n}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{constants} \\ $$ Answered by chengulapetrom last updated on 04/Jun/22 $$\mathrm{if}\:{a}<{b},\:\mathrm{show}\:\mathrm{that}\:{a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$$${a},{b},{m},{n}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{constants} \\…
Question Number 170833 by LEKOUMA last updated on 01/Jun/22 $${Solve}\: \\ $$$$\begin{cases}{{C}_{{x}} ^{{y}} ={C}_{{x}} ^{{y}+\mathrm{1}} }\\{\mathrm{4}{C}_{{x}} ^{{y}} =\mathrm{5}{C}_{{x}} ^{{y}−\mathrm{1}} }\end{cases}\:{or}\:{C}_{{n}} ^{{k}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$ Answered…
Question Number 105289 by Anindita last updated on 27/Jul/20 $$\sqrt{\mathrm{52}}+\sqrt{\mathrm{3}}−\sqrt{\frac{\mathrm{4}}{\mathrm{6}}}\:=\:? \\ $$ Answered by SUMIT007 last updated on 27/Jul/20 $$\Rightarrow\:\mathrm{2}\sqrt{\mathrm{13}}+\sqrt{\mathrm{3}}−\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\Rightarrow\:\mathrm{2}\sqrt{\mathrm{13}}+\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{6}\sqrt{\mathrm{13}}+\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}}}{\mathrm{3}} \\…
Question Number 170824 by 0731619 last updated on 31/May/22 Answered by floor(10²Eta[1]) last updated on 01/Jun/22 $$\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}\sqrt[{\mathrm{3}}]{\mathrm{a}−\mathrm{x}}−\sqrt{\mathrm{a}−\mathrm{x}}\sqrt{\mathrm{a}+\mathrm{x}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}−\mathrm{x}}\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid\sqrt{\mathrm{a}−\mathrm{x}}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid\sqrt[{\mathrm{6}}]{\mathrm{a}−\mathrm{x}}}−\frac{\sqrt{\mathrm{a}+\mathrm{x}}}{\:\sqrt[{\mathrm{6}}]{\mathrm{a}−\mathrm{x}}\sqrt[{\mathrm{3}}]{\mathrm{a}+\mathrm{x}}+\mid\mathrm{a}\mid} \\ $$$$\Rightarrow\mathrm{lim}\left(…\right)=\frac{\sqrt[{\mathrm{3}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2a}}}−\frac{\sqrt{\mathrm{2a}}}{\mid\mathrm{a}\mid}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{a}}} \\ $$ Terms of Service Privacy…
Question Number 170820 by LEKOUMA last updated on 31/May/22 $${Solve} \\ $$$${y}^{\left(\mathrm{4}\right)} +\mathrm{2}{y}^{\left(\mathrm{3}\right)} +{y}^{\left(\mathrm{2}\right)} ={xe}^{−{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170812 by MathsFan last updated on 31/May/22 $$\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{x}}+\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{horizontal}}\:\boldsymbol{\mathrm{asymptote}} \\ $$ Answered by Mathspace last updated on 31/May/22 $${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow−\infty} \frac{{x}+\mathrm{1}}{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{9}}{{x}^{\mathrm{2}}…
Question Number 170811 by mokys last updated on 31/May/22 $${find}\:{minimum}\:{and}\:{maximum}\:{z}\:=\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170809 by thean last updated on 31/May/22 Answered by som(math1967) last updated on 31/May/22 $$\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{−\mathrm{1}} \left({e}^{\mathrm{2}{x}} −\mathrm{1}\right){sin}\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{6}}{{e}}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({e}^{\mathrm{2}{x}} −\mathrm{1}\right)}{\mathrm{2}{x}}×\frac{{sin}\mathrm{3}{x}}{\mathrm{3}{x}}…