Question Number 1499 by 112358 last updated on 14/Aug/15 $${Use}\:{the}\:\epsilon−\delta\:{definition}\:{of}\:{the} \\ $$$${limit}\:{to}\:{show}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{6}−{x}}=\mathrm{1}\:. \\ $$$$\left({I}'{m}\:{hoping}\:{to}\:{better}\:{understand}\right. \\ $$$${this}\:{concept}\:{by}\:{example}\:{so}\:{please} \\ $$$${help}\:{me}\:{by}\:{explaning}\:{the}\: \\ $$$$\left.{reasoning}\:{behind}\:{your}\:{steps}.\right) \\ $$…
Question Number 67035 by rajesh4661kumar@gmail.com last updated on 22/Aug/19 Commented by rajesh4661kumar@gmail.com last updated on 22/Aug/19 $${solve}\:{please}\:{argent}\:{hai} \\ $$ Commented by mathmax by abdo last…
Question Number 1498 by Rasheed Soomro last updated on 14/Aug/15 $$\mathrm{Find}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:{and}\:\:\beta\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{\:{m}} =\beta^{\:\:{n}} \:\:\:\:{and}\:\:\:\beta^{\:\:{m}} =\alpha^{\mathrm{n}} \:\:,\:{m},{n}\:\in\:\mathbb{Z} \\ $$$$\boldsymbol{\mathrm{D}}\mathrm{etermine}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\left(\alpha,\beta\right)\:\mathrm{fulfilling}\:\mathrm{the} \\ $$$$\mathrm{above}\:\mathrm{conditions}.\:\:\left(\:\mathrm{You}\:\mathrm{may}\:\mathrm{ignore}\:\mathrm{this}\:\mathrm{part}\:\mathrm{in}\:\mathrm{your}\:\mathrm{answer}\right) \\ $$ Commented by…
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Question Number 1497 by Rasheed Soomro last updated on 14/Aug/15 $$\mathrm{Find}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{\mathrm{2}} =\beta^{\mathrm{3}} \:\:\mathrm{and}\:\:\beta^{\mathrm{2}} =\alpha^{\mathrm{3}} \\ $$ Commented by 123456 last updated on 14/Aug/15…
Question Number 67033 by TawaTawa last updated on 22/Aug/19 Commented by Tony Lin last updated on 22/Aug/19 $${let}\angle{DBM}=\angle{MBC}=\phi \\ $$$$\:\:\:\:\:\angle{ECM}=\angle{MCB}=\theta \\ $$$$\mathrm{2}\theta+\mathrm{2}\phi=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{45}°−\phi \\…
Question Number 132564 by abony1303 last updated on 15/Feb/21 Answered by abony1303 last updated on 15/Feb/21 $$\mathrm{pls}\:\mathrm{help}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{region} \\ $$ Answered by MJS_new last updated on…
Question Number 67031 by hmamarques1994@gmai.com last updated on 21/Aug/19 $$\: \\ $$$$\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{1}} {\boldsymbol{\mathrm{lim}}}\sqrt[{\mathrm{3}}]{\frac{\sqrt[{\mathrm{7}}]{\boldsymbol{\mathrm{x}}^{\mathrm{5}} }+\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{9}\:}]{\boldsymbol{\mathrm{x}}^{\mathrm{7}} }}}=? \\ $$$$\: \\ $$ Commented by Tony Lin last updated…
Question Number 67026 by TawaTawa last updated on 21/Aug/19 Commented by Tony Lin last updated on 22/Aug/19 $$\because{slope}\:{of}\:{L}_{{BC}} ={tan}\mathrm{150}°=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\therefore{L}_{{BC}} :\:{y}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left({x}+\mathrm{4}\right) \\ $$$$\because{L}_{{BC}} \bot{L}_{{AB}}…
Question Number 67025 by Masumsiddiqui399@gmail.com last updated on 21/Aug/19 Commented by Rasheed.Sindhi last updated on 22/Aug/19 $${x}^{{x}^{{x}+\mathrm{5}} } =\:^{\mathrm{3}} \sqrt{\mathrm{3}} \\ $$$$\left({x}^{{x}^{{x}+\mathrm{5}} } \right)^{\mathrm{3}} =\left(\:^{\mathrm{3}}…