Question Number 126708 by slahadjb last updated on 23/Dec/20 $$\boldsymbol{{solve}}\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{x}}^{\sqrt{\mathrm{2}}} =\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last updated on 23/Dec/20 $$\sim\mathrm{0}.\mathrm{559793} \\ $$ Commented…
Question Number 192241 by yaslm last updated on 12/May/23 Answered by Frix last updated on 12/May/23 $${Z}_{\mathrm{1}} =\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }=−\frac{\mathrm{1}−\mathrm{2cos}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\ $$$${Z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{{ix}} }=\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{x}\right)}{\mathrm{5}−\mathrm{4cos}\:{x}}+\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\…
Question Number 192239 by mathdave last updated on 12/May/23 Commented by mathdave last updated on 12/May/23 $${can}\:{someone}\:{pls}\:{help}\:{me}\:{out}\:{it}\:{very}\:{urgent} \\ $$ Answered by aleks041103 last updated on…
Question Number 192233 by gatocomcirrose last updated on 12/May/23 $$\mathrm{show}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\in\mathrm{N}\:\mathrm{that} \\ $$$$\mathrm{3}\left(\mathrm{1}^{\mathrm{5}} +…+\mathrm{n}^{\mathrm{5}} \right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{1}^{\mathrm{3}} +…+\mathrm{n}^{\mathrm{3}} \\ $$ Commented by Frix last updated on 12/May/23 $${S}_{\mathrm{5}}…
Question Number 192227 by sonukgindia last updated on 12/May/23 Answered by senestro last updated on 12/May/23 $$…\mathrm{76} \\ $$ Commented by text last updated on…
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Question Number 192220 by gatocomcirrose last updated on 12/May/23 $$ \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{n}\in\mathbb{N},\:\mathrm{n}>\mathrm{1}\:\mathrm{and}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{then} \\ $$$$\:\mathrm{1}^{\mathrm{n}} +…+\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{n} \\ $$$$\left(\mathrm{dont}\:\mathrm{use}\:\equiv\left(\mathrm{modn}\right)\right) \\ $$ Commented by AST last updated…
Question Number 126672 by shaker last updated on 23/Dec/20 Answered by liberty last updated on 23/Dec/20 $$\:−\mathrm{1}\leqslant\:\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant\mathrm{1}\:;\:−\left({x}+\mathrm{4}\right)\leqslant\left({x}+\mathrm{4}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant{x}+\mathrm{4} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left({x}+\mathrm{4}\right)\leqslant\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}+\mathrm{4}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}}{{x}}\right)\leqslant\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({x}+\mathrm{4}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\left({x}+\mathrm{4}\right)=−\mathrm{4}…
Question Number 192204 by naka3546 last updated on 11/May/23 Answered by a.lgnaoui last updated on 11/May/23 $$\mathrm{soit}:\:\:\boldsymbol{\mathrm{C}}\left(\boldsymbol{\mathrm{O}},\boldsymbol{\mathrm{R}}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{entre}\boldsymbol{\mathrm{O}}\left(\boldsymbol{\mathrm{a}},\mathrm{b}\right) \\ $$$$\boldsymbol{\mathrm{equatin}}\:\boldsymbol{\mathrm{cercle}}\:\left(\boldsymbol{\mathrm{origine}}\:\boldsymbol{\mathrm{O}}\right) \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} =\boldsymbol{\mathrm{R}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{E}}\right) \\…
Question Number 192203 by naka3546 last updated on 11/May/23 Answered by HeferH last updated on 11/May/23 $$\left(\mathrm{3h}−\mathrm{2x}\right)\mathrm{2h}\:=\:\mathrm{3hx} \\ $$$$\:\mathrm{6h}^{\mathrm{2}} −\mathrm{4hx}\:=\:\mathrm{3hx} \\ $$$$\:\mathrm{6h}^{\mathrm{2}} =\:\mathrm{7hx} \\ $$$$\:\mathrm{6h}\:=\:\mathrm{7x}…