Question Number 165778 by Odhiambojr last updated on 08/Feb/22 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m},\mathrm{n}\:\mathrm{and}\:\mathrm{l} \\ $$$$\mathrm{l}+\mathrm{m}+\mathrm{n}=\mathrm{0} \\ $$$$\mathrm{l}^{\mathrm{2}} +\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} =\mathrm{0} \\ $$ Answered by TheSupreme last updated on…
Question Number 100223 by DGmichael last updated on 25/Jun/20 Answered by maths mind last updated on 25/Jun/20 $${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} }{{k}+\mathrm{1}} \\ $$$${let}\:{f}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{{n}}…
Question Number 165756 by CrispyXYZ last updated on 07/Feb/22 $$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{compact}\:\mathrm{smooth}\:\mathrm{manifold}\:\mathrm{of}\:\mathrm{dimension}\:{d}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{some}\:{n}\:\in\mathbb{Z}^{+} \:\mathrm{such}\:\mathrm{that}\:{M}\:\mathrm{can}\:\mathrm{be}\:\mathrm{regularly}\:\mathrm{embedded}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Euclidean}\:\mathrm{space}\:\mathbb{R}^{{n}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 165731 by mkam last updated on 07/Feb/22 $$\int\:\sqrt{{sinx}}\:{dx} \\ $$ Commented by puissant last updated on 07/Feb/22 Commented by Tawa11 last updated on…
Question Number 165724 by mathocean1 last updated on 06/Feb/22 $$\left({U}_{{n}} \right)_{{n}\in\mathbb{N}^{\ast} } \::\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\: \\ $$$${Show}\:{that}\:\forall\:{n}\in\mathbb{N}^{\ast\:} ,\:{U}_{\mathrm{2}{n}} −{U}_{{n}} \geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$.\:{Deduct}\:{that}\:\underset{{n}\rightarrow+\infty} {{lim}}\:{U}_{{n}} =+\infty…
Question Number 165725 by mathocean1 last updated on 06/Feb/22 $${Show}\:{that}\::\:\forall\:{k}\:\in\:\mathbb{N}^{\ast\:} ,\: \\ $$$$\frac{\mathrm{1}}{{k}+\mathrm{1}}\leqslant{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\leqslant\frac{\mathrm{1}}{{k}} \\ $$ Answered by TheSupreme last updated on 07/Feb/22 $${ln}\left(\mathrm{1}+{x}\right)\leqslant{x}\:\forall\mathbb{R}^{+} \\ $$$${f}\left({x}\right)={ln}\left(\mathrm{1}+{x}\right)…
Question Number 100173 by Tinku Tara last updated on 25/Jun/20 $$\mathrm{Updated}\:\mathrm{apk}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{changes}\:\left(\mathrm{with}\:\mathrm{fixes}\:\mathrm{for}\:\mathrm{all}\:\mathrm{reported}\right. \\ $$$$\left.\mathrm{issues}\:\mathrm{so}\:\mathrm{far}\right)\:\mathrm{is}\:\mathrm{available}\:\mathrm{at} \\ $$$$\mathrm{www}.\mathrm{tinkutara}.\mathrm{com}. \\ $$$$\bullet\:\mathrm{Review}\:\mathrm{a}\:\mathrm{post} \\ $$$$\bullet\:\mathrm{copy}\:\mathrm{all}\:\mathrm{to}\:\mathrm{buffer} \\ $$$$\bullet\:\mathrm{Ability}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{diagrams} \\ $$…
Question Number 165699 by lapache last updated on 06/Feb/22 $${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{{x}−\sqrt{{x}+\mathrm{2}}}{\:\sqrt{\mathrm{4}{x}+\mathrm{1}}\:−\mathrm{3}}=….??? \\ $$ Commented by cortano1 last updated on 07/Feb/22 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{x}+\mathrm{2}}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}−\mathrm{3}}\:×\frac{\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}×\frac{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}}{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}}\:×\underset{{x}\rightarrow\mathrm{2}}…
Question Number 100158 by Algoritm last updated on 25/Jun/20 Answered by MJS last updated on 25/Jun/20 $${u}=\sqrt{{x}+\mathrm{3}}\:\Leftrightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{3}\wedge{u}\geqslant\mathrm{0} \\ $$$$\left.\left(\left.{u}+\mathrm{4}\right)\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}−\mathrm{3}\right)=\mathrm{3}{u}^{\mathrm{2}} −\mathrm{20} \\ $$$$\left({u}+\mathrm{4}\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}=\mathrm{3}{u}^{\mathrm{2}}…