Question Number 62626 by Tawa1 last updated on 23/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\:\:\:\:\frac{\mathrm{n}!}{\mathrm{4}^{\mathrm{n}} }\:\:\:\mathrm{as}\:\:\mathrm{n}\:\:\mathrm{approach}\:\mathrm{infinity} \\ $$ Commented by mathmax by abdo last updated on 23/Jun/19 $${let}\:{u}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\:\:\:{we}\:{have}\:{n}!\:\sim{n}^{{n}}…
Question Number 62624 by Jmasanja last updated on 23/Jun/19 Answered by MJS last updated on 23/Jun/19 $$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{before},\:\mathrm{they}\:\mathrm{all}\:\mathrm{are}\:\mathrm{similar} \\ $$$${P}=\begin{pmatrix}{\mathrm{9cos}\:\mathrm{270}°}\\{\mathrm{9sin}\:\mathrm{270}°}\end{pmatrix}\:\:{Q}=\begin{pmatrix}{\mathrm{10cos}\:\mathrm{45}°}\\{\mathrm{10sin}\:\mathrm{45}°}\end{pmatrix}\:\:{R}=\begin{pmatrix}{\mathrm{10cos}\:\mathrm{135}°}\\{\mathrm{10sin}\:\mathrm{135}°}\end{pmatrix} \\ $$$${P}=\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{9}}\end{pmatrix}\:\:{Q}=\begin{pmatrix}{\mathrm{5}\sqrt{\mathrm{2}}}\\{\mathrm{5}\sqrt{\mathrm{2}}}\end{pmatrix}\:\:{R}=\begin{pmatrix}{−\mathrm{5}\sqrt{\mathrm{2}}}\\{\mathrm{5}\sqrt{\mathrm{2}}}\end{pmatrix} \\ $$$${P}+{Q}+{R}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{10}\sqrt{\mathrm{2}}−\mathrm{9}}\end{pmatrix} \\ $$…
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Question Number 128156 by DomaPeti last updated on 04/Jan/21 $${f}\left({x}\right)''=\frac{\mathrm{1}}{{f}\left({x}\right)^{\mathrm{2}} }\:\:\:\:\: \\ $$$${f}\left({x}\right)=? \\ $$ Answered by mr W last updated on 05/Jan/21 $${let}\:{y}'={u} \\…
Question Number 128157 by naka3546 last updated on 04/Jan/21 $${Let}\:\:{P}\left({x}\right)\:\:{is}\:\:{monic}\:\:{polynomial}\:\:{function}\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{P}\left(\mathrm{2}{x}\right)\right]^{\mathrm{2}} \:=\:{P}\left(\mathrm{4}{x}\right)\:+\:\mathrm{16}{P}\left({x}^{\mathrm{2}} \right) \\ $$$${P}\left({x}\right)\:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128150 by MJS_new last updated on 04/Jan/21 $$\mathrm{question}\:\mathrm{128091},\:\mathrm{trying}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{completely}\:\mathrm{in}\:\mathbb{R} \\ $$$${a}^{\mathrm{3}} =\mathrm{3}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${b}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${c}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{25}…
Question Number 62613 by necx1 last updated on 23/Jun/19 Commented by Prithwish sen last updated on 23/Jun/19 $$=\int\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\:\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \sqrt{\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }}}…
Question Number 128149 by MathSh last updated on 04/Jan/21 $$\underset{{x}\rightarrow−\mathrm{1}} {{lim}}\left(\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}}{{x}+\mathrm{2}−{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{{x}+\mathrm{1}}} =\:? \\ $$ Answered by liberty last updated on 04/Jan/21 $$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{2x}^{\mathrm{3}}…
Question Number 128146 by sahnaz last updated on 04/Jan/21 $$\mathrm{4}<\mathrm{x}<\mathrm{5}\:\mathrm{a}=\sqrt{\mathrm{x}−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{x}−\mathrm{4}}\:\:}\:\:\:\mathrm{b}=\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{x}−\mathrm{a}×\mathrm{b}=? \\ $$ Answered by Fikret last updated on 04/Jan/21 $${a}=\sqrt{{x}−\mathrm{4}}+\mathrm{1} \\ $$$${b}=\sqrt{{x}−\mathrm{4}}−\mathrm{1} \\ $$$${a}.{b}={x}−\mathrm{5} \\…