Question Number 3695 by Rasheed Soomro last updated on 19/Dec/15 $$\mathcal{C}{an}\:{we}\:{say}\:{that} \\ $$$$\mathcal{A}\:{line}\:{is}\:{a}\:{circle}\:{whose}\:{radius}\:{is}\:\infty \\ $$$$\mathcal{O}{r} \\ $$$${A}\:{circle}\:{with}\:\infty\:{radius}\:{is}\:{a}\:{line}\:\:? \\ $$ Commented by Filup last updated on…
Question Number 134764 by bramlexs22 last updated on 07/Mar/21 $$\:\mathbb{Z}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{dx}}{\:\sqrt{\mid\mathrm{x}−\mathrm{1}\mid}}\:? \\ $$ Answered by EDWIN88 last updated on 07/Mar/21 $$\mathbb{Z}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{dx}}{\:\sqrt{\mid\mathrm{x}−\mathrm{1}\mid}}\:;\:\mathrm{let}\:\mathrm{u}=\mid\mathrm{x}−\mathrm{1}\mid \\…
Question Number 69231 by ~ À ® @ 237 ~ last updated on 21/Sep/19 $${Prove}\:{that}\:\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{4}{p}+\mathrm{1}}\:=\:\frac{\pi−{argcoth}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{4}\sqrt{\mathrm{2}}}\:\:{and} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty\:} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{4}{p}+\mathrm{3}}\:=\:\frac{\pi+{argcoth}\left(\sqrt{\mathrm{2}}\:\right)}{\mathrm{4}\sqrt{\mathrm{2}}\:}\:\:\: \\ $$…
Question Number 134767 by jagadharkumar last updated on 07/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 3693 by Rasheed Soomro last updated on 19/Dec/15 $$\mathcal{E}{very}\:{equation}\:{of}\:{x},{y}\:{has}\:{a}\:{curve}\:{in}\:{a}\:{plane}. \\ $$$$\mathcal{D}{oes}\:{every}\:{curve}\:{in}\:{a}\:{plane}\:{has}\:{an}\:{equation}? \\ $$ Commented by 123456 last updated on 19/Dec/15 $$\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{yes} \\ $$$$\mathrm{lets}\:{f}:\mathbb{R}^{\mathrm{2}}…
Question Number 69229 by naka3546 last updated on 21/Sep/19 $${a},\:{b},\:{c}\:\:\in\:\:{nonnegative}\:\:{real}\:\:{numbers} \\ $$$$\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\sqrt{{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\sqrt{{c}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:\geqslant\:\:\mathrm{2}\:+\:\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right)\:+\:\mathrm{1}} \\ $$$${Find}\:\:{all}\:\:{triplets}\:\left({a},\:{b},\:{c}\right)\:\:{so}\:\:{that}\:\:{inequality}\:\:{above}\:\:{hold}\:. \\ $$…
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Question Number 134760 by rs4089 last updated on 07/Mar/21 Answered by mathmax by abdo last updated on 07/Mar/21 $$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left[\frac{\left(\mathrm{n}!\right)}{\mathrm{n}}\right]^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\frac{\mathrm{n}!}{\mathrm{n}}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}}…
Question Number 134763 by bramlexs22 last updated on 07/Mar/21 $$\mathrm{Given}\:{a}=\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{a}^{\mathrm{3}} −\mathrm{9}{a}+\mathrm{1}. \\ $$ Answered by EDWIN88 last updated on 07/Mar/21 $$=\:\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{3}}\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{9}}−\frac{\mathrm{26}\:\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\frac{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{3}}−\frac{\mathrm{26}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{3}}+\mathrm{4} \\ $$…
Question Number 69222 by mustakim420 last updated on 21/Sep/19 Commented by mr W last updated on 21/Sep/19 $${n}=\mathrm{2}\:{or}\:\mathrm{4} \\ $$$$\mathrm{2}+\mathrm{4}=\mathrm{6} \\ $$ Terms of Service…