Question Number 6511 by Yozzii last updated on 30/Jun/16 $${Find}\:{the}\:{function}\:{y}\:{satisfying} \\ $$$${y}'+\frac{{c}_{\mathrm{1}} }{{y}\left({x}−{c}_{\mathrm{2}} \right)^{{n}} }={c}_{\mathrm{3}} \\ $$$${where}\:{n}\in\left(\mathbb{Z}−\left\{\mathrm{0}\right\}\right),\:{c}_{\mathrm{1}} \:{and}\:{c}_{\mathrm{3}} \:{are}\:{nonzero} \\ $$$${constants},\:{and}\:{c}_{\mathrm{2}} \:{is}\:{constant}. \\ $$ Terms…
Question Number 72047 by aliesam last updated on 23/Oct/19 Commented by mind is power last updated on 23/Oct/19 $$\mathrm{x}_{\mathrm{1}} =\mathrm{1},\mathrm{x}_{\mathrm{2}} =\mathrm{4},\mathrm{x}_{\mathrm{3}} =\mathrm{9},\mathrm{x}_{\mathrm{4}} =\mathrm{16} \\ $$$$\forall_{\mathrm{n}}…
Question Number 72044 by 20190927 last updated on 23/Oct/19 $$\mathrm{y}=\mathrm{ln}\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{x}\right)\:\:\:\mathrm{solve}\:\mathrm{y}^{\left(\mathrm{6}\right)} \left(\mathrm{x}\right) \\ $$ Commented by mathmax by abdo last updated on 24/Oct/19 $${we}\:{have}\:{y}^{'} \left({x}\right)=\frac{\mathrm{6}{x}+\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}}…
Question Number 137582 by SOMEDAVONG last updated on 04/Apr/21 $$\mathrm{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Answered by Ñï= last updated on 04/Apr/21 $${A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)}…
Question Number 137579 by liberty last updated on 04/Apr/21 $$\left(−\mathrm{1}\right)×\frac{\mathrm{1}}{\pi.{i}}\:=?\: \\ $$ Answered by Engr_Jidda last updated on 04/Apr/21 $$\left(−\mathrm{1}\right)×\frac{\mathrm{1}}{\pi{i}}=\:{i}^{\mathrm{2}} \frac{\mathrm{1}}{\pi{i}}=\frac{{i}}{\pi} \\ $$ Terms of…
Question Number 6505 by nburiburu last updated on 29/Jun/16 $${Find}\:{general}\:{solution}\:{for} \\ $$$${y}'\:+{P}\left({x}\right)\centerdot{y}^{\mathrm{2}} ={Q}\left({x}\right) \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 72036 by mr W last updated on 23/Oct/19 Commented by mr W last updated on 23/Oct/19 $${the}\:{cup}\:{has}\:{the}\:{shape}\:{of}\:{hemisphere} \\ $$$${with}\:{radius}\:{R}.\:{a}\:{stick}\:{of}\:{mass}\:{m}\:{and} \\ $$$${length}\:{l}\:\left(\mathrm{2}{R}<{l}<\mathrm{4}{R}\right)\:{is}\:{released}\:{at} \\ $$$${blue}\:{position}.\:{there}\:{is}\:{no}\:{friction}…
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Question Number 137568 by JulioCesar last updated on 04/Apr/21 Commented by JulioCesar last updated on 04/Apr/21 $${why}\:{it}\:{has}\:{background}?? \\ $$ Commented by mr W last updated…