Question Number 101291 by Mikael_786 last updated on 01/Jul/20 Answered by mr W last updated on 01/Jul/20 $$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \frac{{dy}}{{dx}}+{ay}+{ax}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{6}{x}−\mathrm{6}{y}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +\left({ax}−\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}}…
Question Number 101258 by bemath last updated on 01/Jul/20 $$\mathrm{minimum}\:\mathrm{value}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{constrain}\:\mathrm{g}\left(\mathrm{x},\mathrm{y}\right)=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{16} \\ $$ Commented by john santu last updated on 01/Jul/20 $$\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)=\mathrm{x}^{\mathrm{2}}…
Question Number 101182 by bemath last updated on 01/Jul/20 $$\mathcal{C}\mathrm{onsider}\:\mathrm{a}\:\mathrm{square}\:−\:\mathrm{based} \\ $$$$\mathrm{pyramid}\:\mathrm{with}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{6}{x}\:\mathrm{cm} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{base}\:\mathrm{length}\:\mathrm{of}\:\left(\mathrm{2}−{x}\right)\mathrm{cm}. \\ $$$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{volume}\:\mathrm{of}\:\mathrm{pyramid}. \\ $$ Answered by bemath last updated…
Question Number 35642 by rahul 19 last updated on 21/May/18 $${If}\:{y}=\:\sqrt{\left({a}−{x}\right)\left({x}−{b}\right)}−\left({a}−{b}\right)\mathrm{tan}^{−\mathrm{1}} \left(\left(\frac{{a}−{x}}{{x}−{b}}\right)^{\mathrm{0}.\mathrm{5}} \right). \\ $$$${Then}\:{find}\:\frac{{dy}}{{dx}}\:? \\ $$ Commented by rahul 19 last updated on 21/May/18…
Question Number 35623 by abdo mathsup 649 cc last updated on 21/May/18 $${solve}\:{the}\:{differencial}\:{system} \\ $$$$\left\{_{{y}^{'} =\:{y}\:+\mathrm{2}{z}\:+{t}^{\mathrm{2}} } ^{{x}^{'} \:\:=\:{y}\:+{t}^{\mathrm{2}} } \right. \\ $$$$\left\{{z}^{'} \:=\mathrm{2}{x}−\mathrm{2}{y}\right. \\…
Question Number 166690 by mnjuly1970 last updated on 25/Feb/22 $$ \\ $$$$\:\:\:\:\:\:\:\:\:{Calculate}\: \\ $$$$\:\:\:\:\:{If}\:\:,\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tanh}^{\:−\mathrm{1}} \left(\:{x}^{\:\mathrm{3}} \:\right)}{{x}}\:{dx}\:=\:\alpha.\zeta\left(\:\mathrm{2}\right)\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{then}\:,\:\:\:\:\:\alpha\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\mathscr{M}.\mathscr{N}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$ \\…
Question Number 35553 by tanmay.chaudhury50@gmail.com last updated on 20/May/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 35537 by Raj Singh last updated on 20/May/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18 $${f}\left({x}\right)={y}={x}^{\mathrm{2}} \:\: \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{2}} =\mathrm{1}\:\:\:\:{and}\:{f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${slope}\:{m}=\frac{{f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{3}…
Question Number 35461 by Raj Singh last updated on 19/May/18 Answered by tanmay.chaudhury50@gmail.com last updated on 19/May/18 $${x}.\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}+{y}\right)^{−\mathrm{1}/\mathrm{2}} ×\frac{{dy}}{{dx}}+{y}×\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} \\ $$$$ \\ $$ Terms of…
Question Number 166437 by mnjuly1970 last updated on 20/Feb/22 $$ \\ $$$$\:\:\:\:\:\:{calculate}\: \\ $$$$\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:{ln}\left(\mathrm{1}−{x}\:\right).{ln}\left(\mathrm{1}+\:{x}\:\right)}{{x}^{\:\mathrm{2}} }\:{dx}=? \\ $$$$\:\:\:\:\:\:−−−−−−− \\ $$ Answered by qaz last…