Question Number 102328 by bemath last updated on 08/Jul/20 Answered by bobhans last updated on 08/Jul/20 $$\left({ii}\right)\:{y}=\:\mathrm{2ln}\left(\frac{\sqrt{\mathrm{8}{x}−\mathrm{4}}}{{x}}\right)\:\Rightarrow{x}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}\:=\:\mathrm{ln}\left(\frac{\mathrm{8}{x}−\mathrm{4}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{e}^{{y}} .{x}^{\mathrm{2}} \:=\:\mathrm{8}{x}−\mathrm{4} \\ $$$${e}^{{y}} .{x}^{\mathrm{2}}…
Question Number 102323 by Learner101 last updated on 08/Jul/20 $${Solve}\:{this}\:{linear}\:{Equation} \\ $$$$\frac{{dy}}{{dx}}\:+\:{y}\:{cos}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\:{x} \\ $$ Answered by bemath last updated on 08/Jul/20 $${IF}\:\:{u}\left({x}\right)={e}^{\int\mathrm{cos}\:{x}\:{dx}} \:=\:{e}^{\mathrm{sin}\:{x}} \\ $$$${y}\left({x}\right)=\frac{\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}\:{e}^{\mathrm{sin}\:{x}}…
Question Number 167838 by cortano1 last updated on 27/Mar/22 $$\:\:\:\:\:\:\frac{{dy}}{{dx}}=\mathrm{8}{x}+\mathrm{4}{y}+\left(\mathrm{2}{x}+{y}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:{y}=? \\ $$ Answered by mr W last updated on 27/Mar/22 $${let}\:{u}=\mathrm{2}{x}+{y}−\mathrm{1} \\ $$$$\frac{{du}}{{dx}}=\mathrm{2}+\frac{{dy}}{{dx}}…
Question Number 36746 by prof Abdo imad last updated on 05/Jun/18 $${solve}\:{the}\:{d}.{e}.\:{y}^{''} \left({t}\right)\:+{y}\left({t}\right)=\sum_{{n}=\mathrm{0}} ^{{N}} \:{a}_{{n}} {cos}\left({nt}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36745 by prof Abdo imad last updated on 05/Jun/18 $${solve}\:{the}\:{d}.{e}\:\:{y}^{''} \left({t}\right)\:+{y}\left({t}\right)={cos}\left({nt}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167709 by mnjuly1970 last updated on 23/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 36590 by ajfour last updated on 03/Jun/18 Commented by ajfour last updated on 03/Jun/18 $${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC} \\ $$$${if}\:{one}\:{vertex}\:{is}\:{on}\:{smaller}\:{circle} \\ $$$${of}\:{radius}\:\boldsymbol{{r}}\:{and}\:{other}\:{two}\:{on}\: \\ $$$${larger}\:{circle}\:{of}\:{radius}\:\boldsymbol{{R}}\:. \\ $$$${The}\:{distance}\:{between}\:{centres}\:{of}…
Question Number 102094 by M±th+et+s last updated on 06/Jul/20 $${good}\:{evenig}\:{for}\:{all} \\ $$$$ \\ $$$${this}\:{is}\:{an}\:{answerd}\:{question}\:{i}\:{will}\:{repost}\:{it} \\ $$$${if}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\sqrt{{x}−\mathrm{2}} \\ $$$${is}\:{there}\:{a}\:{cirtical}\:{point}\:{when}\:{x}=\mathrm{2}\:? \\ $$$$ \\ $$$$ \\…
Question Number 167630 by cortano1 last updated on 21/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167567 by pete last updated on 19/Mar/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\mathrm{y}=\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{5},\:\mathrm{where}\:\mathrm{the}\:\mathrm{tangent} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}−\mathrm{5}=\mathrm{8x}. \\ $$ Commented by greogoury55 last updated on 19/Mar/22 $$\:{y}'=\:\mathrm{8}…