Question Number 121807 by abdelsalamalmukasabe last updated on 11/Nov/20 Answered by sukumara3@gmail.com last updated on 14/Nov/20 $${while}\:{cancelling}\:{step},{it}\:{is}\:{not}\:\mathrm{2}{yy}^{'} {x}.{But}\:{it}\:{is}\:\mathrm{2}{y}^{'} {y}^{'} {x}\:{only}. \\ $$ Terms of Service…
Question Number 121746 by KathleenannLabrador last updated on 11/Nov/20 $${what}\:{are}\:{the}\:{numerical}\:{coefficients}\:{in}\:{the}\:{quadration}\:{equation}\:−\mathrm{10}×\mathrm{2}−\mathrm{7}×+\mathrm{2}=\mathrm{0}?\:\:\:{a}.{a}=\mathrm{10}.{b}=.\mathrm{7}{c}=\mathrm{2}\:\:\:\:\:\:\:\:\:{b}.{a}=−\mathrm{2}.{b}=\mathrm{4}.{c}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:{c}.{a}=\mathrm{2}.{b}=\mathrm{7}.{c}=\mathrm{10}\:\:\:\:\:\:\:\:{d}.{a}=−\mathrm{10}.{b}=−\mathrm{7}.{c}=\mathrm{2} \\ $$ Answered by MJS_new last updated on 11/Nov/20 $$\mathrm{if}\:\mathrm{you}\:\mathrm{mean} \\ $$$$−\mathrm{10}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\left({d}\right)\:{a}=−\mathrm{10};\:{b}=−\mathrm{7};\:{c}=\mathrm{2}…
Question Number 121687 by abdelsalamalmukasabe last updated on 10/Nov/20 Commented by bemath last updated on 11/Nov/20 $$\left(\frac{\mathrm{2}{y}^{\mathrm{4}} +\mathrm{1}}{{y}^{\mathrm{2}} }\right)\:{dx}\:+\:\left(\mathrm{4}{xy}\right)\:{dy}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}{y}^{\mathrm{4}} +\mathrm{1}\right)\:{dx}\:+\:\left(\mathrm{4}{xy}^{\mathrm{3}} \right)\:{dy}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}\:=\:\frac{−\left(\mathrm{2}{y}^{\mathrm{4}}…
Question Number 121633 by liberty last updated on 10/Nov/20 $$\mathrm{Given}\:\mathrm{implicit}\:\mathrm{expression}\:\begin{cases}{\mathrm{x}=\mathrm{tan}\:\alpha}\\{\mathrm{y}=\mathrm{tan}\:\beta}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x},\mathrm{y},\alpha\:\mathrm{and}\:\beta\:\mathrm{is}\:\mathrm{variable}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{0}\:. \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 121609 by benjo_mathlover last updated on 10/Nov/20 $$\:\mathrm{y}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\mathrm{dx}\:−\mathrm{x}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right)\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$$$ \\ $$ Answered by liberty last updated on 10/Nov/20…
Question Number 121517 by liberty last updated on 09/Nov/20 $$\:\mathrm{y}''+\mathrm{4y}'+\mathrm{4y}\:=\:\left(\mathrm{x}+\mathrm{3}\right)\mathrm{e}^{−\mathrm{2x}} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2}\:\wedge\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{5} \\ $$ Answered by benjo_mathlover last updated on 09/Nov/20 Answered by TANMAY PANACEA…
Question Number 55939 by MJS last updated on 06/Mar/19 $$\frac{{y}'}{{y}}=\mathrm{1}−\mathrm{cot}\:{x} \\ $$$${y}=? \\ $$ Answered by kaivan.ahmadi last updated on 06/Mar/19 $$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}=\mathrm{1}−{cotx}\Rightarrow \\ $$$$\frac{\mathrm{1}}{{y}}{dy}=\left(\mathrm{1}−{cotx}\right){dx}\Rightarrow \\…
Question Number 186999 by Mingma last updated on 12/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121461 by bramlexs22 last updated on 08/Nov/20 Answered by Olaf last updated on 08/Nov/20 $$\left(\mathrm{3}{y}^{\mathrm{2}} +\mathrm{10}{xy}^{\mathrm{2}} \right){dx}+\left(\mathrm{6}{xy}−\mathrm{2}+\mathrm{10}{x}^{\mathrm{2}} {y}\right){dy}\:=\:\mathrm{0} \\ $$$$\mathrm{d}{f}\:=\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right){dx}+\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dy}\:=\:\mathrm{0} \\ $$$$ \\…
Question Number 121378 by liberty last updated on 07/Nov/20 $$\:\mathrm{cos}\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{y}.\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}.\mathrm{sin}\:\mathrm{x}\: \\ $$ Answered by Ar Brandon last updated on 07/Nov/20 $$\mathrm{cosx}\centerdot\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\centerdot\mathrm{sinx}=\mathrm{2cos}^{\mathrm{3}} \mathrm{x}\centerdot\mathrm{sinx} \\ $$$$\mathrm{secx}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ysecxtanx}=\mathrm{2cosxsinx}…