Question Number 149274 by BHOOPENDRA last updated on 04/Aug/21 Answered by Olaf_Thorendsen last updated on 05/Aug/21 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}{y}\:=\:{e}^{{x}} \:\:\:\:\left(\mathrm{E}\right) \\ $$$$\mathrm{Let}\:{y}\:=\:{e}^{{x}} {u},\:{y}'\:=\:{e}^{{x}} \left({u}'+{u}\right),\:{y}''\:=\:{e}^{{x}} \left({u}''+\mathrm{2}{u}'+{u}\right)…
Question Number 83737 by jagoll last updated on 05/Mar/20 $$\int\:\:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{4}} +\mathrm{cos}\:{x}}\:{dx}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18199 by Tinkutara last updated on 17/Jul/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equivalent}\:\mathrm{weight}\:\mathrm{of} \\ $$$$\mathrm{KH}\left(\mathrm{IO}_{\mathrm{3}} \right)_{\mathrm{2}} \:\mathrm{as}\:\mathrm{an}\:\mathrm{oxidant}\:\mathrm{in}\:\mathrm{presence}\:\mathrm{of} \\ $$$$\mathrm{4}\:\left(\mathrm{N}\right)\:\mathrm{HCl}\:\mathrm{when}\:\mathrm{ICl}\:\mathrm{becomes}\:\mathrm{the} \\ $$$$\mathrm{reduced}\:\mathrm{form}?\:\left(\mathrm{K}\:=\:\mathrm{39},\:\mathrm{I}\:=\:\mathrm{127}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 149268 by Samimsultani last updated on 04/Aug/21 Answered by dkenechukwu04 last updated on 04/Aug/21 $$\mathrm{Let}\:\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}+…}}}={x} \\ $$$$\mathrm{5}+\frac{\mathrm{2}}{{x}}=\mathrm{5}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{3}}{{x}}} \\ $$$$\mathrm{5}+\frac{\mathrm{2}}{{x}}=\mathrm{5}+\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$$\frac{\mathrm{5}{x}+\mathrm{2}}{{x}}=\frac{\mathrm{12}{x}+\mathrm{15}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$$\mathrm{10}{x}^{\mathrm{2}}…
Question Number 18198 by Tinkutara last updated on 16/Jul/17 $$\mathrm{Mixture}\:\mathrm{X}\:=\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{SO}_{\mathrm{4}} \right]\mathrm{Br}\:\mathrm{and}\:\mathrm{0}.\mathrm{02}\:\mathrm{mol}\:\mathrm{of} \\ $$$$\left[\mathrm{Co}\left(\mathrm{NH}_{\mathrm{3}} \right)_{\mathrm{5}} \mathrm{Br}\right]\mathrm{SO}_{\mathrm{4}} \:\mathrm{was}\:\mathrm{prepared}\:\mathrm{in}\:\mathrm{2} \\ $$$$\mathrm{litre}\:\mathrm{of}\:\mathrm{solution} \\ $$$$\mathrm{1}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{mixture}\:\mathrm{X}\:+\:\mathrm{excess}\:\mathrm{AgNO}_{\mathrm{3}} \:\rightarrow\:\mathrm{Y}…
Question Number 149271 by Samimsultani last updated on 04/Aug/21 Answered by fred last updated on 04/Aug/21 $${x}^{{x}^{\mathrm{4}} } =\mathrm{4} \\ $$$${x}=\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$ Answered by…
Question Number 18197 by Tinkutara last updated on 16/Jul/17 $$\mathrm{Rearrange}\:\mathrm{the}\:\mathrm{following}\:\left(\mathrm{I}\:\mathrm{to}\:\mathrm{IV}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{increasing}\:\mathrm{masses}. \\ $$$$\mathrm{I}.\:\mathrm{1}\:\mathrm{molecule}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{II}.\:\mathrm{1}\:\mathrm{atom}\:\mathrm{of}\:\mathrm{nitrogen} \\ $$$$\mathrm{III}.\:\mathrm{10}^{\mathrm{10}} \:\mathrm{g}\:\mathrm{molecular}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{oxygen} \\ $$$$\mathrm{IV}.\:\mathrm{10}^{−\mathrm{18}} \:\mathrm{g}\:\mathrm{atomic}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{copper} \\ $$ Answered…
Question Number 149266 by mathdanisur last updated on 04/Aug/21 $${Solve}\:{for}\:{natural}\:{numbers}: \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+{x}\:+\:{y}\:=\:\mathrm{3}{xy} \\ $$ Answered by nimnim last updated on 04/Aug/21 $$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} +\left({x}+{y}\right)=\mathrm{5}{xy}…
Question Number 18186 by Tinkutara last updated on 16/Jul/17 $$\mathrm{3}.\mathrm{92}\:\mathrm{g}\:\mathrm{of}\:\mathrm{ferrous}\:\mathrm{ammonium}\:\mathrm{sulphate} \\ $$$$\mathrm{are}\:\mathrm{dissolved}\:\mathrm{in}\:\mathrm{100}\:\mathrm{ml}\:\mathrm{of}\:\mathrm{water}.\:\mathrm{20}\:\mathrm{ml} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{solution}\:\mathrm{requires}\:\mathrm{18}\:\mathrm{ml}\:\mathrm{of} \\ $$$$\mathrm{potassium}\:\mathrm{permanganate}\:\mathrm{during} \\ $$$$\mathrm{titration}\:\mathrm{for}\:\mathrm{complete}\:\mathrm{oxidation}.\:\mathrm{The} \\ $$$$\mathrm{weight}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{present}\:\mathrm{in}\:\mathrm{one}\:\mathrm{litre} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is} \\ $$…
Question Number 149259 by john_santu last updated on 04/Aug/21 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{sin}\:\left(\mathrm{41}°\right)\mathrm{sin}\:\left(\mathrm{19}°\right)+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)=? \\ $$ Answered by john_santu last updated on 04/Aug/21 $$\Leftrightarrow\mathrm{sin}\:\mathrm{41}°\mathrm{sin}\:\mathrm{19}°=\frac{−\mathrm{2sin}\:\mathrm{4l}°\mathrm{sin}\:\mathrm{19}°}{−\mathrm{2}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{60}°−\mathrm{cos}\:\mathrm{22}°}{−\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{4}}…