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Question Number 104383 by abony1303 last updated on 21/Jul/20 $$\mathrm{When}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{differentiable}\:\mathrm{function} \\ $$$$\mathrm{satisfying}\:\:{x}\centerdot{f}\left({x}\right)={x}^{\mathrm{2}} +\int_{\mathrm{0}} ^{\:{x}} \left({x}−{t}\right)\centerdot{f}\:'\left({t}\right){dt} \\ $$$$\mathrm{Find}\:\Rightarrow\:{f}\left(\mathrm{1}\right) \\ $$ Commented by abony1303 last updated on…
Question Number 169915 by daus last updated on 12/May/22 Answered by daus last updated on 13/May/22 $$ \\ $$ Answered by dumitrel last updated on…
Question Number 169900 by pticantor last updated on 12/May/22 $$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\forall\boldsymbol{{z}}\in\mathbb{C}/\:\mid\boldsymbol{{z}}\mid=\mathrm{2}, \\ $$$$\mid\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{4}} −\mathrm{5}\boldsymbol{{z}}+\mathrm{1}}\mid\leqslant\frac{\mathrm{1}}{\mathrm{5}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 169896 by Beginner last updated on 11/May/22 Answered by mr W last updated on 12/May/22 $${x}−\mathrm{3}\geqslant\mathrm{0}\:\wedge\:\mathrm{2}{x}\left({x}+\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant\mathrm{3}\:\wedge\:{x}<−\mathrm{1}\:{or}\:{x}>\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant\mathrm{3}\:\checkmark \\ $$$${x}−\mathrm{3}\leqslant\mathrm{0}\:\wedge\:\mathrm{2}{x}\left({x}+\mathrm{1}\right)<\mathrm{0} \\…
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Question Number 169892 by weltr last updated on 11/May/22 $$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{xe}^{−{y}} \:,\:\:{y}\left(\mathrm{1}\right)\:=\:\mathrm{0} \\ $$ Answered by mahdipoor last updated on 11/May/22 $$\Rightarrow\frac{{dy}}{{e}^{−{y}} }={e}^{{y}} {dy}=\mathrm{2}{xdx}\Rightarrow{e}^{{y}} ={x}^{\mathrm{2}} +{c}…