Question Number 130699 by LYKA last updated on 28/Jan/21 Answered by MJS_new last updated on 28/Jan/21 $${z}=\mathrm{ln}\:\left({x}−\mathrm{2}{y}+\mathrm{3}\right) \\ $$$${x}−\mathrm{2}{y}+\mathrm{3}>\mathrm{0}\:\Leftrightarrow\:{y}<\frac{{x}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\wedge{x}\in\mathbb{R}\:\Rightarrow\:{z}\in\mathbb{R} \\ $$ Terms of Service Privacy…
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Question Number 130670 by EDWIN88 last updated on 27/Jan/21 $$\:{Given}\:{f}\left({x}\right)=\sqrt{{ax}+{b}}\:;\:{g}\left({x}\right)={px}+{q} \\ $$$${and}\:{f}\left({x}\right)<{g}\left({x}\right)\:{has}\:{solution} \\ $$$$\mathrm{1}<{x}<\mathrm{5}.\:{Give}\:{one}\:{example}\:{f}\left({x}\right) \\ $$$${and}\:{g}\left({x}\right)\:{satisfy}\:{the}\:{condition} \\ $$ Commented by MJS_new last updated on 28/Jan/21…
Question Number 65134 by turbo msup by abdo last updated on 25/Jul/19 $${let}\:{U}_{{n}} \:{a}\:{sequence}\:{U}_{\mathrm{0}} ={a}\:{and} \\ $$$${U}_{{n}} ={nU}_{{n}−\mathrm{1}} \:\:\:−\mathrm{2}\:\:\:\left({n}>\mathrm{0}\right) \\ $$$${calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n}. \\ $$ Commented…
Question Number 65044 by mathmax by abdo last updated on 24/Jul/19 $$\left.{solve}\:{x}^{\mathrm{2}} {y}^{''} \:+{xy}^{'} \:+{y}\:=\mathrm{0}\:\:{on}\:\right]\mathrm{0},+\infty\left[\:\:\:\left({put}\:{x}\:={e}^{{t}} \right)\right. \\ $$ Commented by mathmax by abdo last updated…
Question Number 130535 by mathmax by abdo last updated on 26/Jan/21 $$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{nx}} \mathrm{arctan}\left(\mathrm{nx}+\mathrm{2}\right) \\ $$$$\mathrm{calculate}\:\mathrm{g}^{\left(\mathrm{n}\right)} \left(\mathrm{o}\right) \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 64886 by mathmax by abdo last updated on 22/Jul/19 $${let}\:{f}\left({x}\right)\:={cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\:\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:{at}\:{x}_{\mathrm{0}} =\frac{\mathrm{3}}{\pi} \\ $$ Terms of Service Privacy Policy…