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Question Number 97401 by M±th+et+s last updated on 07/Jun/20 Commented by M±th+et+s last updated on 07/Jun/20 $$\mathrm{1}{F}\mathrm{1}\:{is}\:{the}\:{kummer}\:{confluent}\:{hypergeometric} \\ $$$${function} \\ $$$${U}\left({a},{b},{x}\right)\:{the}\:{confluent}\:{hypergeometric} \\ $$$${function}\:{of}\:{the}\:{second}\:{kind} \\ $$…
Question Number 97380 by mathocean1 last updated on 07/Jun/20 $${E}\:{is}\:{a}\:{vectorial}\:{plane}.\:{his}\:{base}\:{is}\: \\ $$$${B}=\left(\overset{\rightarrow} {{i}};\overset{\rightarrow} {{j}}\right).\:{f}\:{is}\:{an}\:{endomorphism}\:{defined} \\ $$$${by}\:{f}\left(\overset{\rightarrow} {{i}}\right)=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{i}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{j}}\:{and}\:{f}\left(\overset{\rightarrow} {{j}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{i}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{j}} \\ $$$$\left.\mathrm{1}\right){Show}\:{that}\:{ker}\:{f}\:{is}\:{a}\:{vectorial}\:{straigh} \\…
Question Number 31820 by mondodotto@gmail.com last updated on 15/Mar/18 Answered by mrW2 last updated on 15/Mar/18 $$\left({i}\right) \\ $$$$\frac{{dx}}{{d}\theta}=−\mathrm{4}\:\mathrm{sin}\:\theta \\ $$$$\frac{{dy}}{{d}\theta}=\mathrm{3}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{dy}}{{d}\theta}×\frac{\mathrm{1}}{\frac{{dx}}{{d}\theta}}=−\frac{\mathrm{3}\:\mathrm{cos}\:\theta}{\mathrm{4}\:\mathrm{sin}\:\theta} \\ $$$$\left({ii}\right)…
Question Number 31794 by mondodotto@gmail.com last updated on 14/Mar/18 Commented by MJS last updated on 14/Mar/18 $${f}\left({x}\right)=\sqrt{{x}^{\mathrm{3}} }={x}\sqrt{{x}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}}}=\frac{\sqrt{{x}}}{{x}} \\ $$$${f}\left({x}\right)−{g}\left({x}\right)={x}\sqrt{{x}}−\frac{\sqrt{{x}}}{{x}}=\left({x}−\frac{\mathrm{1}}{{x}}\right)\sqrt{{x}} \\ $$$$\mathrm{domain}=\left\{{x}\in\mathbb{R}\mid{x}>\mathrm{0}\right\}=\mathbb{R}^{+} \\…
Question Number 97323 by mathocean1 last updated on 07/Jun/20 $${Mr}\:{Peter}\:{has}\:\mathrm{4}\:{children}.\:{x}\:{are}\:{in}\: \\ $$$${class}\:{C}\:{and}\:{y}\:{are}\:{in}\:{class}\:{D}.\:{x}\geqslant\mathrm{1}\:{and} \\ $$$${y}\geqslant\mathrm{1}.\:{Show}\:{that}\:{the}\:{number}\:{of}\:{possibility}\: \\ $$$${to}\:{choose}\:{at}\:{random}\:{and}\:{simultaneous} \\ $$$$\mathrm{2}\:{children}\:{in}\:{same}\:{class}\:{verify}\:{this} \\ $$$${equation}\:{p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6} \\ $$ Answered by…
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Question Number 162847 by mkam last updated on 01/Jan/22 Answered by abdullahhhhh last updated on 01/Jan/22 $$\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{want}}\:\boldsymbol{\mathrm{vaule}}\:\boldsymbol{\mathrm{of}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\Pi}/\mathrm{4}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{what}} \\ $$$$ \\ $$ Commented by mkam last…
Question Number 162845 by mkam last updated on 01/Jan/22 Commented by mkam last updated on 01/Jan/22 $${prove}\:{this} \\ $$ Answered by abdullahhhhh last updated on…
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