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Question Number 217769 by nECxx2 last updated on 20/Mar/25 $${Given}\:{a}\:{consumer}\:{with}\:{the}\:{utility} \\ $$$${function}\:{U}\:=\:{X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +\:{X}_{\mathrm{2}} \:{who}\:{faces} \\ $$$${a}\:{budget}\:{constraint}\:{of}\:{B}={P}_{\mathrm{1}} {X}_{\mathrm{1}} {P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${Show}\:{that}\:{the}\:{expemditure}\:{function} \\ $$$${facing}\:{the}\:{consumer}\:{is} \\…
Question Number 208608 by Tawa11 last updated on 18/Jun/24 Answered by mr W last updated on 18/Jun/24 $${R}^{\mathrm{2}} =\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+{ab}+{b}^{\mathrm{2}} } \\…
Question Number 204469 by DeArtist last updated on 18/Feb/24 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2},\:\mathrm{0}<\:{x}\:<\mathrm{2}}\\{−\mathrm{2},\:−\mathrm{2}\:<{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{of}\:\mathrm{period}\:\mathrm{4} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right)\:,\:\mathrm{for}\:−\mathrm{6}\:<\:{x}\:<\:\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{coefficient}\:{a}_{\mathrm{0}} ,\:{a}_{{n}} ,\:\mathrm{and}\:{b}_{{n}} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{series}.\: \\ $$$$\left(\mathrm{d}\right)\:\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:\frac{\pi}{\mathrm{4}}…
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Question Number 130827 by bemath last updated on 29/Jan/21 $$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\ $$ Answered by EDWIN88 last updated on 29/Jan/21 $${let}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)={t}\:\Rightarrow{x}+\mathrm{1}\:=\:{e}^{{t}} \\ $$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}.\frac{{dy}}{{dt}}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:=\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\frac{{dy}}{{dt}}\:\right]=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 129131 by Boucatchou last updated on 13/Jan/21 $${Solve}\:\:\mathrm{134}^{{x}+\mathrm{1}} =\mathrm{16}^{{x}} −\mathrm{768} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Jan/21 $${No}\:{solution}\:{in}\:\:{x}\in\mathbb{R} \\ $$…
Question Number 129102 by AbdulAzizabir last updated on 12/Jan/21 Answered by Olaf last updated on 12/Jan/21 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:=\:\left\{\mathrm{1},\:\mathrm{5},\:\mathrm{14},\:\mathrm{30},\:\mathrm{55},…\right\}\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{56}}−\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{220}}−…\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{55}}−…\right) \\…
Question Number 63298 by Rio Michael last updated on 02/Jul/19 $${A}\:{particle}\:{P},\:{moves}\:{on}\:{the}\:{curve}\:{with}\:{polar}\:{equation}\:\: \\ $$$${r}\:=\:{e}^{{k}\theta} \:,\:{where}\:\left({r},\theta\right)\:{are}\:{polar}\:{coordinates}\:{referred}\:{to}\:{a}\:{fixed} \\ $$$${pole}\:{and}\:{k}\:{is}\:{a}\:{positive}\:{constant}.\:{Given}\:{that}\:{the}\:{radial}\:{velocity} \\ $$$${of}\:{P}\:{is}\:\frac{{k}}{{r}}\:\:{show}\:{that}\:{the}\:{transverse}\:{acceleration}\:{of}\:{th}\:{particle} \\ $$$${is}\:{zero}. \\ $$$$ \\ $$ Commented…