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Question Number 183766 Answers: 1 Comments: 0
Question Number 183769 Answers: 0 Comments: 4
$${solve}\:{for}\:{x}: \\ $$$${x}^{{x}^{{x}} } =\mathrm{2}^{\mathrm{2048}} \\ $$$${by}\:{using}\:{lambert}\:{function} \\ $$
Question Number 183761 Answers: 1 Comments: 0
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{for}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{given}\:\mathrm{by}\:{U}\left({x},{t}\right). \\ $$$$\begin{cases}{\frac{\partial{U}}{\partial{t}}\:=\:\mathrm{2}\frac{\partial^{\mathrm{2}} {U}}{\partial{x}^{\mathrm{2}} }\:,\:\mathrm{0}\:<\:{x}\:<\:\pi}\\{{U}\left(\mathrm{0},{t}\right)\:=\:\mathrm{0},\:{U}\left(\pi,{t}\right)\:=\:\mathrm{0},\:{t}\:>\:\mathrm{0}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{U}\left({x},\mathrm{0}\right)\:=\:\mathrm{25}{x} \\ $$
Question Number 183795 Answers: 2 Comments: 0
Question Number 183794 Answers: 0 Comments: 1
$$\:{If}\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{cos}\:{x}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:{dx}=\:{T} \\ $$$$\:{then}\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}+\mathrm{1}}\:{dx}\:=\:?\: \\ $$
Question Number 183756 Answers: 2 Comments: 0
$${solve}\:{for}\:{x}\:{by}\:{using}\:{lambert}\:{function} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$
Question Number 183750 Answers: 1 Comments: 0
Question Number 183747 Answers: 1 Comments: 0
$$\:{Find}\:{the}\:{perimeter}\:{of}\:{a}\:{regular}\:{heptagon}\: \\ $$$$\:{ABCDEFG}\:{if}\:\frac{\mathrm{1}}{{AE}}\:+\:\frac{\mathrm{1}}{{AC}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\: \\ $$
Question Number 183746 Answers: 1 Comments: 5
$${find}\:{x} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{19}{x}^{\mathrm{2}} +\mathrm{93}{x}−\mathrm{128}=\mathrm{0} \\ $$
Question Number 183737 Answers: 2 Comments: 0
Question Number 183734 Answers: 1 Comments: 0
Question Number 183728 Answers: 1 Comments: 1
Question Number 183726 Answers: 2 Comments: 1
$${Prove}\:{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{2}^{{n}−\mathrm{1}} ={n}×\mathrm{2}^{{n}−\mathrm{1}} \: \\ $$
Question Number 183725 Answers: 0 Comments: 0
Question Number 183712 Answers: 1 Comments: 0
$${solve}: \\ $$$${W}\left({In}\left(\mathrm{4}{x}\right)\right)=\sqrt{\left({x}−\mathrm{1}\right)} \\ $$
Question Number 183711 Answers: 1 Comments: 1
Question Number 183710 Answers: 0 Comments: 0
Question Number 183709 Answers: 0 Comments: 1
Question Number 183706 Answers: 1 Comments: 0
$$\:\:\:\:{A}=\int\:\frac{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{4}} {x}}\:{dx} \\ $$
Question Number 183693 Answers: 1 Comments: 0
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\left(\mathrm{ln}\:{x}\right)^{{x}} }{{x}^{\mathrm{ln}\:{x}} } \\ $$
Question Number 183690 Answers: 1 Comments: 0
Question Number 183687 Answers: 3 Comments: 0
$$\int\frac{\mathrm{1}}{\mathrm{lnx}}\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}\:\mathrm{out} \\ $$
Question Number 183678 Answers: 0 Comments: 0
$${find}\:{the}\:{laplace}\:{invesrse}\:{for}\:{I}\left({s}\right) \\ $$$${I}\left({s}\right)=\frac{\mathrm{2000}{s}^{\mathrm{2}} }{\left({s}^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} \right)\left({s}^{\mathrm{2}} +\mathrm{400}{s}+\mathrm{2}×\mathrm{10}^{\mathrm{5}} \right)} \\ $$
Question Number 183673 Answers: 1 Comments: 0
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\:\left[\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{{a}\mathrm{x}}}\right)\right]^{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{\mathrm{bx}}}\right)} \\ $$
Question Number 183669 Answers: 4 Comments: 0
$$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\mathrm{1}\:+\:{x}\:+\:{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:{min}_{\:{f}} \:=\:? \\ $$
Question Number 183668 Answers: 1 Comments: 1
$$ \\ $$$$\:\:\:\:\:\:{x}\:,\:{y}\:,\:{z}\:\in\mathbb{R}: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{If}\:\:\:\:\begin{cases}{\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{y}+{z}}\:=\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{{y}}\:+\frac{\mathrm{1}}{{x}+{z}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{1}}{{z}_{\:} }\:+\frac{\mathrm{1}}{{x}+{y}}\:=\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{x}\:,\:{y}\:,\:{z}\:=? \\ $$$$ \\ $$
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