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Question Number 185101 Answers: 2 Comments: 0
Question Number 185090 Answers: 1 Comments: 2
$$\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}\:=\:\:\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=? \\ $$$$ \\ $$$$ \\ $$
Question Number 185087 Answers: 2 Comments: 0
$$ \\ $$$$ \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2004}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}=?\:\:\: \\ $$$$ \\ $$
Question Number 185086 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\:=\:? \\ $$$$ \\ $$$$ \\ $$
Question Number 185085 Answers: 2 Comments: 0
$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{a}+\mathrm{1}\right)=\mathrm{log}_{\mathrm{4}} \left(\mathrm{a}+\mathrm{8}\right) \\ $$$$\mathrm{a}=?? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$
Question Number 185077 Answers: 1 Comments: 0
Question Number 185076 Answers: 2 Comments: 0
$${x}^{\mathrm{2}{x}^{\mathrm{6}} } =\mathrm{3} \\ $$$${x}=? \\ $$
Question Number 185075 Answers: 0 Comments: 5
$$ \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{4x}\:\Rightarrow\:\:\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{x}} }=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}.\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{xln}\left(\mathrm{2}\right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2x}\right)} =−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{W}\left(−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)} \right)=\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)=\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=−\frac{\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right)\:\:}{\mathrm{ln}\left(\mathrm{2}\right)}\:=\frac{\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)\right)}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=−\frac{\mathrm{W}\left(−\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)} \right)}{\mathrm{ln}\left(\mathrm{2}\right)}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=\frac{\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=\mathrm{4} \\ $$
Question Number 185073 Answers: 1 Comments: 0
Question Number 185058 Answers: 1 Comments: 0
Question Number 185054 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\mathrm{I}{f}\:,\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\lfloor\:{x}\:\rfloor}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\mathrm{D}_{\:{f}} \:=?\left({domain}\right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:{f}\:} =\:?\:\left({range}\:\right) \\ $$
Question Number 185053 Answers: 1 Comments: 0
Question Number 185052 Answers: 0 Comments: 3
Question Number 185042 Answers: 0 Comments: 0
$${ABCD}\:{is}\:{a}\:{quadrilateral}\:{inscribed}\:{in}\:{a}\:{circle} \\ $$$$\:{if}\:\:\:\:\overset{\frown} {{AB}}\:+\:\overset{\frown} {{CD}}\:=\:\mathrm{307}°\:{and}\:{AC}\centerdot{BD}\:=\:\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{ABCD}.\: \\ $$
Question Number 185038 Answers: 1 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{series}\:\frac{\mathrm{x}}{\mathrm{27}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{125}}+...+\frac{\mathrm{x}^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }+... \\ $$$$\mathrm{is}\:\mathrm{absolutely}\:\mathrm{convergent}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Question Number 185036 Answers: 1 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{series}\:\mathrm{for}\:\mathrm{cosx}.\:\mathrm{Hence},\: \\ $$$$\mathrm{deduce}\:\mathrm{series}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{and}\:\mathrm{show}\:\mathrm{that}, \\ $$$$\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{small},\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{cosx}}{\mathrm{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{360}} \\ $$$$\mathrm{approximately}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Question Number 185030 Answers: 1 Comments: 1
Question Number 185024 Answers: 1 Comments: 0
$$\mathrm{prove}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}<\mathrm{ln}\left(\mathrm{2}\right) \\ $$
Question Number 185023 Answers: 1 Comments: 0
$$\mathrm{provet}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}>\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$
Question Number 185014 Answers: 1 Comments: 1
$$\mathrm{The}\:\mathrm{relation}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{kx}+\mathrm{c},\:\mathrm{where}\:\mathrm{K} \\ $$$$\mathrm{and}\:\mathrm{C}\:\mathrm{are}\:\mathrm{constant}\:\mathrm{passes}\:\mathrm{through} \\ $$$$\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{1},\:−\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{1},\:\mathrm{8}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{coordinate}\:\mathrm{axes}.\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{C}\:\mathrm{and}\:\mathrm{K}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$
Question Number 185013 Answers: 4 Comments: 3
Question Number 185012 Answers: 0 Comments: 2
$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{sin}\left(\mathrm{kx}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{Im}\left(\mathrm{e}^{\mathrm{ikx}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Im}\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{k}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Im}\left(\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }\right) \\ $$$$\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} \left(\mathrm{e}^{−\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} −\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}} \right)}{\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} \left(\mathrm{e}^{−\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} −\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} ×\frac{−\mathrm{2sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{−\mathrm{2sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} \frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{sin}\left(\mathrm{kx}\right)=\frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\:\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{Im}\left(\mathrm{e}^{\mathrm{in}\frac{\mathrm{x}}{\mathrm{2}}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\left(\left(\mathrm{n}+\mathrm{1}\right)\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{sin}\left(\frac{\mathrm{nx}}{\mathrm{2}}\right) \\ $$
Question Number 185011 Answers: 0 Comments: 0
Question Number 185010 Answers: 0 Comments: 0
Question Number 185009 Answers: 0 Comments: 0
Question Number 185008 Answers: 1 Comments: 0
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