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Question Number 185101    Answers: 2   Comments: 0

Question Number 185090    Answers: 1   Comments: 2

x^3 +1 = 2((2x−1))^(1/3) x=?

$$\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}\:=\:\:\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=? \\ $$$$ \\ $$$$ \\ $$

Question Number 185087    Answers: 2   Comments: 0

(x^2 +x+1)^2 +(y^2 −y+1)^2 =2004^2 x;y∈Z x;y=?

$$ \\ $$$$ \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2004}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}=?\:\:\: \\ $$$$ \\ $$

Question Number 185086    Answers: 1   Comments: 0

y = ln(x+(√(x^2 +1))) (d^n y/dx^n ) = ?

$$ \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\:=\:? \\ $$$$ \\ $$$$ \\ $$

Question Number 185085    Answers: 2   Comments: 0

log_3 (a+1)=log_4 (a+8) a=?? please solution

$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{a}+\mathrm{1}\right)=\mathrm{log}_{\mathrm{4}} \left(\mathrm{a}+\mathrm{8}\right) \\ $$$$\mathrm{a}=?? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$

Question Number 185077    Answers: 1   Comments: 0

Question Number 185076    Answers: 2   Comments: 0

x^(2x^6 ) =3 x=?

$${x}^{\mathrm{2}{x}^{\mathrm{6}} } =\mathrm{3} \\ $$$${x}=? \\ $$

Question Number 185075    Answers: 0   Comments: 5

2^x =4x ⇒ (x/2^x )=(1/4) ⇒x.e^(−ln(2^x )) =(1/4) ⇒−xln(2)e^(−ln(2x)) =−(1/4)ln(2) ⇒W(−ln(2^x )e^(−ln(2^x )) )=W(−(1/4)ln(2)) ⇒−ln(2^x )=W(−(1/4)ln(2)) ⇒x=−((W(−(1/4)ln(2)) )/(ln(2))) =((W(−(1/2^4 )ln(2^4 )))/(ln(2))) ⇒x=−((W(−ln(2^4 )e^(−ln(2^4 )) ))/(ln(2))) ⇒x=((ln(2^4 ))/(ln(2))) ⇒x=4

$$ \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{4x}\:\Rightarrow\:\:\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{x}} }=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}.\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{xln}\left(\mathrm{2}\right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2x}\right)} =−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{W}\left(−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)} \right)=\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}} \right)=\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=−\frac{\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\right)\:\:}{\mathrm{ln}\left(\mathrm{2}\right)}\:=\frac{\mathrm{W}\left(−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)\right)}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=−\frac{\mathrm{W}\left(−\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)\mathrm{e}^{−\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)} \right)}{\mathrm{ln}\left(\mathrm{2}\right)}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=\frac{\mathrm{ln}\left(\mathrm{2}^{\mathrm{4}} \right)}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=\mathrm{4} \\ $$

Question Number 185073    Answers: 1   Comments: 0

Question Number 185058    Answers: 1   Comments: 0

Question Number 185054    Answers: 1   Comments: 0

If , f(x)= (x/(⌊ x ⌋)) ⇒ D_( f) =?(domain) and R_( f ) = ? (range )

$$ \\ $$$$\:\:\mathrm{I}{f}\:,\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\lfloor\:{x}\:\rfloor}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\mathrm{D}_{\:{f}} \:=?\left({domain}\right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:{f}\:} =\:?\:\left({range}\:\right) \\ $$

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 AB^(⌢) + CD^(⌢) = 307° and AC∙BD = 6(√5) find the area of ABCD.

$${ABCD}\:{is}\:{a}\:{quadrilateral}\:{inscribed}\:{in}\:{a}\:{circle} \\ $$$$\:{if}\:\:\:\:\overset{\frown} {{AB}}\:+\:\overset{\frown} {{CD}}\:=\:\mathrm{307}°\:{and}\:{AC}\centerdot{BD}\:=\:\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{ABCD}.\: \\ $$

Question Number 185038    Answers: 1   Comments: 0

Find the range of values of x for which the series (x/(27))+(x^2 /(125))+...+(x^n /((2n+1)^3 ))+... is absolutely convergent. Help!

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{series}\:\frac{\mathrm{x}}{\mathrm{27}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{125}}+...+\frac{\mathrm{x}^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }+... \\ $$$$\mathrm{is}\:\mathrm{absolutely}\:\mathrm{convergent}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 185036    Answers: 1   Comments: 0

Find the series for cosx. Hence, deduce series sin^2 x and show that, if x is small, ((sin^2 x−x^2 cosx)/x^4 )=(1/6)+(x^2 /(360)) approximately. Help!

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{series}\:\mathrm{for}\:\mathrm{cosx}.\:\mathrm{Hence},\: \\ $$$$\mathrm{deduce}\:\mathrm{series}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{and}\:\mathrm{show}\:\mathrm{that}, \\ $$$$\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{small},\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{cosx}}{\mathrm{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{360}} \\ $$$$\mathrm{approximately}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 185030    Answers: 1   Comments: 1

Question Number 185024    Answers: 1   Comments: 0

prove that : Σ_(k=1) ^n (1/(n+k))<ln(2)

$$\mathrm{prove}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}<\mathrm{ln}\left(\mathrm{2}\right) \\ $$

Question Number 185023    Answers: 1   Comments: 0

provet that : Σ_(k=1) ^n (1/(n+k))>ln(((2n+1)/(n+1)))

$$\mathrm{provet}\:\mathrm{that}\::\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}>\mathrm{ln}\left(\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right) \\ $$

Question Number 185014    Answers: 1   Comments: 1

The relation y=x^2 +kx+c, where K and C are constant passes through the points (−1, −2) and (1, 8) in the coordinate axes. calculate the value of C and K. M.m

$$\mathrm{The}\:\mathrm{relation}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{kx}+\mathrm{c},\:\mathrm{where}\:\mathrm{K} \\ $$$$\mathrm{and}\:\mathrm{C}\:\mathrm{are}\:\mathrm{constant}\:\mathrm{passes}\:\mathrm{through} \\ $$$$\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{1},\:−\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{1},\:\mathrm{8}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{coordinate}\:\mathrm{axes}.\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{C}\:\mathrm{and}\:\mathrm{K}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 185013    Answers: 4   Comments: 3

Question Number 185012    Answers: 0   Comments: 2

Σ_(k=0) ^n sin(kx)=Σ_(k=0) ^n Im(e^(ikx) ) =Im(Σ_(k=0) ^n (e^(ix) )^k ) =Im(((1−e^(i(n+1)x) )/(1−e^(ix) ))) ((1−e^(i(n+1)x) )/(1−e^(ix) ))=((e^(i(n+1)(x/2)) (e^(−i(n+1)(x/2)) −e^(i(n+1)(x/2)) ))/(e^(i(x/2)) (e^(−i(x/2)) −e^(i(x/2)) ))) =e^(in(x/2)) ×((−2sin((n+1)(x/2)))/(−2sin((x/2)))) =e^(in(x/2)) ((sin((n+1)(x/2)))/(sin((x/2)))) Σ_(k=0) ^n sin(kx)=((sin((n+1)(x/2) ))/(sin((x/2))))Im(e^(in(x/2)) ) =((sin((n+1)(x/2)))/(sin((x/2))))sin(((nx)/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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