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Question Number 218494    Answers: 4   Comments: 0

Question Number 218493    Answers: 2   Comments: 0

Question Number 218491    Answers: 5   Comments: 1

Question Number 218485    Answers: 1   Comments: 0

solve the equation 1) X^6 −1=0 2) X^4 +X^2 +1=0

$${solve}\:{the}\:{equation} \\ $$$$\left.\mathrm{1}\right)\:\:\:{X}^{\mathrm{6}} −\mathrm{1}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\:{X}^{\mathrm{4}} +{X}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$

Question Number 218482    Answers: 0   Comments: 1

solve for real x; _( 3(√(((x^5 −5x^4 +10x^3 −10x^2 +5x)/(x^2 −2x+1))+(√(x^4 +4x^2 +4))))=(√(3 )) )

$$ \\ $$$$\:\:\:\:{solve}\:{for}\:{real}\:\boldsymbol{{x}}; \\ $$$$ \\ $$$$\underset{\:\:\:\mathrm{3}\sqrt{\frac{\boldsymbol{{x}}^{\mathrm{5}} −\mathrm{5}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{10}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{10}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}}}=\sqrt{\mathrm{3}\:}\:\:\:} {\:} \\ $$$$ \\ $$

Question Number 218476    Answers: 1   Comments: 0

Question Number 218475    Answers: 5   Comments: 0

Question Number 218474    Answers: 1   Comments: 0

solve for real x (√(x+(√(x^2 +1)))) = (√2)

$$ \\ $$$$\:\:\:\:{solve}\:{for}\:{real}\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:=\:\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Question Number 218473    Answers: 2   Comments: 0

Solve the following equation for the real x value; x^4 −4x^3 +6x^2 −4x+2 = (√(2x^4 −8x^3 +12x^2 −8x+5))

$$ \\ $$$$\:{Solve}\:{the}\:{following}\:{equation}\:{for}\:{the}\:{real}\:\boldsymbol{{x}}\:{value}; \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:=\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{5}}\:\:\:\: \\ $$$$ \\ $$

Question Number 218462    Answers: 2   Comments: 0

∫(dx/( (√(2x−x^2 +3))))

$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$

Question Number 218459    Answers: 0   Comments: 2

F^→ (x,y)=−(1/2)ye_1 ^→ +(1/2)xe_2 ^→ ▽^→ ×F^→ (x,y)= determinant ((( e_1 ^→ ),e_2 ^→ ,e_3 ^→ ),(( ∂_x ),( ∂_y ),∂_z ),((−(1/2)y),((1/2)x),0))=0e_1 ^→ −0e_2 ^→ +((1/2)−(−(1/2)))e_3 ^→ ∮_( C) F^→ ∙dl=∮_( C) −(1/2)ydx+(1/2)xdx=∫∫_( R^2 ) e_3 ^→ ∙n^→ dS ∴∮_( C) −(1/2)ydx+(1/2)xdx=∫∫_( R^2 ) dS.....is right...??

$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=\begin{vmatrix}{\:\:\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} }\\{\:\:\:\:\:\partial_{{x}} }&{\:\partial_{{y}} }&{\partial_{{z}} }\\{−\frac{\mathrm{1}}{\mathrm{2}}{y}}&{\frac{\mathrm{1}}{\mathrm{2}}{x}}&{\mathrm{0}}\end{vmatrix}=\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\oint_{\:{C}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \centerdot\overset{\rightarrow} {\boldsymbol{\mathrm{n}}}\:\mathrm{dS} \\ $$$$\therefore\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \:\mathrm{dS}.....\mathrm{is}\:\mathrm{right}...?? \\ $$

Question Number 218456    Answers: 5   Comments: 0

Question Number 218445    Answers: 3   Comments: 0

there are 100 students in a school. it is found out that each student should select at least 4 courses, so that no two students have the same selection. how many different courses does the school offer?

$${there}\:{are}\:\mathrm{100}\:{students}\:{in}\:{a}\:{school}. \\ $$$${it}\:{is}\:{found}\:{out}\:{that}\:{each}\:{student}\: \\ $$$${should}\:{select}\:{at}\:{least}\:\mathrm{4}\:{courses},\:{so}\: \\ $$$${that}\:{no}\:{two}\:{students}\:{have}\:{the}\:{same}\: \\ $$$${selection}.\: \\ $$$${how}\:{many}\:{different}\:{courses}\:{does}\: \\ $$$${the}\:{school}\:{offer}? \\ $$

Question Number 218438    Answers: 1   Comments: 0

An amazing thing i saw S = 1 + 2 + 3 + 4 + 5 + 6... = 1 + 2(2/2 + 3/2 + 4/2 + 5/2 +6/2....) = 1 + 2(1 + 3/2 + 2 + 5/2 + 3...) = 1 + 2(1+ 2 + 3 ... + 3/2 + 5/2...) = 1 + 2S + 2Σ_(n= 1) ^∞ ((2n + 1)/2) or,S − 2S = 1 + Σ_(n=1) ^∞ 2n + 1 ∴ −S = Σ_(n=0) ^∞ 2n + 1 Sum of all odd numbers! I know the step S−2S = −S is not allowed

$${An}\:{amazing}\:{thing}\:{i}\:{saw} \\ $$$${S}\:=\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:\mathrm{5}\:+\:\mathrm{6}... \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{2}/\mathrm{2}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{4}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\mathrm{6}/\mathrm{2}....\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\:\mathrm{3}...\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}+\:\mathrm{2}\:+\:\mathrm{3}\:...\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}...\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}{S}\:+\:\mathrm{2}\underset{{n}=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$${or},{S}\:−\:\mathrm{2}{S}\:=\:\mathrm{1}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$$ \\ $$$$\therefore\:−{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$${Sum}\:{of}\:{all}\:{odd}\:{numbers}! \\ $$$${I}\:{know}\:{the}\:{step}\:{S}−\mathrm{2}{S}\:=\:−{S}\:{is}\:{not}\:{allowed} \\ $$

Question Number 218437    Answers: 0   Comments: 0

Question Number 218428    Answers: 3   Comments: 0

Question Number 218427    Answers: 1   Comments: 0

give a recurrence relation for I_n . I_n =∫_0 ^1 (1/((1+x^2 )^n ))dx, ∀n ∈ N.

$$ \\ $$$${give}\:{a}\:{recurrence}\:{relation}\:{for}\:{I}_{{n}} . \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx},\:\forall{n}\:\in\:\mathbb{N}. \\ $$

Question Number 218421    Answers: 1   Comments: 3

prove ∫_( ∂𝚺) 𝛚=∫_( 𝚺 ) d𝛚

$$\mathrm{prove}\: \\ $$$$\int_{\:\partial\boldsymbol{\Sigma}} \:\boldsymbol{\omega}=\int_{\:\boldsymbol{\Sigma}\:} \mathrm{d}\boldsymbol{\omega} \\ $$

Question Number 218420    Answers: 0   Comments: 0

Question Number 218418    Answers: 1   Comments: 0

Question Number 218411    Answers: 2   Comments: 0

Question Number 218410    Answers: 3   Comments: 0

Question Number 218400    Answers: 4   Comments: 0

Hard problem..... prove. for all α∈Z α^(37) ≡α Mod(1729) pls help :(

$$\mathrm{Hard}\:\mathrm{problem}..... \\ $$$$\:\mathrm{prove}. \\ $$$$\:\mathrm{for}\:\mathrm{all}\:\alpha\in\mathbb{Z} \\ $$$$\alpha^{\mathrm{37}} \equiv\alpha\:\mathrm{Mod}\left(\mathrm{1729}\right) \\ $$$$\mathrm{pls}\:\mathrm{help}\::\left(\right. \\ $$

Question Number 218399    Answers: 1   Comments: 0

Question Number 218398    Answers: 1   Comments: 0

Question Number 218397    Answers: 0   Comments: 0

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