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Question Number 158907 Answers: 1 Comments: 0

Question Number 158906 Answers: 1 Comments: 0

determiner le reste de la division eucludienne de: 10^(100) par 105

$${determiner}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{eucludienne}\:{de}: \\ $$$$\mathrm{10}^{\mathrm{100}} \:{par}\:\mathrm{105} \\ $$

Question Number 158903 Answers: 0 Comments: 0

# solve # Φ:=∫_(−∞) ^( ∞) (( xsin(x))/(( 2+ x +x^( 2) )^( 2) )) dx =? −−−−−−−−

$$ \\ $$$$\:\:\:#\:{solve}\:# \\ $$$$\:\:\:\:\Phi:=\int_{−\infty} ^{\:\infty} \frac{\:{xsin}\left({x}\right)}{\left(\:\mathrm{2}+\:{x}\:+{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=? \\ $$$$−−−−−−−− \\ $$

Question Number 158902 Answers: 0 Comments: 0

nice mathematics # calculate # Ω := Σ_(n=0) ^∞ (( 1)/(( 6n + 1 )^( 3) )) = ? −−−−−−−−−−−−

$$ \\ $$$$\:\:\:\:\:\:{nice}\:\:{mathematics} \\ $$$$\:\:\:\:\:\:\:#\:{calculate}\:# \\ $$$$\:\:\:\:\:\:\:\:\Omega\::=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{\left(\:\mathrm{6}{n}\:+\:\mathrm{1}\:\right)^{\:\mathrm{3}} }\:=\:? \\ $$$$\:\:\:\:\:\:−−−−−−−−−−−− \\ $$$$ \\ $$

Question Number 158899 Answers: 1 Comments: 0

(1/2)+(3/4)=

$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}= \\ $$

Question Number 158894 Answers: 0 Comments: 1

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Question Number 158884 Answers: 1 Comments: 0

Question Number 158883 Answers: 1 Comments: 4

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The roots of the equation 2x^2 +px+p=0 are 2α+β and α+2β. Calculate the value of p

$$\:{The}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\:\mathrm{2}{x}^{\mathrm{2}} +{px}+{p}=\mathrm{0}\:{are}\:\mathrm{2}\alpha+\beta\:{and} \\ $$$$\:\alpha+\mathrm{2}\beta.\:{Calculate}\:{the}\:{value}\:{of}\:{p} \\ $$

Question Number 158862 Answers: 1 Comments: 0

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Question Number 158858 Answers: 0 Comments: 0

I_n =∫_(−1) ^1 (1−x^2 )^n cos ((a/(2b))x)dx to integrating by piece for n≥2 proven (a^2 /(4b^2 ))I_(n ) =2n(2n−1)I_(n−1) −4(n−1)I_(n−2) proven by rearring that ((a/(2b)))^(2n+1) I_n =n![p((q/(2b)))sin ((a/(2b)))+Q((a/(2b)))cos ((a/(2b)))]

$${I}_{{n}} =\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} \mathrm{cos}\:\left(\frac{{a}}{\mathrm{2}{b}}{x}\right){dx} \\ $$$${to}\:{integrating}\:{by}\:{piece}\:{for}\:{n}\geqslant\mathrm{2}\: \\ $$$${proven}\: \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }{I}_{{n}\:} =\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right){I}_{{n}−\mathrm{1}} −\mathrm{4}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$${proven}\:{by}\:{rearring}\:{that}\: \\ $$$$\left(\frac{{a}}{\mathrm{2}{b}}\right)^{\mathrm{2}{n}+\mathrm{1}} {I}_{{n}} ={n}!\left[{p}\left(\frac{{q}}{\mathrm{2}{b}}\right)\mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}{b}}\right)+{Q}\left(\frac{{a}}{\mathrm{2}{b}}\right)\mathrm{cos}\:\left(\frac{{a}}{\mathrm{2}{b}}\right)\right] \\ $$

Question Number 158855 Answers: 1 Comments: 0

Resolve the system d′ unknow (x, y,z) ∈ ⊂^3 x+y+z=1 x^2 +y^2 +z^2 =1 x^3 +y^3 +z^3 =−5

$${Resolve}\:{the}\:{system}\:{d}'\:{unknow}\:\:\left({x},\:{y},{z}\right)\:\in\:\subset^{\mathrm{3}} \\ $$$${x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =−\mathrm{5} \\ $$

Question Number 158847 Answers: 1 Comments: 0