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Question Number 219619    Answers: 2   Comments: 0

∫_0 ^∞ (x^2 +1)^(−1/2) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$ \\ $$

Question Number 219618    Answers: 0   Comments: 0

Prove; ∫_0 ^( 1) ((ln ln (1/x))/((1+x)^2 )) dx = (1/2)(ln((π/2))−γ)

$$ \\ $$$$\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{{ln}\:{ln}\:\frac{\mathrm{1}}{{x}}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{\pi}{\mathrm{2}}\right)−\gamma\right) \\ $$$$ \\ $$

Question Number 219617    Answers: 0   Comments: 0

ok, let′s all answer questions from anywhere on the www using the given results from the sources or wolframalpha or any AI available. this promises great fun!

$$\mathrm{ok},\:\mathrm{let}'\mathrm{s}\:\mathrm{all}\:\mathrm{answer}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{anywhere} \\ $$$$\mathrm{on}\:\mathrm{the}\:{www}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{results}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{sources}\:\mathrm{or}\:{wolframalpha}\:\mathrm{or}\:\mathrm{any}\:\mathrm{AI} \\ $$$$\mathrm{available}.\:\mathrm{this}\:\mathrm{promises}\:\mathrm{great}\:\mathrm{fun}! \\ $$

Question Number 219606    Answers: 2   Comments: 0

prove that for positive real numbers a,b,c, the following inequality holds; (a^2 /(b + c)) + (b^2 /(c + a)) + (c^2 /(a + b)) ≥ ((a + b + c)/2)

$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 219602    Answers: 1   Comments: 2

∫_0 ^( ∞) dt e^(−ikt) J_(−(2/3)) (t)−∫_0 ^( ∞) dt e^(−ikt) Y_(−(2/3)) (t)=??

$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {J}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)−\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {Y}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)=?? \\ $$

Question Number 222520    Answers: 0   Comments: 3

Question Number 222519    Answers: 0   Comments: 0

Prove: ∫_0 ^(+∞) (√(cosh x))−(√(sinh x))dx=((2−(√2))/( (√π)))Γ^2 ((3/4))

$$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{+\infty} \sqrt{\mathrm{cosh}\:{x}}−\sqrt{\mathrm{sinh}\:{x}}{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\pi}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$

Question Number 219597    Answers: 1   Comments: 0

Evaluate integral by Complex integral method ∫_0 ^( 2π) (1/(a+b∙cos(nθ))) dθ

$$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$

Question Number 219591    Answers: 0   Comments: 0

LT{((Ai^((1)) (−((3/2))^(2/3) z^(2/3) )+(√3)Bi^((1)) (−((3/2))^(2/3) z^(2/3) )/(^3 (√2)∙^6 (√3)z^(2/3) ))}=??? LT{∗}=∫_0 ^( ∞) e^(−zt) ∗ Ai(x) and Bi(x) Airy Function f^((1)) (z) is ((d )/dz)f(z)

$$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\ $$$$\mathrm{Ai}\left({x}\right)\:\mathrm{and}\:\mathrm{Bi}\left({x}\right)\:\mathrm{Airy}\:\mathrm{Function} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({z}\right)\:\mathrm{is}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}{f}\left({z}\right) \\ $$

Question Number 219589    Answers: 0   Comments: 2

Evaluate; L(tan^(−1) (t−(1/t))) solution; ⇒F(s)= L(tan^(−1) (t−(1/t))) ⇔ sF(s)+(π/2)=L(((t^2 +1)/(t^4 −t^2 +1)))(s) ⇒ sF(s)+(π/2)=L(((1/2)/(t^2 −(√3) t +1))+((1/2)/(t^2 −(√3) t+1)))(s) ⇒ sF(s)+(π/2)=(1/2)L((1/((t−((√3)/2))^2 +(1/4))))(s)+(1/2)L((1/((t+((√3)/2))^2 +(1/4))))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s)+(1/2)e^(((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) (1/(1/2)) sin ((1/2) s)∗(1/s)+(1/2)e^(((√3)/2) s) (1/(1/2))sin((1/2) s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) L(sin((1/2) t))(s)∗(1/s)+e^(((√3)/2) s) L(sin((1/2) t))(s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/(2 ))s) ((1/2)/(s^2 +1/4))∗(1/s) Final Answer; F(s)=(1/s)(e^(−((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s))−(π/(2s))

$${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\:\:\:\:\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:{F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\left({e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\right)−\frac{\pi}{\mathrm{2}{s}}\:\: \\ $$$$ \\ $$

Question Number 219586    Answers: 1   Comments: 0

Integrate : ∫(x^2 /( (√(x^2 − x)) + 1))dx.

$${Integrate}\:: \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:−\:{x}}\:+\:\mathrm{1}}{dx}. \\ $$

Question Number 219587    Answers: 1   Comments: 1

Question Number 219600    Answers: 1   Comments: 0

∫_( 1) ^( 2) [x]^x dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \left[{x}\right]^{{x}} \:{dx} \\ $$$$ \\ $$

Question Number 219581    Answers: 0   Comments: 0

Question Number 219580    Answers: 0   Comments: 0

∫_0 ^( ∞) J_ν (kt)e^(−t) dt

$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({kt}\right){e}^{−{t}} \:\mathrm{d}{t} \\ $$

Question Number 219579    Answers: 0   Comments: 0

∫_0 ^( ∞) sin(t)J_ν (kt) dt ∫_0 ^( ∞) t∙J_ν (at)J_ν (kt) dt

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\mathrm{sin}\left({t}\right){J}_{\nu} \left({kt}\right)\:\mathrm{d}{t} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:{t}\centerdot{J}_{\nu} \left({at}\right){J}_{\nu} \left({kt}\right)\:\mathrm{d}{t} \\ $$

Question Number 219577    Answers: 0   Comments: 1

prove ∫_0 ^( 1) ∫_0 ^( 1) ...∫_0 ^( 1) _(n times) x_1 ^α x_2 ^α ....x_n ^α ln(x_1 )ln(x_2 )....ln(x_n )dx_1 dx_2 ..dx_n =^(Equal) (((−)^n )/((α+1)^(2n) ))

$$\mathrm{prove} \\ $$$$\underset{{n}\:\mathrm{times}} {\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} ...\int_{\mathrm{0}} ^{\:\mathrm{1}} }\:\:{x}_{\mathrm{1}} ^{\alpha} {x}_{\mathrm{2}} ^{\alpha} ....{x}_{{n}} ^{\alpha} \mathrm{ln}\left({x}_{\mathrm{1}} \right)\mathrm{ln}\left({x}_{\mathrm{2}} \right)....\mathrm{ln}\left({x}_{{n}} \right)\mathrm{d}{x}_{\mathrm{1}} \mathrm{d}{x}_{\mathrm{2}} ..\mathrm{d}{x}_{{n}} \\ $$$$\overset{\mathrm{Equal}} {=}\:\:\:\frac{\left(−\right)^{{n}} }{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}{n}} } \\ $$

Question Number 219574    Answers: 0   Comments: 0

∫_0 ^( ∞) ((sin(z))/(z(z^2 +4))) dz=(1/4)∫_0 ^( ∞) (((sin(z))/z)−((sin(z))/(2z+4i))−((sin(z))/(2z−4i))) dz and next....??? 2πiΣ_(j=1) ^M Res_(h=a_j ) {f(h)}....

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({z}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{4}\right)}\:\mathrm{d}{z}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \:\:\left(\frac{\mathrm{sin}\left({z}\right)}{{z}}−\frac{\mathrm{sin}\left({z}\right)}{\mathrm{2}{z}+\mathrm{4}\boldsymbol{{i}}}−\frac{\mathrm{sin}\left({z}\right)}{\mathrm{2}{z}−\mathrm{4}\boldsymbol{{i}}}\right)\:\mathrm{d}{z} \\ $$$$\mathrm{and}\:\mathrm{next}....??? \\ $$$$\mathrm{2}\pi\boldsymbol{{i}}\underset{{j}=\mathrm{1}} {\overset{{M}} {\sum}}\:\:\mathrm{Res}_{{h}={a}_{{j}} } \left\{{f}\left({h}\right)\right\}.... \\ $$

Question Number 219571    Answers: 1   Comments: 0

pls Help.....! prove ∫∫_( S) g^→ ∙dS^→ =0 ⇆ div g^→ =0

$$\mathrm{pls}\:\mathrm{Help}.....! \\ $$$$\mathrm{prove} \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{g}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}=\mathrm{0}\:\leftrightarrows\:\mathrm{div}\:\overset{\rightarrow} {\boldsymbol{\mathrm{g}}}=\mathrm{0} \\ $$

Question Number 219570    Answers: 2   Comments: 0

8x^2 +9y^2 =36(x+y), x,y∈R, find maximum (x+y)

$$\:\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} =\mathrm{36}\left({x}+{y}\right),\: \\ $$$$\:\:{x},{y}\in{R},\:{find}\:{maximum}\:\left({x}+{y}\right) \\ $$$$\: \\ $$

Question Number 219564    Answers: 0   Comments: 0

solve for real numbers: { ((((xy(x^3 −y^3 )+yz(y^3 −z^3 )+zx(z^3 −x^3 ))/(xy(x−y)+yz(y−z)+zx(z−x))) = 55)),((x^3 + y^3 + z^3 = 99)),((((xy(x^3 −y^3 )+yz(y^3 −z^3 )+zx(z^3 −x^3 ))/(xy(x^2 −y^2 )+yz(y^2 −z^2 )+zx(z^2 −y^2 ))) = ((55)/9))) :}

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{xy}\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{yz}\left(\mathrm{y}−\mathrm{z}\right)+\mathrm{zx}\left(\mathrm{z}−\mathrm{x}\right)}\:=\:\mathrm{55}}\\{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{99}}\\{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{xy}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{55}}{\mathrm{9}}}\end{cases} \\ $$

Question Number 219563    Answers: 2   Comments: 0

find all n ∈ N^∗ such that ∫_0 ^( 1) (sinx)^(2n−2) ∙ (cosx)^(2n) dx ≥ (1/4^(1011) )

$$\mathrm{find}\:\mathrm{all}\:\:\:\mathrm{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{sinx}\right)^{\mathrm{2n}−\mathrm{2}} \:\centerdot\:\left(\mathrm{cosx}\right)^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$

Question Number 219562    Answers: 1   Comments: 0

prove that exists X ∈ M_(2,3) (R) Y ∈ M_(3,2) (R) such that X∙Y = ((1,1),(1,1) ) Y∙X = ((2,6,6),(3,9,9),((-3),(-9),(-9)) )

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{exists}\:\:\:\mathrm{X}\:\in\:\mathrm{M}_{\mathrm{2},\mathrm{3}} \:\left(\mathbb{R}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Y}\:\in\:\mathrm{M}_{\mathrm{3},\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\:\mathrm{X}\centerdot\mathrm{Y}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Y}\centerdot\mathrm{X}\:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{6}}&{\mathrm{6}}\\{\mathrm{3}}&{\mathrm{9}}&{\mathrm{9}}\\{-\mathrm{3}}&{-\mathrm{9}}&{-\mathrm{9}}\end{pmatrix}\: \\ $$

Question Number 219561    Answers: 0   Comments: 3

let be the sequence (x_n )n ≥ 1 defined by x_1 = 1 x_(n+2) = 3x_(n+1) − x_n ∀n ∈ N find L =lim_(n→∞) ((Σ_(k=0) ^0 (x_(2k+1) /(x_k + x_(k+1) ))))^(1/n) = ?

$$\mathrm{let}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sequence}\:\:\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)\mathrm{n}\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\:\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}+\mathrm{2}} \:=\:\mathrm{3x}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} −\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$$$\forall\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{find}\:\:\:\boldsymbol{\mathrm{L}}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\mathrm{0}} {\sum}}\:\:\frac{\mathrm{x}_{\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}} }{\mathrm{x}_{\boldsymbol{\mathrm{k}}} \:+\:\mathrm{x}_{\boldsymbol{\mathrm{k}}+\mathrm{1}} }}\:=\:? \\ $$

Question Number 219560    Answers: 0   Comments: 0

Suppose that an urn contains 100,000 marbles and 120 are red . If a random sample of 1000 is drawn, what are the probabilities that 0,1,2,3 and 4 respectively will be red. What is the mean and variance?

Suppose that an urn contains 100,000 marbles and 120 are red . If a random sample of 1000 is drawn, what are the probabilities that 0,1,2,3 and 4 respectively will be red. What is the mean and variance?

Question Number 219556    Answers: 0   Comments: 0

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