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Question Number 220540    Answers: 0   Comments: 0

Find: 𝛀 = Σ_(n=1) ^∞ (((−1)^(n+1) )/(n^3 ∙(n + 1)^3 ∙(2n + 1)^2 )) = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{n}^{\mathrm{3}} \centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)^{\mathrm{3}} \centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$

Question Number 220538    Answers: 0   Comments: 0

Question Number 220526    Answers: 2   Comments: 0

Question Number 220519    Answers: 0   Comments: 1

Solve in R ((15)/(x^2 - 3x + 4)) + (7/(x^2 + 7x)) + ((10)/(x^2 + 4x - 21)) + 1 = 0

$$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 220511    Answers: 2   Comments: 0

AB=2CE & DE=2(√3)+4 CE ⊥AB & AD⊥BC & AB=AC & EF ⊥BC BF=?

$${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$

Question Number 220486    Answers: 1   Comments: 2

z ∈ C and λ > 0 Then prove that: ∣z + 2λ∣ + ∣z + λ∣ ≥ ∣z + ((3λ − λ(√3)i)/2)∣

$$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$

Question Number 220480    Answers: 1   Comments: 0

Can you guys teach me about Weber function E_ν (z) and Anger function J_ν (z)?? Let′s Consider n-dimensional Euclidean Space and function f , f;R^n →R Helmholtz Equation defined as (▽^2 +k^2 )f=0 and in 2-dimensional Solution is f(r,θ)=Σ_(ℓ=0) ^∞ (a_ℓ ^ cos(ℓθ)+b_ℓ sin(ℓθ))(c_ℓ ^ J_ℓ (kr)+d_ℓ Y_ℓ (kr)) When I solved this equation I knew from the separation of variables that each of the Bessel functions J_ν (z) and Y_ν (z) comes out as a basis for solution But, When Bessel Equation not equal to 0 in other word x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=(((x−ν)sin(νπ))/π) (A) and x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=−((x+ν+(x−ν)sin(πν))/π) (B) and Each Solution as Follows Solution (A) {Weber}=C_1 J_ν (x)+C_2 Y_ν (x)+E_ν (x) Solution (B) {Anger}=C_1 J_ν (x)+C_2 Y_ν (x)+J_ν (x) I know how the Bessel function works aka J_ν (z) and Y_ν (z) but I don′t know How these two functions (each Weber function and Anger function) work... I′d like to know what its for or is it just a nonlinear differential equation thats been create by these weirdo mathematicians for their intellectual play???

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teach}\:\mathrm{me}\:\mathrm{about} \\ $$$$\mathrm{Weber}\:\mathrm{function}\:\boldsymbol{\mathrm{E}}_{\nu} \left({z}\right)\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\:\boldsymbol{\mathrm{J}}_{\nu} \left({z}\right)?? \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Consider}\:{n}-\mathrm{dimensional}\:\mathrm{Euclidean}\:\mathrm{Space} \\ $$$$\mathrm{and}\:\mathrm{function}\:{f}\:,\:{f};\mathbb{R}^{{n}} \rightarrow\mathbb{R} \\ $$$$\mathrm{Helmholt}{z}\:\mathrm{Equation}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\left(\bigtriangledown^{\mathrm{2}} +{k}^{\mathrm{2}} \right){f}=\mathrm{0}\:\mathrm{and}\:\mathrm{in}\:\:\mathrm{2}-\mathrm{dimensional}\:\mathrm{Solution}\:\mathrm{is} \\ $$$${f}\left({r},\theta\right)=\underset{\ell=\mathrm{0}} {\overset{\infty} {\sum}}\left({a}_{\ell} ^{\:} \mathrm{cos}\left(\ell\theta\right)+{b}_{\ell} \mathrm{sin}\left(\ell\theta\right)\right)\left({c}_{\ell} ^{\:} {J}_{\ell} \left({kr}\right)+{d}_{\ell} {Y}_{\ell} \left({kr}\right)\right) \\ $$$$\: \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{variables}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{functions} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{comes}\:\mathrm{out}\:\mathrm{as}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{for}\:\mathrm{solution} \\ $$$$\mathrm{But}, \\ $$$$\mathrm{When}\:\mathrm{Bessel}\:\mathrm{Equation}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{other}\:\mathrm{word} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=\frac{\left({x}−\nu\right)\mathrm{sin}\left(\nu\pi\right)}{\pi} \\ $$$$\left(\mathrm{A}\right) \\ $$$$\mathrm{and} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=−\frac{{x}+\nu+\left({x}−\nu\right)\mathrm{sin}\left(\pi\nu\right)}{\pi} \\ $$$$\left(\mathrm{B}\right)\: \\ $$$$\mathrm{and}\:\mathrm{Each}\:\mathrm{Solution}\:\mathrm{as}\:\mathrm{Follows} \\ $$$$\mathrm{Solution}\:\left(\mathrm{A}\right)\:\left\{\mathrm{Weber}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{E}}_{\nu} \left({x}\right) \\ $$$$\mathrm{Solution}\:\left(\mathrm{B}\right)\:\left\{\mathrm{Anger}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{J}}_{\nu} \left({x}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{know}\:\mathrm{how}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{works}\:\mathrm{aka}\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{How}\:\mathrm{these}\:\mathrm{two}\:\mathrm{functions} \\ $$$$\left(\mathrm{each}\:\mathrm{Weber}\:\mathrm{function}\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\right)\:\mathrm{work}... \\ $$$$\:\mathrm{I}'\mathrm{d}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{its}\:\mathrm{for}\:\mathrm{or}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{just}\:\mathrm{a}\:\mathrm{nonlinear}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{thats}\:\mathrm{been} \\ $$$$\mathrm{create}\:\mathrm{by}\:\mathrm{these}\:\mathrm{weirdo}\:\mathrm{mathematicians}\:\mathrm{for}\:\mathrm{their} \\ $$$$\mathrm{intellectual}\:\mathrm{play}??? \\ $$

Question Number 220474    Answers: 0   Comments: 1

Question Number 220473    Answers: 0   Comments: 0

lim_(n→∞) (tan((π/4)+(1/n)))^n →^(t=(1/n)) =lim_(t→0) [tan((π/4)+t)]^(1/t) ⇒lim_(t→0) [tan((π/4)+t)−1]×(1/t) =lim_(t→0) (((1+tant)/(1−tant))−1)×(1/t) =lim_(t→0) (((2tant)/(1−tant)))×(1/t)=2 ⇒Ans=e^2 ✓

$${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Question Number 220469    Answers: 0   Comments: 0

∫_0 ^( 1) ((6x(1 − x))/((x + 1)(x^2 + 1) ln (x + 1))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$

Question Number 220468    Answers: 3   Comments: 0

a+b+c=1, a^2 +b^2 +c^2 =1 (a,b,c ∈R) a^(10) +b^(10) +c^(10) =1, a^4 +b^4 +c^4 =?

$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$

Question Number 220459    Answers: 3   Comments: 0

find Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=? (a∈R)

$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=?\:\:\:\:\:\:\:\:\:\:\left({a}\in{R}\right) \\ $$

Question Number 220457    Answers: 0   Comments: 0

Find: 𝛀 = ∫_0 ^( ∞) ∫_0 ^( (𝛑/2)) (((x+1)sin^2 (x)ln(y^3 +1))/(xy(y^2 +1))) dxdy = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{xy}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{dxdy}\:=\:? \\ $$

Question Number 220456    Answers: 0   Comments: 0

a, b, c, d ≥ 1 ; a + b + c = d show that; ab + bc + ca + (1/a) + (1/b) + (1/c) ≥ 2d − 3 + (9/d)

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{a},\:{b},\:{c},\:{d}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:;\:\:\:{a}\:+\:{b}\:+\:{c}\:=\:{d} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that};\: \\ $$$$\:\:\:{ab}\:+\:{bc}\:+\:{ca}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:\geqslant\:\mathrm{2}{d}\:−\:\mathrm{3}\:+\:\frac{\mathrm{9}}{{d}}\:\:\:\: \\ $$

Question Number 220454    Answers: 0   Comments: 0

In the CO_2 molecule, each oxygen atom forms a double bond with central carbon atom. Given that the Bohr radius (a_o ) is approximately 0.529 A^o and the experimental C=O bond length is about 1.16 A^(o ) , calculate : a) The approximate region (in the terms of Bohr radii) where the shared electrons are most likely to be found between C and O. b) Compare the calculated bond region to the sum of the covalent radii of carbon and oxygen, and comment on the effect of π bonding on the contraction of bond length.

$$\:\mathrm{In}\:\mathrm{the}\:{CO}_{\mathrm{2}} \:\mathrm{molecule},\:\mathrm{each}\:\mathrm{oxygen}\:\mathrm{atom}\:\mathrm{forms}\:\mathrm{a}\:\mathrm{double}\:\mathrm{bond}\:\mathrm{with}\:\mathrm{central}\:\mathrm{carbon}\:\mathrm{atom}. \\ $$$$\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{Bohr}\:\mathrm{radius}\:\left({a}_{{o}} \right)\:\mathrm{is}\:\mathrm{approximately}\:\mathrm{0}.\mathrm{529}\:\overset{\mathrm{o}} {\mathrm{A}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{experimental}\:{C}={O}\: \\ $$$$\:\:\mathrm{bond}\:\mathrm{length}\:\mathrm{is}\:\mathrm{about}\:\mathrm{1}.\mathrm{16}\:\overset{\mathrm{o}\:} {\mathrm{A}},\:\mathrm{calculate}\:: \\ $$$$\: \\ $$$$\left.\:\mathrm{a}\right)\:\mathrm{The}\:\mathrm{approximate}\:\mathrm{region}\:\left(\mathrm{in}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{Bohr}\:\mathrm{radii}\right)\:\mathrm{where}\:\mathrm{the}\:\mathrm{shared}\:\mathrm{electrons}\:\mathrm{are} \\ $$$$\:\mathrm{most}\:\mathrm{likely}\:\mathrm{to}\:\mathrm{be}\:\mathrm{found}\:\mathrm{between}\:{C}\:\mathrm{and}\:{O}. \\ $$$$\left.\:\mathrm{b}\right)\:\mathrm{Compare}\:\mathrm{the}\:\mathrm{calculated}\:\mathrm{bond}\:\mathrm{region}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{covalent}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{carbon}\:\mathrm{and} \\ $$$$\:\mathrm{oxygen},\:\mathrm{and}\:\mathrm{comment}\:\mathrm{on}\:\mathrm{the}\:\mathrm{effect}\:\mathrm{of}\:\pi\:\mathrm{bonding}\:\mathrm{on}\:\mathrm{the}\:\mathrm{contraction}\:\mathrm{of}\:\mathrm{bond}\:\mathrm{length}. \\ $$

Question Number 220447    Answers: 0   Comments: 1

Question Number 220499    Answers: 3   Comments: 1

Question Number 220405    Answers: 2   Comments: 0

9^x^2 −3^(x+1) =0

$$\mathrm{9}^{{x}^{\mathrm{2}} } −\mathrm{3}^{{x}+\mathrm{1}} =\mathrm{0} \\ $$

Question Number 220403    Answers: 1   Comments: 0

Question Number 220493    Answers: 0   Comments: 1

Let ABC be a triangle such CosA+cosB+cosC=(√2) SinA+sinB+sinC=1+(√2) Find A,B,C

$${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such} \\ $$$${CosA}+{cosB}+{cosC}=\sqrt{\mathrm{2}} \\ $$$${SinA}+{sinB}+{sinC}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${Find}\:{A},{B},{C} \\ $$

Question Number 220502    Answers: 1   Comments: 0

each J_ν (z),Y_ν (z) are linear independent....?? W_(Ronskian) {J_ν ^ (z),Y_ν (z)}= determinant (((J_ν (z)),( Y_ν (z))),((J_ν ′(z)),(Y_ν ′(z)))) =J_ν ^((1)) (z)Y_ν (z)−J_ν (z)Y_ν ^((1)) (z) J_ν ^((1)) (z)Y_ν (z)=J_(ν−1) (z)Y_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_ν (z)Y_ν ^((1)) (z)=Y_(ν−1) (z)J_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_(ν−1) (z)Y_ν (z)−J_ν (z)Y_(ν−1) (z).... .....damn..... Result is (2/(πz)) ......

$$\mathrm{each}\:{J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\mathrm{linear}\:\mathrm{independent}....?? \\ $$$${W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\begin{vmatrix}{{J}_{\nu} \left({z}\right)}&{\:{Y}_{\nu} \left({z}\right)}\\{{J}_{\nu} '\left({z}\right)}&{{Y}_{\nu} '\left({z}\right)}\end{vmatrix} \\ $$$$={J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right) \\ $$$${J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)={J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right)={Y}_{\nu−\mathrm{1}} \left({z}\right){J}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right).... \\ $$$$.....\mathrm{damn}..... \\ $$$$\mathrm{Result}\:\mathrm{is}\:\frac{\mathrm{2}}{\pi{z}}\:...... \\ $$

Question Number 220495    Answers: 0   Comments: 5

Question Number 220396    Answers: 0   Comments: 2

Question Number 220395    Answers: 2   Comments: 0

Find: Ω =Σ_(x=1) ^∞ Σ_(y=1) ^∞ (1/(x^2 y^3 (x^2 + 1)(y + 2))) = ?

$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\:=\:? \\ $$

Question Number 220393    Answers: 1   Comments: 1

Question Number 220391    Answers: 3   Comments: 0

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