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Question Number 220375 Answers: 1 Comments: 2

Question Number 220366 Answers: 2 Comments: 0

solve the system of equation using gaussian elimination method x+2y+3z=10 2x−3y+z=1 3x+y−2z=9

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{gaussian}}\:\boldsymbol{\mathrm{elimination}}\:\boldsymbol{\mathrm{method}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{10} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{9} \\ $$

Question Number 220365 Answers: 3 Comments: 0

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Question Number 220377 Answers: 1 Comments: 0

Prove equation ∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=(1/(2πi)) ∫_(−∞i+𝛄) ^( +∞i+𝛄) F(u)G(u−ρ)du F(u)=∫_0 ^( ∞) f(t)e^(−ut) dt G(u)=∫_0 ^( ∞) g(t)e^(−ut) dt

$$\mathrm{Prove}\:\mathrm{equation} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$${F}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$$${G}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$

Question Number 220340 Answers: 2 Comments: 0

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Question Number 220353 Answers: 1 Comments: 3

Question Number 220307 Answers: 1 Comments: 0

Question Number 220286 Answers: 3 Comments: 5

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Question Number 220269 Answers: 1 Comments: 0

lim_(t→0) ((C_1 J_ν (t)+C_2 Y_ν (t)+H_ν (t))/(C_1 J_ν (t)+C_2 Y_ν (t)))=?? ν∈R J_ν (z) Bessel function First kind Y_ν (z) Bessel function Second Kind H_ν (z) Struve H function

$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)+\boldsymbol{\mathrm{H}}_{\nu} \left({t}\right)}{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)}=?? \\ $$$$\nu\in\mathbb{R} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{First}\:\mathrm{kind} \\ $$$${Y}_{\nu} \left({z}\right)\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{Second}\:\mathrm{Kind} \\ $$$$\boldsymbol{\mathrm{H}}_{\nu} \left({z}\right)\:\mathrm{Struve}\:\mathrm{H}\:\mathrm{function} \\ $$

Question Number 220266 Answers: 1 Comments: 0

2^a + 2^b + 2^c = 148

$$\mathrm{2}^{\mathrm{a}} \:\:+\:\:\mathrm{2}^{\mathrm{b}} \:\:+\:\:\mathrm{2}^{\mathrm{c}} \:\:=\:\:\mathrm{148} \\ $$

Question Number 220264 Answers: 1 Comments: 0

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Question Number 220262 Answers: 7 Comments: 0

Question Number 220257 Answers: 2 Comments: 0

proof that volume of frustum of circular cone is (1/3)h[A1+A2+(√(A1A2)) A_1 and A_2 are areas of base

$${proof}\:{that}\:{volume}\:{of}\:{frustum}\:{of} \\ $$$$\:{circular}\:{cone}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left[{A}\mathrm{1}+{A}\mathrm{2}+\sqrt{{A}\mathrm{1}{A}\mathrm{2}}\right. \\ $$$${A}_{\mathrm{1}} {and}\:{A}_{\mathrm{2}} \:{are}\:\:{areas}\:{of}\:{base} \\ $$

Question Number 220253 Answers: 0 Comments: 0

Question Number 220250 Answers: 3 Comments: 0

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