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Question Number 220396 Answers: 0 Comments: 2
Question Number 220395 Answers: 2 Comments: 0
$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\:=\:? \\ $$
Question Number 220393 Answers: 1 Comments: 1
Question Number 220391 Answers: 3 Comments: 0
Question Number 220390 Answers: 1 Comments: 0
$$\mathrm{sin}\theta\:+\:\mathrm{sin}\left(\pi\:+\:\theta\right)\:+\:\mathrm{sin}\left(\mathrm{2}\pi\:+\:\theta\right)\:+\:...\: \\ $$$$+\:\mathrm{sin}\left({n}\pi\:+\:\theta\right)\:=\:?\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{integer}. \\ $$
Question Number 220388 Answers: 0 Comments: 1
Question Number 220380 Answers: 3 Comments: 0
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\left[\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right]^{{n}} =? \\ $$
Question Number 220378 Answers: 0 Comments: 0
Question Number 220375 Answers: 1 Comments: 2
Question Number 220366 Answers: 2 Comments: 0
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{gaussian}}\:\boldsymbol{\mathrm{elimination}}\:\boldsymbol{\mathrm{method}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{10} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{9} \\ $$
Question Number 220365 Answers: 3 Comments: 0
Question Number 220362 Answers: 0 Comments: 0
Question Number 220377 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{equation} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$${F}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$$${G}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$
Question Number 220340 Answers: 2 Comments: 0
Question Number 220320 Answers: 1 Comments: 3
Question Number 220353 Answers: 1 Comments: 3
Question Number 220307 Answers: 1 Comments: 0
Question Number 220286 Answers: 3 Comments: 5
Question Number 220278 Answers: 0 Comments: 0
Question Number 220269 Answers: 1 Comments: 0
$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)+\boldsymbol{\mathrm{H}}_{\nu} \left({t}\right)}{{C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right)}=?? \\ $$$$\nu\in\mathbb{R} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{First}\:\mathrm{kind} \\ $$$${Y}_{\nu} \left({z}\right)\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{Second}\:\mathrm{Kind} \\ $$$$\boldsymbol{\mathrm{H}}_{\nu} \left({z}\right)\:\mathrm{Struve}\:\mathrm{H}\:\mathrm{function} \\ $$
Question Number 220266 Answers: 1 Comments: 0
$$\mathrm{2}^{\mathrm{a}} \:\:+\:\:\mathrm{2}^{\mathrm{b}} \:\:+\:\:\mathrm{2}^{\mathrm{c}} \:\:=\:\:\mathrm{148} \\ $$
Question Number 220264 Answers: 1 Comments: 0
Question Number 220263 Answers: 3 Comments: 0
Question Number 220262 Answers: 7 Comments: 0
Question Number 220257 Answers: 2 Comments: 0
$${proof}\:{that}\:{volume}\:{of}\:{frustum}\:{of} \\ $$$$\:{circular}\:{cone}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left[{A}\mathrm{1}+{A}\mathrm{2}+\sqrt{{A}\mathrm{1}{A}\mathrm{2}}\right. \\ $$$${A}_{\mathrm{1}} {and}\:{A}_{\mathrm{2}} \:{are}\:\:{areas}\:{of}\:{base} \\ $$
Question Number 220253 Answers: 0 Comments: 0
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