Question Number 212801 by Spillover last updated on 24/Oct/24 Answered by som(math1967) last updated on 24/Oct/24 Commented by som(math1967) last updated on 24/Oct/24 $$\frac{{OL}}{{PL}}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 212802 by Spillover last updated on 24/Oct/24 Answered by mr W last updated on 24/Oct/24 Commented by mr W last updated on 24/Oct/24…
Question Number 212803 by Spillover last updated on 24/Oct/24 Answered by Spillover last updated on 24/Oct/24 Answered by Spillover last updated on 24/Oct/24 Answered by…
Question Number 212765 by issac last updated on 23/Oct/24 $$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu}…
Question Number 212760 by Spillover last updated on 23/Oct/24 Answered by mr W last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24…
Question Number 212761 by Spillover last updated on 23/Oct/24 Commented by mr W last updated on 23/Oct/24 $${AB}=\mathrm{2}×{GE}=\mathrm{8}\sqrt{\mathrm{7}} \\ $$ Answered by som(math1967) last updated…
Question Number 212762 by universe last updated on 23/Oct/24 Commented by MrGaster last updated on 23/Oct/24 $$\mathrm{Rewrite}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{function}\:\mathrm{as}: \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{Computational}\:\mathrm{integral}: \\ $$$$\int_{\mathrm{0}} ^{{n}}…
Question Number 212784 by Spillover last updated on 23/Oct/24 Answered by A5T last updated on 24/Oct/24 $${d}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\left({r}−{a}\right){cos}\theta \\ $$$${c}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{r}\left({r}−{a}\right){cos}\theta…
Question Number 212783 by Spillover last updated on 23/Oct/24 Commented by York12 last updated on 25/Oct/24 $$ \\ $$$$ \\ $$$$ \\ $$ Answered by…
Question Number 212777 by Ghisom last updated on 23/Oct/24 $$\mathrm{question}\:\mathrm{212462} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{160}}}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{95}{x}^{\mathrm{2}} +\mathrm{2560}}} \\ $$$$ \\…