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Question Number 221415 by wewji12 last updated on 04/Jun/25 $$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$ Answered…
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Question Number 221406 by ajfour last updated on 03/Jun/25 Commented by ajfour last updated on 03/Jun/25 $${Find}\:\mathrm{tan}\:\phi\:\:\:{if}\:\mathrm{tan}\:\theta=\mathrm{1}/\mathrm{4}. \\ $$ Answered by mr W last updated…
Question Number 221407 by fantastic last updated on 03/Jun/25 $${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221411 by ajfour last updated on 03/Jun/25 Answered by mr W last updated on 04/Jun/25 $${r}=\frac{\theta}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\pi}{\mathrm{6}} \\ $$ Commented by mr W last…
Question Number 221400 by Nicholas666 last updated on 02/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\int\int\int_{\mathrm{0}\leqslant{x}\leqslant{y}\leqslant{z}\leqslant\mathrm{1}} \:\left[\left({y}\:−\:{x}\right)^{\mathrm{2}} \left({z}\:−\:{y}\right)^{\mathrm{2}} \left({z}\:−\:{x}\right)^{\mathrm{2}} \right]\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$ \\ $$ Answered by MrGaster last updated…
Question Number 221397 by Nicholas666 last updated on 02/Jun/25 Commented by Nicholas666 last updated on 02/Jun/25 $$\:\:{find}\:{x} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221399 by Nicholas666 last updated on 02/Jun/25 $$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\:\geqslant\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{ab}}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}; \\ $$$$\:\:\frac{\mathrm{38}}{\mathrm{55}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{2}} \:+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:,\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{3}}…