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Question Number 219844 by cherokeesay last updated on 02/May/25 Answered by mr W last updated on 02/May/25 Commented by mr W last updated on 02/May/25…
Question Number 219846 by hardmath last updated on 02/May/25 $$\mathrm{If}\:\:\:\mathrm{f}:\left[\mathrm{a},\mathrm{b}\right]\rightarrow\mathbb{R} \\ $$$$\:\:\:\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b} \\ $$$$\mathrm{f}\:-\:\mathrm{continuous} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{b}^{\mathrm{4047}} \:−\:\mathrm{a}^{\mathrm{4047}} }{\mathrm{4047}}\:+\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\mathrm{f}\:^{\mathrm{2}} \:\left(\mathrm{x}^{\mathrm{2024}} \right)\:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{1012}}\:\int_{\boldsymbol{\mathrm{a}}^{\mathrm{2024}} }…
Question Number 219843 by hardmath last updated on 02/May/25 $$\mathrm{If}\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:>\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{ab}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{ac}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{ad}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{bc}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{bd}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{cd}\right)^{\mathrm{3}}…
Question Number 219709 by SdC355 last updated on 01/May/25 $$\mathrm{solve}\:\mathrm{Differantial}\:\mathrm{Equation} \\ $$$$\left({y}'\left({t}\right)\right)^{\mathrm{2}} =\mathrm{4}\left({y}\left({t}\right)\right)^{\mathrm{3}} −{ay}\left({t}\right)−{b}\:,\:\left\{{a},{b}\in\mathbb{C}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219710 by SdC355 last updated on 01/May/25 $$\int_{\rho} ^{\:\infty} \:\frac{{e}^{{r}} \centerdot\Gamma\left(\mathrm{0},{r}\right)}{{r}}\:\mathrm{d}{r}=?? \\ $$$$\Gamma\left(\alpha,{r}\right)=\frac{\mathrm{1}}{\Gamma\left(\mathrm{1}−\alpha\right)}\centerdot\int_{\mathrm{0}} ^{\:\infty} \:\frac{\theta\left({s}−\mathrm{1}\right)}{{s}\left({s}−\mathrm{1}\right)^{\alpha} }{e}^{−{sr}} \mathrm{d}{s} \\ $$ Terms of Service Privacy…
Question Number 219704 by Spillover last updated on 01/May/25 Answered by SdC355 last updated on 01/May/25 $$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\mathrm{ln}\left(\frac{\mathrm{1}}{{z}}−\mathrm{1}\right)\:\mathrm{d}{z}=\mathrm{0}\: \\ $$$$\int_{\mathrm{0}} ^{\:\rho} \:+\int_{\:\rho} ^{\:\mathrm{1}} =\mathrm{0}\:,\:\left(\rho<\mathrm{1}\right)…
Question Number 219769 by mr W last updated on 01/May/25 Commented by mr W last updated on 01/May/25 $${a}\:{ball}\:{with}\:{mass}\:\boldsymbol{{m}}\:{and}\:{an}\:{uniform} \\ $$$${rod}\:{with}\:{mass}\:\boldsymbol{{M}}\:{and}\:{length}\:\boldsymbol{{L}}\:{are} \\ $$$${released}\:{from}\:{the}\:{rest}\:{at}\:{the}\:{same} \\ $$$${heigth}\:\boldsymbol{{h}}\:{as}\:{shown}.\:\left({L}>{h}\right)…
Question Number 219770 by ajfour last updated on 01/May/25 Commented by ajfour last updated on 01/May/25 $${A}\:{rolling}\:{cylinder}\:{of}\:{length}\:{L}\:{rolls} \\ $$$$\:{gathering}\:{moss}\:{at}\:{the}\:{rate}\:\frac{{dm}}{{dt}}=\rho_{\mathrm{0}} {Lv}.\: \\ $$$${Find}\:{radius}\:{r}\left({t}\right){and}\:{speed}\:{v}\left({t}\right)\:{of}\: \\ $$$${center}\:{of}\:{mass}. \\…
Question Number 219696 by Spillover last updated on 01/May/25 Answered by SdC355 last updated on 01/May/25 $$\boldsymbol{\mathrm{E}}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{cus}\:\mathrm{A}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\: \\ $$$$\mathrm{swap}\:\mathrm{row}\:\mathrm{1}\:\mathrm{and}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\:\mathrm{and}\:\mathrm{Subtract}\:\frac{\mathrm{3}}{\mathrm{5}}×\:\mathrm{row}\:\mathrm{1}\:\mathrm{from}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{4}}&{\:\:\mathrm{2}}&{\:\:\:\:\:\mathrm{7}}\end{pmatrix}\:\:…