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} \\ $$$$\mathrm{cus}\:\mathrm{A}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\: \\ $$$$\mathrm{swap}\:\mathrm{row}\:\mathrm{1}\:\mathrm{and}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\:\mathrm{and}\:\mathrm{Subtract}\:\frac{\mathrm{3}}{\mathrm{5}}×\:\mathrm{row}\:\mathrm{1}\:\mathrm{from}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{4}}&{\:\:\mathrm{2}}&{\:\:\:\:\:\mathrm{7}}\end{pmatrix}\:\:…
Question Number 219761 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by mr W last updated on 01/May/25 $$\mathrm{1}+{x}+{x}^{\mathrm{2}}…
Question Number 219697 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 2 2…
Question Number 219763 by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by Spillover last updated on 01/May/25 Answered by…
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Question Number 219733 by Spillover last updated on 01/May/25 Answered by SdC355 last updated on 01/May/25 $$\mid\mathrm{g}\left({x}\right)−\mathrm{2}\mid\leq\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$−\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\leq\mathrm{g}\left({x}\right)\leq\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\leq\underset{{x}\rightarrow\mathrm{1}}…
Question Number 219730 by SdC355 last updated on 01/May/25 $${f}\left({w}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\theta\left({s}−\mathrm{1}\right)}{{s}\left({s}−\mathrm{1}\right)^{\alpha} }{e}^{−{sw}} \:\mathrm{d}{s} \\ $$$$\hat {\theta}\left({s}\right)=\begin{cases}{\mathrm{0}\:\:{s}<\mathrm{0}}\\{\mathrm{1}\:\:{s}>\mathrm{0}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact:…