Question Number 226486 by ajfour last updated on 30/Nov/25 Commented by ajfour last updated on 30/Nov/25 yes Commented by ajfour last updated on 30/Nov/25 $${Length}\:{OB}=\theta\:\:{as}\:{well}.…
Question Number 226471 by Rojarani last updated on 30/Nov/25 $$\:{If},\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \infty{xy}\: \\ $$$$\:\:{then}\:{prove}\:{that},\:\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \infty{xy} \\ $$ Commented by fantastic2 last updated on 30/Nov/25…
Question Number 226467 by Lara2440 last updated on 30/Nov/25 Answered by Lara2440 last updated on 30/Nov/25 $$\:\mathrm{let}\:\mathrm{differantable}\:\mathrm{Smooth}\:\mathrm{curve}\:\phi;\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{3}} \\ $$$$\:\phi\left({u},{v}\right)=\begin{cases}{−\mathrm{sin}\left({u}\right)−\mathrm{3sin}\left({v}\right)}\\{\mathrm{cos}\left({u}\right)+\mathrm{3cos}\left({v}\right)}\\{\mathrm{4}{v}}\end{cases}\:\:\:,\:−\mathrm{2}\pi\leq{u}\leq\mathrm{2}\pi\:,\:−\mathrm{2}\pi\leq{v}\leq\mathrm{2}\pi \\ $$$$\mathrm{Find}\:\mathrm{Normal}\:\mathrm{curvature}\:,\:\mathrm{Principal}\:\mathrm{curvature}\:,\:\mathrm{Principal}\:\mathrm{dirction} \\ $$$$\: \\…
Question Number 226447 by Tawa11 last updated on 29/Nov/25 Commented by Tawa11 last updated on 29/Nov/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}. \\ $$ Answered by mr W last updated…
Question Number 226453 by ajfour last updated on 29/Nov/25 Commented by ajfour last updated on 29/Nov/25 $${Find}\:{radii}\:{of}\:{smaller}\:{circles} \\ $$$${a}\left({top}\right),\:{b}\left({left}\right),\:{c}\left({right}\right). \\ $$ Commented by mr W…
Question Number 226455 by Spillover last updated on 29/Nov/25 Answered by Ghisom_ last updated on 29/Nov/25 $$\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}}= \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }+\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}}…
Question Number 226464 by Spillover last updated on 29/Nov/25 Commented by Frix last updated on 29/Nov/25 $$\mathrm{By}\:\mathrm{parts} \\ $$$${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{cot}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:\rightarrow\:{v}'=−\frac{\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}…
Question Number 226442 by mr W last updated on 28/Nov/25 Commented by mahdipoor last updated on 29/Nov/25 $$\mathrm{for}\:\mathrm{mass}\:\mathrm{M} \\ $$$$\Sigma\mathrm{F}=\mathrm{ma}\:\:\left(\mathrm{for}\:\mathrm{CM}\right) \\ $$$$\begin{cases}{\mathrm{A}=\left(\overset{..} {\mathrm{x}}−\alpha.\mathrm{l}.\mathrm{cos}\theta−\omega^{\mathrm{2}} .\mathrm{l}.\mathrm{sin}\theta\right)\mathrm{M}}\\{\mathrm{B}=\left(−\alpha.\mathrm{l}.\mathrm{sin}\theta+\omega^{\mathrm{2}} .\mathrm{l}.\mathrm{cos}\theta\right)\mathrm{M}}\end{cases}…
Question Number 226409 by hardmath last updated on 27/Nov/25 Answered by fantastic2 last updated on 27/Nov/25 Commented by fantastic2 last updated on 27/Nov/25 $$\alpha+\mathrm{90}^{\mathrm{0}} +\mathrm{70}^{\mathrm{0}}…
Question Number 226400 by Lara2440 last updated on 27/Nov/25 $$\mathrm{Prove}\:\mathrm{klein}\:\mathrm{bottle}\:\mathrm{is}\:\mathrm{Immersion} \\ $$$$\mathrm{but}\:\mathrm{klein}\:\mathrm{bottle}\:\mathrm{can}'\mathrm{t}\:\mathrm{Imbedding}\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \:\mathrm{Space}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com