Question Number 227085 by Rojarani last updated on 29/Dec/25 Commented by mr W last updated on 30/Dec/25 $${to}\:{wabi}\:{sir}: \\ $$$${please}\:{don}'{t}\:{put}\:{your}\:{new}\:{question} \\ $$$${as}\:{comment}\:{to}\:{an}\:{other}\:{question}! \\ $$$${please}\:{post}\:{it}\:{as}\:{new}\:{question}! \\…
Question Number 227080 by Spillover last updated on 29/Dec/25 Answered by Kassista last updated on 29/Dec/25 $$\left.{a}\right)\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:\mathrm{7}−\mathrm{4}{x}\:=\:\mathrm{7}−\mathrm{4}\left(−\mathrm{6}\right)=\:\mathrm{31} \\ $$$$ \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}^{−}…
Question Number 227067 by Spillover last updated on 29/Dec/25 Answered by peace2 last updated on 29/Dec/25 $$\frac{\mathrm{1}}{{x}}\rightarrow{y} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{y}}.\frac{\left[{y}\right]}{{y}^{\mathrm{2}} }{dy}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}}…
Question Number 227063 by fantastic2 last updated on 29/Dec/25 $${is}\:\left(−{e}\right)^{\pi} \:\:{real}\:{number}? \\ $$ Answered by Frix last updated on 29/Dec/25 $$\left.\begin{matrix}{\mathrm{abs}\:\left(−\mathrm{e}\right)\:=\mathrm{e}}\\{\mathrm{arg}\:\left(−\mathrm{e}\right)\:=\pi}\end{matrix}\right\}\:\Rightarrow\:−\mathrm{e}=\mathrm{e}^{\mathrm{i}\pi} \mathrm{e} \\ $$$$\left(−\mathrm{e}\right)^{\pi} =\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}}…
Question Number 227073 by Spillover last updated on 29/Dec/25 Answered by Kassista last updated on 29/Dec/25 $$ \\ $$$$\frac{{dy}}{{dx}}=\:\frac{−\left({x}−{b}\right)+\left({a}−{x}\right)}{\mathrm{2}\sqrt{\left({a}−{x}\right)\left({x}−{b}\right)}}−\left({a}−{b}\right).\frac{\mathrm{1}}{\mathrm{1}+\left(\sqrt{\frac{{a}−{x}}{{x}−{b}}}\right)^{\mathrm{2}} }.\frac{\frac{−\left({x}−{b}\right)−\left({a}−{x}\right)}{\left({x}−{b}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{\frac{{a}−{x}}{{x}−{b}}}} \\ $$$$ \\ $$$$…
Question Number 227074 by Spillover last updated on 29/Dec/25 Answered by Spillover last updated on 04/Jan/26 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 227075 by Spillover last updated on 29/Dec/25 Answered by Spillover last updated on 04/Jan/26 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 227054 by Spillover last updated on 28/Dec/25 $${A}\:{Segment}\:{of}\:{a}\:{sphere}\:{has}\:{radius}\:{r} \\ $$$${and}\:{maximum}\:{height}\:{h}.{Prove}\:{that} \\ $$$${its}\:{volume}\:\frac{\boldsymbol{\pi{h}}}{\mathrm{6}}\left(\boldsymbol{{h}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{r}}^{\mathrm{2}} \right) \\ $$ Answered by fantastic2 last updated on 29/Dec/25…
Question Number 227055 by Spillover last updated on 28/Dec/25 $${A}\:{parabolic}\:{refector}\:{is}\:{formed}\:{by} \\ $$$${revolving}\:{the}\:{arc}\:{of}\:{the}\:{parabala} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax}\:\:{from}\:{x}=\mathrm{0}\:\:\:\:{to}\:\:{x}={h} \\ $$$${about}\:{the}\:{axis}.{If}\:{the}\:\:{diameter} \\ $$$${of}\:{the}\:{reflector}\:{is}\:\mathrm{2}{l}.{Show}\:{that} \\ $$$${the}\:{area}\:{of}\:{the}\:{reflecting}\:{surface}\:{is} \\ $$$$\frac{\pi{l}}{\mathrm{6}{h}^{\mathrm{2}} }\left\{\left({l}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}}…
Question Number 227051 by mnjuly1970 last updated on 28/Dec/25 $$ \\ $$$$\:\:\:\:\:\underset{{n}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}} {\sum}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}\:+\:\mathrm{2}}\:\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by Spillover last updated on…