Question Number 226910 by Estevao last updated on 18/Dec/25 Answered by Estevao last updated on 18/Dec/25 $$>{Ashed}\:{Area} \\ $$ Answered by fantastic2 last updated on…
Question Number 226907 by fantastic2 last updated on 18/Dec/25 $${two}\:{small}\:{balls}\:{are}\:{hung}\:{from}\:{a}\:{point} \\ $$$$\left({same}\:{mass},\:{same}\:{charge}\:{and}\:{rope}\:{length}\:{are}\:{same}\right) \\ $$$${the}\:{two}\:{strings}\:{make}\:{an}\:{angle}\:\mathrm{30}^{\mathrm{0}} \\ $$$${when}\:{immersed}\:{in}\:{a}\:{liquid}\:{of}\:\rho=\mathrm{0}.\mathrm{8}{g}/{cc} \\ $$$${the}\:{angle}\:{remains}\:{same}.\rho_{{ball}} =\mathrm{1}.\mathrm{6}{g}/{cc} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\:\kappa\left({dielectric}\:{const}.\right){of} \\ $$$${the}\:{liquid} \\ $$…
Question Number 226901 by mr W last updated on 18/Dec/25 $${if}\:{x}+{y}=\mathrm{2}\:{with}\:{x},\:{y}\:>\mathrm{0},\:{find}\:{the} \\ $$$${minimum}\:{of}\:{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }. \\ $$ Commented by fantastic2 last updated on 18/Dec/25 $$\mathrm{3}…
Question Number 226898 by TonyCWX last updated on 18/Dec/25 $$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{canonical}\:\mathrm{form}: \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}^{\mathrm{2}} }+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}\partial\mathrm{y}}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{y}^{\mathrm{2}} }=\mathrm{0} \\ $$ Terms of Service…
Question Number 226912 by Spillover last updated on 18/Dec/25 Answered by Raphael254 last updated on 19/Dec/25 $$ \\ $$$${b}+{c}\:=\:\mathrm{16} \\ $$$$ \\ $$$${a}+{b}+{c}=\mathrm{24} \\ $$$${a}+\mathrm{16}\:=\:\mathrm{24}…
Question Number 226913 by Spillover last updated on 18/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226914 by Estevao last updated on 18/Dec/25 Answered by breniam last updated on 19/Dec/25 $$\mathrm{We}\:\mathrm{can}\:\mathrm{concider}\:\mathrm{it}\:\mathrm{only}\:\mathrm{for}\:\mathrm{odd}\:{n},\:\mathrm{because} \\ $$$$\mathrm{real}\:\mathrm{even}\:\mathrm{root}\:\mathrm{of}\:\mathrm{negative}\:\mathrm{number}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}. \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{2}\right)!\left(\mathrm{2}−\mathrm{2}{n}\right)}= \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{1}\right)!}×\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\frac{\mathrm{2}−\mathrm{2}{n}}{\mathrm{2}{n}−\mathrm{1}}}…
Question Number 226877 by Spillover last updated on 17/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226878 by Spillover last updated on 17/Dec/25 Answered by gregori last updated on 18/Dec/25 $$\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\sqrt{{x}+{h}}}{\left(\sqrt{{x}+{h}}\:.\sqrt{{x}}\:\right){h}} \\ $$$$\:=\:\frac{\mathrm{1}}{{x}}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{h}}{\left(\sqrt{{x}}+\sqrt{{x}+{h}}\:\right){h}} \\ $$$$\:=\:−\frac{\mathrm{1}}{{x}.\:\mathrm{2}\sqrt{{x}}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}} \\ $$…
Question Number 226879 by Spillover last updated on 17/Dec/25 Answered by gregori last updated on 18/Dec/25 $$\left({a}\right)\:{f}\:'\left({x}\right)=\:−\mathrm{sin}\:{x}\:.\:{e}^{\mathrm{cos}\:{x}} \: \\ $$$$\:\left({b}\right)\:{f}\:'\left({x}\right)\:=\:\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:{x}\:.{e}^{\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$ \\…