Question Number 162088 by mathlove last updated on 26/Dec/21 $${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$${solve}\:\:\:{for}\:\:\:\:{x}=? \\ $$ Answered by Ar Brandon last updated on 26/Dec/21 $${x}^{\mathrm{2}} =\mathrm{2}^{{x}}…
Question Number 31018 by jasno91 last updated on 02/Mar/18 Commented by rahul 19 last updated on 02/Mar/18 $${sum}\:{of}\:{roots}\:=−\:\frac{{coefficient}\:{of}\:{b}}{{coefficient}\:{of}\:{a}} \\ $$$$\:{let}\:\mathrm{3}{rd}\:{root}=\:\zeta \\ $$$$\Rightarrow\:\sqrt{\mathrm{3}}\:+\left(\:−\sqrt{\mathrm{3}}\:\right)+\zeta\:=\:\frac{−\mathrm{1}}{\mathrm{2}}. \\ $$$${hence}\:\zeta\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:. \\…
Question Number 162083 by HongKing last updated on 26/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31003 by rahul 19 last updated on 01/Mar/18 $${Number}\:{of}\:{positive}\:{integers}\:{x}\:{for} \\ $$$${which}\:{f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{13}\:{is}\:\:{a}\: \\ $$$${prime}\:{number}\:{are}\:? \\ $$ Answered by MJS last updated on…
Question Number 162068 by HongKing last updated on 25/Dec/21 Answered by Lordose last updated on 26/Dec/21 $$ \\ $$$$\Omega\:=\:\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{sin}^{−\mathrm{1}} \left(\epsilon\right)} ^{\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−\epsilon\right)} \mathrm{log}\left(\left(\mathrm{cos}\left(\mathrm{x}\right)\right)^{\mathrm{cot}\left(\mathrm{x}\right)} \centerdot\left(\mathrm{sin}\left(\mathrm{x}\right)\right)^{\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{sin}\left(\mathrm{x}\right)}}…
Question Number 30990 by rahul 19 last updated on 01/Mar/18 Answered by MJS last updated on 01/Mar/18 $$−\mathrm{2}{x}^{\mathrm{2}} +{kx}+\left({k}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{{k}}{\mathrm{4}}−\frac{\sqrt{\mathrm{9}{k}^{\mathrm{2}} +\mathrm{40}}}{\mathrm{4}} \\…
Question Number 96527 by student work last updated on 02/Jun/20 $$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +….+\mathrm{n}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$ Commented by 06122004 last updated on 02/Jun/20 Answered…
Question Number 162062 by HongKing last updated on 25/Dec/21 $$\Omega\left(\alpha;\beta\right)\:=\underset{\:-\mathrm{1}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\alpha}-\mathrm{1}} \:\left(\mathrm{1}-\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\beta}-\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }\:\mathrm{dx}\:;\:\alpha;\beta>\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Omega\left(\mathrm{3},\mathrm{5}\right)\:>\:\sqrt{\Omega\left(\mathrm{4},\mathrm{5}\right)\centerdot\Omega\left(\mathrm{3},\mathrm{6}\right)} \\ $$ Answered by mindispower…
Question Number 162042 by HongKing last updated on 25/Dec/21 $$\mathrm{let}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\in\mathbb{R}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:\geqslant\:\mathrm{a}^{\mathrm{3}} \mathrm{b}\:+\:\mathrm{b}^{\mathrm{3}} \mathrm{c}\:+\:\mathrm{c}^{\mathrm{3}} \mathrm{a} \\ $$ Terms of Service…
Question Number 162043 by HongKing last updated on 25/Dec/21 $$\mathrm{Find}\:\mathrm{valu}\:\mathrm{of}\:\:\boldsymbol{\mathrm{x}}\:\:\mathrm{if}\:\:\mathrm{x}\in\mathbb{R}\: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{9x}\:-\:\mathrm{1}}\:+\:\sqrt{\mathrm{8x}\:-\:\mathrm{1}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{8x}\:+\:\mathrm{15}}\:-\:\frac{\mathrm{5}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$ Commented by mr W last updated on 25/Dec/21 $${with}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{8}}−\mathrm{1}}+\sqrt{\frac{\mathrm{8}}{\mathrm{8}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{15}}−\frac{\mathrm{5}}{\mathrm{2}}…