Question Number 181221 by cortano1 last updated on 23/Nov/22 Answered by mr W last updated on 23/Nov/22 Commented by cortano1 last updated on 23/Nov/22 $$\mathrm{nicesolution}…
Question Number 181196 by depressiveshrek last updated on 22/Nov/22 $${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:…{a}_{\mathrm{2022}} \:{be}\:{numbers} \\ $$$${ranging}\:{from}\:\left(\mathrm{0},\:+\infty\right)\:\backslash\:\left\{\mathrm{1}\right\},\:{for}\:{which} \\ $$$${the}\:{function}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:{is}\:{defined}\:{as} \\ $$$${f}\left({x}\right)={a}_{\mathrm{1}} ^{{x}} +{a}_{\mathrm{2}} ^{{x}} +{a}_{\mathrm{3}} ^{{x}}…
Question Number 181182 by depressiveshrek last updated on 22/Nov/22 $${For}\:{what}\:{values}\:{of}\:{a}\:{does}\:{the}\:{system} \\ $$$${of}\:{equations}\:{only}\:{have}\:{one}\:{solution}: \\ $$$$\begin{cases}{{a}\left({x}^{\mathrm{4}} +\mathrm{1}\right)={y}+\mathrm{2}−\mid{x}\mid}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$ Answered by mr W last updated…
Question Number 181183 by Shrinava last updated on 22/Nov/22 $$\Omega_{\boldsymbol{\mathrm{n}}} \:=\:\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{…}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}^{\mathrm{2}} }&{\mathrm{2}^{\mathrm{3}} }&{…}&{\mathrm{2}^{\boldsymbol{\mathrm{n}}} }\\{\mathrm{1}}&{\mathrm{3}^{\mathrm{2}} }&{\mathrm{3}^{\mathrm{3}} }&{…}&{\mathrm{3}^{\boldsymbol{\mathrm{n}}} }\\{…}&{…}&{…}&{…}&{…}\\{\mathrm{1}}&{\mathrm{n}^{\mathrm{2}} }&{\mathrm{n}^{\mathrm{3}} }&{…}&{\mathrm{n}^{\boldsymbol{\mathrm{n}}} }\end{vmatrix}\:\:,\:\:\:\mathrm{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\frac{\Omega_{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\Omega_{\boldsymbol{\mathrm{n}}} }}\:…
Question Number 50089 by Cheyboy last updated on 13/Dec/18 $$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{16}}{\mid\mathrm{x}−\mathrm{8}\mid} \\ $$ Commented by Abdo msup. last updated on 14/Dec/18 $${let}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{16}}{\mid{x}−\mathrm{8}\mid}\:\Rightarrow{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{2}{x}−\mathrm{16}}{\mid{x}−\mathrm{8}\mid}…
Question Number 181144 by Agnibhoo98 last updated on 22/Nov/22 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left(\frac{\mathrm{3}{x}\:−\:\mathrm{28}}{\mathrm{3}{x}\:−\:\mathrm{26}}\right)^{\mathrm{3}} \:=\:\frac{{x}\:−\:\mathrm{10}}{{x}\:−\:\mathrm{8}} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Nov/22 $$\left(\frac{\mathrm{3}{x}\:−\:\mathrm{28}}{\mathrm{3}{x}\:−\:\mathrm{26}}\right)^{\mathrm{3}} \:=\:\frac{{x}\:−\:\mathrm{10}}{{x}\:−\:\mathrm{8}} \\…
Question Number 181138 by Agnibhoo98 last updated on 22/Nov/22 $$\mathrm{If}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}, \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} \:+\:{bc}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{b}^{\mathrm{2}} \:+\:{ca}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{c}^{\mathrm{2}} \:+\:{ab}}\:=\:? \\ $$ Answered by som(math1967) last updated on 22/Nov/22 $$\:{a}+{b}+{c}=\mathrm{0}…
Question Number 181130 by Shrinava last updated on 21/Nov/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115564 by zakirullah last updated on 26/Sep/20 Answered by Olaf last updated on 26/Sep/20 $$\mathrm{A}\:\mathrm{is}\:\mathrm{symetrix} \\ $$$$\Rightarrow\:\mathrm{2}{b}\:=\:\mathrm{3}\:\mathrm{and}\:\mathrm{3}{a}\:=\:−\mathrm{2} \\ $$$${a}\:=\:−\frac{\mathrm{2}}{\mathrm{3}},\:{b}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Commented by…
Question Number 181099 by Ari last updated on 21/Nov/22 $$ \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\: \\ $$$$\mathrm{positivenumber}\:\mathrm{p}\:\mathrm{e}\:\mathrm{q}\:\mathrm{wee} \\ $$$$\mathrm{hav}{e}: \\ $$$${p}+{q}\geqslant\sqrt{\mathrm{4}{pq}} \\ $$ Answered by Agnibhoo98 last updated…