Category: Algebra

Question Number 181196 by depressiveshrek last updated on 22/Nov/22 $$Let{a}_1,a_2,a_3,\dots a_{2022}benumbers$$$$rangingfrom(0,+\mathrm{\infty })\mathrm{\setminus }\left\{1\right\},forwhich$$$$thefunctionf:\mathbb{R}\to \mathbb{R}isdefinedas$$$${f}\left({x}\right)={a}_{\mathrm{1}} ^{{x}} +{a}_{\mathrm{2}} ^{{x}} +{a}_{\mathrm{3}} ^{{x}}…

Question Number 181183 by Shrinava last updated on 22/Nov/22 $${\mathrm{\Omega }}_{\mathbf{n}}=\left|\begin{array}{ccccc}1& 1& 1& \dots & 1\\ 1& 2^2& 2^3& \dots & 2^{\mathbf{n}}\\ 1& 3^2& 3^3& \dots & 3^{\mathbf{n}}\\ \dots & \dots & \dots & \dots & \dots \\ 1& {\mathrm{n}}^2& {\mathrm{n}}^3& \dots & {\mathrm{n}}^{\mathbf{n}}\end{array}\right|,\mathrm{n}\in {\mathbb{N}}^{\ast }$$$$\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 181144 by Agnibhoo98 last updated on 22/Nov/22 $$\mathrm{Solve}\mathrm{for}x:$$$${\left(\frac{3x-28}{3x-26}\right)}^3=\frac{x-10}{x-8}$$ 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 181130 by Shrinava last updated on 21/Nov/22 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 181099 by Ari last updated on 21/Nov/22 $$$$$$\mathrm{prove}\mathrm{that}\mathrm{for}\mathrm{every}$$$$\mathrm{positivenumber}\mathrm{p}\mathrm{e}\mathrm{q}\mathrm{wee}$$$$\mathrm{hav}e:$$$$p+q⩾\sqrt{4pq}$$ Answered by Agnibhoo98 last updated…