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Find-the-sum-of-n-terms-of-the-series-S-n-1-22-333-4444-

Question Number 121235 by ZiYangLee last updated on 06/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$${S}_{{n}} =\mathrm{1}+\mathrm{22}+\mathrm{333}+\mathrm{4444}+\ldots\ldots \\ $$ Answered by Ar Brandon last updated on 06/Nov/20 $$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{22}+\mathrm{333}+\mathrm{4444}+\centerdot\centerdot\centerdot…

Given-function-f-R-R-satisfy-that-f-f-x-x-x-1-f-x-Find-the-value-of-f-1-

Question Number 186769 by naka3546 last updated on 10/Feb/23 $$\mathrm{Given}\:\:\mathrm{function}\:\:\:{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\:\:\mathrm{satisfy}\:\:\mathrm{that} \\ $$$$\:\:\:\:\left({f}\:\circ\:{f}\right)\left({x}\right)\:+\:{x}\:=\:\left({x}+\mathrm{1}\right)\:{f}\left({x}\right)\:. \\ $$$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\:{f}\left(\mathrm{1}\right)\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

If-x-1-2-find-the-value-of-x-4-1-x-4-

Question Number 121218 by ZiYangLee last updated on 06/Nov/20 $$\mathrm{If}\:{x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }. \\ $$ Answered by liberty last updated on 06/Nov/20 $$\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{x}^{\mathrm{4}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\…

Given-that-a-b-and-c-are-three-consecutive-numbers-where-a-gt-b-gt-c-such-that-its-product-is-the-same-as-its-sum-that-is-abc-a-b-c-how-many-such-a-b-c-

Question Number 121219 by ZiYangLee last updated on 06/Nov/20 $$\mathrm{Given}\:\mathrm{that}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{numbers},\:\mathrm{where}\:{a}>{b}>{c},\:\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{product} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{its}\:\mathrm{sum},\:\mathrm{that}\:\mathrm{is}\:{abc}={a}+{b}+{c}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{such}\:\left({a},{b},{c}\right)? \\ $$ Commented by liberty last updated on 06/Nov/20…

Question-186750

Question Number 186750 by pascal889 last updated on 09/Feb/23 Answered by Frix last updated on 09/Feb/23 $$\frac{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{8}}}{\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}}= \\…

Question-186751

Question Number 186751 by 073 last updated on 09/Feb/23 Answered by Rasheed.Sindhi last updated on 09/Feb/23 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=? \\ $$$$\:…

5-5-5-1-3-1-3-1-3-

Question Number 186739 by Davidtim last updated on 09/Feb/23 $$\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+…}}}=? \\ $$ Answered by pablo1234523 last updated on 09/Feb/23 $${x}=\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+…}}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{5}+{x}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{x}−\mathrm{5}=\mathrm{0}…

find-the-difference-of-the-roots-of-the-following-quadratic-equation-3-2-2-x-2-1-2-x-2-

Question Number 55658 by otchereabdullai@gmail.com last updated on 01/Mar/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{following}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:}\right)\mathrm{x}^{\mathrm{2}} \:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}\:=\mathrm{2} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19 $$\alpha+\beta=−\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}\:\:\:\alpha\beta=\frac{−\mathrm{2}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}…