Question Number 119835 by ZiYangLee last updated on 27/Oct/20 $$\mathrm{If}\:{M}\:\mathrm{and}\:{m}\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{smallest}\:\mathrm{integers}\:\mathrm{that}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\mathrm{6}{n}^{\mathrm{2}} −\mathrm{5}{n}\leqslant\mathrm{99},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${M}−{m}. \\ $$ Commented by talminator2856791 last updated on…
Question Number 185361 by yaslm last updated on 20/Jan/23 Answered by Rasheed.Sindhi last updated on 20/Jan/23 $$\mathrm{2}{x}+{y}>\mathrm{5}….\left({i}\right) \\ $$$$−\mathrm{1}>{y}……\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right) \\ $$$$\mathrm{2}{x}+{y}−\mathrm{1}>{y}+\mathrm{5} \\ $$$$\mathrm{2}{x}−\mathrm{1}>\mathrm{5}…
Question Number 185354 by Ml last updated on 20/Jan/23 Answered by HeferH last updated on 20/Jan/23 $$\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:−\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:{k} \\ $$$$\:\mathrm{2}\:+\:\sqrt{\mathrm{3}}\:+\:\mathrm{2}\:−\:\sqrt{\mathrm{3}}\:−\:\mathrm{2}\left(\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\:\centerdot\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)={k}^{\mathrm{2}} \\ $$$$\:\:\mathrm{4}\:−\mathrm{2}\left(\sqrt{\mathrm{2}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right)=\:{k}^{\mathrm{2}} \\ $$$$\:\sqrt{\mathrm{2}}\:=\:{k}\:…
Question Number 185348 by yaga7 last updated on 20/Jan/23 Answered by Frix last updated on 20/Jan/23 $${a}={b}={t}\wedge{c}=−\mathrm{2}{t} \\ $$$${abc}=−\mathrm{2}{t}^{\mathrm{3}} \\ $$$${a}+{b}+{c}=\mathrm{0} \\ $$ Terms of…
Question Number 119808 by Khalmohmmad last updated on 27/Oct/20 $${f}^{−\mathrm{1}} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${f}\left(\mathrm{8}\right)=? \\ $$ Answered by Ar Brandon last updated on 27/Oct/20 $${f}^{−\mathrm{1}}…
Question Number 185346 by KONE last updated on 20/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119803 by repentence last updated on 27/Oct/20 Answered by Dwaipayan Shikari last updated on 27/Oct/20 $${T}_{{n}} =\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$=\sqrt{\frac{\left({n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}}…
Question Number 185329 by liuxinnan last updated on 20/Jan/23 $${lim}_{{k}\rightarrow+\infty} \underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{i}−\mathrm{1}}+\sqrt{{i}}}= \\ $$ Answered by aleks041103 last updated on 20/Jan/23 $$\sqrt{{i}−\mathrm{1}}<\sqrt{{i}}\Rightarrow\sqrt{{i}−\mathrm{1}}+\sqrt{{i}}<\mathrm{2}\sqrt{{i}}\leqslant\mathrm{2}{i} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}}…
Question Number 185328 by liuxinnan last updated on 20/Jan/23 $${lim}_{{k}\rightarrow+\infty} \underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{i}}}= \\ $$ Answered by aleks041103 last updated on 20/Jan/23 $${since}\:\sqrt{{i}}\leqslant{i}\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{i}}}\geqslant\frac{\mathrm{1}}{{i}} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}}…
Question Number 185325 by Ml last updated on 20/Jan/23 Answered by aleks041103 last updated on 20/Jan/23 $$\mathrm{14}+\mathrm{8}\sqrt{\mathrm{3}}=\mathrm{14}+\mathrm{2}.\mathrm{2}\sqrt{\mathrm{2}}.\sqrt{\mathrm{6}}= \\ $$$$=\left(\sqrt{\mathrm{6}}\right)^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{6}}\right)= \\ $$$$=\left(\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{14}+\mathrm{8}\sqrt{\mathrm{3}}}=\sqrt{\left(\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…