Question Number 141004 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 75468 by aliesam last updated on 11/Dec/19 $$\underset{{n}\rightarrow\infty} {{lim}}\sqrt{\frac{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{3}}{{n}^{\mathrm{5}} +\mathrm{1}}} \\ $$ Commented by MJS last updated on 11/Dec/19 $${P}_{{n}} \left({x}\right)={c}_{{n}} {x}^{{n}}…
Question Number 75469 by aliesam last updated on 11/Dec/19 $$\underset{{n}\rightarrow\infty} {{lim}}\left(\:\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}\:\right) \\ $$ Commented by MJS last updated on 11/Dec/19 $$\mathrm{0} \\ $$$$\mathrm{because}\:{n}+\mathrm{1}\sim{n}\:\mathrm{for}\:\mathrm{large}\:{n} \\ $$…
Question Number 9929 by ridwan balatif last updated on 16/Jan/17 Answered by mrW1 last updated on 18/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{4}{x}\centerdot\mathrm{tan}^{\mathrm{2}} \:\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{sin}\:\mathrm{3}{x}\centerdot\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}\left(\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}{x}}\right)\mathrm{tan}^{\mathrm{2}}…
Question Number 141002 by EnterUsername last updated on 14/May/21 $$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$ Answered by…
Question Number 75465 by necxxx last updated on 11/Dec/19 Commented by necxxx last updated on 12/Dec/19 $${Find}\:{the}\:{yellow}\:{area} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 9928 by Tawakalitu ayo mi last updated on 16/Jan/17 Commented by RasheedSoomro last updated on 16/Jan/17 $$\mathrm{Picture}\:\mathrm{not}\:\mathrm{visible}. \\ $$ Terms of Service Privacy…
Question Number 140997 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Commented by MJS_new last updated on 15/May/21 $$\mathrm{as}\:\mathrm{easy}\:\mathrm{as}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\…
Question Number 140996 by mnjuly1970 last updated on 14/May/21 $$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}}…
Question Number 75463 by indalecioneves last updated on 11/Dec/19 Answered by mr W last updated on 12/Dec/19 $${shape}\:{of}\:{cable}\:{is}\:{catenary}: \\ $$$$\frac{{L}}{{d}}=\frac{\mathrm{2}\:\mathrm{sinh}\:\frac{{d}}{{x}}}{\frac{{d}}{{x}}} \\ $$$$\frac{{f}}{{d}}=\frac{\mathrm{cosh}\:\frac{{d}}{{x}}−\mathrm{1}}{\frac{{d}}{{x}}} \\ $$$${with}\:{t}=\frac{{d}}{{x}}\:{as}\:{parameter}: \\…