Question Number 10667 by Gaurav3651 last updated on 22/Feb/17 $${prove}\:{that}\:{the}\:{quadrilateral}\:{formed} \\ $$$${by}\:{angle}\:{bisectors}\:{of}\:{a}\:{cyclic}\: \\ $$$${quadrilateral}\:{is}\:{also}\:{cyclic}. \\ $$ Answered by mrW1 last updated on 22/Feb/17 $${ABCD}\:{is}\:{a}\:{convex}\:\:{quadrilateral}. \\…
Question Number 76203 by Master last updated on 25/Dec/19 Commented by Master last updated on 25/Dec/19 $$\mathrm{example}: \\ $$$$\sqrt[{\mathrm{3}}]{−\mathrm{8}}=? \\ $$$$\left(−\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =? \\ $$ Commented…
Question Number 76200 by Joel578 last updated on 25/Dec/19 Commented by benjo last updated on 25/Dec/19 $$\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{merry}\:\mathrm{chrismastto}\:\mathrm{all} \\ $$ Commented by Rio Michael last updated…
Question Number 10665 by FilupS last updated on 22/Feb/17 $${n}\mathrm{th}\:\mathrm{prime}\:=\:{p}_{{n}} \\ $$$${n}\mathrm{th}\:\mathrm{non}−\mathrm{prime}\:=\:{q}_{{n}} \\ $$$$\: \\ $$$$\mathrm{Determine}\:\mathrm{if}\:{q}_{{n}} >{p}_{{n}} \:\mathrm{for}\:\forall{n}\geqslant\mathrm{2} \\ $$ Commented by FilupS last updated…
Question Number 10664 by FilupS last updated on 22/Feb/17 $${S}=\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\notin\mathbb{P}}} {\overset{\infty} {\sum}}{n} \\ $$$${Q}=\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\in\mathbb{P}}} {\overset{\infty} {\sum}}{n} \\ $$$$\: \\ $$$$\mathrm{Prove}\:\mathrm{if}\:\mathrm{true}: \\ $$$${S}>{Q} \\…
Question Number 141733 by BHOOPENDRA last updated on 23/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10662 by FilupS last updated on 22/Feb/17 $$\mathrm{determine}\:\mathrm{if}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\frac{\mathrm{1}}{\mathrm{5}^{{s}} }+…\geqslant\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+\frac{\mathrm{1}}{\mathrm{6}^{{s}} }+… \\ $$$$\mathrm{or}: \\ $$$$\underset{\underset{{n}\geqslant\mathrm{1}} {{n}\in\mathbb{P}}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}}…
Question Number 10660 by okhema last updated on 22/Feb/17 $${ind}\:{the}\:{centre}\:{and}\:{radius}\:{of}\:{these}: \\ $$$$\left({a}\right)\:\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{y}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}{y}−\mathrm{2}=\mathrm{0} \\ $$$$\left({b}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}{y}−\mathrm{1}=\mathrm{0} \\ $$$$\left({c}\right)\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$ Answered…
Question Number 76194 by abdomathmax last updated on 25/Dec/19 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$ Commented by abdomathmax last updated on 25/Dec/19 $${let}\:{A}\:=\int_{\mathrm{0}}…
Question Number 76192 by abdomathmax last updated on 25/Dec/19 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{arctan}\left({sin}\left(\mathrm{2}{x}\right)\right)−{sin}\left({arctan}\left(\mathrm{2}{x}\right)\right)}{{x}^{\mathrm{2}} } \\ $$ Answered by benjo last updated on 25/Dec/19 Commented by benjo last…