Question Number 195279 by edwinpollito last updated on 28/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195232 by CrispyXYZ last updated on 28/Jul/23 $${a},\:{b},\:{c}>\mathrm{0},\:\:{a}+{b}+{c}\geqslant\mathrm{1}.\:\mathrm{Find} \\ $$$$\mathrm{max}\:\underset{\mathrm{cyc}} {\sum}\:\frac{{a}−{bc}}{{a}+{bc}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195206 by otchereabdullai@gmail.com last updated on 27/Jul/23 Answered by MM42 last updated on 27/Jul/23 $${p}\left({a}\right)={p}\left({b}\right)={p}\left({c}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:;\:{probability}\:{of}\:{winning}\:{each}\:{race} \\ $$$$\left({i}\right)\:{p}\left({a}'\cap{b}\cap{c}'\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{64}} \\ $$$$\left({ii}\right)\:{p}\left({a}\cap{b}\cap{c}\right)=\frac{\mathrm{27}}{\mathrm{64}} \\ $$$$\left({iii}\right)\:{p}\left(\Sigma{abc}'\right)=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{27}}{\mathrm{64}} \\ $$…
Question Number 195223 by Rodier97 last updated on 27/Jul/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{In}=\int_{\mathrm{1}} ^{\mathrm{e}} \frac{\left({lnx}\right)^{{n}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\mathrm{using}\:\mathrm{an}\:\mathrm{enclosing}\:{lnx}\:\mathrm{on}\:\mathrm{interval}\:\left[\mathrm{1};\mathrm{e}\right]\:\mathrm{show}\:\mathrm{that}\:\forall{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{0}\:\leq\mathrm{In}\leq\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$…
Question Number 195202 by sonukgindia last updated on 27/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195135 by 073 last updated on 25/Jul/23 Answered by som(math1967) last updated on 25/Jul/23 $$\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}{x}.\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 195120 by sonukgindia last updated on 25/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195118 by York12 last updated on 25/Jul/23 $${a},{b},{c}>\mathrm{0}\:\&\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{{a}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}+{b}+{c}}{{ab}+{bc}+{ac}}\right)^{\mathrm{2}} \\ $$ Commented by York12…
Question Number 195171 by azirali0198316 last updated on 25/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195114 by mustafazaheen last updated on 24/Jul/23 $$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\underset{\mathrm{k}=\mathrm{3}} {\overset{\mathrm{55}} {\sum}}\mathrm{k}=? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{5}\left.\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{9k}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{53k}\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right)\mathrm{45} \\ $$$$ \\ $$$$ \\ $$ Answered…