Question Number 18363 by 314159 last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18361 by diofanto last updated on 19/Jul/17 $${N}\:\mathrm{propositions}\:\mathrm{are}\:\mathrm{judged}\:\mathrm{by}\:\mathrm{2}{k}−\mathrm{1}\:\mathrm{people}. \\ $$$$\mathrm{Each}\:\mathrm{person}\:\mathrm{assigns}\:“\mathrm{true}''\:\mathrm{to} \\ $$$$\mathrm{exactly}\:{M}\:\mathrm{propositions}\:\mathrm{and}\:“\mathrm{false}'' \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:{N}−{M}\:\left({M}\:\leqslant\:{N}\right). \\ $$$$\mathrm{To}\:\mathrm{say}\:\mathrm{a}\:\mathrm{proposition}\:\mathrm{is}\:“\mathrm{approved}''\:\mathrm{means} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{according}\:\mathrm{to}\:\mathrm{at}\:\mathrm{least}\:{k}\:\mathrm{judges}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{numbers} \\ $$$$\mathrm{of}\:\mathrm{approved}\:\mathrm{propositions}\:\mathrm{given}\:{N},\:{M}\:\mathrm{and}\:{k}. \\…
Question Number 18362 by 314159 last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149425 by help last updated on 05/Aug/21 Answered by dumitrel last updated on 05/Aug/21 $${x}>\mathrm{0};{x}\neq\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{log}_{\mathrm{3}} {x}}={y}>\mathrm{0}\Rightarrow{x}=\mathrm{3}^{{y}^{\mathrm{4}} } \\ $$$$\mathrm{9}^{{y}} −\mathrm{4}{x}^{\frac{\mathrm{1}}{{y}^{\mathrm{3}} }}…
Question Number 83886 by mr W last updated on 07/Mar/20 $${To}\:{the}\:{developers}\:{of}\:{TinkuTara}: \\ $$$${problem}\:\mathrm{1}: \\ $$$${i}\:{get}\:{no}\:{notifications}\:{when}\:{my}\:{posts} \\ $$$${are}\:{updated}. \\ $$$$ \\ $$$${problem}\:\mathrm{2}: \\ $$$${i}\:{can}\:{edit}\:{my}\:{post},\:{see}\:{picture}\:\mathrm{1},\:{but} \\ $$$${the}\:{content}\:{is}\:{not}\:{visiable},\:{see}\:{picture}\:\mathrm{2}.…
Question Number 18342 by mondodotto@gmail.com last updated on 19/Jul/17 Answered by alex041103 last updated on 19/Jul/17 $$\frac{{x}+\mathrm{4}}{{x}+\mathrm{1}}=\frac{{x}+\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{3}}{{x}+\mathrm{1}}=\mathrm{1}+\frac{\mathrm{3}}{{x}+\mathrm{1}} \\ $$$$\frac{{x}−\mathrm{2}}{{x}−\mathrm{4}}=\frac{{x}−\mathrm{4}}{{x}−\mathrm{4}}+\frac{\mathrm{2}}{{x}−\mathrm{4}}=\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{{x}+\mathrm{1}}<\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$${The}\:{inequality}\:{makes}\:{sense}\:{when} \\ $$$${x}\neq−\mathrm{1};\mathrm{4}…
Question Number 149415 by abdurehime last updated on 05/Aug/21 $$\mathrm{show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{sint}−\mathrm{tcost}}{\mathrm{t}^{\mathrm{3}} }\right)^{\mathrm{2}} \mathrm{dt}=\frac{\Pi}{\mathrm{15}} \\ $$ Commented by abdurehime last updated on 05/Aug/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}???? \\…
Question Number 18327 by mondodotto@gmail.com last updated on 18/Jul/17 Commented by mondodotto@gmail.com last updated on 18/Jul/17 $$\mathrm{please}\:\mathrm{help}!! \\ $$ Answered by ajfour last updated on…
Question Number 18307 by mondodotto@gmail.com last updated on 18/Jul/17 Answered by ajfour last updated on 18/Jul/17 $$\mathrm{Q}.\mathrm{2} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\:\pi} \frac{\mathrm{ysin}\:\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}=\int_{\mathrm{0}} ^{\:\:\pi} \frac{\left(\pi−\mathrm{y}\right)\mathrm{sin}\:\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}…
Question Number 18306 by mondodotto@gmail.com last updated on 18/Jul/17 Answered by diofanto last updated on 18/Jul/17 $$\frac{\mathrm{log}\:\mathrm{2}^{\mathrm{4}} \:−\:\mathrm{log}\:\mathrm{3}^{\mathrm{4}} }{\mathrm{log}\:\mathrm{3}^{\mathrm{3}} \:+\:\mathrm{log}\:\mathrm{2}^{\mathrm{3}} }\:=\:\frac{\mathrm{4}\:\mathrm{log}\:\mathrm{2}\:−\:\mathrm{4}\:\mathrm{log}\:\mathrm{3}}{\mathrm{3}\:\mathrm{log}\:\mathrm{3}\:+\:\mathrm{3}\:\mathrm{log}\:\mathrm{2}}\:= \\ $$$$=\:\frac{\mathrm{4}\left(\mathrm{log}\:\mathrm{2}\:−\:\mathrm{log}\:\mathrm{3}\right)}{\mathrm{3}\left(\mathrm{log}\:\mathrm{2}\:+\:\mathrm{log}\:\mathrm{3}\right)}\:=\:\frac{\mathrm{4}\:\mathrm{log}\:\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{3}\:\mathrm{log}\:\mathrm{6}}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\mathrm{log}_{\mathrm{6}} \:\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\…