Question Number 168131 by daus last updated on 04/Apr/22 Commented by daus last updated on 04/Apr/22 $${help}\:{me}\:{to}\:{solve}\:{above}\:{equation} \\ $$ Answered by benhamimed last updated on…
Question Number 37048 by MJS last updated on 08/Jun/18 $$\mathrm{to}\:\mathrm{everybody}: \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{solved}\:\mathrm{an}\:\mathrm{integral},\:\mathrm{please}\:\mathrm{test}\:\mathrm{your} \\ $$$$\mathrm{solution}\:\mathrm{by}\:\mathrm{differentiating}.\:\mathrm{I}'\mathrm{m}\:\mathrm{doing}\:\mathrm{the} \\ $$$$\mathrm{same}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}! \\ $$$${MJS} \\ $$ Commented by rahul 19 last…
Question Number 168119 by Beginner last updated on 03/Apr/22 Answered by mr W last updated on 04/Apr/22 $${l}={length}\:{of}\:{rod}=\mathrm{24}\:{cm} \\ $$$${I}=\frac{{Ml}^{\mathrm{2}} }{\mathrm{12}}+{M}\left(\frac{{l}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} \\ $$$${I}\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−{Mg}\left(\frac{{l}}{\mathrm{2}}−{x}\right)\mathrm{sin}\:\theta\:\approx−{Mg}\left(\frac{{l}}{\mathrm{2}}−{x}\right)\theta…
Question Number 102576 by MAB last updated on 10/Jul/20 $${let}\:{A}_{\mathrm{1}} {A}_{\mathrm{2}} …{A}_{{n}} \:\:{a}\:{regular}\:{polygon}\:{of}\:{n}\:{sides} \\ $$$${with}\:{length}\:{l}\:{each},\:{let}\:{X}\:{a}\:{point}\:{such} \\ $$$${that}\:{A}_{\mathrm{1}} {X}={x}\centerdot{l}\:{where}\:\mathrm{0}<{x}<\mathrm{1}\:\left({as}\:{shown}\right) \\ $$$${what}\:{is}\:{the}\:{length}\:{of}\:{the}\:{shortest}\:{path} \\ $$$${begins}\:{fromX}\:{touching}\:{all}\:{sides}\:{once} \\ $$$${and}\:{ends}\:{at}\:{X} \\…
Question Number 168115 by Beginner last updated on 03/Apr/22 Commented by Beginner last updated on 04/Apr/22 Commented by Beginner last updated on 04/Apr/22 $${Not}\:{surd}\:{buh}\:{that}\:{was}\:{my}\:{thought}\:{process}….{Any}\:{mistake}? \\…
Question Number 102572 by Tinku Tara last updated on 10/Jul/20 $$\mathrm{App}\:\mathrm{updates}:\: \\ $$$$\mathrm{Version}\:\mathrm{2}.\mathrm{094}\:\mathrm{has}\:\mathrm{been}\:\mathrm{uploaded} \\ $$$$\mathrm{to}\:\mathrm{playstore}: \\ $$$$\mathrm{Enhancements}: \\ $$$$\bullet\:\mathrm{paste}\:\mathrm{plain}\:\mathrm{text} \\ $$$$\bullet\:\mathrm{Auto}\:\mathrm{wrapping}\:\mathrm{of}\:\mathrm{lines}\:\left(\mathrm{settings}\right) \\ $$$$\bullet\:\mathrm{New}\:\mathrm{construct}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{below}. \\ $$$$\:\:\:\:\mathrm{strike}\:\mathrm{thru}…
Question Number 168114 by Beginner last updated on 03/Apr/22 Answered by MJS_new last updated on 04/Apr/22 $$\mathrm{7} \\ $$ Commented by Beginner last updated on…
Question Number 168102 by MathsFan last updated on 03/Apr/22 $${solve}\:{the}\:{differential}\:{equation} \\ $$$${sin}^{\mathrm{3}} {x}\:{y}''+{sinxcosx}\:{y}'+\mathrm{4}{y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168093 by leicianocosta last updated on 02/Apr/22 Answered by nurtani last updated on 03/Apr/22 $${sin}\:\mathrm{60}°\:=\:\frac{\mathrm{2}}{{d}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\frac{\mathrm{2}}{{d}}\: \\ $$$$\sqrt{\mathrm{3}}\:{d}\:=\:\mathrm{4}\Rightarrow\:{d}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:\centerdot\:\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\mathrm{3}}\:{m}\:\left({b}\right) \\ $$ Terms of…
Question Number 168092 by leicianocosta last updated on 02/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com