Question Number 87105 by Chi Mes Try last updated on 02/Apr/20 Commented by mathmax by abdo last updated on 03/Apr/20 $${let}\:{f}\left({x}\right)=\frac{{x}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{lnx}}\:\Rightarrow{f}\left({x}\right)=\frac{{xln}\left({x}\right)−{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right){ln}\left({x}\right)}\:{we}\:{do}\:{the}\:{changement} \\ $$$${x}−\mathrm{1}\:={t}\:\Rightarrow{f}\left({x}\right)=\frac{\left({t}+\mathrm{1}\right){ln}\left(\mathrm{1}+{t}\right)+\mathrm{1}−\left({t}+\mathrm{1}\right)}{{tln}\left(\mathrm{1}+{t}\right)}={g}\left({t}\right) \\ $$$$=\frac{\left({t}+\mathrm{1}\right){ln}\left(\mathrm{1}+{t}\right)−{t}}{{tln}\left(\mathrm{1}+{t}\right)}\:\:\left({x}\rightarrow\mathrm{1}\:\Rightarrow{t}\rightarrow\mathrm{0}\right)\:{so}…
Question Number 87103 by hamdhan last updated on 02/Apr/20 $$\int\frac{{dx}}{\left(\mathrm{1}+{x}\right)\sqrt{{x}−{x}^{\mathrm{2}} }} \\ $$ Commented by abdomathmax last updated on 02/Apr/20 $${I}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{−\left({x}^{\mathrm{2}} −{x}\right)}}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{−\left({x}^{\mathrm{2}} −{x}\:+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\right)}} \\ $$$$=\int\:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 21566 by mondodotto@gmail.com last updated on 27/Sep/17 Commented by mrW1 last updated on 27/Sep/17 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{2} \\ $$ Commented by…
Question Number 21564 by Tinkutara last updated on 27/Sep/17 $$\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{smooth} \\ $$$$\mathrm{ground}.\:\mathrm{Its}\:\mathrm{upper}\:\mathrm{surface}\:\mathrm{is}\:\mathrm{smooth}\:\mathrm{and} \\ $$$$\mathrm{vertical}\:\mathrm{surface}\:\mathrm{is}\:\mathrm{rough}\:\mathrm{with}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:\mathrm{friction}\:\mu.\:\mathrm{A}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{m}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{placed} \\ $$$$\mathrm{on}\:\mathrm{its}\:\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{and}\:\mathrm{tied}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{massless}\:\mathrm{inextensible}\:\mathrm{string}\:\mathrm{passing} \\ $$$$\mathrm{over}\:\mathrm{smooth}\:\mathrm{pulley}.\:\mathrm{Its}\:\mathrm{other}\:\mathrm{end}\:\mathrm{is} \\ $$$$\mathrm{connected}\:\mathrm{to}\:\mathrm{another}\:\mathrm{block}\:\mathrm{of}\:\mathrm{mass}\:{m}_{\mathrm{2}}…
Question Number 152631 by mnjuly1970 last updated on 30/Aug/21 Answered by Olaf_Thorendsen last updated on 30/Aug/21 $$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\left[−\frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{4}}…
Question Number 21557 by Tinkutara last updated on 27/Sep/17 $$\mathrm{A}\:\mathrm{man}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{85}\:\mathrm{kg}\:\mathrm{stands}\:\mathrm{on}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{30}\:\mathrm{kg}.\:\mathrm{When}\:\mathrm{he}\:\mathrm{pulls}\:\mathrm{on}\:\mathrm{the}\:\mathrm{rope}, \\ $$$$\mathrm{he}\:\mathrm{exerts}\:\mathrm{a}\:\mathrm{force}\:\mathrm{of}\:\mathrm{400}\:\mathrm{N}\:\mathrm{on}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{lift}.\:\mathrm{Calculate}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{lift}.\:\mathrm{Given}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} . \\ $$ Terms of Service Privacy…
Question Number 87093 by M±th+et£s last updated on 02/Apr/20 $$\lfloor\frac{{x}−\mathrm{1}}{\mathrm{4}}\rfloor+\lfloor\frac{{x}−\mathrm{2}}{\mathrm{3}}\rfloor=\lfloor\frac{{x}−\mathrm{3}}{\mathrm{2}}\rfloor \\ $$ Commented by M±th+et£s last updated on 02/Apr/20 $${slove}\:{the}\:{equation} \\ $$ Commented by MJS…
Question Number 152625 by Dandelion last updated on 30/Aug/21 $${x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{for}\:{c}_{{n}} \in\mathbb{R}\:\mathrm{this}\:\mathrm{can}\:\mathrm{have} \\ $$$$\mathrm{4}\:\mathrm{unique}\:\mathrm{zeros}\:\in\mathbb{R} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{zeros}\:+\:\mathrm{1}\:\mathrm{double}\:\mathrm{zero}\:\in\mathbb{R} \\…
Question Number 87088 by TawaTawa1 last updated on 02/Apr/20 Answered by mr W last updated on 02/Apr/20 Commented by mr W last updated on 02/Apr/20…
Question Number 152626 by mathdanisur last updated on 30/Aug/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{x}-\mathrm{1}}\:+\:\frac{\mathrm{2}}{\mathrm{x}-\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{x}-\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{x}-\mathrm{4}}\:=\:\mathrm{2x}^{\mathrm{2}} -\mathrm{5x}-\mathrm{4} \\ $$ Answered by MJS_new last updated on 30/Aug/21 $${x}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}}…