Question Number 8846 by Rasheed Soomro last updated on 31/Oct/16 $$\mathrm{Let}\:\mathrm{by}\:\left(\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…\mathrm{a}_{\mathrm{n}} \right)\:\mathrm{we}\:\mathrm{mean}\:\mathrm{LCM} \\ $$$$\mathrm{of}\:\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…\mathrm{a}_{\mathrm{n}} \:,\mathrm{where}\:\mathrm{a}_{\mathrm{i}} \in\mathbb{N}. \\ $$$$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}\:\left(\:\left(\mathrm{a},\mathrm{b}\right),\left(\mathrm{b},\mathrm{c}\right)\:\:\right)=\left(\mathrm{a},\mathrm{b},\mathrm{c}\right). \\ $$ Answered…
Question Number 8845 by tawakalitu last updated on 31/Oct/16 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{figure}\:\mathrm{out}\:\mathrm{this}\: \\ $$$$\mathrm{Quantitative}\:\mathrm{reasoning} \\ $$$$\mathrm{How}\:\mathrm{did}\:\mathrm{they}\:\mathrm{get}\:\mathrm{this}\:\mathrm{answers}. \\ $$$$ \\ $$$$\mathrm{2},\mathrm{613},\mathrm{400}\:=\:\mathrm{3} \\ $$$$\mathrm{2},\mathrm{451},\mathrm{100}\:=\:\mathrm{1} \\ $$$$\mathrm{2},\mathrm{541},\mathrm{100}\:=\:\mathrm{2} \\ $$$$\mathrm{3},\mathrm{000},\mathrm{001}\:=\:\mathrm{1} \\…
Question Number 8829 by tawakalitu last updated on 30/Oct/16 $$\mathrm{The}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{three}\:\mathrm{whole}\:\mathrm{number}\:\mathrm{is}\:\mathrm{144}.\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{their}\:\mathrm{third}\:\mathrm{common}\:\mathrm{multiple}. \\ $$ Commented by Rasheed Soomro last updated on 31/Oct/16 $$\mathrm{All}\:\mathrm{the}\:\mathrm{common}\:\mathrm{multiples}\:\mathrm{are}\:\mathrm{multiples}\:\mathrm{of}\:\mathrm{LCM}. \\ $$$$\mathrm{First}\:\mathrm{common}\:\mathrm{multiple}\:\mathrm{is}\:\mathrm{LCM}\:\mathrm{itself}.\:\mathrm{The}\:\mathrm{second}…
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Question Number 8764 by tawakalitu last updated on 26/Oct/16 $$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{given}\:\mathrm{by} \\ $$$$\mathrm{S}_{\mathrm{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}}\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{2n}\:−\:\mathrm{1}\right)\left(\mathrm{2n}\:+\:\mathrm{1}\right)} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\:\:\mathrm{S}_{\mathrm{n}} \:\:\mathrm{as}\:\:\mathrm{n}\:\rightarrow\:\infty \\ $$ Commented by sou1618 last updated on 26/Oct/16…
Question Number 139814 by help last updated on 01/May/21 Answered by physicstutes last updated on 01/May/21 $${PV}\:=\:{nRT} \\ $$$$\mathrm{and}\:{P}\:\mathrm{increases}\:\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{0}.\mathrm{10}\:\mathrm{atm}/\mathrm{min}\:\mathrm{and}\:{V}\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{at} \\ $$$$\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{0}.\mathrm{15}\:\mathrm{L}/\mathrm{min}\:\mathrm{when}\:{n}\:=\:\mathrm{10}\: \\ $$$${T}\:=\:\frac{{PV}}{{nR}}\:\Rightarrow\:\frac{{dT}}{{dt}}\:=\:\frac{\mathrm{1}}{{nR}}\left({V}_{\mathrm{0}} \frac{{dP}}{{dt}}+{P}_{\mathrm{0}} \frac{{dV}}{{dt}}\right)…
Question Number 139771 by Engr_Jidda last updated on 01/May/21 Commented by mohammad17 last updated on 01/May/21 $${yes}\:{sir}\:{realy}\:{im}\:{sory}\:{can}\:{you}\:{solve}\:{this}? \\ $$ Commented by mr W last updated…
Question Number 8554 by Sopheak last updated on 16/Oct/16 $${Problem}\:.\mathrm{15} \\ $$$${Find}\:{the}\:{sum}\:{of} \\ $$$${S}=\:\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}+\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+\frac{\mathrm{5}}{\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}+…+\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!} \\ $$$$\: \\ $$ Commented by Yozzias last updated on 16/Oct/16…
Question Number 8558 by Sopheak last updated on 16/Oct/16 Commented by FilupSmith last updated on 16/Oct/16 $${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{4}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}=\frac{{A}}{\mathrm{2}{k}−\mathrm{1}}+\frac{{B}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\mathrm{1}={A}\left(\mathrm{2}{k}+\mathrm{1}\right)+{B}\left(\mathrm{2}{k}−\mathrm{1}\right)…
Question Number 8553 by Sopheak last updated on 16/Oct/16 Commented by Yozzias last updated on 16/Oct/16 $$\frac{\mathrm{1}}{\mathrm{n}!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{\mathrm{n}!}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)=\frac{\mathrm{n}}{\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!} \\ $$ Answered by Yozzias last updated on…