Question Number 29222 by imamu222 last updated on 05/Feb/18 $$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{12},\:{xy}=\mathrm{2} \\ $$ Answered by ajfour last updated on 05/Feb/18 $$\mathrm{3}{x}−\frac{\mathrm{8}}{{x}}=\mathrm{12} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}=\mathrm{8} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}}…
Question Number 29209 by tawa tawa last updated on 05/Feb/18 Commented by tawa tawa last updated on 05/Feb/18 $$\mathrm{please}\:\mathrm{help}\:\mathrm{with}.\:\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7} \\ $$ Commented by ajfour last…
Question Number 94739 by 174 last updated on 20/May/20 Commented by i jagooll last updated on 20/May/20 $$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{3sin}\:\mathrm{x}−\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)}= \\ $$$$\int\:\frac{\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{3sin}\:\mathrm{x}−\mathrm{2}}\:=\:\int\:\frac{\mathrm{du}}{\left(\mathrm{2u}−\mathrm{1}\right)\left(\mathrm{u}+\mathrm{2}\right)} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\…
Question Number 29172 by byaw last updated on 04/Feb/18 Answered by mrW2 last updated on 04/Feb/18 $${mv}_{\mathrm{0}} =\left({m}+{M}\right){v}_{\mathrm{1}} \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\frac{\left({m}+{M}\right){v}_{\mathrm{1}} }{{m}}=\frac{\mathrm{0}.\mathrm{084}+\mathrm{20}}{\mathrm{0}.\mathrm{084}}×\mathrm{0}.\mathrm{24}=\mathrm{57}.\mathrm{38}\:{m}/{s} \\ $$ Terms…
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Question Number 29168 by byaw last updated on 04/Feb/18 Commented by byaw last updated on 04/Feb/18 $${please}\:{help}\:{me}\:{solve}\:{all}\:{the}\:{questions} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 29144 by tawa tawa last updated on 04/Feb/18 $$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{orbit}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{4R}\:\mathrm{round}\:\mathrm{the}\:\mathrm{earth}.\:\:\mathrm{Express}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{free}\:\mathrm{fall}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{g} \\ $$$$\mathrm{R}\:=\:\mathrm{radius}\:\mathrm{if}\:\mathrm{the}\:\mathrm{earth} \\ $$$$\mathrm{g}\:=\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity} \\ $$ Answered by ajfour last updated on…
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Question Number 29048 by $@ty@m last updated on 03/Feb/18 Commented by ajfour last updated on 04/Feb/18 $${thanks}\:{for}\:{verifying}\:{Sir}! \\ $$ Commented by $@ty@m last updated on…
Question Number 28991 by abdo imad last updated on 02/Feb/18 $${prove}\:{that}\:{L}\left(\mathrm{1}\right)\left({s}\right)=\:\frac{\mathrm{1}}{{s}}\:\:{and}\:{L}\left({t}^{{n}} \right)\left({s}\right)=\:\frac{{n}!}{{s}^{{n}+\mathrm{1}} }\:.{L}\:{means} \\ $$$${laplace}\:{transform}. \\ $$ Commented by abdo imad last updated on 03/Feb/18…