Question Number 168722 by LEKOUMA last updated on 16/Apr/22 $${Resolve}\: \\ $$$${x}^{\mathrm{2}} {y}^{''} +{xy}^{'} +{y}=\mathrm{1} \\ $$ Commented by mokys last updated on 16/Apr/22 $${solution}…
Question Number 37645 by Raj Singh last updated on 16/Jun/18 Answered by Rasheed.Sindhi last updated on 16/Jun/18 $$\mathrm{Both}\:\mathrm{did}\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{of}\:\mathrm{work}\:\mathrm{in}\:\mathrm{1}\:\mathrm{day}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{work}\:\mathrm{in}\:\mathrm{4}\:\mathrm{days} \\ $$$$\mathrm{Remaining}\:\mathrm{part}\:\mathrm{of}\:\mathrm{work}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Ram}\:\mathrm{did}\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{work}\:\mathrm{in}\:\mathrm{20}\:\mathrm{days} \\…
Question Number 37632 by Raj Singh last updated on 16/Jun/18 Commented by Rasheed.Sindhi last updated on 16/Jun/18 $$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{word}\:\mathrm{problem}.\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{a}\:“\mathrm{puzzle}''. \\ $$ Answered by Rasheed.Sindhi…
Question Number 168679 by Tawa11 last updated on 15/Apr/22 Answered by mr W last updated on 15/Apr/22 $${N}={total}\:{force}\:{in}\:{A}−{frame} \\ $$$${N}_{{x}} =\mathrm{450}\:\mathrm{cos}\:\mathrm{20}° \\ $$$${N}_{{y}} =\mathrm{450}\:\mathrm{sin}\:\mathrm{20}°+\mathrm{10000} \\…
Question Number 103119 by Dwaipayan Shikari last updated on 12/Jul/20 $$\int\frac{{log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\sqrt{{x}}−\mathrm{1}\right)}{{x}^{\sqrt{{x}}} {log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\sqrt{{x}}+\mathrm{1}\right)−\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168616 by Tawa11 last updated on 14/Apr/22 Answered by mr W last updated on 14/Apr/22 $${M}={M}_{\mathrm{1}} +{M}_{\mathrm{2}} \\ $$$${M}_{\mathrm{1}} {v}=\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){V} \\…
Question Number 168617 by Tawa11 last updated on 14/Apr/22 Commented by mr W last updated on 14/Apr/22 $${since}\:{the}\:{position}\:{of}\:{particle}\:{at}\:{t}=\mathrm{0}\:{is} \\ $$$${not}\:{specified},\:{you}\:{can}'{t}\:{determine}\:{its} \\ $$$${position}\:{at}\:{t}=\mathrm{3}.\mathrm{5}\:{s}. \\ $$$${assume}\:{at}\:{t}=\mathrm{0}\:{it}\:{is}\:{at}\:{the}\:{mean} \\…
Question Number 168613 by LEKOUMA last updated on 14/Apr/22 $${Resolve}\: \\ $$$$\left.\mathrm{1}\right)\:{x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$ Commented by Mastermind last updated on…
Question Number 168608 by Mastermind last updated on 14/Apr/22 Answered by infinityaction last updated on 14/Apr/22 Commented by SLVR last updated on 14/Apr/22 $${nice}\:{sir} \\…
Question Number 168549 by LEKOUMA last updated on 13/Apr/22 $${Resolve} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} {y}^{''} −\mathrm{3}\left({x}−\mathrm{2}\right){y}'+{y}={x} \\ $$ Answered by mindispower last updated on 13/Apr/22 $${y}={ax}+{b} \\…