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Sir-Aifour-I-just-answered-your-question-number-37209-greetings-from-a-rainy-Saturday-evening-in-Vienna-Austria-

Question Number 37691 by MJS last updated on 16/Jun/18 $$\mathrm{Sir}\:\mathrm{Aifour},\:\mathrm{I}\:\mathrm{just}\:\mathrm{answered}\:\mathrm{your}\:\mathrm{question} \\ $$$$\mathrm{number}\:\mathrm{37209}…\:\mathrm{greetings}\:\mathrm{from}\:\mathrm{a}\:\mathrm{rainy} \\ $$$$\mathrm{Saturday}\:\mathrm{evening}\:\mathrm{in}\:\mathrm{Vienna},\:\mathrm{Austria}! \\ $$ Commented by ajfour last updated on 16/Jun/18 $${rainy}\:{even}\:{up}\:{here},\:{Sir} \\…

Question-103205

Question Number 103205 by DGmichael last updated on 13/Jul/20 Answered by OlafThorendsen last updated on 13/Jul/20 $${f}\left(\lambda\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\lambda^{{k}} }{{k}!} \\ $$$$\Rightarrow\:{f}'\left(\lambda\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}\lambda^{{k}−\mathrm{1}} }{{k}!}\:=\:\underset{{k}=\mathrm{1}}…

A-car-of-mass-1000kg-has-an-engine-capable-of-developing-power-of-15Kw-against-a-constand-Resistance-R-N-The-maximum-speed-of-the-car-on-level-road-is-100-3-ms-1-Calculate-the-value-of-R-Gi

Question Number 37652 by Rio Mike last updated on 16/Jun/18 $$\mathrm{A}\:\mathrm{car},\mathrm{of}\:\mathrm{mass}\:\mathrm{1000kg},\mathrm{has}\:\mathrm{an}\:\mathrm{engine} \\ $$$$\mathrm{capable}\:\mathrm{of}\:\mathrm{developing}\:\mathrm{power}\:\mathrm{of}\: \\ $$$$\mathrm{15Kw}\:\mathrm{against}\:\mathrm{a}\:\mathrm{constand}\:\mathrm{Resistance} \\ $$$$\mathrm{R}\:\mathrm{N}.\mathrm{The}\:\mathrm{maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{on}\:\mathrm{level}\:\mathrm{road}\:\mathrm{is}\:\frac{\mathrm{100}}{\mathrm{3}}\:{ms}^{−\mathrm{1}} \: \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{R}. \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{resistance}\:\mathrm{and}\:\mathrm{the}\:\mathrm{power} \\…

convert-the-intigeral-1-2-0-1-4xy-3-dxdy-to-polar-cordinaite-and-it-solve-

Question Number 168669 by mokys last updated on 15/Apr/22 $$\boldsymbol{{convert}}\:\boldsymbol{{the}}\:\boldsymbol{{intigeral}}\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{4}\boldsymbol{{xy}}^{\mathrm{3}} \boldsymbol{{dxdy}}\:\boldsymbol{{to}}\:\boldsymbol{{polar}}\:\boldsymbol{{cordinaite}}\:\boldsymbol{{and}}\:\boldsymbol{{it}}\:\boldsymbol{{solve}}\:? \\ $$ Commented by mokys last updated on 15/Apr/22 $${false}…

Question-168665

Question Number 168665 by mokys last updated on 15/Apr/22 Commented by cortano1 last updated on 15/Apr/22 $$\:{A}_{\mathrm{1}} =\:\underset{−\mathrm{2}} {\overset{\mathrm{0}} {\int}}\left(\mathrm{8}−{x}^{\mathrm{3}} \right){dx}=\:\left[\mathrm{8}{x}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{−\mathrm{2}} ^{\mathrm{0}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{8}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{16}\right)=\mathrm{16}+\mathrm{4}=\mathrm{20}…