Question Number 103205 by DGmichael last updated on 13/Jul/20 Answered by OlafThorendsen last updated on 13/Jul/20 $${f}\left(\lambda\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\lambda^{{k}} }{{k}!} \\ $$$$\Rightarrow\:{f}'\left(\lambda\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{k}\lambda^{{k}−\mathrm{1}} }{{k}!}\:=\:\underset{{k}=\mathrm{1}}…
Question Number 37652 by Rio Mike last updated on 16/Jun/18 $$\mathrm{A}\:\mathrm{car},\mathrm{of}\:\mathrm{mass}\:\mathrm{1000kg},\mathrm{has}\:\mathrm{an}\:\mathrm{engine} \\ $$$$\mathrm{capable}\:\mathrm{of}\:\mathrm{developing}\:\mathrm{power}\:\mathrm{of}\: \\ $$$$\mathrm{15Kw}\:\mathrm{against}\:\mathrm{a}\:\mathrm{constand}\:\mathrm{Resistance} \\ $$$$\mathrm{R}\:\mathrm{N}.\mathrm{The}\:\mathrm{maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{on}\:\mathrm{level}\:\mathrm{road}\:\mathrm{is}\:\frac{\mathrm{100}}{\mathrm{3}}\:{ms}^{−\mathrm{1}} \: \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{R}. \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{resistance}\:\mathrm{and}\:\mathrm{the}\:\mathrm{power} \\…
Question Number 103186 by mohammad17 last updated on 13/Jul/20 $${Discuss}\:{whether}\:{the}\:{mean}\:{value}\:{theorem}\: \\ $$$${applies}\:{to}\:{the}\:{function}\:{f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103180 by mohammad17 last updated on 13/Jul/20 $${Let}\:{S}\:{be}\:{a}\:{nonempty}\:{subset}\:{of}\:{R}\:{that}\:{is}\:{bounded} \\ $$$${below}\:.\:{prove}\:{that}\:\left({inf}\left({S}\right)=−{sup}\left\{−{s}:{s}\in{S}\right\}\right)? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103177 by mohammad17 last updated on 13/Jul/20 $${Find}\:{the}\:{gineral}\:{form}\:{of}\:{the}\:{sequence}\:\langle\mathrm{2},−\mathrm{2},\mathrm{2},−\mathrm{2},…..\rangle? \\ $$ Answered by Dwaipayan Shikari last updated on 13/Jul/20 $$\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$…
Question Number 168710 by bekzodjumayev last updated on 16/Apr/22 Commented by bekzodjumayev last updated on 16/Apr/22 $${Please}\:{help} \\ $$ Commented by mather last updated on…
Question Number 168680 by mokys last updated on 15/Apr/22 Commented by mokys last updated on 15/Apr/22 $${help}\:{me}\:{sir}\:{please} \\ $$ Terms of Service Privacy Policy Contact:…
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 $${false}…
Question Number 168665 by mokys last updated on 15/Apr/22 Commented by cortano1 last updated on 15/Apr/22 $$\:{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}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{8}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{16}\right)=\mathrm{16}+\mathrm{4}=\mathrm{20}…
Question Number 37591 by mondodotto@gmail.com last updated on 15/Jun/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{y}}−\mathrm{8}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{24}\boldsymbol{{x}}+\boldsymbol{{hy}}=\mathrm{0}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{orthogonally}}, \\ $$$$\boldsymbol{\mathrm{determine}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{h}}. \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com…