Question Number 56749 by pete last updated on 22/Mar/19 $$\mathrm{The}\:\mathrm{curve}\:\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\:\mathrm{crosses}\:\mathrm{the} \\ $$$$\mathrm{y}−\mathrm{axis}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{0},\mathrm{3}\right)\:\mathrm{and}\:\mathrm{has} \\ $$$$\mathrm{stationary}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{1},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{values}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$ Answered by ajfour last updated on…
Question Number 56738 by pieroo last updated on 22/Mar/19 $$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{bx},\:\mathrm{where} \\ $$$$\mathrm{b}\:\mathrm{is}\:\mathrm{constant},\:\mathrm{at}\:\mathrm{x}=\mathrm{1}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the} \\ $$$$\mathrm{points}\:\mathrm{A}\left(−\mathrm{1},\mathrm{6}\right)\:\mathrm{and}\:\left(\mathrm{2},−\mathrm{15}\right).\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{b}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 122264 by mnjuly1970 last updated on 15/Nov/20 Commented by mr W last updated on 15/Nov/20 $$\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{{k}} ^{{n}} \right){x}^{{k}} \\ $$$$\left(\mathrm{1}−\mathrm{1}\right)^{{n}}…
Question Number 122185 by mnjuly1970 last updated on 14/Nov/20 $$\:\:\:\:\:\:…\:{challanging}\:\:{math}… \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}−\mathrm{1}} {n}}\overset{???} {=}\gamma\:+{ln}\left(\frac{\pi}{\mathrm{4}}\right)\: \\ $$$$\:\:{hint}\:::\:\zeta\left({s}\right)=\frac{\mathrm{1}}{\Gamma\left({s}\right)}\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} −\mathrm{1}}{dx}…
Question Number 121956 by bemath last updated on 13/Nov/20 $${Find}\:{concave}\:{up}\:{and}\:{concove}\:{down} \\ $$$${of}\:{function}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:\mid{x}−\mathrm{3}\mid\: \\ $$ Answered by MJS_new last updated on 13/Nov/20 $${f}\left({x}\right)=\begin{cases}{\mathrm{3}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} ;\:{x}<\mathrm{3}}\\{{x}^{\mathrm{3}}…
Question Number 187458 by mnjuly1970 last updated on 17/Feb/23 $$ \\ $$$$ \\ $$$$\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:\:\mathrm{2}{sin}\left({x}\right)\:−\:{cos}\left({x}\right)}{{sin}\left({x}\right)\:+\:{cos}\:\left({x}\right)}\:{dx}\:=\:? \\ $$$$−−−− \\ $$ Answered by Frix last updated…
Question Number 121882 by oustmuchiya@gmail.com last updated on 12/Nov/20 Commented by oustmuchiya@gmail.com last updated on 12/Nov/20 Differentiate the above functions Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121869 by mnjuly1970 last updated on 12/Nov/20 $$\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\frac{{ln}\left({tan}\left({x}\right)\right)}{{sin}\left({x}\right)−{cos}\left({x}\right)}\right)^{\mathrm{2}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$ Answered by mindispower…
Question Number 121729 by oustmuchiya@gmail.com last updated on 11/Nov/20 Answered by bemath last updated on 11/Nov/20 $$\left({i}\right)\:\mathrm{2}{yy}'+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} {y}'=\mathrm{3}{y}' \\ $$$$\Leftrightarrow\:\mathrm{3}{x}^{\mathrm{2}} \:=\:{y}'\left(\mathrm{3}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{2}{y}\right) \\ $$$$\Leftrightarrow\:{y}'\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}}…
Question Number 121713 by mnjuly1970 last updated on 11/Nov/20 $$\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:{please}\:\:{prove}: \\ $$$$\Phi=\int_{\mathrm{0}} ^{\:\infty} {arctan}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\frac{{dx}}{{x}}\:=\pi{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}… \\ $$ Terms of Service Privacy…