Question Number 134040 by benjo_mathlover last updated on 27/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$$\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}−\mathrm{1and}\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}−\mathrm{1} \\ $$ Commented by benjo_mathlover last updated on 27/Feb/21…
Question Number 133994 by benjo_mathlover last updated on 26/Feb/21 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}\:\mathrm{satisfies} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{{x}+{a}\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}\:;\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{4}}}\\{\mathrm{2}{x}\:\mathrm{cot}\:{x}\:+{b}\:;\:\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}}\\{{a}\:\mathrm{cos}\:\mathrm{2}{x}−{b}\mathrm{sin}\:{x}\:;\:\frac{\pi}{\mathrm{2}}<{x}\leqslant\pi}\end{cases} \\ $$$$\:\mathrm{continuous}\:\mathrm{in}\:\left[\:\mathrm{0},\pi\:\right],\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b}.\: \\ $$ Answered by EDWIN88 last updated on 26/Feb/21…
Question Number 133928 by mnjuly1970 last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{2}\right)\:\mathrm{u}_{\mathrm{n}} =\left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:=\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}}…
Question Number 133915 by nherly last updated on 25/Feb/21 Commented by TheSupreme last updated on 25/Feb/21 $${nice}\:{photo} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 68350 by mhmd last updated on 09/Sep/19 Commented by MJS last updated on 09/Sep/19 $$\left({x}−\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\left({x}−\mathrm{2cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\left({x}−\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=\mathrm{0} \\ $$$$\mathrm{approximating}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}…
Question Number 68331 by Peculiar last updated on 08/Sep/19 $${Differentiate}\:{y}={ln}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \underset{} {\right)} \\ $$ Answered by MJS last updated on 08/Sep/19 $${y}={h}\left({g}\left({f}\left({x}\right)\right)\right) \\ $$$${y}'={h}'\left({g}\left({f}\left({x}\right)\right)\right)×{g}'\left({f}\left({x}\right)\right)×{f}'\left({x}\right)…
Question Number 133859 by mnjuly1970 last updated on 24/Feb/21 Answered by Ñï= last updated on 24/Feb/21 $$\underset{{n}=−\infty} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}−{n}\pi\right)^{\mathrm{2}}…
Question Number 68270 by ~ À ® @ 237 ~ last updated on 08/Sep/19 $$\:{Prove}\:{that}\:\:{if}\:\:{Li}_{\mathrm{2}} \left({x}\right)=\underset{{n}=\mathrm{1}} {\sum}\:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:\:\:{then} \\ $$$$\forall\:{x}\:\:{Li}_{\mathrm{2}} \left({x}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\:\: \\…
Question Number 68257 by ~ À ® @ 237 ~ last updated on 07/Sep/19 $$\:\:{Prove}\:{that}\:\:\frac{\mathrm{6}}{\mathrm{673}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2019}}\: \\ $$ Terms of Service Privacy…
Question Number 68232 by mhmd last updated on 07/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com