Question Number 133851 by physicstutes last updated on 24/Feb/21 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:{r}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{subtends}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\alpha\:\mathrm{at}\:\mathrm{its}\:\mathrm{centre}.\:\mathrm{A}\:\mathrm{chord}\:\mathrm{cuts}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{it}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{into}\:\mathrm{two}\:\mathrm{segments}\:\mathrm{of}\:\mathrm{areas}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:\mathrm{5}.\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\mathrm{sin}\:\alpha\:=\:\alpha−\frac{\pi}{\mathrm{3}} \\ $$ Terms of Service Privacy Policy…
Question Number 2777 by Filup last updated on 27/Nov/15 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\zeta\left(−\mathrm{2}{n}\right)=\mathrm{0} \\ $$ Answered by prakash jain last updated on 27/Nov/15 $$\mathrm{Functional}\:\mathrm{Equation}\:\mathrm{for}\:\zeta\left({s}\right) \\ $$$$\zeta\left({s}\right)=\mathrm{2}^{{s}}…
Question Number 68313 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\mathrm{1}−\frac{\mathrm{6}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }+\sqrt{{x}}\right) \\ $$ Commented by mathmax by abdo last updated on 10/Sep/19 $$\int\:\left(\mathrm{1}−\frac{\mathrm{6}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:+\sqrt{{x}}\right){dx}\:={x}−\mathrm{6}{ln}\mid{x}\mid\:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}\:}} \:+{c}…
Question Number 2776 by Filup last updated on 27/Nov/15 $$\mathrm{Is}: \\ $$$$\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}\geqslant\underset{{x}={a}} {\overset{{b}} {\sum}}{f}\left({x}\right) \\ $$$${a}<{x}<{b} \\ $$ Commented by Filup last updated…
Question Number 68308 by mr W last updated on 08/Sep/19 $${solve}\:{y}'''={y}''{y}' \\ $$ Answered by mind is power last updated on 08/Sep/19 $$\frac{{y}^{'''} }{{y}'^{'} }={y}'…
Question Number 68309 by ajfour last updated on 08/Sep/19 Commented by ajfour last updated on 08/Sep/19 $${Find}\:{AB}\:{and}\:{AC}\:{in}\:{terms}\:{of}\:{a}\: \\ $$$${and}\:\theta. \\ $$ Commented by mr W…
Question Number 2771 by prakash jain last updated on 26/Nov/15 $$\mathrm{64}\:\mathrm{points}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}: \\ $$$$\left({x},{y}\right),\:{x}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\},\:{y}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\} \\ $$$$\mathrm{4}\:\mathrm{points}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{proabability}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{connecting} \\ $$$$\mathrm{them}\:\mathrm{do}\:\mathrm{not}\:\mathrm{form}\:\mathrm{a}\:\mathrm{square}\:\mathrm{or}\:\mathrm{rectangle}? \\ $$ Answered by Rasheed Soomro…
Question Number 133840 by Algoritm last updated on 24/Feb/21 Answered by Dwaipayan Shikari last updated on 24/Feb/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \Gamma\left({x}+\mathrm{1}\right)−\mathrm{4}\Gamma\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}=\frac{{x}^{{x}} \left({logx}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)+\Gamma'\left({x}+\mathrm{1}\right){x}^{{x}} −\mathrm{12}\Gamma'\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma'\left({x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)+\mathrm{4}\Gamma'\left(\mathrm{3}\right)−\mathrm{12}\Gamma'\left(\mathrm{3}\right)}{\mathrm{3}\Gamma'\left(\mathrm{3}\right)}=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)}{\mathrm{6}\left(\frac{\mathrm{3}}{\mathrm{2}}−\gamma\right)}−\frac{\mathrm{8}}{\mathrm{3}} \\…
Question Number 2770 by Yozzi last updated on 26/Nov/15 $$\int_{\mathrm{2}} ^{\infty} \frac{{dx}}{{x}\sqrt{{x}−\mathrm{1}}}=? \\ $$ Answered by prakash jain last updated on 27/Nov/15 $${x}−\mathrm{1}=\mathrm{tan}^{\mathrm{2}} {u} \\…
Question Number 68305 by TawaTawa last updated on 08/Sep/19 Commented by ajfour last updated on 10/Sep/19 Commented by ajfour last updated on 10/Sep/19 $${Area}\:\:{S}_{{req}} =\underset{\smile}…