Question Number 137946 by otchereabdullai@gmail.com last updated on 08/Apr/21 $$\mathrm{From}\:\mathrm{a}\:\mathrm{poin}\:\mathrm{P}\:\mathrm{outside}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{a} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{T}. \\ $$$$\mathrm{from}\:\mathrm{P}\:\mathrm{a}\:\mathrm{line}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{circle}\:\mathrm{at} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:.\:\mathrm{Find}\:\mathrm{lenth}\:\mathrm{PT}\:\mathrm{giving}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\mathrm{PA}\:=\mathrm{6} \\ $$$$\:\:\:\:\mathrm{AB}=\mathrm{8} \\ $$$$\mathrm{ii}.\:\mathrm{PB}=\mathrm{18} \\ $$$$\:\:\:\:\:\mathrm{PT}=\mathrm{12} \\…
Question Number 137941 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}{ln}\:\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{dx}=\frac{\pi}{\mathrm{2}{ln}^{\mathrm{2}} \mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 137936 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{4}{n}}\\{\mathrm{2}{n}}\end{pmatrix}^{−\mathrm{1}} =\frac{\mathrm{16}}{\mathrm{15}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{27}}\pi−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{25}}{ln}\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$ Commented by Dwaipayan Shikari last updated on 08/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 137939 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left(\mathrm{1}+{x}+{y}\right){dxdy}=\frac{\mathrm{5}}{\mathrm{2}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\pi−\frac{\mathrm{9}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 137938 by Ñï= last updated on 09/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{12}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{3}/\mathrm{2}} }{dx}=? \\ $$$${A}\:{very}\:{nice}\:{solution}:: \\ $$$$\int\frac{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{12}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{3}/\mathrm{2}} }{dx}=\frac{{f}\left({x}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}}+{C}…
Question Number 72362 by Maclaurin Stickker last updated on 27/Oct/19 $${Calculate}\:{the}\:{sides}\:{of}\:{a}\:{triangle} \\ $$$${knowing}\:{the}\:{heights}\:{h}_{\mathrm{a}\:} =\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${h}_{\mathrm{b}} =\frac{\mathrm{1}}{\mathrm{7}}\:{and}\:{h}_{\mathrm{c}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Answered by mr W last…
Question Number 6825 by madscientist last updated on 30/Jul/16 $${is}\:{there}\:{a}\:{triple}\:{integral}\:{that}\:{equals} \\ $$$${to}\:{the}\:{golden}\:{ratio}\:\phi\:\mathrm{1}.\mathrm{618}? \\ $$$$ \\ $$$$ \\ $$ Answered by FilupSmith last updated on 30/Jul/16…
Question Number 72339 by naka3546 last updated on 27/Oct/19 $$\sqrt{{x}\:+\:\sqrt{\mathrm{4}{x}\:+\:\sqrt{\mathrm{16}{x}\:+\:…\:+\:\sqrt{\mathrm{4}^{\mathrm{2019}} {x}\:+\:\mathrm{3}}}}}\:\:=\:\:\sqrt{{x}}\:+\:\mathrm{1} \\ $$ Commented by naka3546 last updated on 27/Oct/19 $${x}\:\:=\:\:? \\ $$ Terms of…
Question Number 137869 by mohammad17 last updated on 07/Apr/21 $${prove}\:{that}\:{Arg}\left({z}\right)\:{is}\:{harmonic}\:{function} \\ $$$${and}\:{find}\:{the}\:{conjecate}\:{this}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 137871 by pete last updated on 07/Apr/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{standard} \\ $$$$\mathrm{normal}\:\mathrm{distribution}\:\mathrm{p}\left(−\mathrm{1}.\mathrm{96}\leqslant\mathrm{Z}\leqslant\mathrm{2}.\mathrm{5}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com