Question Number 71233 by mr W last updated on 13/Oct/19 Commented by ajfour last updated on 13/Oct/19 Commented by ajfour last updated on 13/Oct/19 $$\mathrm{cos}\:\mathrm{60}°=\frac{{a}^{\mathrm{2}}…
Question Number 136765 by Ñï= last updated on 25/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$ Answered by snipers237 last updated on 25/Mar/21 $${let}\:{named}\:{it}\:{A} \\…
Question Number 5695 by FilupSmith last updated on 24/May/16 Commented by FilupSmith last updated on 24/May/16 $$\mathrm{What}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{methods}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$$$\left(\mathrm{Both}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{identical}\right) \\ $$ Commented by…
Question Number 71229 by naka3546 last updated on 13/Oct/19 Commented by naka3546 last updated on 13/Oct/19 $${Difference}\:\:{of}\:\:{red}\:\:{area}\:\:{and}\:\:{blue}\:\:{area}\:\:{is}\:\:… \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 5692 by FilupSmith last updated on 24/May/16 $${f}\left({x}\right)={e}^{{x}} \\ $$$${g}\left({x}\right)=\mathrm{ln}\:{x} \\ $$$${h}\left({x}\right)={x} \\ $$$$ \\ $$$$\mathrm{A}\:\mathrm{line}\:{L}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:{h}\left({x}\right)\:\mathrm{at}\:\mathrm{point} \\ $$$${P}\left({x},{y}\right)\:\mathrm{and}\:\mathrm{extends}\:\mathrm{and}\:\mathrm{disects}\:{f}\left({x}\right)\:\mathrm{and}\:{g}\left({x}\right). \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:{L}\:\mathrm{between}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right) \\ $$$$\mathrm{is}\:{r}.\:\mathrm{When}\:\mathrm{is}\:{r}\:\mathrm{minimum}? \\…
Question Number 136761 by I want to learn more last updated on 25/Mar/21 Commented by I want to learn more last updated on 25/Mar/21 $$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.…
Question Number 136760 by I want to learn more last updated on 25/Mar/21 Answered by mr W last updated on 25/Mar/21 $$\frac{{BC}}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\angle{BDA}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{{DA}}\:\:\:…\left({ii}\right) \\…
Question Number 136762 by snipers237 last updated on 25/Mar/21 $$\:{let}\:\:{C}\left({a},{r}\right)=\left\{{z}\in\mathbb{C},\:\:\mid{z}−{a}\mid={r}\:\right\} \\ $$$${Let}\:{u},{v},{w}\:\in{C}\left({a},{r}\right)\:{such}\:{as}\:\:\:{u}+{v}=\mathrm{2}{w} \\ $$$${Prove}\:{that}\:\:\frac{{u}−{a}}{{v}−{a}}\:=\mathrm{1}\:,\:{u}={v}={w} \\ $$$$ \\ $$$${It}\:{shows}\:{that}\:{the}\:{middle}\:{of}\:{a}\:{segment}\: \\ $$$${joining}\:{two}\:{points}\:{in}\:{a}\:{circle}\:\:{is}\:{not}\:{in}\:{that}\:{circle} \\ $$ Terms of Service…
Question Number 136757 by Dwaipayan Shikari last updated on 25/Mar/21 $$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}+\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right).\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+….=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{2}\pi^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{4}} }\right)^{\mathrm{2}} \\ $$ Answered by mindispower last…
Question Number 71220 by ozodbek last updated on 13/Oct/19 Terms of Service Privacy Policy Contact: info@tinkutara.com