Question Number 84381 by behi83417@gmail.com last updated on 12/Mar/20 Commented by behi83417@gmail.com last updated on 12/Mar/20 $$\mathrm{AB}=\mathrm{9},\mathrm{BC}=\mathrm{10},\mathrm{CA}=\mathrm{12} \\ $$$$\mathrm{FI},\mathrm{divides}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{two}\:\mathrm{parts}\:\mathrm{that}\:\mathrm{equail}\:\mathrm{in}\: \\ $$$$\mathrm{area}\:\mathrm{and}\:\mathrm{perimeter}. \\ $$$$\mathrm{find}:\mathrm{CI},\mathrm{CF},\mathrm{FI}. \\ $$…
Question Number 84379 by Power last updated on 12/Mar/20 Answered by mind is power last updated on 12/Mar/20 $$=\int\frac{{x}\mathrm{2}{i}}{{e}^{{ix}} −{e}^{−{ix}} }{dx} \\ $$$$=\int\frac{{x}\mathrm{2}{ie}^{{ix}} }{\left({e}^{\mathrm{2}{ix}} −\mathrm{1}\right)}{dx}={f}\left({x}\right)…
Question Number 18841 by mondodotto@gmail.com last updated on 30/Jul/17 Answered by Tinkutara last updated on 31/Jul/17 $$\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{8}\theta\right)}}} \\ $$$$=\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{4}\theta\right)}}\:=\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{2}\theta\right)} \\ $$$$=\:\mathrm{2}\:\mathrm{cos}\:\theta \\ $$ Terms of…
Question Number 149914 by 7770 last updated on 08/Aug/21 $$\boldsymbol{{show}}\:\boldsymbol{{that}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{2}+\mathrm{6}\theta^{\mathrm{2}} −\mathrm{2}\boldsymbol{\theta}^{\mathrm{3}} }{\boldsymbol{\theta}^{\mathrm{2}} \left(\boldsymbol{\theta}^{\mathrm{2}} +\mathrm{1}\right)}\right)\boldsymbol{{d}\theta}=\mathrm{1}.\mathrm{606} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84377 by john santu last updated on 12/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{x}} \\ $$ Answered by john santu last updated on 12/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}\:×\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 84370 by Roland Mbunwe last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:<\mathrm{5} \\ $$$$\left.\mathrm{2}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:+\:\mid\mathrm{2}−{x}\mid\:\leqslant\mathrm{8} \\ $$ Answered by john santu last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\left(\mathrm{i}\right)\:\mathrm{x}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{2x}\:<\:\mathrm{3}\:,\:\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}}\: \\…
Question Number 18834 by mondodotto@gmail.com last updated on 30/Jul/17 Answered by Tinkutara last updated on 30/Jul/17 $$\mathrm{tan}\:\mathrm{2A}\:=\:\mathrm{cot}\:\left(\mathrm{90}°\:−\:\mathrm{2A}\right)\:=\:\mathrm{cot}\:\left(\mathrm{A}\:−\:\mathrm{18}°\right) \\ $$$$\mathrm{Since}\:\mathrm{A}\:\mathrm{is}\:\mathrm{acute},\:\mathrm{so}\:\mathrm{cot}\:\mathrm{A}\:=\:\mathrm{cot}\:\mathrm{B}\:\Leftrightarrow\:\mathrm{A}\:=\:\mathrm{B} \\ $$$$\mathrm{90}°\:−\:\mathrm{2A}\:=\:\mathrm{A}\:−\:\mathrm{18}° \\ $$$$\mathrm{A}\:=\:\mathrm{36}° \\ $$…
Question Number 84367 by bagjamath last updated on 12/Mar/20 $$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Given}\:\mathrm{the}\:\mathrm{squence}\:\mathrm{of}\:\mathbb{R} \\ $$$$\left({x}_{{n}} \right).\:{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}_{{k}} <\infty, \\ $$$${Find} \\…
Question Number 18830 by Joel577 last updated on 30/Jul/17 $$\mathrm{If}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:{a} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{simplest}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{cos}\:{x} \\ $$ Commented by mrW1 last updated on 30/Jul/17 $$\mathrm{cos}\:\mathrm{2x}=\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{2cos}^{\mathrm{2}}…
Question Number 149903 by bramlexs22 last updated on 08/Aug/21 $$\:\Omega\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}}\:\mathrm{dx}\: \\ $$ Answered by Ar Brandon last updated on 08/Aug/21 $$\Omega=\int_{\mathrm{0}}…