Question Number 84834 by jagoll last updated on 16/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}}{\mathrm{sin}\:\mathrm{3x}} \\ $$ Commented by john santu last updated on 16/Mar/20 $$\mathrm{do}\:\mathrm{not}\:\mathrm{exist} \\ $$ Answered…
Question Number 84835 by Power last updated on 16/Mar/20 Commented by abdomathmax last updated on 16/Mar/20 $${S}=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \:\frac{{k}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \:\frac{{k}+\mathrm{1}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \left(\mathrm{1}\right)−\sum_{{k}=\mathrm{1}}…
Question Number 84830 by jagoll last updated on 16/Mar/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}−\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}} \\ $$ Answered by john santu last updated on 16/Mar/20 Terms of Service Privacy Policy…
Question Number 19294 by mondodotto@gmail.com last updated on 08/Aug/17 Answered by mrW1 last updated on 08/Aug/17 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{3} \\ $$$$=\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{5}}{\mathrm{4}}\mathrm{x}+\frac{\mathrm{25}}{\mathrm{16}}\right)+\mathrm{3}−\frac{\mathrm{25}}{\mathrm{8}} \\ $$$$=\mathrm{2}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{8}} \\…
Question Number 84828 by john santu last updated on 17/Mar/20 $$\mathrm{log}_{\mathrm{3}} \left(\mathrm{25x}^{\mathrm{2}} −\mathrm{4}\right)−\mathrm{log}_{\mathrm{3}} \left(\mathrm{x}\right)\:\leqslant\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{26x}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{x}}−\mathrm{10}\right) \\ $$ Commented by john santu last updated on…
Question Number 19293 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Let}\:{AC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{B}\:\mathrm{a}\:\mathrm{point}\:\mathrm{between}\:{A}\:\mathrm{and}\:{C}. \\ $$$$\mathrm{Construct}\:\mathrm{isosceles}\:\mathrm{triangles}\:{PAB}\:\mathrm{and} \\ $$$${QBC}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}\:{AC} \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle{APB}\:=\:\angle{BQC}\:=\:\mathrm{120}°\:\mathrm{and} \\ $$$$\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:{RAC}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{of}\:{AC}\:\mathrm{such}\:\mathrm{that}\:\angle{ARC}\:=\:\mathrm{120}°. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{PQR}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\…
Question Number 150366 by SLVR last updated on 11/Aug/21 Commented by SLVR last updated on 11/Aug/21 $${kindly}\:{provide}\:{solution}..{already} \\ $$$${given}\:{in}\:{group} \\ $$ Terms of Service Privacy…
Question Number 19292 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:…\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$ Commented by prakash jain last updated on 09/Aug/17 $$\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following} \\…
Question Number 84826 by jagoll last updated on 16/Mar/20 $$\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{2}}\:+\:\mathrm{log}_{\mathrm{3}} \:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{10}}\:=\:\mathrm{2} \\ $$ Commented by john santu last updated on 16/Mar/20 $$\mathrm{x}\:=\:\mathrm{1} \\…
Question Number 19291 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:. \\ $$ Answered by Tinkutara last updated on 16/Aug/17 $$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{sin}\:\mathrm{48}°\:+\:\mathrm{sin}\:\mathrm{24}°}{\mathrm{sin}\:\mathrm{48}°\:\mathrm{sin}\:\mathrm{24}°} \\…