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Question Number 51768 by hassentimol last updated on 30/Dec/18 Commented by hassentimol last updated on 30/Dec/18 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{in}\:\mathrm{purple}\:\mathrm{given}\:\mathrm{that}\:\mathrm{1}\:\mathrm{small} \\ $$$$\mathrm{square}\:\mathrm{is}\:\mathrm{1}\:\mathrm{cm}^{\mathrm{2}} \:? \\ $$$$\mathrm{Thanks} \\ $$ Commented…

Question Number 182829 by yaslm last updated on 14/Dec/22 Answered by cortano1 last updated on 15/Dec/22 $$\:{L}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}\:\mathrm{sec}\:{x}−\mathrm{2}} \\ $$$$\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}\left(\mathrm{2sin}\:{x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\:{x}} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2cos}\:{x}} \\…

Question Number 117295 by mohammad17 last updated on 10/Oct/20 Answered by mr W last updated on 10/Oct/20 $${y}={r}\:\mathrm{sin}\:\theta \\ $$$${x}={r}\:\mathrm{cos}\:\theta \\ $$$${r}=\mathrm{1}+\mathrm{cos}\:\theta \\ $$$$\frac{{dy}}{{d}\theta}=\frac{{dr}}{{d}\theta}\:\mathrm{sin}\:\theta+{r}\:\mathrm{cos}\:\theta=−\mathrm{sin}^{\mathrm{2}} \:\theta+\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\mathrm{cos}\:\theta…

Question Number 182809 by leicianocosta last updated on 14/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 182791 by Matica last updated on 14/Dec/22 $$\:{We}\:{have}\:\mathrm{0}<{x}<{a}\:{and}\:{m},{n}\:\in\mathbb{N}. \\ $$$$\:{Prove}\:{x}^{{m}} \left({a}−{x}\right)^{{n}} \leqslant\:\frac{{m}^{{m}} {n}^{{n}} }{\left({m}+{n}\right)^{{m}+{n}} }\centerdot{a}^{{m}+{n}} \\ $$ Answered by mahdipoor last updated on…