Question Number 208741 by Tawa11 last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…
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Question Number 208743 by Tawa11 last updated on 22/Jun/24 Commented by Tawa11 last updated on 22/Jun/24 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle}. \\ $$ Commented by Tawa11 last updated on…
Question Number 208738 by Tawa11 last updated on 22/Jun/24 $$\mathrm{sum}: \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{4}.\mathrm{7}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{7}.\mathrm{10}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{10}.\mathrm{13}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{10}.\mathrm{13}.\mathrm{16}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{13}.\mathrm{16}.\mathrm{19}}\:\:+\:\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{94}.\mathrm{97}.\mathrm{100}} \\ $$ Answered by BaliramKumar last updated on 22/Jun/24 $$\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{\mathrm{1}}{\mathrm{1}×\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{97}×\mathrm{100}}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{d}=\:\mathrm{7}−\mathrm{1}\:=\:\mathrm{10}−\mathrm{4}\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\frac{\left(\mathrm{97}×\mathrm{25}−\mathrm{1}\right)}{\mathrm{97}×\mathrm{100}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{2424}}{\mathrm{9700}}\: \\…
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Question Number 208692 by Ghisom last updated on 21/Jun/24 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{x}\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}=? \\ $$ Answered by Berbere last updated on 21/Jun/24 $${ln}\left({sin}\left({x}\right)\right)=−{ln}\left(\mathrm{2}\right)−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{kx}\right)}{{k}} \\ $$$${proof}\:{ln}\left({sin}\left({x}\right)\right)={Re}\left({ln}\left({sin}\left({x}\right)\right)\right.…
Question Number 208693 by Frix last updated on 21/Jun/24 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\left(\frac{\mathrm{1}}{{x}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Answered by mr W last updated on 21/Jun/24 $$\frac{\mathrm{1}}{{x}}−\mathrm{2}={u}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}+{u}}…
Question Number 208690 by Tawa11 last updated on 21/Jun/24 Answered by mr W last updated on 21/Jun/24 $$\left(\mathrm{2}{r}−\mathrm{3}\right)×\mathrm{3}=\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$${shaded}\:{area}=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{13}}{\mathrm{6}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{13}}{\mathrm{6}}\right)×\mathrm{2}\approx\mathrm{3}.\mathrm{04} \\ $$…
Question Number 208685 by hardmath last updated on 21/Jun/24 $$\mathrm{25}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{3} \\ $$$$\mathrm{x}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/24 $$\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{3}}{\mathrm{25}} \\ $$$$\mathrm{x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{25}}\right)…
Question Number 208686 by hardmath last updated on 21/Jun/24 $$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2x}\:=\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2x}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$ Answered by MM42 last updated on 21/Jun/24 $${cos}^{\mathrm{2}} \mathrm{2}{x}−{sin}^{\mathrm{2}}…