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Question-201890

Question Number 201890 by sonukgindia last updated on 15/Dec/23 Answered by mr W last updated on 15/Dec/23 $${B}'=\left(−\mathrm{2},\mathrm{1}\right) \\ $$$$\frac{{y}_{{P}} −\mathrm{1}}{\mathrm{0}−\left(−\mathrm{2}\right)}=\frac{\mathrm{5}−\mathrm{1}}{\mathrm{5}−\left(−\mathrm{2}\right)} \\ $$$$\Rightarrow{y}_{{P}} =\frac{\mathrm{15}}{\mathrm{7}}\:\Rightarrow{P}\left(\mathrm{0},\:\frac{\mathrm{15}}{\mathrm{7}}\right) \\…

Question-201853

Question Number 201853 by sonukgindia last updated on 14/Dec/23 Answered by AST last updated on 14/Dec/23 $${Let}\:{green}\:{segment}={g};{orange}\:{segment}={r} \\ $$$${and}\:{blue}\:{segment}={b};{side}\:{of}\:{square}={s} \\ $$$$\frac{{b}}{{r}}=\frac{\mathrm{7}−\mathrm{5}}{\mathrm{7}−\mathrm{3}}\Rightarrow{r}=\mathrm{2}{b};\frac{{g}}{{g}+{s}}=\frac{\mathrm{2}}{\mathrm{4}}\Rightarrow{g}={s} \\ $$$$\frac{\mathrm{2}{b}}{\mathrm{2}{b}+{s}}=\frac{\mathrm{7}−\mathrm{3}}{\mathrm{13}−\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{5}}\Rightarrow\mathrm{5}{r}=\mathrm{2}{r}+\mathrm{2}{s}\Rightarrow{r}=\frac{\mathrm{2}{s}}{\mathrm{3}} \\ $$$${r}^{\mathrm{2}}…

Question-201873

Question Number 201873 by sonukgindia last updated on 14/Dec/23 Answered by witcher3 last updated on 14/Dec/23 $$\mathrm{Re}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\right)=\frac{\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} \right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\theta\right)}{\mathrm{2}+\mathrm{2cos}\left(\theta\right)}=\frac{\mathrm{1}}{\mathrm{2}};\forall\theta\in\mathbb{R}−\left\{\left(\mathrm{1}+\mathrm{2k}\right)\pi\right\} \\ $$$$\mathrm{I}==\frac{\varphi^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\…

n-1-1-n-H-n-n-1-

Question Number 201860 by MrGHK last updated on 14/Dec/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}=?? \\ $$ Answered by mnjuly1970 last updated on 14/Dec/23 $$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty}…

Do-Not-Use-sin-is-small-Enough-g-sin-0-y-t-g-sin-y-t-0-y-t-y-t-g-sin-y-t-y-t-0-y-t-y-t-1-2-d-dt-y-t-2-g-sin-y-t-y-t-g-d-dt-cos-y

Question Number 201822 by MathedUp last updated on 13/Dec/23 $$\mathrm{Do}\:\mathrm{Not}\:\mathrm{Use}\:\mathrm{sin}\left(\theta\right)\sim\theta\:\left(\theta\:\:\mathrm{is}\:\mathrm{small}\:\mathrm{Enough}\right) \\ $$$$\ddot {\theta}+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left(\theta\right)=\mathrm{0} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\:\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0} \\ $$$${y}''\left({t}\right){y}'\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=\mathrm{0} \\ $$$${y}'\left({t}\right){y}''\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=−\frac{\mathrm{g}}{\ell}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{cos}\left({y}\left({t}\right)\right) \\ $$$$\therefore\:\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)\right]=\mathrm{0} \\…

Question-201832

Question Number 201832 by sonukgindia last updated on 13/Dec/23 Answered by aleks041103 last updated on 13/Dec/23 $${x}={e}^{−{t}} \:\Rightarrow\:{dx}=−{e}^{−{t}} {dt} \\ $$$$\Rightarrow{I}=\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{5}} \left(−{t}\right)}{−{t}}{e}^{−{t}} {dt}=\mathrm{32}\int_{\mathrm{0}}…

Question-201806

Question Number 201806 by sonukgindia last updated on 12/Dec/23 Commented by mr W last updated on 30/Dec/23 $${see}\:{Q}\mathrm{202543} \\ $$$${answer}\:{is}\:\mathrm{2}\left(\mathrm{4}^{\mathrm{3}} −\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−…−\sqrt{\mathrm{15}}\right) \\ $$ Commented by…