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Question-226855

Question Number 226855 by Kassista last updated on 17/Dec/25 Answered by mehdee7396 last updated on 17/Dec/25 $${x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{4}\Rightarrow{x}=\mathrm{0},\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\Rightarrow{P}\left(\mathrm{0},,\mathrm{2}\right)\:\&\:\:{Q}\left(\frac{\mathrm{8}}{\mathrm{5}},−\frac{\mathrm{6}}{\mathrm{5}}\right)\Rightarrow{m}_{{PQ}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}}\backsimeq\mathrm{36}^{\mathrm{0}} \Rightarrow\angle{POQ}\backsimeq\mathrm{126}^{\mathrm{0}}…

Question-226785

Question Number 226785 by fantastic2 last updated on 14/Dec/25 Commented by fantastic2 last updated on 15/Dec/25 $${there}\:{are}\:{three}\:{small}\:{balls}\:{of}\:{mass}\:{m}_{\mathrm{1},} {m}_{\mathrm{2}} \:{and}\:{m}_{\mathrm{3}} \\ $$$${all}\:{are}\:{hanging}\:{from}\:{a}\:{point}\:{O}\:{by}\:{a}\:{string} \\ $$$${length}\:{l}\:. \\ $$$${what}\:{charge}\:{should}\:{be}\:{given}\:{to}\:{them}\:{so}…

Question-226755

Question Number 226755 by Hanuda354 last updated on 13/Dec/25 Answered by TonyCWX last updated on 13/Dec/25 $${A}_{{Green}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{40}}{\mathrm{3}}\pi−\mathrm{16}\sqrt{\mathrm{3}} \\ $$$${A}_{{Blue}\:} =\:\left(\frac{\mathrm{12}−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2}\pi}{\mathrm{12}}\right)\left(\mathrm{8}^{\mathrm{2}} \right)=\mathrm{64}−\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{32}}{\mathrm{3}}\pi…

Prove-lim-v-n-1-2-e-n-k-1-2n-2n-k-1-2n-k-0-n-n-k-k-1-pi-e-2-2-3-

Question Number 226738 by MrAjder last updated on 12/Dec/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}: \\ $$$$\underset{{v}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{e}^{−{n}} \left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}^{\frac{\mathrm{1}}{\mathrm{2}{n}}} −\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}^{{k}} }{{k}!}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\left(\frac{{e}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$ Answered by TonyCWX…

Question-226651

Question Number 226651 by Hanuda354 last updated on 08/Dec/25 Commented by Hanuda354 last updated on 08/Dec/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\left(\mathrm{without}\:\mathrm{using}\:\mathrm{integration}\right). \\ $$ Answered by fantastic2 last updated on…