Question Number 109305 by peter frank last updated on 22/Aug/20 Answered by Aziztisffola last updated on 22/Aug/20 $$\:\mathrm{let}\:\mathrm{A}\left(\mathrm{4};\mathrm{0}\right)\:\mathrm{and}\:\mathrm{P}\left({x};{y}\right)\in\mathscr{C}\left(\mathrm{o};\mathrm{2}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow\:{y}=\underset{−} {+}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} } \\…
Question Number 109302 by peter frank last updated on 22/Aug/20 Answered by Dwaipayan Shikari last updated on 22/Aug/20 $${x}=\left(\mathrm{10}−{y}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={y}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{20}{y}+\mathrm{100}=\mathrm{2}\left({y}−\mathrm{5}\right)^{\mathrm{2}}…
Question Number 109303 by peter frank last updated on 22/Aug/20 Commented by som(math1967) last updated on 23/Aug/20 $$\mathrm{Acute}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{y}=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:\mathrm{istan}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore\mathrm{angle}\:\mathrm{between}\:\mathrm{bisector} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:=\frac{\pi}{\mathrm{8}}…
Question Number 109271 by 675480065 last updated on 22/Aug/20 Commented by bobhans last updated on 22/Aug/20 can't be read the writing in your photo Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 109246 by peter frank last updated on 22/Aug/20 Answered by Ar Brandon last updated on 22/Aug/20 $$\mathrm{a}.\:\mathrm{I}=\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{2}}}=\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{2}=\sqrt{\mathrm{2}}\mathrm{ch}\theta\Rightarrow\mathrm{dx}=\sqrt{\mathrm{2}}\mathrm{sh}\theta\mathrm{d}\theta \\ $$$$\:\:\:\mathrm{I}=\int\frac{\mathrm{sh}\theta\mathrm{d}\theta}{\left(\sqrt{\mathrm{2}}\mathrm{ch}\theta−\mathrm{1}\right)\sqrt{\mathrm{sh}^{\mathrm{2}}…
Question Number 174677 by pluton last updated on 08/Aug/22 Commented by Rasheed.Sindhi last updated on 08/Aug/22 $$\mathrm{See}\:\mathrm{Q}#\mathrm{174591} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 109075 by Rio Michael last updated on 20/Aug/20 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{twe}\:\mathrm{circles}\: \\ $$$${C}_{\mathrm{1}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{6}{x}−\mathrm{4}{y}\:+\:\mathrm{9}\:=\:\mathrm{0}\:\mathrm{and}\:{C}_{\mathrm{2}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{6}{y}\:+\:\mathrm{9}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:{C}_{\mathrm{3}} \:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre} \\ $$$$\mathrm{of}\:{C}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{C}_{\mathrm{1}}…
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