Question Number 175103 by Tawa11 last updated on 19/Aug/22 $$\mathrm{If}\:\:\mathrm{20}\:\left(\mathrm{mod}\:\mathrm{9}\right)\:\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\:\:\mathrm{y}\:\left(\mathrm{mod}\:\mathrm{6}\right).\:\:\mathrm{Find}\:\:\mathrm{y} \\ $$ Answered by Rasheed.Sindhi last updated on 19/Aug/22 $$\mathrm{20}\left(\mathrm{mod}\:\mathrm{9}\right)=\mathrm{11} \\ $$$$\mathrm{y}\equiv\mathrm{11}\left(\mathrm{mod}\:\mathrm{6}\right) \\ $$$$\mathrm{y}\equiv\mathrm{11}−\mathrm{6}=\mathrm{5}\left(\mathrm{mod}\:\mathrm{6}\right) \\…
Question Number 175004 by saboorhalimi last updated on 15/Aug/22 Commented by kaivan.ahmadi last updated on 15/Aug/22 $${s}_{\mathrm{9}} =\mathrm{6}{s}_{\mathrm{6}} \Rightarrow\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{8}{d}\right)=\mathrm{6}×\mathrm{3}\left(\mathrm{2}{a}+\mathrm{5}{d}\right) \\ $$$$\Rightarrow\mathrm{9}{a}+\mathrm{36}{d}=\mathrm{36}{a}+\mathrm{90}{d} \\ $$$$\Rightarrow−\mathrm{27}{a}=\mathrm{54}{d}\Rightarrow\mathrm{d}=\frac{−\mathrm{1}}{\mathrm{2}}\mathrm{a} \\ $$$$\mathrm{so}…
Question Number 174939 by BaliramKumar last updated on 14/Aug/22 $$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{all}\:\mathrm{proper}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{360}\:? \\ $$ Answered by JDamian last updated on 14/Aug/22 $$\mathrm{360}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5} \\ $$$$\frac{\mathrm{2}^{\mathrm{3}+\mathrm{1}} −\mathrm{1}}{\mathrm{2}−\mathrm{1}}×\frac{\mathrm{3}^{\mathrm{2}+\mathrm{1}}…
Question Number 43793 by gunawan last updated on 15/Sep/18 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:^{\mathrm{3}} \sqrt{\mathrm{20}+\mathrm{14}\sqrt{\mathrm{2}}}+^{\mathrm{3}} \sqrt{\mathrm{20}−\mathrm{14}\sqrt{\mathrm{2}}}\:\mathrm{is}\ldots \\ $$ Answered by Joel578 last updated on 15/Sep/18 $$\mathrm{Let}\:{a}\:=\:\mathrm{20}\:+\:\mathrm{14}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}\:=\:\mathrm{20}\:−\:\mathrm{14}\sqrt{\mathrm{2}} \\…
Question Number 109305 by peter frank last updated on 22/Aug/20 Answered by Aziztisffola last updated on 22/Aug/20 $$\:\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} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:\mathrm{istan}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore\mathrm{angle}\:\mathrm{between}\:\mathrm{bisector} \\ $$$$\mathrm{and}\:\mathrm{x}−\mathrm{axis}\:=\frac{\pi}{\mathrm{8}}…
Question Number 109271 by 675480065 last updated on 22/Aug/20 Commented by bobhans last updated on 22/Aug/20 can't be read the writing in your photo Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109248 by peter frank last updated on 22/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 109246 by peter frank last updated on 22/Aug/20 Answered by Ar Brandon last updated on 22/Aug/20 $$\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}}…