Question Number 201526 by 281981 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $${WLOG},{let}\:{O}\:{be}\:{the}\:{origin};\:{g}=\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$$\mid{o}−{a}\mid=\mid{o}−{b}\mid=\mid{o}−{c}\mid={R} \\ $$$$\Rightarrow\left({o}−{a}\right)\left(\overset{−} {{o}}−\overset{−} {{a}}\right)={R}^{\mathrm{2}} \Rightarrow{a}\overset{−} {{a}}={R}^{\mathrm{2}}…
Question Number 201527 by cortano12 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $$#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=#\left(\mathrm{6}\right)+#\left(\mathrm{8}\right)−#\left(\mathrm{6}{n}\mathrm{8}\right) \\ $$$$#\left(\mathrm{6}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{6}}\rfloor=\mathrm{333};#\left(\mathrm{8}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{8}}\rfloor=\mathrm{250} \\ $$$$#\left(\mathrm{6}{n}\mathrm{8}\right)=#\left(\mathrm{24}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{24}}\rfloor=\mathrm{83} \\ $$$$\Rightarrow#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=\mathrm{500} \\ $$$$\Rightarrow{Probability}=\frac{\mathrm{2000}−\mathrm{500}}{\mathrm{2000}}=\frac{\mathrm{3}}{\mathrm{4}}…
Question Number 201553 by Calculusboy last updated on 08/Dec/23 Answered by som(math1967) last updated on 09/Dec/23 $$\mathrm{1}.\:\int\frac{\mathrm{1}+{logx}−\mathrm{1}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)}\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}−\int\left\{\frac{{d}}{{dx}}×\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}}…
Question Number 201555 by Simurdiera last updated on 08/Dec/23 Answered by mr W last updated on 09/Dec/23 $${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}}\:\Rightarrow\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\Rightarrow\sqrt{{u}+\mathrm{1}}\left(\frac{{du}}{{dx}}−\mathrm{1}\right)=\sqrt{{u}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\frac{\sqrt{{u}−\mathrm{1}}}{\:\sqrt{{u}+\mathrm{1}}}+\mathrm{1} \\…
Question Number 201548 by Calculusboy last updated on 08/Dec/23 Answered by MathematicalUser2357 last updated on 20/Jan/24 $$\left[−\left({x}−\mathrm{1}\right)\left\{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}}{x}−\mathrm{2}{x}+\sqrt{\mathrm{1}−{x}}−\mathrm{2}\sqrt{{x}+\mathrm{1}}\mathrm{log}\left(\sqrt{\mathrm{1}−{x}}+\mathrm{1}\right)−\sqrt{{x}+\mathrm{1}}\mathrm{log}\:{x}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\mathrm{log}\left(\sqrt{{x}+\mathrm{1}}+\mathrm{1}\right)−\mathrm{2}\right\}−\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\mathrm{tanh}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right]+{C} \\ $$ Terms of…
Question Number 201516 by sonukgindia last updated on 08/Dec/23 Answered by Calculusboy last updated on 08/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$${when}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\boldsymbol{{n}}} }\left(−\boldsymbol{{dy}}\right)\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…
Question Number 201517 by sonukgindia last updated on 08/Dec/23 Answered by Calculusboy last updated on 08/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\boldsymbol{{d}\theta}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{\theta}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…
Question Number 201519 by vahid last updated on 08/Dec/23 Answered by cortano12 last updated on 08/Dec/23 $$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{dx}\:=\:−\int\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right) \\ $$ Answered by…
Question Number 201544 by sonukgindia last updated on 08/Dec/23 Answered by aleks041103 last updated on 09/Dec/23 $${J}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left({ln}\left({x}\right)\right)}{{ln}\left(\mathrm{1}/{x}\right)}{dx}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{1}.{ln}\left({x}\right)\right)}{{ln}\left({x}\right)}{dx} \\ $$$${I}\left({s}\right)=−\int_{\mathrm{0}}…
Question Number 201545 by Calculusboy last updated on 08/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com