Question Number 201604 by ajfour last updated on 09/Dec/23 Commented by ajfour last updated on 09/Dec/23 $${Both}\:{circles}\:{in}\:{blue}\:{have}\:{equal} \\ $$$${radii}.\:{Find}. \\ $$ Answered by mr W…
Question Number 201573 by sonukgindia last updated on 09/Dec/23 Answered by witcher3 last updated on 09/Dec/23 $$=\int_{−\infty} ^{\infty} \frac{\mathrm{e}^{−\mathrm{2024x}} +\mathrm{e}^{−\mathrm{2020}} }{\left(\mathrm{e}^{−\mathrm{2025x}} +\mathrm{e}^{−\mathrm{2019x}} \right)\left(\left(−\mathrm{4x}^{\mathrm{3}} +\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}}…
Question Number 201533 by Rodier97 last updated on 08/Dec/23 $$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Un}\:=\:{ln}\:\left({cos}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\right) \\ $$$$\:\:\:\:{show}\:\:{that}\:{Un}\:\leqslant\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$…
Question Number 201534 by Mathspace last updated on 08/Dec/23 $${let}\:{f}\left({x}\right)={tanx} \\ $$$${find}\:{f}^{\left({n}\right)} \left({x}\right)\:{with}\:{n}\:{integr} \\ $$$${natural} \\ $$ Commented by Frix last updated on 09/Dec/23 $$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{very}\:\mathrm{complicated}\:\mathrm{formula},\:\mathrm{you}\:\mathrm{must}…
Question Number 201557 by hardmath last updated on 08/Dec/23 $$\mathrm{5}\:\centerdot\:\underset{\:\mathrm{50}} {\underbrace{\mathrm{555}…\mathrm{5}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$ Answered by aleks041103 last updated on 09/Dec/23 $$\mathrm{5}.\mathrm{5}=\mathrm{25}…
Question Number 201526 by 281981 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $${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…