Question Number 172050 by henderson last updated on 23/Jun/22 Commented by henderson last updated on 23/Jun/22 $$\mathrm{any}\:\mathrm{possibility}\:\mathrm{for}\:\mathrm{this}\:\mathrm{situation}\:??? \\ $$ Commented by greg_ed last updated on…
Question Number 106493 by mohammad17 last updated on 05/Aug/20 $${Solve}? \\ $$$$\left(\mathrm{1}\right){cos}\left(\mathrm{70}\right) \\ $$$$\left(\mathrm{2}\right){cos}\left(\mathrm{10}\right) \\ $$ Commented by malwaan last updated on 05/Aug/20 $${cos}\:\mathrm{10}\:=\:{cos}\:\mathrm{70}\:+\:{cos}\:\mathrm{50} \\…
Question Number 106488 by Algoritm last updated on 05/Aug/20 Commented by Dwaipayan Shikari last updated on 05/Aug/20 $$\:{Q}\mathrm{106379} \\ $$$$\frac{\mathrm{1}.\mathrm{2}^{\mathrm{2}} }{\mathrm{10}}+\frac{\mathrm{2}.\mathrm{3}^{\mathrm{2}} }{\mathrm{10}}+…..=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{n}}…
Question Number 106479 by mohammad17 last updated on 05/Aug/20 Answered by Rio Michael last updated on 05/Aug/20 $$\mathrm{c}.\:\mathrm{Let}\:\:{S}\:=\:{a}\:+\:{ar}\:+\:{ar}^{\mathrm{2}} \:+\:…+\:{ar}^{{n}−\mathrm{1}} \:….\left({i}\right) \\ $$$$\:\:\:\mathrm{now}\:{rS}\:=\:{ra}\:+\:{ar}^{\mathrm{2}} \:+\:{ar}^{\mathrm{3}} \:+…+{ar}^{{n}} …..\left({ii}\right)…
Question Number 106469 by ZiYangLee last updated on 05/Aug/20 $$\mathrm{Solve}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{3cos}^{\mathrm{2}} \mathrm{2x}=\mathrm{cos}^{\mathrm{2}} \mathrm{3x} \\ $$ Answered by mahdi last updated on 05/Aug/20 $$\mathrm{cos}\left(\mathrm{3x}\right)=\mathrm{cos}\left(\mathrm{2x}\right).\mathrm{cosx}−\mathrm{sin}\left(\mathrm{2x}\right).\mathrm{sinx}= \\ $$$$\left(\mathrm{2cos}^{\mathrm{2}}…
Question Number 106465 by mohammad17 last updated on 05/Aug/20 Answered by Dwaipayan Shikari last updated on 05/Aug/20 $$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 106449 by mohammad17 last updated on 05/Aug/20 Answered by Dwaipayan Shikari last updated on 05/Aug/20 $${y}\sqrt{{x}}−{x}\sqrt{{y}}=\mathrm{9} \\ $$$${y}\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}+\sqrt{{x}}.\frac{{dy}}{{dx}}−\frac{{x}}{\mathrm{2}\sqrt{{y}}}.\frac{{dy}}{{dx}}−\sqrt{{y}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\left(\sqrt{{x}}−\frac{{x}}{\mathrm{2}\sqrt{{y}}}\right)=\sqrt{{y}}−\frac{{y}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}\sqrt{{xy}}−{y}}{\mathrm{2}\sqrt{{x}}}.\frac{\mathrm{2}\sqrt{{y}}}{\mathrm{2}\sqrt{{xy}}−{x}} \\…
Question Number 106444 by mathdave last updated on 05/Aug/20 Answered by nimnim last updated on 05/Aug/20 $${LHS}:\:\mathrm{3}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{9}}{\mathrm{13}}\right) \\ $$$$=\:\pi+{tan}\left(\frac{\mathrm{6}−\mathrm{8}}{\mathrm{1}−\mathrm{12}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{9}}{\mathrm{13}}\right)\:\:\:\:\:\:\:\left({as}\:\mathrm{2}>\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\pi+{tan}^{\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{11}}\right)+{tan}^{−\mathrm{1}}…
Question Number 106424 by pandiyan last updated on 05/Aug/20 $$\int{x}^{\mathrm{2}} \\ $$ Answered by Her_Majesty last updated on 05/Aug/20 $${syntax}\:{error};\:{it}\:{must}\:{be}\:\int{x}^{\mathrm{2}} {dx} \\ $$$$\int{x}^{{n}} {dx}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+{C}…
Question Number 106422 by Algoritm last updated on 05/Aug/20 Commented by mr W last updated on 05/Aug/20 $$\mathrm{73}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×…×\mathrm{49}×\mathrm{50}×…×\mathrm{73} \\ $$$$\mathrm{73}!=\mathrm{0}\:\left({mod}\:\mathrm{2}\right) \\ $$$$\mathrm{73}!=\mathrm{0}\:\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{73}!=\mathrm{0}\:\left({mod}\:\mathrm{4}\right) \\…