Question Number 199911 by mr W last updated on 11/Nov/23 Commented by mr W last updated on 11/Nov/23 $${three}\:{mirrors}\:{build}\:{the}\:{sides}\:{of}\:{an} \\ $$$${equilateral}\:{triangle}.\:{a}\:{laser}\:{ray} \\ $$$${emitted}\:{from}\:{the}\:{midpoint}\:{of}\:{a}\:{side} \\ $$$${should}\:{reach}\:{the}\:{opposite}\:{vertex}\:{after}…
Question Number 199968 by a.lgnaoui last updated on 11/Nov/23 $$\mathrm{determiner}\:\boldsymbol{\mathrm{x}}\:\:? \\ $$ Commented by a.lgnaoui last updated on 11/Nov/23 Answered by mr W last updated…
Question Number 199907 by cortano12 last updated on 11/Nov/23 $$\:\:\:\:\:\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:=?\: \\ $$ Answered by Frix last updated on 11/Nov/23 $${t}=\sqrt[{\mathrm{6}}]{{x}} \\ $$$${x}={t}^{\mathrm{6}} \:\:\:{dx}=\mathrm{6}{x}^{\frac{\mathrm{5}}{\mathrm{6}}} {dt} \\…
Question Number 199900 by Calculusboy last updated on 11/Nov/23 Answered by Rasheed.Sindhi last updated on 11/Nov/23 $$\frac{\sqrt[{{x}}]{\mathrm{3}^{{x}} +\mathrm{7}^{−{x}} }\:+\sqrt[{{x}}]{\mathrm{3}^{−{x}} +\mathrm{7}^{{x}} }\:}{\:\sqrt[{{x}}]{\mathrm{21}^{{x}} +\mathrm{1}}} \\ $$$$=\frac{\sqrt[{{x}}]{\frac{\mathrm{21}^{{x}} +\mathrm{1}}{\mathrm{7}^{{x}}…
Question Number 199903 by frankpenredpen last updated on 11/Nov/23 $$\int\frac{{x}^{\mathrm{2}} {dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{16}}}\:=\:? \\ $$ Commented by cortano12 last updated on 11/Nov/23 $$\mathrm{why}\:\partial\mathrm{x}\:? \\ $$ Answered…
Question Number 199898 by Calculusboy last updated on 11/Nov/23 Answered by Frix last updated on 11/Nov/23 $${P}=\mathrm{50}×\mathrm{2}^{\frac{{n}}{\mathrm{12}}} \\ $$ Commented by Calculusboy last updated on…
Question Number 199899 by sonukgindia last updated on 11/Nov/23 Answered by Rajpurohith last updated on 11/Nov/23 $${cos}\left({z}\right)+{sin}\left({z}\right)={k}\: \\ $$$$\Rightarrow\left(\frac{{e}^{{iz}} +{e}^{−{iz}} }{\mathrm{2}}\right)+\left(\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}}\right)={k} \\ $$$$\Rightarrow{e}^{{iz}}…
Question Number 199956 by cortano12 last updated on 11/Nov/23 Answered by Frix last updated on 11/Nov/23 $$\left(\mathrm{1}\right)\:\Rightarrow\:{x}={y}\left({y}^{\mathrm{2}} +{y}+\mathrm{2}\right)\wedge{y}>\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:\frac{{x}}{{y}}=\mathrm{4}\vee\frac{{x}}{{y}}=\mathrm{8} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{4}\wedge{y}=\mathrm{1}\:\vee\:{x}=\mathrm{16}\wedge{y}=\mathrm{2} \\…
Question Number 199959 by Yakubu last updated on 11/Nov/23 $$\:\boldsymbol{{how}}\:\boldsymbol{{do}}\:\boldsymbol{{I}}\:\boldsymbol{{calculate}}\:\boldsymbol{{for}} \\ $$$$\:\mathrm{7}.\mathrm{86}!\:\boldsymbol{{without}}\:\boldsymbol{{calculator}}? \\ $$ Commented by mr W last updated on 11/Nov/23 $$\mathrm{7}.\mathrm{86}!=\Gamma\left(\mathrm{8}.\mathrm{86}\right) \\ $$$${i}\:{think}\:{you}\:{can}'{t}\:{calculate}\:{it}\:{even}\:{with}…
Question Number 199952 by Abdullahrussell last updated on 11/Nov/23 Answered by cortano12 last updated on 11/Nov/23 $$\:=\:\frac{\mathrm{2sin}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:+\:\mathrm{2sin}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{6}\theta}{\mathrm{2cos}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:+\:\mathrm{2cos}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{6}\theta} \\ $$$$\:=\:\frac{\mathrm{sin}\:\mathrm{7}\theta\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{6}\theta\right)}{\mathrm{cos}\:\mathrm{7}\theta\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{6}\theta\right)} \\ $$$$\:=\:\mathrm{tan}\:\mathrm{7}\theta \\ $$ Terms of…