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Author: Tinku Tara

Question-199915

Question Number 199915 by lealiegalendez last updated on 11/Nov/23 Answered by Rasheed.Sindhi last updated on 11/Nov/23 $$\mathrm{5}{x}+\mathrm{30}−\mathrm{3}{x}−\mathrm{24}=\mathrm{0} \\ $$$$\mathrm{2}{x}=−\mathrm{6}\Rightarrow{x}=−\mathrm{6}/\mathrm{2}=−\mathrm{3} \\ $$ Terms of Service Privacy…

lim-x-0-2-x-3-x-5-x-3-3-x-

Question Number 199908 by cortano12 last updated on 11/Nov/23 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} +\mathrm{5}^{\mathrm{x}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{x}}} =?\: \\ $$ Answered by cortano12 last updated on 11/Nov/23 $$\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}}…

Question-199911

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}…

dx-x-x-1-3-

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-199900

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-199899

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}}…