Question Number 149853 by 0731619 last updated on 07/Aug/21 Answered by mindispower last updated on 08/Aug/21 $$\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{{izcos}\left(\theta\right)} {d}\theta={J}_{\mathrm{0}} \left({z}\right)\:\:{bassel}\:{function} \\ $$$${cos}^{\mathrm{2}} \left(\theta\right)=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}} \\…
Question Number 149830 by DonQuichote last updated on 07/Aug/21 $${lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−…}}}}{{sin}\:\left({sin}\:\left({sin}\:\left(….{x}\right)\right.\right.}=? \\ $$ Answered by mathmax by abdo last updated on 08/Aug/21 $$\mathrm{y}=\sqrt{\mathrm{x}−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}−}..}}=\sqrt{\mathrm{x}−\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\mathrm{x}−\mathrm{y}\:\Rightarrow\mathrm{y}^{\mathrm{2}} +\mathrm{y}−\mathrm{x}=\mathrm{0}…
Question Number 18748 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by Joel577 last updated on 29/Jul/17 $${f}\left({x}\right)\:=\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}} \:−\:\mathrm{6}{x} \\ $$$${f}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)\:=\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{\mathrm{3}} \:−\:\mathrm{6}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{2}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{16}}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{32}}\:+\:\mathrm{4}\right)\:−\:\left(\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{6}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}…
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Question Number 18747 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by Joel577 last updated on 29/Jul/17 $$\left(−\mathrm{2}^{−\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:\left(−\mathrm{2}\right)^{−\mathrm{6}} \:=\:\mathrm{0},\mathrm{015625} \\ $$ Commented by Joel577…
Question Number 18746 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by daffa22 last updated on 30/Jul/17 $$\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{q}}\right)+{p}}+\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{r}}\right)+{q}}+\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)+{r}} \\ $$$${misal}, \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{q}}\right)={a} \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{r}}\right)={b} \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)={c}…
Question Number 18744 by aplus last updated on 29/Jul/17 Commented by aplus last updated on 29/Jul/17 $$\mathrm{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 149809 by 0731619 last updated on 07/Aug/21 Answered by Ar Brandon last updated on 07/Aug/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \mathrm{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy}…
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Question Number 149789 by Mathfinity last updated on 07/Aug/21 $${form} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com