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Question-19615

Question Number 19615 by icyfalcon999 last updated on 13/Aug/17 Answered by Tinkutara last updated on 13/Aug/17 $$\frac{\mathrm{1}}{\mathrm{4}}\underset{−\infty} {\overset{\infty} {\int}}\frac{{dx}}{{x}^{\mathrm{2}} \:+\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{tan}^{−\mathrm{1}} \:\mathrm{2}{x}\right]_{−\infty} ^{\infty} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\mathrm{2}}\:−\:\left(−\frac{\pi}{\mathrm{2}}\right)\right]\:=\:\frac{\pi}{\mathrm{2}}…

For-x-R-solve-the-equation-below-2-x-4-3-4-x-2-3-4-x-2-x-6-3-

Question Number 19595 by khamizan833@yahoo.com last updated on 13/Aug/17 $$\mathrm{For}\:{x}\:\in\:\mathrm{R},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{below}! \\ $$$$\left(\mathrm{2}^{{x}} \:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left(\mathrm{4}^{{x}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:=\:\left(\mathrm{4}^{{x}} \:+\:\mathrm{2}^{{x}} \:−\:\mathrm{6}\right)^{\mathrm{3}} \\ $$ Commented by khamizan833@yahoo.com last updated…

sin9x-sim5x-sin3x-help-

Question Number 150661 by Jamshidbek last updated on 14/Aug/21 $$\mathrm{sin9x}=\mathrm{sim5x}+\mathrm{sin3x}\:\:\:\mathrm{help}\: \\ $$ Commented by MJS_new last updated on 15/Aug/21 $$\mathrm{what}\:\mathrm{are}\:\mathrm{we}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{do}? \\ $$$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{are}\:{x}_{\mathrm{1}} =\mathrm{0}\:\left(\mathrm{obviously}\right) \\ $$$$\mathrm{and}\:{x}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{30}}\:\mathrm{for}\:\mathrm{some}\:{n},\:\mathrm{not}\:\mathrm{for}\:\mathrm{all}.…

Question-85105

Question Number 85105 by naka3546 last updated on 19/Mar/20 Answered by MJS last updated on 19/Mar/20 $${x}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{9}}+\mathrm{1}+\frac{\mathrm{64}}{\mathrm{81}}+\frac{\mathrm{125}}{\mathrm{243}}+…={S}_{\infty} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{k}^{\mathrm{3}} }{\mathrm{3}^{{k}} }\:=\frac{\mathrm{33}}{\mathrm{8}}−\left(\frac{{n}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{9}{n}^{\mathrm{2}}…

Prove-by-mathematical-induction-that-2002-n-2-2003-2n-1-is-divisible-by-4005-

Question Number 85073 by Roland Mbunwe last updated on 18/Mar/20 $${Prove}\:{by}\:{mathematical}\:{induction}\:{that} \\ $$$$\mathrm{2002}^{{n}+\mathrm{2}} +\mathrm{2003}^{\mathrm{2}{n}+\mathrm{1}} \:\:\:\:{is}\:{divisible}\:{by}\:\mathrm{4005} \\ $$ Commented by MJS last updated on 18/Mar/20 $$\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:{n}=\mathrm{1}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{for}\:{n}=\mathrm{2},\:\mathrm{3},\:\mathrm{4}…

Question-19531

Question Number 19531 by mondodotto@gmail.com last updated on 12/Aug/17 Answered by Tinkutara last updated on 12/Aug/17 Commented by Tinkutara last updated on 12/Aug/17 $$\mathrm{Total}\:\mathrm{who}\:\mathrm{have}\:\mathrm{taken}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{of} \\…

If-A-x-x-x-4-2-2-3-1-4-findX-if-p-A-3-

Question Number 85036 by gopikrishnan last updated on 18/Mar/20 $${If}\:{A}=\begin{bmatrix}{{x}\:\:\:\:{x}\:\:\:{x}\:\:}\\{\underset{\mathrm{2}\:\:} {\mathrm{4}}\:−\underset{\mathrm{3}} {\mathrm{2}}\:\:\:\underset{\mathrm{4}} {\mathrm{1}}}\end{bmatrix}{findX}\:{if}\:{p}\left({A}\right)=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$ Commented by jagoll last updated on…

Question-150560

Question Number 150560 by DELETED last updated on 13/Aug/21 Answered by DELETED last updated on 13/Aug/21 $$\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{12}\right)\mathrm{sin}\:\left(\mathrm{x}−\mathrm{3}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} }\:\:\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\:\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}−\mathrm{3}\right)\mathrm{sin}\:\left(\mathrm{x}−\mathrm{3}\right)}{\left\{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)\right\}^{\mathrm{2}} }…