Question Number 72628 by Rio Michael last updated on 30/Oct/19 $${solve}\:{the}\:{inequality}\: \\ $$$$\:\:{log}_{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:+\:\mathrm{9}\right)\:<\:\mathrm{0} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 7093 by FilupSmith last updated on 10/Aug/16 $$\mathrm{If}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{vector}\:\boldsymbol{{v}}=\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix} \\ $$$$\mathrm{and}\:\mathrm{apply}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{tranformation}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{new}\:\mathrm{vector}\:\boldsymbol{{v}}^{'} =\begin{bmatrix}{{x}'}\\{{y}'}\end{bmatrix}\:\mathrm{is}\:\mathrm{made}, \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{change}\:\mathrm{of}\:\mathrm{area}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\hat {\imath}\:\mathrm{unit}\:\mathrm{vector}? \\ $$ Commented by FilupSmith…
Question Number 138163 by mnjuly1970 last updated on 10/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..{nice}\:\:…\:….\:….\:{calculus}….. \\ $$$$\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}}\right)\overset{???} {=}\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:…………. \\ $$ Answered…
Question Number 7087 by Tawakalitu. last updated on 10/Aug/16 Commented by Yozzii last updated on 10/Aug/16 $${W}=\mathrm{2}^{\mathrm{129}} \mathrm{3}^{\mathrm{81}} \mathrm{5}^{\mathrm{131}} \\ $$$${X}=\mathrm{2}^{\mathrm{127}} \mathrm{3}^{\mathrm{81}} \mathrm{5}^{\mathrm{131}} \\ $$$${Y}=\mathrm{2}^{\mathrm{126}}…
Question Number 138159 by BHOOPENDRA last updated on 10/Apr/21 Answered by Dwaipayan Shikari last updated on 10/Apr/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−{x}} {dx}=\aleph \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{xlog}\left({x}\right)}…
Question Number 138153 by mathlove last updated on 10/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{x}−\mid{tanx}\mid}{\mid{sinx}\mid−{x}}=? \\ $$ Answered by TheSupreme last updated on 10/Apr/21 $${for}\:{x}\rightarrow\mathrm{0}^{+} \:\mid{sin}\left({x}\right)\mid={sin}\left({x}\right)\:{and}\:\mid{tan}\left({x}\right)\mid={tan}\left({x}\right) \\ $$$${lim}\:\frac{{x}−{tan}\left({x}\right)}{{sin}\left({x}\right)−{x}}=…
Question Number 138154 by JulioCesar last updated on 10/Apr/21 Answered by TheSupreme last updated on 10/Apr/21 $$\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}}{{x}}\geqslant\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \\ $$$$\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}\geqslant\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}−\left(\mathrm{1}−{x}\right)\right]\geqslant\mathrm{0} \\…
Question Number 7077 by Tawakalitu. last updated on 09/Aug/16 Commented by Yozzii last updated on 09/Aug/16 $${a}^{\mathrm{2}} +{a}=\mathrm{5} \\ $$$${Let}\:{u}={a}+\mathrm{3}\Rightarrow{a}={u}−\mathrm{3} \\ $$$$\Rightarrow\left({u}−\mathrm{3}\right)^{\mathrm{2}} +{u}−\mathrm{3}=\mathrm{5} \\ $$$${u}^{\mathrm{2}}…
Question Number 138150 by liberty last updated on 10/Apr/21 $${Given}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+\:{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:. \\ $$ Answered by EDWIN88 last updated…
Question Number 7076 by Tawakalitu. last updated on 09/Aug/16 Commented by Yozzii last updated on 09/Aug/16 $$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\left\{\sqrt{\mathrm{5}}\left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right\}^{\mathrm{1}/\mathrm{3}} =\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{1}+{u}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${u}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}<\mathrm{1}. \\ $$$${Also}\:\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =−\mathrm{5}^{\mathrm{1}/\mathrm{6}}…