Question Number 142349 by mnjuly1970 last updated on 30/May/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}^{{n}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 142348 by mathdanisur last updated on 30/May/21 $$\frac{\sqrt[{\mathrm{6}}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{15}}}\:\centerdot\:\sqrt[{\mathrm{3}}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:=\:? \\ $$ Answered by bramlexs22 last updated on 30/May/21 $$\:\frac{\sqrt[{\mathrm{6}\:}]{\mathrm{4}−\sqrt{\mathrm{15}}}}{\:\sqrt[{\mathrm{6}\:}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{3}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\ $$$$\:\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{6}}]{\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }.\sqrt[{\mathrm{3}\:}]{\mathrm{4}+\sqrt{\mathrm{15}}}}\:= \\…
Question Number 76815 by Rio Michael last updated on 30/Dec/19 $$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{is}\:\boldsymbol{\mathrm{N}}\mathrm{o}\boldsymbol{\mathrm{t}}\:\mathrm{convergent}? \\ $$$$\mathrm{A}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} } \\ $$$$\mathrm{B}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{C}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}}…
Question Number 11278 by malwaan last updated on 19/Mar/17 $${please} \\ $$$${what}\:{is}\:{the}\:{meaning}\:{of} \\ $$$${I}\:{go}\:{go}\:{the}\:{go} \\ $$ Commented by Joel576 last updated on 19/Mar/17 $$??? \\…
Question Number 142351 by bramlexs22 last updated on 30/May/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{27}+{x}}−\sqrt[{\mathrm{3}\:}]{\mathrm{27}−{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}^{\mathrm{2}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{3}} }}\:=? \\ $$ Answered by mathmax by abdo last updated on 30/May/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{27}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 76812 by Rio Michael last updated on 30/Dec/19 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{3}\:×\:\mathrm{3}\:\mathrm{invertible}\:\mathrm{matrices}, \\ $$$$\:\mathrm{then}\:\left(\mathrm{A}^{−\mathrm{1}} \mathrm{B}\right)^{−\mathrm{1}} \:= \\ $$$$\mathrm{A}.\:\mathrm{AB}^{−\mathrm{1}} \\ $$$$\mathrm{B}.\mathrm{B}^{−\mathrm{1}} \mathrm{A} \\ $$$$\mathrm{C}.\:\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \\ $$$$\mathrm{D}.\:\mathrm{BA}^{−\mathrm{1}}…
Question Number 76813 by Rio Michael last updated on 30/Dec/19 $$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \:=\: \\ $$$$\mathrm{A}.\:\infty \\ $$$$\mathrm{B}.\:\mathrm{1} \\ $$$$\mathrm{C}.\:−\mathrm{1} \\ $$$$\mathrm{D}.\:\mathrm{0} \\ $$ Commented…
Question Number 142344 by mnjuly1970 last updated on 30/May/21 $$ \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}:=\underset{\frac{\mathrm{1}}{{e}}} {\int}^{\:{e}} \left\{\frac{\mathrm{1}}{{ln}\left({x}\right)}+{ln}\left({ln}\left({x}\right)\right)\right\}{dx} \\ $$ Commented by Dwaipayan Shikari last updated on 30/May/21 $${log}\left({x}\right)={t}…
Question Number 76811 by Rio Michael last updated on 30/Dec/19 $$\mathrm{The}\:\mathrm{eccentricity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{64}}\:−\:\frac{{y}^{\mathrm{2}} }{\mathrm{36}}\:=\:\mathrm{1}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}.\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$…
Question Number 11274 by uni last updated on 18/Mar/17 $$\frac{{sin}\mathrm{20}+\sqrt{\mathrm{3}}×{cos}\mathrm{20}}{{cos}\mathrm{10}}=? \\ $$ Answered by bahmanfeshki1 last updated on 18/Mar/17 $${sin}\:\mathrm{20}+\sqrt{\mathrm{3}}×{cos}\mathrm{20}=\mathrm{2}×{cos}\left(\mathrm{30}−\mathrm{20}\right)=\mathrm{2}×{cos}\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{2}{cos}\mathrm{10}}{{cos}\mathrm{10}}=\mathrm{2} \\ $$ Answered…