Question Number 203933 by liuxinnan last updated on 02/Feb/24 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} =\mathrm{22} \\ $$$${prove}\:{a}+\mathrm{2}{b}+{c}\leqslant\mathrm{11} \\ $$ Answered by York12 last updated on 02/Feb/24 $$\left({a}+\mathrm{2}{b}+{c}\right)^{\mathrm{2}}…
Question Number 203935 by mathlove last updated on 02/Feb/24 $$\frac{{cos}\frac{{x}}{\mathrm{3}}+{sin}\:\frac{{x}}{\mathrm{3}}}{{cos}\:\frac{{x}}{\mathrm{3}}−{sin}\:\frac{{x}}{\mathrm{3}}}=? \\ $$ Answered by esmaeil last updated on 02/Feb/24 $$=\frac{\sqrt{\mathrm{2}}{sin}\left(\frac{{x}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}{cos}\left(\frac{{x}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right)}={tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by AST…
Question Number 203921 by Davidtim last updated on 02/Feb/24 $${prove}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1} \\ $$ Commented by malwan last updated on 02/Feb/24 $${this}\:{is}\:{indeterminate}\:{form} \\ $$$${but}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\:{x}^{{x}}…
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Question Number 203919 by Samuel12 last updated on 02/Feb/24 $$\:\: \\ $$ Answered by Mathspace last updated on 02/Feb/24 $$\mathrm{1}={e}^{\mathrm{2}{ik}\pi} \:\:{let}\:{z}={re}^{{i}\theta} \\ $$$${z}^{\mathrm{5}} =\mathrm{1}\:\Leftrightarrow{r}^{\mathrm{5}} {e}^{{i}\mathrm{5}\theta}…
Question Number 203941 by Panav last updated on 02/Feb/24 $$\boldsymbol{{If}}\:\boldsymbol{{a}}^{\mathrm{2}\:} −\boldsymbol{{a}}+\mathrm{2}=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{6}} +\mathrm{2}\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{a}}^{\mathrm{2}} \:\boldsymbol{{then}}\:\boldsymbol{{find}}\:\boldsymbol{{x}}? \\ $$$$\boldsymbol{{IIT}}−\boldsymbol{{JEE}}\:\boldsymbol{{based}}\:\boldsymbol{{question}}.\:\boldsymbol{{Find}}\:\boldsymbol{{sol}}^{\boldsymbol{{n}}} . \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 203900 by necx122 last updated on 01/Feb/24 $${Let}\:{A}\:\in\:{R}^{{N}×{N}} \:{be}\:{a}\:{symmetric}\:{positive} \\ $$$${definite}\:{matrix}\:{and}\:{b}\:\in\:{R}^{{N}} \:{a}\:{vector}. \\ $$$${If}\:{x}\:\in\:{R}^{{N}} ,\:{evaluate}\:{the}\:{integral} \\ $$$${Z}\left({A},{b}\right)\:=\:\int{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{{T}} {Ax}\:+\:{b}^{{T}} {x}} {dx}\:{as}\:{a}\:{function} \\ $$$${of}\:{A}\:{and}\:{b}. \\…
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Question Number 203897 by NasaSara last updated on 01/Feb/24 Answered by AST last updated on 01/Feb/24 $${cos}^{\mathrm{2}} \left({x}\right)−{cos}\left({x}\right)=\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}} \\ $$$$\Rightarrow{cos}^{\mathrm{4}} \left({x}\right)+\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{2}{cos}^{\mathrm{3}} \left({x}\right)−\mathrm{1}=\mathrm{0} \\…
Question Number 203898 by necx122 last updated on 02/Feb/24 $${Let}\:{f}\left({W}\right)\:{be}\:{a}\:{function}\:{of}\:{vector}\:{W}\:\in\: {R}^{{N}} , \\ $$$${i}.{e}.\:{f}\left({W}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{e}^{−{W}^{{T}} {x}} } \\ $$$${Determine}\:{the}\:{first}\:{derivative}\:{and} \\ $$$${matrix}\:{of}\:{second}\:{derivatives}\:{of}\:{f}\:{with} \\ $$$${respect}\:{to}\:{W} \\ $$$$ \\…