Question Number 85864 by john santu last updated on 26/Mar/20 $${simplify}\:{the}\:{expression} \\ $$$$\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{10}}}\:−\:\sqrt{\mathrm{7}−\sqrt{\mathrm{3}}}\:\:{in} \\ $$$${the}\:{form}\:\sqrt{\sqrt{{a}}+{b}}\:? \\ $$ Answered by jagoll last updated on 26/Mar/20 $$\mathrm{by}\:\mathrm{trick}\:!\:…
Question Number 85865 by subhankar10 last updated on 25/Mar/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{curl}\left(\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}×\overset{\rightarrow} {\mathrm{r}}\right)=\left(\mathrm{n}+\mathrm{2}\right)\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}−\mathrm{nr}^{\mathrm{n}−\mathrm{2}} \left(\overset{\rightarrow} {\mathrm{r}}.\overset{\rightarrow} {\mathrm{c}}\right)\:\:. \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{vector}. \\ $$ Answered…
Question Number 151393 by Tawa11 last updated on 20/Aug/21 Answered by mr W last updated on 20/Aug/21 $$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} +{x}^{\mathrm{15}} +…\right) \\ $$$$=\left(\mathrm{1}+{x}^{\mathrm{5}}…
Question Number 85858 by jagoll last updated on 25/Mar/20 $$\mathrm{Is}\:\mathrm{a}\:\mathrm{matrix} \\ $$$$\mathrm{A}^{\mathrm{T}} \mathrm{A}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{definite}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85859 by jagoll last updated on 25/Mar/20 $$\:\mathrm{Is}\:\mathrm{the}\:\mathrm{Var}\left(\mathrm{aX}+\mathrm{b}\right)\:=\:\mathrm{a}^{\mathrm{2}} \:\mathrm{Var}\left(\mathrm{X}\right)\:+\:\mathrm{b}? \\ $$ Answered by Joel578 last updated on 25/Mar/20 $${nope} \\ $$$$\mathrm{Assume}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is}\:\mathrm{constant},\:\mathrm{then} \\ $$$$\mathrm{var}\left({b}\right)\:=\:{E}\left[{b}^{\mathrm{2}}…
Question Number 85857 by M±th+et£s last updated on 25/Mar/20 Answered by mind is power last updated on 26/Mar/20 $$\left.{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{1}\right)\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}}…
Question Number 85854 by jagoll last updated on 25/Mar/20 $$\mathrm{If}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:+\:\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{4xyz}\:−\mathrm{1}\: \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\: \\ $$ Commented by mr W…
Question Number 151391 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\in\mathbb{R}\:\:\mathrm{and} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{9}=\mathrm{6}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{find}\:\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =? \\ $$ Answered by dumitrel last updated…
Question Number 20316 by Tinkutara last updated on 25/Aug/17 $$\mathrm{If}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{40}\:\mathrm{m},\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile}\:\mathrm{makes}\:\mathrm{an}\:\mathrm{angle} \\ $$$$\pi/\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal},\:\mathrm{then}\:\mathrm{its}\:\mathrm{initial} \\ $$$$\mathrm{velocity}\:\mathrm{and}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{are}, \\ $$$$\mathrm{respectively} \\ $$$$\left({a}\right)\:\mathrm{30},\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\left({b}\right)\:\mathrm{30},\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\…
Question Number 151383 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y}>\mathrm{0}\:\:\mathrm{peove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\:\centerdot\:\frac{\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\centerdot\:\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\:\geqslant\:\mathrm{xy} \\ $$ Commented by mathdanisur last updated on 20/Aug/21 $$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\geqslant\sqrt[{\mathrm{3}}]{\mathrm{xy}}\:\mathrm{for}\:\mathrm{solution}\:\mathrm{please} \\…