Question Number 39612 by Tinkutara last updated on 08/Jul/18 Commented by MJS last updated on 08/Jul/18 $$\mathrm{true}.\:\mathrm{that}'\mathrm{s}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{the}\:\mathrm{other}\:\mathrm{solutions} \\ $$ Commented by MJS last updated on…
Question Number 105071 by dw last updated on 25/Jul/20 Commented by mr W last updated on 25/Jul/20 $$…\:{that}\:{satisfy}\:{the}\:{equation}\:… \\ $$$${which}\:{equation}? \\ $$ Commented by dw…
Question Number 38960 by Tinkutara last updated on 01/Jul/18 Answered by ajfour last updated on 01/Jul/18 $$\begin{vmatrix}{{xa}}&{{yb}}&{{zc}}\\{{yc}}&{{za}}&{{xb}}\\{{zb}}&{{xc}}&{{ya}}\end{vmatrix}=\:{xa}\left({a}^{\mathrm{2}} {yz}−{bcx}^{\mathrm{2}} \right) \\ $$$$\:+{by}\left({b}^{\mathrm{2}} {zx}−{acy}^{\mathrm{2}} \right)+{cz}\left({c}^{\mathrm{2}} {xy}−{abz}^{\mathrm{2}} \right)…
Question Number 170022 by Moytea last updated on 14/May/22 $$\mathrm{2}.\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$$${Soln}:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A} \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{5}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{−\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow{R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow\left(−\frac{\mathrm{1}}{\mathrm{5}}\right){R}_{\mathrm{2}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −\mathrm{2}{R}_{\mathrm{2}}…
Question Number 38679 by Tinkutara last updated on 28/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18 $$\left({a}−{x}\right)\left({bc}−{bx}−{xc}+{x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−{c}\left({c}^{\mathrm{2}} −{cx}−{ab}\right) \\ $$$$+{b}\left({ac}−{b}^{\mathrm{2}} +{bx}\right)=\mathrm{0} \\ $$$${abc}−{abx}−{acx}+{ax}^{\mathrm{2}}…
Question Number 104141 by bobhans last updated on 19/Jul/20 Answered by bramlex last updated on 19/Jul/20 $$\left[\:\mathrm{1}\:\:\:{x}\:\:\:\mathrm{2}\:\right]\:\begin{bmatrix}{\mathrm{2}{x}+\mathrm{2}}\\{\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}{x}+\mathrm{2}\:+\mathrm{2}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{4}{x}\:=\:−\mathrm{4}\:\Rightarrow{x}\:=\:−\mathrm{1}\: \\ $$ Terms of…
Question Number 38282 by Tinkutara last updated on 23/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103622 by bobhans last updated on 16/Jul/20 $${Given}\:{a}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{24}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}+\sqrt{{n}}}\:{then}\:{the}\:{value}\:{of} \\ $$$${a}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{a}} \left({bc}\right)+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{b}} \left({ac}\right)+\mathrm{1}}\:+ \\ $$$$\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} \left({ab}\right)+\mathrm{1}}\:=\:? \\ $$ Answered by OlafThorendsen last…
Question Number 168339 by MikeH last updated on 09/Apr/22 $$\mathrm{A}\:\mathrm{point}\:\mathrm{in}\:\mathrm{rectangular}\:\mathrm{coordinates}\: \\ $$$$\left({x},{y},{z}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{represented}\:\mathrm{in}\:\mathrm{spherical} \\ $$$$\mathrm{coordinates}\:\left({r},\theta,\varphi\right)\:\mathrm{by}: \\ $$$$\:{x}\:=\:{r}\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi,\:{y}\:=\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi,\: \\ $$$${z}\:=\:\mathrm{sin}\:\varphi,\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\mathrm{2}\pi\:,\:\mathrm{0}\:\leqslant\:\varphi\:\leqslant\:\pi \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{Jacobian}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{transformation}\:\frac{\partial\left({x},{y},{z}\right)}{\partial\left({r},\theta,\varphi\right)} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region} \\…
Question Number 167956 by peter frank last updated on 30/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com