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 $$\mathrm{A}\:\mathrm{point}\:\mathrm{in}\:\mathrm{rectangular}\:\mathrm{coordinates}\: \\ $$$$\left({x},{y},{z}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{represented}\:\mathrm{in}\:\mathrm{spherical} \\ $$$$\mathrm{coordinates}\:\left({r},\theta,\varphi\right)\:\mathrm{by}: \\ $$$$\:{x}\:=\:{r}\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi,\:{y}\:=\:\mathrm{sin}\:\theta\:\mathrm{sin}\:\varphi,\: \\ $$$${z}\:=\:\mathrm{sin}\:\varphi,\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\mathrm{2}\pi\:,\:\mathrm{0}\:\leqslant\:\varphi\:\leqslant\:\pi \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{Jacobian}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{transformation}\:\frac{\partial\left({x},{y},{z}\right)}{\partial\left({r},\theta,\varphi\right)} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region} \\…
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Question Number 167183 by alcohol last updated on 09/Mar/22 Commented by ArielVyny last updated on 09/Mar/22 $${i}\:{let}\:{question}\:\left({f}\right)\:{for}\:{your}\:{attention} \\ $$ Answered by ArielVyny last updated on…
Question Number 100873 by EquationMaker2305 last updated on 29/Jun/20 $$\begin{bmatrix}{\mathrm{1}}&{\mathrm{5}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{6}}&{\mathrm{7}}\\{\mathrm{9}}&{\mathrm{2}}&{\mathrm{8}}\end{bmatrix}+\begin{bmatrix}{\mathrm{9}}&{\mathrm{5}}&{\mathrm{7}}\\{\mathrm{6}}&{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{8}}&{\mathrm{2}}\end{bmatrix} \\ $$ Commented by Tinku Tara last updated on 29/Jun/20 $$\mathrm{Please}\:\mathrm{avoid}\:\mathrm{posting}\:\mathrm{question}\:\mathrm{for} \\ $$$$\mathrm{unless}\:\mathrm{you}\:\mathrm{are}\:\mathrm{really}\:\mathrm{looking}\:\mathrm{for} \\ $$$$\mathrm{inputs}\:\mathrm{or}\:\mathrm{some}\:\mathrm{discussion}.…
Question Number 35319 by Tinkutara last updated on 17/May/18 Answered by MJS last updated on 17/May/18 $$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{multiplicating}\:\mathrm{and}\:\mathrm{factorising} \\ $$$$\mathrm{det}\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{vmatrix}={aei}+{bfg}+{cdh}−{afh}−{bdi}−{ceg} \\ $$ Commented by Tinkutara last…
Question Number 166377 by cortano1 last updated on 19/Feb/22 $$\:\:\mathrm{Given}\:\mathrm{A}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\mathrm{a}\:\:\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:.\:\mathrm{If}\:\mathrm{A}^{\mathrm{3}} =\:\mathrm{A}^{\mathrm{2}} \\ $$$$\:\mathrm{then}\:\mathrm{2a}−\mathrm{3b}=? \\ $$ Answered by bobhans last updated on 19/Feb/22 $$\:\:\mathrm{A}^{\mathrm{2}} =\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\mathrm{a}\:\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\mathrm{a}\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}\:\:\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}}\\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}+\mathrm{ab}\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}+\mathrm{b}^{\mathrm{2}} }\end{pmatrix}…