Question Number 139992 by ajfour last updated on 02/May/21 Commented by ajfour last updated on 03/May/21 $${Let}\:\:{role}\:{of}\:{new}\:{cross}\: \\ $$$${multiplication}\:{by}\:\hat {{k}} \\ $$$$\:{be}\:{to}\:{rotate}\:{the}\:{component} \\ $$$${of}\:{a}\:{vector}\:\bot\:{to}\:{z}\:{axis}\:{by}\:\mathrm{90}° \\…
Question Number 73940 by smartsmith459@gmail.com last updated on 16/Nov/19 Answered by Rio Michael last updated on 16/Nov/19 $$\left.{Q}\mathrm{4}\right)\:{is}\:{equivalent}\:{to}\:{solving}\: \\ $$$$\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:{M}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{1}}\\{\mathrm{23}}\end{pmatrix} \\ $$$${where}\:{M}\:=\:\begin{pmatrix}{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{7}}&{\mathrm{1}}\end{pmatrix} \\ $$$${M}^{−\mathrm{1}}…
Question Number 8390 by tawakalitu last updated on 09/Oct/16 Answered by fernandodantas1996 last updated on 12/Oct/16 $$ \\ $$$$\: \\ $$$$\left.\mathrm{i}\right)\:\overset{\rightarrow} {\mathrm{F}}+\overset{\rightarrow} {\mathrm{P}}\:=\:\mathrm{6i}\:+\:\mathrm{4j}\:−\:\mathrm{k}\:\Rightarrow\: \\ $$$$\Rightarrow\:\parallel\overset{\rightarrow}…
Question Number 139455 by mnjuly1970 last updated on 27/Apr/21 $$\:\:\:\:\:\:\:#\:{calculus}# \\ $$$$\:\:{evaluate}: \\ $$$$\:\:\boldsymbol{\phi}:=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:\Gamma\:\left(\frac{{k}}{\mathrm{2}}\right)}{{k}\:\Gamma\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)}\:=? \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 138976 by I want to learn more last updated on 20/Apr/21 Answered by mr W last updated on 20/Apr/21 $${n}+\mathrm{5}{s}=\begin{pmatrix}{−\mathrm{12}+\mathrm{5}}\\{\mathrm{5}−\mathrm{5}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{7}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${q}=\begin{pmatrix}{−\mathrm{35}}\\{\mathrm{0}}\end{pmatrix} \\…
Question Number 73084 by rajesh4661kumar@gmail.com last updated on 06/Nov/19 Answered by Tanmay chaudhury last updated on 06/Nov/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7372 by FilupSmith last updated on 25/Aug/16 $$\boldsymbol{{v}}=\begin{bmatrix}{{x}}\\{{f}\left({x}\right)}\end{bmatrix},\:\:{f}\left({x}\right)={mx}+{b} \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{rotate}\:\boldsymbol{{v}}\:\mathrm{counter}\:\mathrm{clockwise} \\ $$$$\mathrm{by}\:\theta\:\mathrm{degrees},\:\mathrm{how}\:\mathrm{does}\:\mathrm{one}\:\mathrm{do}\:\mathrm{so}\:\mathrm{where}: \\ $$$$\bullet\:\:\:\boldsymbol{{v}}\:\mathrm{is}\:\mathrm{rotated}\:\mathrm{about}\:\mathrm{the}\:\mathrm{origin} \\ $$$$\bullet\:\:\:\boldsymbol{{v}}\:\mathrm{is}\:\mathrm{rotated}\:\mathrm{about}\:\mathrm{the}\:\mathrm{point}\:\left({x}_{\mathrm{1}} ,\:{f}\left({x}_{\mathrm{1}} \right)\right) \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{new}\:\mathrm{vectors}? \\ $$ Answered…
Question Number 72832 by necxxx last updated on 03/Nov/19 $${In}\:{a}\:{parallelogram}\:{OABC},\:{O}\overset{\rightharpoondown} {{A}}=\overset{−\rightharpoondown} {{a}}, \\ $$$${O}\overset{\rightarrow} {{C}}=\overset{\rightarrow} {{c}},\:{D}\:{is}\:{a}\:{point}\:{such}\:{that}\:{A}\overset{\rightarrow} {{D}}:{D}\overset{\rightarrow} {{B}}=\mathrm{1}:\mathrm{2} \\ $$$${Express}\:{the}\:{following}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{c} \\ $$$$\left({i}\right){C}\overset{\rightarrow} {{B}}\:\left({ii}\right){B}\overset{\rightarrow} {{C}}\:\left({iii}\right){A}\overset{\rightarrow} {{B}}\:\left({iv}\right)\:{A}\overset{\rightarrow}…
Question Number 138366 by KwesiDerek last updated on 12/Apr/21 $$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{−\mathrm{2}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{16}} +\boldsymbol{\mathrm{x}}^{−\mathrm{16}} =? \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}} \\ $$ Answered by TheSupreme last updated…
Question Number 72773 by ajfour last updated on 02/Nov/19 Commented by ajfour last updated on 02/Nov/19 $${If}\:{parabola}\:{in}\:{xz}\:{plane}\:{has}\:{eq}. \\ $$$${z}={ax}^{\mathrm{2}} \:\:\:{find}\:{eq}.\:{of}\:{shadow}\:{of} \\ $$$${this}\:{parabola}\:{on}\:{ground}\:\left({xy}\:{plane}\right). \\ $$ Commented…