Question Number 212176 by MrGHK last updated on 05/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212159 by boblosh last updated on 04/Oct/24 Answered by Rasheed.Sindhi last updated on 04/Oct/24 $$\begin{bmatrix}{\:\:\:\:\:\mathrm{2}}&{\:\:\:\:\mathrm{3}}\\{−\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}+{a}\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{4}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}{a}+\mathrm{2}}&{{a}+\mathrm{3}}\\{\mathrm{0}{a}−\mathrm{1}}&{\mathrm{4}{a}−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\mathrm{2}{a}+\mathrm{2}=\mathrm{8}\Rightarrow{a}=\mathrm{3} \\ $$$${c}=−\mathrm{1} \\ $$$${b}={a}+\mathrm{3}=\mathrm{3}+\mathrm{3}=\mathrm{6}…
Question Number 212160 by boblosh last updated on 04/Oct/24 Answered by A5T last updated on 04/Oct/24 $$\frac{{x}}{\mathrm{2}}+\mathrm{4}{y}=\mathrm{4}\Rightarrow{x}+\mathrm{8}{y}=\mathrm{8}…\left({i}\right) \\ $$$$\frac{{x}}{\mathrm{4}}−\frac{\mathrm{2}{y}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\mathrm{3}{x}−\mathrm{8}{y}=\mathrm{8}…\left({ii}\right) \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}\\{\mathrm{8}}\end{bmatrix} \\ $$$${By}\:{Cramer}'{s}\:{rule}:\:{x}=\frac{\begin{vmatrix}{\mathrm{8}}&{\mathrm{8}}\\{\mathrm{8}}&{−\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{128}}{−\mathrm{32}}=\mathrm{4} \\ $$$${and}\:{y}=\frac{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{\mathrm{8}}\end{vmatrix}}{\begin{vmatrix}{\mathrm{1}}&{\mathrm{8}}\\{\mathrm{3}}&{−\mathrm{8}}\end{vmatrix}}=\frac{−\mathrm{16}}{−\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 212137 by MrGaster last updated on 03/Oct/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{{n}}]{{n}}−\mathrm{1}\right)\sqrt{{n}}=? \\ $$$$ \\ $$ Answered by Frix last updated on 03/Oct/24 $$\underset{{n}\rightarrow\infty}…
Question Number 212146 by CrispyXYZ last updated on 03/Oct/24 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{without}\:\mathrm{derivative} \\ $$$${x}\left(\mathrm{6}−{x}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:\left(\mathrm{3}<{x}<\mathrm{6}\right) \\ $$ Answered by Frix last updated on 03/Oct/24 $$\mathrm{Let}\:{x}=\mathrm{3}+\mathrm{3sin}\:{t} \\ $$$${x}\left(\mathrm{6}−{x}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}}…
Question Number 212148 by boblosh last updated on 03/Oct/24 $$ \\ $$ Answered by Frix last updated on 03/Oct/24 $$\mathrm{42} \\ $$ Terms of Service…
Question Number 212130 by Ismoiljon_008 last updated on 02/Oct/24 $$\:\:\:\: \\ $$ Answered by a.lgnaoui last updated on 03/Oct/24 $$\:\:\:\:\frac{\mathrm{AD}}{\mathrm{BC}}=\mathrm{2},\mathrm{48}\:\left(\mathrm{detail}\:\mathrm{en}\:\mathrm{commentaire}\right) \\ $$ Commented by a.lgnaoui…
Question Number 212108 by Ismoiljon_008 last updated on 01/Oct/24 $$ \\ $$$$\:\:\:\int_{\:−\mathrm{1}} ^{\:\:\mathrm{1}} \mid\:{x}\:\mid\:\centerdot\:{ln}\left({x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)\:{dx}\:=\:? \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:{please} \\ $$$$ \\ $$ Answered by Frix last…
Question Number 212114 by Ismoiljon_008 last updated on 01/Oct/24 $$ \\ $$$$\:\:\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$$\:\:\:\mathcal{F}{ind}\:{that}:\:\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:? \\ $$$$ \\ $$ Answered by Frix…
Question Number 212028 by siva12345 last updated on 27/Sep/24 $${DEFINATION}\:\:\:\:{OF}\:\:\:{QUADRATIC}\:\:{FORM}:\: \\ $$$$\:\:\:\:\:{A}\:\:{Quadratic}\:\:{form}\:\:{is}\:\:{a}\:{homogeneous}\:\:{polynomial}\:\:{of}\:\:{degree}\:{two}\:\:{in}\:\:{multiple}\:\:{variable}.\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}={X}^{{T}} {AX} \\ $$$${Here}\:\:{Q}={Quadratic}\:{form}. \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +{cz}^{\mathrm{2}} +\mathrm{2}{hxy}+\mathrm{2}{fyz}+\mathrm{2}{gzx}=\mathrm{0} \\ $$$${By}\:\:{using}\:\:{these}\:\:{Q}={X}^{{T}} {AX}\:\:\left[{we}\:\:{can}\:\:{write}\:{matrix}\:{A}\right]…