Question Number 9392 by tawakalitu last updated on 04/Dec/16 $$\mathrm{x}^{\mathrm{x}} \:=\:\mathrm{16} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings}. \\ $$ Answered by mrW last updated on 04/Dec/16 $$\mathrm{using}\:\mathrm{Lambert}\:\boldsymbol{\mathrm{W}}\:\mathrm{function}…
Question Number 140459 by EDWIN88 last updated on 08/May/21 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{2}×\mathrm{2}\:\mathrm{matrices}\:\mathrm{A}\:\mathrm{with}\: \\ $$$$\:\mathrm{A}^{\mathrm{3}} −\mathrm{3A}^{\mathrm{2}} \:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix}\:.\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9380 by FilupSmith last updated on 03/Dec/16 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{n}} \\ $$$${S}=?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74912 by behi83417@gmail.com last updated on 03/Dec/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\boldsymbol{\mathrm{yz}}+\mathrm{1}}\\{\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\boldsymbol{\mathrm{xz}}+\mathrm{2}}\\{\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\boldsymbol{\mathrm{xy}}+\mathrm{3}}\end{cases}\:\:\:\:\Rightarrow\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=? \\ $$ Answered by MJS last updated on 03/Dec/19 $${y}={px}\wedge{z}={qx} \\ $$$$\left(\mathrm{1}\right)\:\:\left(\mathrm{1}−{pq}\right){x}^{\mathrm{2}}…
Question Number 9378 by tawakalitu last updated on 03/Dec/16 $$\mathrm{Solve}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{18x}\:−\:\mathrm{32}\:=\:\mathrm{0} \\ $$ Answered by mrW last updated on 03/Dec/16 $$\mathrm{a}=\mathrm{1},\:\mathrm{b}=\mathrm{0},\:\mathrm{c}=−\mathrm{18},\:\mathrm{d}=−\mathrm{32} \\ $$$$\Delta=\mathrm{b}^{\mathrm{2}}…
Question Number 74910 by vishalbhardwaj last updated on 03/Dec/19 $$\mathrm{Explain}\:\mathrm{a}\:\mathrm{function}\:\mathrm{with}\:\mathrm{examples}\:\mathrm{based} \\ $$$$\mathrm{on}\:\mathrm{our}\:\mathrm{daily}\:\mathrm{life}\:? \\ $$ Answered by MJS last updated on 03/Dec/19 $$\mathrm{renting}\:\mathrm{a}\:\mathrm{car} \\ $$$$\mathrm{deal}\:\mathrm{1} \\…
Question Number 140445 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{lines}\:\bar {{a}}_{{i}} {z}+{a}_{{i}} \bar {{z}}+{b}_{{i}} =\mathrm{0}\left({i}=\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right),\:\mathrm{where} \\ $$$${b}_{{i}} \:\mathrm{are}\:\mathrm{real},\:\mathrm{are}\:\mathrm{concurrent},\:\mathrm{then}\:\Sigma{b}_{{i}} \left({a}_{\mathrm{2}} \bar {{a}}_{\mathrm{3}} −\bar {{a}}_{\mathrm{2}} {a}_{\mathrm{3}} \right)\:\mathrm{is}…
Question Number 140444 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{points}\:\mathrm{1}+\mathrm{2}{i}\:\mathrm{and}\:−\mathrm{1}+\mathrm{4}{i}\:\mathrm{are}\:\mathrm{reflections}\:\mathrm{of} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{z}\left(\mathrm{1}+{i}\right)+\bar {{z}}\left(\mathrm{1}−{i}\right)+{K}=\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{K}\:\mathrm{is}\:\_\_\_\_\_. \\ $$ Answered by mr W last updated on 08/May/21…
Question Number 140446 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{vertices}\:{Z}_{\mathrm{1}} ,\:{Z}_{\mathrm{2}} \:\mathrm{and}\:{Z}_{\mathrm{3}} \:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lambda{i}\:\:\begin{vmatrix}{{Z}_{\mathrm{1}} }&{\bar {{Z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{Z}_{\mathrm{2}} }&{\bar {{Z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{Z}_{\mathrm{3}} }&{\bar {{Z}}_{\mathrm{3}}…
Question Number 140443 by EnterUsername last updated on 07/May/21 $$\mathrm{If}\:{z}_{\mathrm{2}} /{z}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{and}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{non}-\mathrm{zero}\:\mathrm{real} \\ $$$$\mathrm{numbers},\:\mathrm{then}\:\mid\left({az}_{\mathrm{1}} +{bz}_{\mathrm{2}} \right)/\left({az}_{\mathrm{1}} −{bz}_{\mathrm{2}} \right)\mid\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_\_. \\ $$ Answered by mr W last…