Question Number 61895 by maxmathsup by imad last updated on 10/Jun/19 $${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:\:,{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{e}^{{iA}} \:\:\:{then}\:\:{cosA}\:\:{and}\:{sinA}\:. \\ $$ Commented by maxmathsup…
Question Number 61892 by hhghg last updated on 10/Jun/19 $$\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$ Commented by Rasheed.Sindhi last updated on 11/Jun/19 $$\frac{\mathrm{2}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$ Terms of Service…
Question Number 127429 by Study last updated on 29/Dec/20 $$\mathrm{6}\boldsymbol{\div}\mathrm{3}\left(\mathrm{2}\right)=??? \\ $$ Commented by talminator2856791 last updated on 29/Dec/20 $$\:\mathrm{why}\:\mathrm{is}\:\mathrm{your}\:\mathrm{name}\:\mathrm{in}\:\mathrm{red}? \\ $$ Terms of Service…
Question Number 192957 by York12 last updated on 31/May/23 $$ \\ $$$${bx}^{\mathrm{3}} =\mathrm{10}{a}^{\mathrm{2}} {bx}\:+\:\mathrm{3}{a}^{\mathrm{3}} {y}\:,\:{ay}^{\mathrm{3}} =\:\mathrm{10}{ab}^{\mathrm{2}} {y}\:+\:\mathrm{3}{b}^{\mathrm{3}} {x} \\ $$$${solve}\:{for}\:{x}\:{and}\:{y}\:{in}\:{terms}\:{of}\:\left({a}\:,\:{b}\right) \\ $$$${and}\:{solve}\:{for}\:{a}\:{and}\:{b}\:{in}\:{terms}\:{of}\:\:\left({x}\:,\:{y}\:\right) \\ $$ Commented…
Question Number 192958 by Mingma last updated on 31/May/23 Answered by Frix last updated on 01/Jun/23 $$\mathrm{Question}\:\mathrm{191675} \\ $$$${x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{4}} =\left(\sqrt{\mathrm{5}}\right)^{\mathrm{4}} =\mathrm{25} \\ $$…
Question Number 61873 by lalitchand last updated on 10/Jun/19 $$\sqrt[{\mathrm{3}}]{\mathrm{30x}^{\mathrm{8}} \mathrm{y}^{\mathrm{12}} }/^{\mathrm{4}} \sqrt{\mathrm{6x}^{\mathrm{2}} \mathrm{y}^{\mathrm{9}} \mathrm{z}}\:\:\:\:\mathrm{simplifh}\:\mathrm{this}\:\mathrm{question} \\ $$ Commented by MJS last updated on 10/Jun/19 $$\mathrm{please}\:\mathrm{make}\:\mathrm{clear}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{really}\:{y}\mathrm{12}…
Question Number 192942 by depressiveshrek last updated on 31/May/23 Commented by depressiveshrek last updated on 31/May/23 $${A}\:{survey}\:{was}\:{conducted}\:{on}\:{a}\:{number}\:{of}\:{familes} \\ $$$${about}\:{the}\:{number}\:{of}\:{children}\:{they}\:{have}. \\ $$$${The}\:{families}\:{had}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:{or}\:\mathrm{5}\:{children}. \\ $$$${If}\:{the}\:{y}−{axis}\:{represents}\:{the}\:{number}\:{of}\:{families}, \\ $$$${and}\:{the}\:{x}−{axis}\:{represents}\:{the}\:{number}\:{of}\:{children},…
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Question Number 192936 by Abdullahrussell last updated on 31/May/23 Answered by Frix last updated on 31/May/23 $$\mathrm{Obviously}\:{x}^{\mathrm{2}} =\mathrm{3}\vee{x}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow \\ $$$${x}=\pm\sqrt{\mathrm{3}}\vee{x}=\pm\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{use}\:\sqrt[{\mathrm{3}}]{−{z}}=−\sqrt[{\mathrm{3}}]{{z}}\:\mathrm{also}\:{x}^{\mathrm{2}} =\mathrm{12}\:\Rightarrow\:{x}=\pm\mathrm{2}\sqrt{\mathrm{3}} \\…
Question Number 192928 by BaliramKumar last updated on 31/May/23 $$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+{k}\:=\:\mathrm{0}\:{where}\:{k}<\mathrm{0}\: \\ $$$$\left(\frac{\alpha}{\beta}\:+\:\frac{\beta}{\alpha}\right)_{\mathrm{max}} \:=\:? \\ $$ Answered by MM42 last updated on 31/May/23 $${s}=\mathrm{3}\:\:\&\:\:{p}=\frac{{k}}{\mathrm{2}} \\…