Question Number 192187 by alcohol last updated on 10/May/23 $$\begin{cases}{{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:=\:{a}}\\{{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:=\:{b}}\\{{z}_{\mathrm{3}} \:+\:{z}_{\mathrm{4}} \:=\:{c}}\\{{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{4}} \:=\:{d}}\end{cases} \\ $$$${solve}\:{using}\:{gaussian}\:{elimination} \\ $$ Terms of Service…
Question Number 61096 by Arthur El-bomart last updated on 29/May/19 $$\forall\:{a},\:{n}\:\in\:{N}\::\:\mid{a}−{n}\mid=\mathrm{1}\:{pour}\:{a},\:{n}\:\geqslant\mathrm{3} \\ $$$${a}^{{m}} \equiv\mathrm{1}{modn}\:\left(\ast\right) \\ $$$${posons}\::\:{m}={n}−\mathrm{1}\:\left(\ast'\right) \\ $$$${subtituons}\:{cette}\:{valeur}\:{dans}\:\left(\ast\right). \\ $$$${on}\:{a}:\:{a}^{{n}−\mathrm{1}} \equiv\mathrm{1}{modn}.\:{Mais}\:{n}\:{n}'{est}\:{pas}\:{forcement}\:{premier}. \\ $$$${Test}\:{de}\:{primalite} \\ $$$$\forall\:{n}\:\in\:{N},\:{n}\:\geqslant\mathrm{3}.…
Question Number 191966 by Shlock last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $${y}={x}^{{x}^{…} } ={x}^{{y}} \\ $$$${y}={e}^{{y}\mathrm{ln}\:{x}} \\ $$$$\left(−{y}\mathrm{ln}\:{x}\right){e}^{−{y}\mathrm{ln}\:{x}} =−\mathrm{ln}\:{x}…
Question Number 60853 by Tony Lin last updated on 26/May/19 $$\sqrt{\mathrm{5}−\mathrm{12}{i}}+\sqrt{\mathrm{5}+\mathrm{12}{i}}=? \\ $$ Answered by tanmay last updated on 26/May/19 $$\sqrt{\mathrm{9}−\mathrm{4}−\mathrm{12}{i}}\:+\sqrt{\mathrm{9}−\mathrm{4}+\mathrm{12}{i}}\: \\ $$$$\sqrt{\left(\mathrm{3}−\mathrm{2}{i}\right)^{\mathrm{2}} }\:+\sqrt{\left(\mathrm{3}+\mathrm{2}{i}\right)^{\mathrm{2}} }\:…
Question Number 191831 by Shlock last updated on 01/May/23 Answered by mehdee42 last updated on 02/May/23 $${suppose}\:,\:\:{n}=<{abcdefghij}>\:,{is}\:{an}\:<{i}.{n}> \\ $$$${a}+{b}+{c}+…+{i}+{j}\overset{\mathrm{9}} {\equiv}\mathrm{0}\Rightarrow\mathrm{9}\mid{n} \\ $$$$\mathrm{9}\mid{n}\:,\:\mathrm{11111}\mid{n}\:\:,\:\left(\mathrm{9},\mathrm{11111}\right)=\mathrm{1}\Rightarrow\mathrm{99999}\mid{n} \\ $$$${let}\:\:{x}=<{abcde}>\:\:\&\:\:{y}={f}<{fghij}> \\…
Question Number 191821 by mehdee42 last updated on 01/May/23 $$\:{Q}\:\blacktriangleright\:{Show}\:{that}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{{i}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{i}+{n}} \\ $$ Answered by mehdee42 last updated on…
Question Number 191758 by Mingma last updated on 30/Apr/23 Answered by AST last updated on 30/Apr/23 $$\mathrm{521}^{\mathrm{1000}} \equiv\mathrm{0001}\left({mod}\:\mathrm{10000}\right)\: \\ $$$$\left(\mathrm{521},\mathrm{10000}\right)=\mathrm{1}\:\Rightarrow\:\:\mathrm{521}^{\mathrm{999}} \equiv\frac{\mathrm{1}}{\mathrm{521}}\left({mod}\:\mathrm{10000}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{521}}\equiv{x}\left({mod}\mathrm{10000}\right)\Rightarrow\mathrm{521}{x}\equiv\mathrm{1}\left({mod}\:\mathrm{10000}\right) \\ $$$$\mathrm{10000}{q}−\mathrm{521}{x}=−\mathrm{1}…
Question Number 60441 by ajfour last updated on 21/May/19 $$\mathrm{If}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\:\mathrm{money}\:\mathrm{doubles}\:\mathrm{itself} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{time}\:\mathrm{T},\:\mathrm{when}\:\mathrm{compounded} \\ $$$$\mathrm{continuously},\:\mathrm{find}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of} \\ $$$$\mathrm{interest},\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{T}. \\ $$ Answered by tanmay last updated on 21/May/19…
Question Number 125977 by liberty last updated on 16/Dec/20 $$\:{Suppose}\:{you}\:{put}\:\$\mathrm{3000}\:{in}\:{a}\:{savings}\:{account} \\ $$$${with}\:{a}\:\mathrm{5\%}\:{annual}\:{interesrate}\:,\:{compounded} \\ $$$${continously}\:.\:{How}\:{long}\:{would}\:{it}\:{take}\:{for} \\ $$$${your}\:{money}\:{tl}\:{double}\:?\: \\ $$ Answered by bramlexs22 last updated on 16/Dec/20…
Question Number 125918 by I want to learn more last updated on 15/Dec/20 $$\underset{\mathrm{n}\:\:=\:\:−\:\:\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$ Answered by mindispower last updated on…