Question Number 94095 by Jidda28 last updated on 16/May/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{subgroups}\:\mathrm{do}\:\mathrm{Z}_{\mathrm{3}} \oplus\mathrm{Z}_{\mathrm{16}\:} \:\mathrm{has}?\:\mathrm{Justify}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28554 by naka3546 last updated on 27/Jan/18 $$\frac{\left({a}\:−\:{b}\right)}{\left({c}\:−\:{d}\right)}\:\:=\:\:\mathrm{3} \\ $$$$\frac{\left({a}\:−\:{c}\right)}{\left({b}\:−\:{d}\right)}\:\:=\:\:\mathrm{4} \\ $$$$\frac{\left({a}\:−\:{d}\right)}{\left({b}\:−\:{c}\right)}\:\:=\:\:? \\ $$$$ \\ $$ Answered by ajfour last updated on 27/Jan/18…
Question Number 28546 by abdo imad last updated on 26/Jan/18 $${let}\:{give}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:{and}\:\:{S}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{w}^{{k}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:{S}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{w}^{\left({q}+{k}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:\mid{S}\mid. \\ $$ Terms…
Question Number 28544 by abdo imad last updated on 26/Jan/18 $${if}\:\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,…{a}_{\mathrm{14}\:} {are}\:{roots}\:{of}\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)={x}^{\mathrm{14}} +{x}^{\mathrm{8}} \:\mathrm{2}{x}+\mathrm{1}\:\:\:{calculate}\:\:\sum_{{i}=\mathrm{1}} ^{\mathrm{14}} \:\:\frac{\mathrm{1}}{\left({a}_{{i}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$ Commented…
Question Number 28533 by abdo imad last updated on 26/Jan/18 $${let}\:{give}\:{the}\:{matrice}\:\:{A}=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${calculate}\:\:{A}^{{n}} \:\:{then}\:{find}\:\:{e}^{{A}} \:. \\ $$ Commented by abdo imad last updated on 28/Jan/18…
Question Number 28534 by abdo imad last updated on 26/Jan/18 $${find}\:{n}\:{from}\:{N}\:\:{in}\:{ordre}\:{tohave}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:{divide} \\ $$$$\left({x}+\mathrm{1}\right)^{{n}} −{x}^{{n}} −\mathrm{1}. \\ $$ Commented by abdo imad last updated on…
Question Number 28532 by abdo imad last updated on 26/Jan/18 $${let}\:{give}\:\:{A}_{{n}} =\:\left(\:{C}_{{n}} ^{\mathrm{0}} \:.{C}_{{n}} ^{\mathrm{1}} \:….{C}_{{n}} ^{{n}} \right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \:\:\:{find}\:^{{n}} \sqrt{{A}}\:_{{n}} . \\ $$ Commented by…
Question Number 159587 by cortano last updated on 19/Nov/21 $$\:\sqrt{\mathrm{2}−{x}}\:\sqrt{\mathrm{3}−{x}}\:+\:\sqrt{\mathrm{3}−{x}}\:\sqrt{\mathrm{4}−{x}}\:+\:\sqrt{\mathrm{2}−{x}}\:\sqrt{\mathrm{4}−{x}}\:=\:{x}+\mathrm{2} \\ $$$$ \\ $$ Commented by mr W last updated on 19/Nov/21 $${we}\:{can}\:{use}\:{my}\:{method}\:{in}\:{Q}\mathrm{159527} \\ $$$${for}\:{solving}\:{this}\:{kind}\:{of}\:{equations}.…
Question Number 28506 by Tinkutara last updated on 26/Jan/18 Commented by Tinkutara last updated on 27/Jan/18 Anyone? Commented by ajfour last updated on 27/Jan/18 $${post}\:{the}\:{answer},\:{please}.…
Question Number 159578 by HongKing last updated on 18/Nov/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\left(\mathrm{log}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{n}\:+\:\mathrm{1}}\right)\right)^{\mathrm{2}} }{\mathrm{log}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\right)}\right) \\ $$$$\mathrm{Answer}:\:\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com