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Category: Algebra

a-b-c-d-3-a-c-b-d-4-a-d-b-c-

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…

let-give-w-e-i-2pi-n-and-S-k-0-n-1-w-k-2-1-prove-that-S-k-0-n-1-w-q-k-2-2-find-S-

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…

if-a-1-a-2-a-14-are-roots-of-the-polynomial-p-x-x-14-x-8-2x-1-calculate-i-1-14-1-a-i-1-2-

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…

2-x-3-x-3-x-4-x-2-x-4-x-x-2-

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}.…

Find-lim-n-log-1-1-n-1-2-log-1-1-n-2-Answer-0-

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