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Category: Algebra

Question-210629

Question Number 210629 by peter frank last updated on 14/Aug/24 Answered by Rasheed.Sindhi last updated on 14/Aug/24 $$\begin{cases}{{x}={a}+\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)+,…+\left({a}+\left({m}−\mathrm{1}\right){d}\right)}\\{{y}=\left({a}+{md}\right)+\left({a}+\left({m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}=\left({a}+\mathrm{2}{md}\right)+\left({a}+\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\: \\ $$$$\begin{cases}{{x}−{ma}=\left({d}\right)+\left(\mathrm{2}{d}\right)+,…+\left(\left({m}−\mathrm{1}\right){d}\right)}\\{{y}−{ma}=\left({md}\right)+\left(\left({m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}−{ma}=\left(\mathrm{2}{md}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\frac{{x}−{ma}}{{d}}=\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+,…+\left(\left({m}−\mathrm{1}\right)\right)}\\{\frac{{y}−{ma}}{{d}}=\left({m}\right)+\left(\left({m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right)\right)\:}\\{\frac{{z}−{ma}}{{d}}=\left(\mathrm{2}{m}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right)\right)}\end{cases}\: \\ $$$$…\:\: \\…

x-2-3x-x-2-3x-1-7-2-2-sin-y-x-x-and-y-

Question Number 210581 by alusto22 last updated on 13/Aug/24 $$\:\:\begin{cases}{{x}^{\mathrm{2}} +\mathrm{3}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}\:=\:\mathrm{7}}}\\{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:{y}\:=\:{x}}\end{cases} \\ $$$$\:\:\:{x}=?\:\:\:\:{and}\:\:\:\:{y}=? \\ $$ Answered by Rasheed.Sindhi last updated on 14/Aug/24 $${x}^{\mathrm{2}} +\mathrm{3}{x}−\sqrt{{x}^{\mathrm{2}}…

if-the-roots-of-the-equation-x-2-k-1-x-k-0-are-and-find-the-value-of-the-real-constant-k-for-which-2-

Question Number 210574 by ChantalYah last updated on 13/Aug/24 $${if}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$${x}^{\mathrm{2}} +\left({k}+\mathrm{1}\right){x}+{k}=\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta, \\ $$$$\:{find}\:{the}\:{value}\:{of}\:{the} \\ $$$$\:{real}\:{constant}\:{k}\:{for} \\ $$$${which}\:\alpha=\mathrm{2}\beta \\ $$ Answered by…

Question-210606

Question Number 210606 by peter frank last updated on 13/Aug/24 Answered by A5T last updated on 14/Aug/24 $$\mathrm{3}\left({y}^{{log}_{\mathrm{5}} \mathrm{2}} \right)+\left({y}^{{log}_{{y}} \mathrm{2}×{log}_{\mathrm{5}} {y}} \right)=\mathrm{3}\left({y}^{{log}_{\mathrm{5}} \mathrm{2}} \right)+{y}^{{log}_{\mathrm{5}}…