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Category: Algebra

Question-210643

Question Number 210643 by ChantalYah last updated on 14/Aug/24 Answered by mahdipoor last updated on 14/Aug/24 $$\left.\mathrm{1}\right){e}^{{x}} ={t} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}−\frac{\mathrm{4}}{{t}}=\mathrm{0}\Rightarrow{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{2}}…

given-that-the-roots-of-the-equation-3x-2-4-2k-x-2k-0-are-and-find-the-value-of-k-for-which-3-

Question Number 210639 by ChantalYah last updated on 14/Aug/24 $${given}\:{that}\:{the}\:{roots} \\ $$$$\:{of}\:{the}\:{equation} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right){x}+\mathrm{2}{k}=\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta \\ $$$${find}\:{the}\:{value}\:{of}\:{k} \\ $$$${for}\:{which}\:\beta=\mathrm{3}\alpha \\ $$ Answered by…

Question-210629

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