Category: Algebra

Question Number 151100 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}0⩽\mathrm{x};\mathrm{y};\mathrm{z}⩽\mathrm{k}\mathrm{and}\mathrm{k}>0\mathrm{then}:$$$$\mathrm{y}(\mathrm{x}-\mathrm{z})-\mathrm{z}(\mathrm{x}-\mathrm{k})⩽{\mathrm{k}}^2$$ Answered by dumitrel last updated on 18/Aug/21 $$\iff y(x-z)+z(k-x)⩽k^2$$$${I}.\:\:{If}\:{x}\leqslant{z}\Rightarrow{y}\left({x}−{z}\right)\leqslant\mathrm{0} \…

Question Number 151085 by mathdanisur last updated on 18/Aug/21 Answered by talminator2856791 last updated on 18/Aug/21 $$$$$$\:\frac{{a}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}} +\left({a}+\mathrm{3}\right)^{\mathrm{2}} }{\left({a}+\mathrm{4}\right)^{\mathrm{2}} −\left({a}+\mathrm{3}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}\right)^{\mathrm{2}}…

Question Number 20013 by Tinkutara last updated on 20/Aug/17 $$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{equation}\frac{1}{x-a}+\frac{1}{x-b}$$$$+\frac{1}{x-c}=0\mathrm{can}\mathrm{have}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{equal}$$$$\mathrm{roots}\mathrm{if}a=b=c.$$ Commented by ajfour last updated on 20/Aug/17 $${how}\:{did}\:{u}\:{solve}\:{it}\:,\:{please}\:{let}\:{me} \…

Question Number 151080 by malwan last updated on 18/Aug/21 $$if(f\circ f\circ f\circ f)\left(x\right)=16x+15$$$$findf\left(x\right)$$ Answered by Mokmokhi last updated on 18/Aug/21 $$\mathrm{It}\mathrm{is}\mathrm{reasonable}\mathrm{to}\mathrm{say}f\left(x\right)\mathrm{is}\mathrm{linear}\mathrm{by}\mathrm{rejecting}\mathrm{other}\mathrm{possibilities}.$$$$\mathrm{Then}\:{f}\left({x}\right)={ax}+{b}\:\mathrm{for}\:\mathrm{some}\:\mathrm{unknowns}\:{a}\:\mathrm{and}\:{b}. \…

Question Number 20001 by Tinkutara last updated on 20/Aug/17 $$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}$$$$\mathrm{equation}{8\sec }^2\theta -6\sec \theta +1=0\mathrm{is}$$ Answered by mrW1 last updated on 20/Aug/17 $$\mathrm{8sec}^{\mathrm{2}} \theta\:−\:\mathrm{6sec}\theta\:+\:\mathrm{1}\:=\:\mathrm{0} \…