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Author: Tinku Tara

Question-11139

Question Number 11139 by Joel576 last updated on 13/Mar/17 Commented by Joel576 last updated on 13/Mar/17 $$\mathrm{A}\:\mathrm{real}\:\mathrm{number}\:{t},\:\mathrm{so}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{three}−\mathrm{ordered}−\mathrm{pair} \\ $$$$\mathrm{solution}\:\left\{{x},\:{y},\:{z}\right\}\:\mathrm{that}\:\mathrm{fulfilled} \\ $$$${x}^{\mathrm{2}\:} \:+\:\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{3}{z}\:\:\:\mathrm{and}\:\:\:{x}\:+\:{y}\:+\:{z}\:=\:{t} \\ $$$$\mathrm{Find}\:{t}…

Find-the-equation-of-the-circle-which-is-orthogonal-to-the-circles-x-2-y-2-7x-y-0-and-x-2-y-2-3x-6y-5-0-and-which-passes-through-the-point-3-0-

Question Number 142208 by ZiYangLee last updated on 27/May/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{7}{x}−{y}=\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}{y}+\mathrm{5}=\mathrm{0}\:\mathrm{and}\:\mathrm{which}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{3},\mathrm{0}\right). \\ $$ Answered by mr…

Question-142204

Question Number 142204 by mnjuly1970 last updated on 27/May/21 Answered by MJS_new last updated on 28/May/21 $$\mathrm{what}\:\mathrm{are}\:\mathrm{you}\:\mathrm{waiting}\:\mathrm{for}? \\ $$$${A}^{\mathrm{2}} =\begin{bmatrix}{\mathrm{7}}&{\mathrm{2}{m}−\mathrm{5}}&{\mathrm{5}}\\{\mathrm{2}}&{{m}^{\mathrm{2}} −\mathrm{1}}&{{m}+\mathrm{2}}\\{\mathrm{2}}&{\mathrm{2}−{m}}&{\mathrm{9}}\end{bmatrix} \\ $$$${A}^{\mathrm{4}} =\begin{bmatrix}{\mathrm{4}{m}+\mathrm{49}}&{\mathrm{2}{m}^{\mathrm{3}} −\mathrm{5}{m}^{\mathrm{2}}…

Question-142207

Question Number 142207 by 777316 last updated on 27/May/21 Answered by mr W last updated on 28/May/21 $${say}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} ={L} \\ $$$${L}=\frac{\mathrm{3}{L}+\mathrm{4}}{\mathrm{2}{L}+\mathrm{3}} \\ $$$$\mathrm{2}{L}^{\mathrm{2}} +\mathrm{3}{L}=\mathrm{3}{L}+\mathrm{4}…