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Question-143602

Question Number 143602 by meetbhavsar25 last updated on 16/Jun/21 Commented by TheHoneyCat last updated on 16/Jun/21 $$\mathrm{Are}\:\mathrm{you}\:\mathrm{asking}\:\mathrm{for}\:\underset{{a}\in\mathbb{C}} {\prod}{g}\left({a}\right)^{\mathrm{val}_{{f}} \left({a}\right)} \\ $$$$ \\ $$$$\mathrm{or}\:{z}\:\mathrm{such}\:\mathrm{that} \\ $$$$\forall\left({a}_{{i}}…

Given-that-the-series-x-3-y-3-2-x-3-3-y-3-4-x-3-n-1-y-3-n-8-x-y-R-Find-the-value-of-3x-y-

Question Number 143596 by ZiYangLee last updated on 16/Jun/21 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{{x}}{\mathrm{3}^{} }+\frac{{y}}{\mathrm{3}^{\mathrm{2}} }+\frac{{x}}{\mathrm{3}^{\mathrm{3}} }+\frac{{y}}{\mathrm{3}^{\mathrm{4}} }+\ldots+\frac{{x}}{\mathrm{3}^{{n}−\mathrm{1}} }+\frac{{y}}{\mathrm{3}^{{n}} }=\mathrm{8},\:{x},{y}\in\mathbb{R}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{x}+{y}. \\ $$ Answered by bobhans…

old-activity-shown-in-my-recent-activity-page-ssems-like-someone-is-hindering-me-from-being-able-to-see-the-recent-activity-page-i-reinstalled-the-app-problem-remains-

Question Number 78045 by ajfour last updated on 13/Jan/20 $${old}\:{activity}\:{shown}\:{in}\:{my} \\ $$$${recent}\:{activity}\:{page},\:{ssems} \\ $$$${like}\:{someone}\:{is}\:{hindering} \\ $$$${me}\:{from}\:{being}\:{able}\:{to}\:{see}\:{the} \\ $$$${recent}\:{activity}\:{page},\:{i}\:{reinstalled} \\ $$$${the}\:{app},\:{problem}\:{remains}.. \\ $$ Commented by MJS…

lim-x-1-2-x-1-x-2-x-lim-n-n-n-1-n-1-

Question Number 143561 by ZiYangLee last updated on 15/Jun/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{2}}{{x}^{} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{x}} =? \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}}−\sqrt{{n}−\mathrm{1}}\right)\sqrt{{n}+\mathrm{1}}\:=? \\ $$ Answered by mr W last updated…

Question-143537

Question Number 143537 by 0731619 last updated on 15/Jun/21 Answered by mr W last updated on 15/Jun/21 $$\frac{{AB}+{BE}}{{BE}}=\frac{{AD}}{{BC}}=\frac{{a}}{{b}} \\ $$$$\frac{{AB}}{{BE}}=\frac{{a}−{b}}{{b}} \\ $$$$\frac{{GE}}{{AD}}=\frac{{BE}}{{AB}}=\frac{{b}}{{a}−{b}} \\ $$$$\Rightarrow{GE}=\frac{{ab}}{{a}−{b}} \\…