Question Number 104859 by qwertyu last updated on 24/Jul/20 Answered by Dwaipayan Shikari last updated on 24/Jul/20 $${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+….+\frac{{n}}{\mathrm{2}^{{n}} } \\ $$$$\frac{{S}_{{n}}…
Question Number 170389 by mr W last updated on 22/May/22 $${solve}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{3}} +\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}−\mathrm{3}=\mathrm{0} \\ $$ Commented by Mastermind last updated on 22/May/22 $${We}\:{can}\:{actually}\:{solve}\:{this}\:{by}\:{graph} \\…
Question Number 104832 by bemath last updated on 24/Jul/20 $$\begin{cases}{{x}+{y}+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{7}}\\{\frac{\left({x}−{y}\right){x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{12}\:}\end{cases} \\ $$ Answered by john santu last updated on 24/Jul/20 $${let}\:{x}\:=\:\lambda{y}\:\rightarrow\begin{cases}{\lambda{y}+{y}+\frac{\lambda^{\mathrm{2}}…
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Question Number 104811 by bobhans last updated on 24/Jul/20 $${Given}\:\begin{cases}{{a}+{b}\sqrt{\mathrm{3}}−\mathrm{2}{c}\:=\:\mathrm{1}}\\{\mathrm{3}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{2}{a}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\mathrm{4}{ac}\:=\:\mathrm{5}{c}^{\mathrm{2}} }\end{cases} \\ $$$${find}\:{b} \\ $$ Answered by maths mind last updated…
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Question Number 104782 by qwertyu last updated on 23/Jul/20 Answered by abdomsup last updated on 23/Jul/20 $${this}\:{question}\:{is}\:{solved}\:{see}\:{the}\: \\ $$$${platform} \\ $$ Commented by qwertyu last…
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Question Number 39218 by behi83417@gmail.com last updated on 03/Jul/18 Answered by MJS last updated on 12/Aug/18 $${x}_{\mathrm{1}} =\mathrm{1};\:{y}_{\mathrm{1}} =\sqrt{\mathrm{2}}\:\left(\mathrm{plain}\:\mathrm{to}\:\mathrm{see}\:\mathrm{but}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{more}\:\mathrm{real}\:\mathrm{solutions}\right) \\ $$$${y}=−{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}} \\…