Question Number 5119 by sanusihammed last updated on 14/Apr/16 Commented by 123456 last updated on 15/Apr/16 $${x}=\sqrt{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${x}=\sqrt{{a}}+\sqrt{{b}}\wedge{a}>\mathrm{0}\wedge{b}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} ={a}+{b}+\mathrm{2}\sqrt{{ab}}…
Question Number 5113 by FilupSmith last updated on 14/Apr/16 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}: \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{Focal}\:\mathrm{point} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{Directrix} \\ $$$$\mathrm{for}\:{y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{simplicity},\:\mathrm{lets}\:\mathrm{assume}\:\mathrm{it}\:\mathrm{goes}\:\mathrm{throigh} \\ $$$$\mathrm{point}\:\left(\mathrm{0},\:\mathrm{0}\right). \\ $$…
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Question Number 70638 by abdusalamyussif@gmail.com last updated on 06/Oct/19 Commented by MJS last updated on 06/Oct/19 $$\mathrm{the}\:\mathrm{first}\:\mathrm{question}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{complete} \\ $$ Answered by mr W last updated…
Question Number 5091 by FilupSmith last updated on 11/Apr/16 $$\mathrm{Part}\:\mathrm{1} \\ $$$${S}=\mathrm{ln}\left({a}+{bi}\right),\:\:\:\:\:\:{a},\:{b}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{is}\:{S}\in\mathbb{C}?\:\:\:\:\:\:\:\left({b}\neq\mathrm{0}\right) \\ $$$$ \\ $$$$\mathrm{Part}\:\mathrm{2} \\ $$$${t}=\mathrm{ln}\left(\mathrm{ln}\left(…\left(\mathrm{ln}\left({k}\right)\right)\right)\right),\:\:\:\:{k}\in\mathbb{R},\:{k}>\mathrm{1} \\ $$$$\therefore{t}\in\mathbb{C}?…
Question Number 136161 by nimnim last updated on 19/Mar/21 $${How}\:{do}\:{I}\:{find}\:{the}\:{sum}\:{of} \\ $$$$\:\:\mathrm{1}+\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{4}} +…….\infty,\:\:{where}\:−\mathrm{1}<{x}<\mathrm{1}\:? \\ $$$$\:\:{Please}\:{Help}.. \\ $$ Answered by bramlexs22 last updated on…
Question Number 70617 by Rio Michael last updated on 06/Oct/19 $${given}\:{that}\:\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{2}} −\mathrm{5}{x}\:+\:\mathrm{4}\:=\mathrm{0}\: \\ $$$$\alpha>\mathrm{0}\:{and}\:\beta\:>\mathrm{0} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\:\sqrt{\alpha}\:{and}\:\sqrt{\beta}\: \\ $$$$ \\ $$$${how}\:{do}\:{i}\:{find}\:\:\sqrt{\alpha\:}\:+\:\sqrt{\beta}\: \\ $$ Answered by mr…
Question Number 70598 by Shamim last updated on 06/Oct/19 $$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\:\Re\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{12}}{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$$$,\:\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$ Answered by som(math1967) last updated on 06/Oct/19 $$\frac{{a}^{\mathrm{2}}…
Question Number 136123 by bramlexs22 last updated on 18/Mar/21 $$\frac{\left\{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−{x}\right)^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−{x}\right)\left({x}−\mathrm{3}\right)}\:+\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:\right\}^{\mathrm{2}} }{\:\sqrt[{\mathrm{3}}]{\mathrm{12}−{x}}\:+\sqrt[{\mathrm{3}}]{{x}−\mathrm{3}}}\:=\:\frac{\mathrm{49}}{\mathrm{3}} \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\sqrt[{\mathrm{3}}]{\left(\mathrm{12}−\mathrm{x}\right)}\:=\:\mathrm{p}\:\&\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:=\:\mathrm{q} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{p}^{\mathrm{2}}…
Question Number 5029 by love math last updated on 04/Apr/16 $${a}\:−\:\frac{\mathrm{2}\:{log}_{\mathrm{12}} \left(\mathrm{8}−{a}\right)+\frac{\mathrm{1}}{{log}_{\mathrm{2}{a}} \mathrm{12}}+\:{log}_{\mathrm{7}+{a}} \mathrm{1}.\mathrm{6}}{{log}_{\mathrm{15}−{a}} \mathrm{2}\:+\:{log}_{\mathrm{10}} {a}} \\ $$$$ \\ $$$${Simplify} \\ $$ Terms of Service…