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…
Question Number 5023 by love math last updated on 03/Apr/16 $$\frac{\mathrm{25}^{−\mathrm{1}} +\:{log}_{\mathrm{2}} {a}^{−\mathrm{1}} }{\mathrm{5}^{−\mathrm{1}} −{log}_{\mathrm{4}} {a}}−\frac{\mathrm{3}}{{a}^{−\mathrm{0}.\mathrm{5}} } \\ $$$$ \\ $$$${Simplify} \\ $$ Answered by…
Question Number 136063 by liberty last updated on 18/Mar/21 $$ \\ $$A and B can do a job in 10 days and 5 days, respectively.…
Question Number 70518 by Shamim last updated on 05/Oct/19 $$\mathrm{If}\:\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} =\:\mathrm{513}\:\mathrm{and}\:\mathrm{ab}=\:\mathrm{54}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\mathrm{a}−\mathrm{b}=\:? \\ $$ Commented by Prithwish sen last updated on 05/Oct/19 $$\boldsymbol{\mathrm{use}}\:\:\:\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)^{\mathrm{3}}…
Question Number 70512 by abdusalamyussif@gmail.com last updated on 04/Oct/19 $$\mathrm{The}\:\mathrm{faculty}\:\mathrm{of}\:\mathrm{science}\:\mathrm{news}\:\mathrm{paper}\:\mathrm{reports}\:\mathrm{that}\:\mathrm{the}\:\mathrm{combined}\:\mathrm{membership}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{mathematics}\:\mathrm{club}\:\mathrm{and}\:\mathrm{science}\:\:\mathrm{club}\:\mathrm{is}\:\mathrm{122}\:\mathrm{students}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{membership}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chemistry}\:\mathrm{club}? \\ $$ Commented by tw000001 last updated on 14/Oct/19 $$\mathrm{To}\:\mathrm{put}\:\mathrm{it}\:\mathrm{simply}, \\ $$$${n}\left({C}_{{M}} \:\cup{C}_{{S}} \right)=\mathrm{122} \\…
Question Number 136054 by azadsir last updated on 18/Mar/21 Commented by mr W last updated on 18/Mar/21 $${a}=\mathrm{log}_{{x}} \:{xyz}=\frac{\mathrm{1}}{\mathrm{log}_{{xyz}} \:{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\mathrm{log}_{{xyz}} \:{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{log}_{{xyz}}…