Question Number 129687 by ajfour last updated on 17/Jan/21 $$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$$\left({solve}\:{for}\:{x}\:\:{even}\:{if}\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$ Answered by ajfour last updated on 18/Jan/21 $${Just}\:{now}\:{I}\:{developed}\:{a}\:{new} \\ $$$${approximate}\:{formula}:…
Question Number 129677 by Adel last updated on 17/Jan/21 Commented by som(math1967) last updated on 17/Jan/21 $$\mathrm{3}^{\frac{\mathrm{1}}{{cscx}}×\frac{\mathrm{1}}{{sinx}}×\frac{\mathrm{1}}{{tanx}}×\frac{\mathrm{1}}{{cotx}}} =\mathrm{3}^{\mathrm{1}} =\mathrm{3} \\ $$ Commented by Adel last…
Question Number 64130 by ANTARES VY last updated on 14/Jul/19 $$\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}+\frac{\mathrm{12}}{\boldsymbol{\mathrm{x}}}}=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{7}}{\boldsymbol{\mathrm{x}}+\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{x}}=? \\ $$ Commented by Prithwish sen last updated on 14/Jul/19 $$\mathrm{x}=\:\mathrm{1},\mathrm{1},\mathrm{3}\:\mathrm{and}\:\frac{−\mathrm{1}\pm\mathrm{15i}}{\mathrm{2}}…
Question Number 64126 by Cheyboy last updated on 13/Jul/19 Answered by ajfour last updated on 13/Jul/19 $$\:\:\mathrm{32}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:−\mathrm{32}\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24}\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{18}\:\:\:\:\:\:\:−\mathrm{18} \\ $$$${d}_{{n}}…
Question Number 129663 by liberty last updated on 17/Jan/21 $$\:\sqrt{\mathrm{14}+\sqrt{\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{5}\sqrt{\mathrm{3}}}}\:=?\: \\ $$ Commented by MathSh last updated on 17/Jan/21 $$\sqrt[{\mathrm{3}}]{\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }=\sqrt[{\mathrm{3}}]{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\sqrt{\mathrm{14}+\sqrt{\mathrm{3}}+\mathrm{2}−\sqrt{\mathrm{3}}}=\sqrt{\mathrm{16}}=\mathrm{4} \\…
Question Number 64112 by mr W last updated on 13/Jul/19 Commented by mr W last updated on 13/Jul/19 $${Find}\:{the}\:{radius}\:{of}\:{the}\:{n}−{th}\:{inscribed} \\ $$$${circle}\:{within}\:{the}\:{parabola}. \\ $$ Answered by…
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Question Number 129622 by Ar Brandon last updated on 16/Jan/21 $$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\geqslant−\mathrm{1}.\:\mathrm{If}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{function}\:\mathrm{whose}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{x},\:\mathrm{then}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}… \\ $$ Commented by mr W last…
Question Number 129607 by HHHHH last updated on 16/Jan/21 $${salom} \\ $$ Answered by prakash jain last updated on 16/Jan/21 $$\mathrm{welcome} \\ $$ Answered by…
Question Number 64066 by mathmax by abdo last updated on 12/Jul/19 $${let}\:\alpha\:,\beta\:{and}\:\lambda\:{the}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\:=\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \:{and}\:\:{B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \:+\lambda^{\mathrm{3}} \:. \\ $$ Answered by…