Question Number 105322 by qwertyu last updated on 27/Jul/20 Commented by qwertyu last updated on 27/Jul/20 $$\boldsymbol{{please}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 170856 by mnjuly1970 last updated on 01/Jun/22 $$ \\ $$$$\:\:{in}\:{A}\overset{\Delta} {{B}C}\::\:\:{cos}\:\left({A}\right)+{cos}\left({B}\right)+{cos}\left({C}\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$$\:\:\:\frac{{R}}{{r}}\:=? \\ $$ Commented by mr W last updated on 01/Jun/22…
Question Number 39783 by ajfour last updated on 10/Jul/18 $${Find}\:{roots}\:{of} \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{35}{x}^{\mathrm{2}} −\mathrm{50}{x}+\mathrm{24} \\ $$ Answered by ajfour last updated on 10/Jul/18 $${let}\:{f}\left({x}\right)=\left({x}^{\mathrm{2}}…
Question Number 105320 by Josias last updated on 27/Jul/20 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{C}} \\ $$$${kx}^{\mathrm{2}} +{ky}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{C} \\ $$$${x},{y},{z},{k}\in\mathbb{R}\:\:{such}\:{that} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 170849 by mathlove last updated on 01/Jun/22 $$\mathrm{2}{x}+\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{8}{x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=? \\ $$ Answered by MJS_new last updated on 01/Jun/22 $$\sqrt{{x}}\in\mathbb{R}\:\Rightarrow\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{let}\:\sqrt{{x}}={t}\geqslant\mathrm{0} \\…
Question Number 39779 by ajfour last updated on 10/Jul/18 $${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5} \\ $$$${Find}\:{the}\:{roots}. \\ $$ Answered by MJS last updated on 11/Jul/18 $$\mathrm{approximate}:…
Question Number 170844 by mr W last updated on 01/Jun/22 $${solve}\:{x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$ Answered by balirampatel last updated on 01/Jun/22 $${solution}:\:\: \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}\:=\:\mathrm{3}…
Question Number 105306 by bobhans last updated on 27/Jul/20 $$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\:}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}+…+\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}} \\ $$ Commented by bemath last updated on 28/Jul/20 $$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}}+\mathrm{1}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}+…+\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}+{n}\sqrt{{n}+\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{n}+\mathrm{1}}\left(\sqrt{{n}^{\mathrm{2}} +{n}}+{n}\right)} \\…
Question Number 170836 by balirampatel last updated on 01/Jun/22 $${x}^{\mathrm{3}} \:−\:\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \:+\:\gamma^{\mathrm{3}} \:=\:? \\ $$ Commented by cortano1 last updated on 01/Jun/22 $$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}}…
Question Number 105301 by Don08q last updated on 27/Jul/20 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{tables}\:\mathrm{or}\:\mathrm{calculator}, \\ $$$$\:{compare}\:\:\:\mathrm{6}^{\mathrm{7}} \:\mathrm{and}\:\:\:\mathrm{7}^{\mathrm{6}} \\ $$ Answered by 1549442205PVT last updated on 28/Jul/20 $$\frac{\mathrm{6}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{6}} }=\mathrm{6}×\left(\frac{\mathrm{6}}{\mathrm{7}}\right)^{\mathrm{6}}…