Question Number 81519 by ajfour last updated on 13/Feb/20 $$\mid\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}\mid=\mathrm{1} \\ $$$${solve}\:{for}\:{real}\:{x}. \\ $$ Commented by mr W last updated on 13/Feb/20 $$\mid\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}\mid=\mathrm{1} \\ $$$$\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}=\pm\mathrm{1}…
Question Number 81523 by Power last updated on 13/Feb/20 Commented by mr W last updated on 14/Feb/20 $$=\mathrm{100} \\ $$ Commented by Power last updated…
Question Number 81506 by Power last updated on 13/Feb/20 Commented by Power last updated on 13/Feb/20 $$\left.\mathrm{A}\left.\right)\left.\mathrm{1}\left.−\frac{\mathrm{1}}{\mathrm{50}!}\:\:\:\:\:\:\:\mathrm{B}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\right)\mathrm{1}+\frac{\mathrm{1}}{\mathrm{49}!}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{2} \\ $$ Commented by mr W last updated…
Question Number 81507 by Power last updated on 13/Feb/20 Commented by mr W last updated on 13/Feb/20 $${we}\:{have}\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\…
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Question Number 147011 by mathdanisur last updated on 17/Jul/21 $$\mathrm{2}{x}\:-\:\sqrt{\mathrm{2}{x}\:-\:\mathrm{3}}\:-\:\mathrm{9}\:=\:\mathrm{0} \\ $$$${if}\:{there}'{s}\:\boldsymbol{{a}}\:{solution}\:{to}\:{equation}, \\ $$$${find}\:\:\mathrm{4}\boldsymbol{{a}}\:+\:\mathrm{3}\:=\:? \\ $$ Commented by 7770 last updated on 17/Jul/21 $$\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{36}\boldsymbol{{x}}+\mathrm{81}−\mathrm{2}\boldsymbol{{x}}+\mathrm{3}=\mathrm{0}…
Question Number 15927 by ajfour last updated on 15/Jun/17 Commented by ajfour last updated on 15/Jun/17 $${Q}.\mathrm{15917}\:\left({sorry}\:{it}\:{got}\:{uploaded}\right. \\ $$$$\left.{as}\:{new}\:{question}\right) \\ $$$${mid}\:{points}\:{of}\:{the}\:{sides}\:{of}\: \\ $$$${quadrilateral}\:{ABCD}\:{are}\: \\ $$$${E},\:{F},\:{G},\:{and}\:{H}.\:{Midpoints}\:{of}…
Question Number 146988 by mathdanisur last updated on 16/Jul/21 $${if}\:\:\:\frac{\boldsymbol{{z}}^{\mathrm{2}} \:+\:\mathrm{4}}{\boldsymbol{{z}}\:-\:\mathrm{2}\boldsymbol{{i}}}\:=\:−\mathrm{5}\:+\:\mathrm{10}\boldsymbol{{i}} \\ $$$${find}\:\:\:{Re}\left({Z}\right)\:+\:{Im}\left({Z}\right)\:=\:? \\ $$ Commented by aliibrahim1 last updated on 17/Jul/21 $${z}^{\mathrm{2}} +\mathrm{4}=\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right) \\…
Question Number 146989 by mathdanisur last updated on 16/Jul/21 $${Simplify}: \\ $$$$\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:{y}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:+\:{y}^{\frac{\mathrm{1}}{\mathrm{6}}} }\:\:-\:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \:-\:{y}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \:-\:{y}^{\frac{\mathrm{1}}{\mathrm{6}}} }\:=\:? \\ $$ Commented by 7770…
Question Number 15916 by tawa tawa last updated on 15/Jun/17 Commented by tawa tawa last updated on 15/Jun/17 $$\mathrm{6a}\:\mathrm{and}\:\mathrm{b} \\ $$ Commented by tawa tawa…