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Category: Arithmetic

2-3-2-2-3-2-3-2-2-3-simplify-

Question Number 59763 by ANTARES VY last updated on 14/May/19 $$\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\ $$$$\boldsymbol{\mathrm{simplify}}. \\ $$ Answered by Kunal12588 last updated on 14/May/19 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}=\sqrt{\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}…

f-x-f-x-x-f-5-1-f-1-

Question Number 190618 by mustafazaheen last updated on 07/Apr/23 $$\mathrm{f}\left(\mathrm{x}−\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\:\:,\mathrm{f}\left(\mathrm{5}\right)=−\mathrm{1} \\ $$$$\mathrm{f}\left(−\mathrm{1}\right)=? \\ $$ Answered by aba last updated on 07/Apr/23 $$\mathrm{f}\left(\mathrm{5}−\mathrm{f}\left(\mathrm{5}\right)\right)=\mathrm{5}\Rightarrow\:\mathrm{f}\left(\mathrm{4}\right)=\mathrm{5} \\ $$$$\mathrm{f}\left(\mathrm{4}−\mathrm{f}\left(\mathrm{4}\right)\right)=\mathrm{4}\Rightarrow\mathrm{f}\left(−\mathrm{1}\right)=\mathrm{4} \\…

Question-59316

Question Number 59316 by mr W last updated on 07/May/19 Answered by MJS last updated on 07/May/19 $${a}+{b}+{c}=\mathrm{25}\:\wedge\:{a},\:{b},\:{c}\:\in\mathbb{N}^{\bigstar} \:\Rightarrow\:\mathrm{1}\leqslant{a},\:{b},\:{c}\leqslant\mathrm{23} \\ $$$$\mathbb{S}=\left\{{n}\in\mathbb{N}^{\bigstar} \mid\mathrm{1}\leqslant{n}\leqslant\mathrm{23}\:\wedge\:\mathrm{mod}\left(\mathrm{360},\:{n}\right)=\mathrm{0}\right\}= \\ $$$$=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{8},\:\mathrm{9},\:\mathrm{10},\:\mathrm{12},\:\mathrm{15},\:\mathrm{18},\:\mathrm{20}\right\} \\…

solve-for-x-R-x-1-3x-5-4x-7-4x-5-

Question Number 190339 by mr W last updated on 01/Apr/23 $${solve}\:{for}\:{x}\:\in\mathbb{R} \\ $$$$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}}=\mathrm{4}{x}−\mathrm{5} \\ $$ Answered by ajfour last updated on 01/Apr/23 $$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}+\mathrm{3}{x}−\mathrm{5}+\mathrm{4}{x}−\mathrm{7}\right)+\frac{\mathrm{3}}{\mathrm{2}}…