Category: Algebra

Question Number 193484 by Frix last updated on 15/Jun/23 $$\mathrm{Nice}\mathrm{problem}:$$$$\mathrm{Find}8\mathrm{distinctive}\mathrm{numbers}\in \mathbb{N}\mathrm{\setminus }\left\{0\right\}\mathrm{such}$$$$\mathrm{that}\mathrm{these}\mathrm{are}\mathrm{simultaniously}\mathrm{true}:$$$$\left(1\right)a+b+c+d=e+f+g+h$$$$\left(\mathrm{2}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{e}^{\mathrm{2}} +{f}^{\mathrm{2}} +{g}^{\mathrm{2}} +{h}^{\mathrm{2}}…

Question Number 193410 by cortano12 last updated on 13/Jun/23 $$\{\begin{array}{l}\mathrm{x}=\sqrt{3-\sqrt{5+2\sqrt{3}}}\\ \mathrm{y}=\sqrt{3+\sqrt{5+2\sqrt{3}}}\end{array}$$$$\underbrace{\mathit{x}}$$ Answered by aba last updated on 13/Jun/23 $$\mathrm{xy}=\sqrt{\mathrm{9}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}−\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{6} \…

Question Number 193363 by MATHEMATICSAM last updated on 11/Jun/23 $$y=4\times {10}^{2x}$$$$\mathrm{Express}x\mathrm{in}\mathrm{terms}\mathrm{of}y,\mathrm{giving}\mathrm{an}\mathrm{exact}$$$$\mathrm{simplified}\mathrm{answer}\mathrm{in}\mathrm{terms}\mathrm{of}\log \mathrm{base}10.$$ Answered by aba last updated on 11/Jun/23 $$\mathrm{y}=\mathrm{4}×\mathrm{10}^{\mathrm{2x}} \:\Rightarrow\:\mathrm{10}^{\mathrm{2x}}…

Question Number 193346 by pascal889 last updated on 11/Jun/23 $$\sqrt{4\mathrm{x}-3}-\sqrt{2\mathrm{x}-5}=2$$$$\mathrm{solve}\mathrm{for}\mathrm{x}$$ Answered by AST last updated on 11/Jun/23 $${u}−{v}=\mathrm{2};\:{u}^{\mathrm{2}} −\mathrm{2}{v}^{\mathrm{2}} =\mathrm{7} \…

Question Number 193314 by York12 last updated on 10/Jun/23 $$$$$$\mathit{a},\mathit{b},\mathit{c}>0\mathrm{\&}{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2=3\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\sqrt[3]{{(1+\frac{3}{\mathit{a}\mathit{b}+\mathit{b}\mathit{c}+\mathit{c}\mathit{a}})}^{{(\mathit{a}+\mathit{b}+\mathit{c})}^2}}⩽(1+\frac{\mathit{a}}{\mathit{b}})(1+\frac{\mathit{b}}{\mathit{c}})(1+\frac{\mathit{c}}{\mathit{a}})$$ Answered by witcher3 last…