answer:The automorphism groups of 3-dimensional complex Lie algebras have been studied in the literature, and one reference you can consult is the following paper: Harvey, A., 1979, "Automorphisms of the Bianchi model Lie groups," Journal of Mathematical Physics, 20: 251–253. Additionally, a paper by Glad and collaborators classifies all such complex Lie algebras along with their automorphism groups. You can find this article for further information. For higher dimensions, the automorphism groups have also been investigated, including dimensions 4, 5, and 6, not only over the complex numbers but also over the real numbers. You can refer to the relevant sources in those studies.

answer:To solve for x in the equation 20x + 2 = 142, we can follow these steps: 1. Subtract 2 from both sides of the equation: 20x + 2 - 2 = 142 - 2 20x = 140 2. Divide both sides of the equation by 20: 20x / 20 = 140 / 20 x = 7 Therefore, the solution for x in the equation 20x + 2 = 142 is x = 7.

answer:Underapplied by 5,000 Explanation: 1. Calculate the predetermined overhead rate: 550,000 / 25,000 = 22 per direct labor hour 2. Calculate the applied manufacturing overhead: 22 x 24,500 = 539,000 3. Determine the difference between actual and applied overhead: 535,000 - 539,000 = -5,000 (underapplied)

answer:1. To rewrite the equation in standard form, we complete the square for both x and y terms: 2 x^2-6 x-6 y^2-10 y+1=0 2 left(x^2-3 xright)-6 left(y^2+frac{5}{3} yright)+1=0 2 left(x^2-3 x+frac{9}{4}right)-6 left(y^2+frac{5}{3} y+frac{25}{36}right)+1+frac{9}{2}-frac{25}{6}=0 2 left(x-frac{3}{2}right)^2-6 left(y+frac{5}{6}right)^2=-frac{2}{3} 2. The equation is now in standard form for a hyperbola: frac{(x-h)^2}{a^2}-frac{(y-k)^2}{b^2}=1 where (h, k) is the center, a^2 is the horizontal semi-axis length squared, and b^2 is the vertical semi-axis length squared. 3. From the equation, we can identify the following properties: - Center: left(frac{3}{2},-frac{5}{6}right) - Horizontal semi-axis length: a=sqrt{2} - Vertical semi-axis length: b=frac{1}{sqrt{6}} - Eccentricity: e=sqrt{1+frac{b^2}{a^2}}=sqrt{2} - Foci: left(frac{3}{2} pm frac{sqrt{2}}{2},-frac{5}{6}right) - Asymptotes: y=pm frac{b}{a} (x-h)+k, which gives y=frac{1}{6} left(3 sqrt{3}-5right)-frac{x}{sqrt{3}},y=frac{x}{sqrt{3}}+frac{1}{6} left(-5-3 sqrt{3}right) Classification: Hyperbola Equation: 2 left(x-frac{3}{2}right)^2-6 left(y+frac{5}{6}right)^2=-frac{2}{3} Foci: left( begin{array}{cc} frac{3}{2} pm frac{sqrt{2}}{2} & -frac{5}{6} end{array} right) Eccentricity: sqrt{2} Center: left(frac{3}{2},-frac{5}{6}right) Asymptotes: y=frac{1}{6} left(3 sqrt{3}-5right)-frac{x}{sqrt{3}},y=frac{x}{sqrt{3}}+frac{1}{6} left(-5-3 sqrt{3}right)