answer:left( begin{array}{cccc} frac{4}{3} & -frac{23}{9} & -frac{1}{9} & frac{20}{9} frac{20}{9} & -frac{20}{9} & 0 & frac{35}{9} -frac{4}{9} & frac{70}{9} & frac{2}{3} & -frac{1}{9} frac{4}{3} & frac{65}{9} & frac{7}{9} & frac{28}{9} frac{32}{9} & frac{1}{9} & frac{1}{3} & frac{59}{9} end{array} right)

answer:The (pseudo)inverse of a matrix A, denoted by A^+, is a matrix that satisfies the following equation: AA^+A = A. To compute the (pseudo)inverse of A, we can use the following formula: A^+ = (A^TA)^{-1}A^T, where A^T is the transpose of A. First, we compute the transpose of A: A^T = left( begin{array}{ccc} -5 & -2 & 1 -3 & -4 & 3 -1 & 4 & 3 end{array} right). Next, we compute the product of A and A^T: A^TA = left( begin{array}{ccc} -5 & -3 & -1 -2 & -4 & 4 1 & 3 & 3 end{array} right) left( begin{array}{ccc} -5 & -2 & 1 -3 & -4 & 3 -1 & 4 & 3 end{array} right) = left( begin{array}{ccc} 35 & 19 & -5 19 & 26 & -6 -5 & -6 & 11 end{array} right). Then, we compute the inverse of A^TA: (A^TA)^{-1} = frac{1}{23} left( begin{array}{ccc} 26 & -6 & 6 -6 & 11 & -5 6 & -5 & 35 end{array} right). Finally, we compute the product of (A^TA)^{-1} and A^T: (A^TA)^{-1}A^T = frac{1}{23} left( begin{array}{ccc} 26 & -6 & 6 -6 & 11 & -5 6 & -5 & 35 end{array} right) left( begin{array}{ccc} -5 & -2 & 1 -3 & -4 & 3 -1 & 4 & 3 end{array} right) = left( begin{array}{ccc} -frac{6}{23} & frac{3}{46} & -frac{4}{23} frac{5}{46} & -frac{7}{46} & frac{11}{46} -frac{1}{46} & frac{3}{23} & frac{7}{46} end{array} right). Therefore, the (pseudo)inverse of A is A^+ = left( begin{array}{ccc} -frac{6}{23} & frac{3}{46} & -frac{4}{23} frac{5}{46} & -frac{7}{46} & frac{11}{46} -frac{1}{46} & frac{3}{23} & frac{7}{46} end{array} right). The answer is A^+ = left( begin{array}{ccc} -frac{6}{23} & frac{3}{46} & -frac{4}{23} frac{5}{46} & -frac{7}{46} & frac{11}{46} -frac{1}{46} & frac{3}{23} & frac{7}{46} end{array} right).

answer:The intersection of open balls with arbitrarily small radii around a point only contains the point itself. However, if each individual ball intersects a set, it only implies that there are points from the set that are arbitrarily close to the point, not that the point itself is in the set.

answer:1. To rewrite the equation in standard form, we complete the square for both x and y terms. 2. The classification of the conic is determined by the coefficients of x^2 and y^2. Since the coefficients have opposite signs, it is a hyperbola. 3. The foci, eccentricity, center, and asymptotes are calculated using the standard form of the equation. Classification: Hyperbola Equation: 3 left(y+frac{1}{2}right)^2-5 left(x+frac{2}{5}right)^2=-frac{61}{20} Foci: left( begin{array}{cc} frac{1}{15} left(-6-sqrt{366}right) & -frac{1}{2} frac{1}{15} left(sqrt{366}-6right) & -frac{1}{2} end{array} right) Eccentricity: 2 sqrt{frac{2}{3}} Center: left{frac{1}{2} left(frac{1}{15} left(-6-sqrt{366}right)+frac{1}{15} left(sqrt{366}-6right)right),-frac{1}{2}right} Asymptotes: left{y=sqrt{frac{5}{3}} x+frac{1}{30} left(4 sqrt{15}-15right),y=frac{1}{30} left(-15-4 sqrt{15}right)-sqrt{frac{5}{3}} xright}