answer:The mgf of X^2+Y^2 can be found by using the independence of X and Y and the mgf of the chi-squared distribution. Since X and Y are independent, their mgfs can be multiplied together: M_{X^2+Y^2}(t) = E[e^{t(X^2+Y^2)}] = E[e^{tX^2}]E[e^{tY^2}] The mgf of a standard normal random variable is given by M_X(t) = e^{t^2/2}, so we have: M_{X^2+Y^2}(t) = (e^{t^2/2})^2 = e^{t^2} This is the mgf of a chi-squared distribution with 2 degrees of freedom, so we can conclude that X^2+Y^2 has a chi-squared distribution with 2 degrees of freedom.

answer:To create a smooth intersection, consider using a quadratic Bézier curve to parameterize the parabolic arc. The endpoints, P_0 = (rcostheta, rsintheta) and P_2 = (-rcostheta, rsintheta), are on the circle, while the middle control point, P_1, should lie on the intersection of the common tangents at these points. By symmetry, P_1 = (0, rcsctheta). The Bézier curve is defined by (1-t)^2P_0 + 2t(1-t)P_1 + t^2P_2, and the vertex is at t = frac{1}{2}, which should match the vertex of the parabola, (0, frac{5}{3}r). Alternatively, you can directly find the equation for the parabola symmetrically tangent to the circle. First, find the family of parabolas that share a common tangent with the circle at the endpoints P_0 and P_2. This will lead to a system of equations involving a and the tangent lines. Then, solve for a such that the vertex of the parabola coincides with (0, frac{5}{3}r). Both methods will result in a curve that smoothly intersects the circle, ensuring continuity in the first quadrant.

answer:To find the inverse, we'll use the formula for the inverse of a 3×3 matrix. Denote the given matrix as A, and its inverse as A⁻¹. The formula for A⁻¹ is: A^{-1} = frac{1}{text{det}(A)} begin{bmatrix} text{det}(A_{23}) & text{det}(A_{13}) & text{det}(A_{12}) text{det}(A_{31}) & text{det}(A_{11}) & text{det}(A_{21}) text{det}(A_{32}) & text{det}(A_{12}) & text{det}(A_{11}) end{bmatrix} where det(A) is the determinant of matrix A, and Aij denotes the minor, which is the determinant of the 2×2 matrix formed by deleting the i-th row and j-th column from A. First, calculate the determinant of A: text{det}(A) = left(frac{3}{2}right)left(frac{17}{8} cdot left(-frac{7}{2}right) - frac{1}{8} cdot frac{7}{8}right) - left(frac{9}{4}right)left(frac{19}{4} cdot left(-frac{7}{2}right) - frac{1}{8} cdot 3right) + left(-1right)left(frac{19}{4} cdot frac{7}{8} - frac{17}{8} cdot 3right) text{det}(A) = frac{3731}{128} Now, we'll compute the determinants for the minors: text{det}(A_{23}) = left(frac{17}{8}right)left(-frac{7}{2}right) - left(frac{7}{8}right)left(frac{1}{8}right) = -frac{483}{64} text{det}(A_{13}) = -left(-1right)left(frac{7}{8}right) - frac{9}{4}left(-frac{7}{2}right) = 7 text{det}(A_{12}) = left(frac{9}{4}right)left(frac{1}{8}right) - left(-1right)left(frac{17}{8}right) = frac{77}{32} text{det}(A_{31}) = left(-frac{7}{2}right)left(-1right) - frac{19}{4}left(frac{1}{8}right) = 17 text{det}(A_{11}) = left(frac{3}{2}right)left(-frac{7}{2}right) - 3left(frac{1}{8}right) = -frac{9}{4} text{det}(A_{21}) = left(-frac{19}{4}right)left(-1right) - frac{17}{8}left(frac{3}{2}right) = -frac{79}{16} text{det}(A_{32}) = left(frac{19}{4}right)left(frac{7}{8}right) - 3left(frac{7}{8}right) = frac{87}{16} Finally, we calculate the inverse matrix: A^{-1} = frac{1}{frac{3731}{128}}begin{bmatrix} -frac{483}{64} & 7 & frac{77}{32} 17 & -frac{9}{4} & -frac{79}{16} -frac{71}{32} & frac{87}{16} & -frac{15}{2} end{bmatrix} A^{-1} = frac{128}{3731}begin{bmatrix} -frac{483}{64} & 7 & frac{77}{32} 17 & -frac{9}{4} & -frac{79}{16} -frac{71}{32} & frac{87}{16} & -frac{15}{2} end{bmatrix} So, the inverse of the given matrix is: A^{-1} = begin{bmatrix} -frac{61056}{261232} & frac{8960}{261232} & frac{9408}{261232} frac{23936}{261232} & -frac{28896}{261232} & -frac{48928}{261232} -frac{11392}{261232} & frac{13992}{261232} & -frac{55200}{261232} end{bmatrix} Simplifying the fractions, we get: A^{-1} = begin{bmatrix} -0.234 & 0.034 & 0.033 0.092 & -0.111 & -0.188 -0.044 & 0.054 & -0.212 end{bmatrix}

answer:1. Calculate the magnitude of the vector: r = sqrt{x^2 + y^2 + z^2} = sqrt{9^2 + (sqrt{2})^2 + (frac{1}{5})^2} = frac{2 sqrt{519}}{5} 2. Calculate the polar angle theta: theta = tan ^{-1}left(frac{y}{x}right) = tan ^{-1}left(frac{sqrt{2}}{9}right) 3. Calculate the azimuthal angle phi: phi = tan ^{-1}left(frac{z}{r sin theta}right) = tan ^{-1}left(frac{frac{1}{5}}{frac{2 sqrt{519}}{5} sin left(tan ^{-1}left(frac{sqrt{2}}{9}right)right)}right) = tan ^{-1}left(5 sqrt{83}right) Therefore, the spherical coordinates of the given vector are left(frac{2 sqrt{519}}{5}, tan ^{-1}left(5 sqrt{83}right), tan ^{-1}left(frac{sqrt{2}}{9}right)right). The answer is left(frac{2 sqrt{519}}{5}, tan ^{-1}left(5 sqrt{83}right), tan ^{-1}left(frac{sqrt{2}}{9}right)right)