answer:To evaluate the given double integral, we can use polar coordinates. Let x=rcostheta quad text{and} quad y=rsintheta. Then, x^2+y^2=r^2 and dxdy=r,dr,dtheta. The region D is the first quadrant of the annulus bounded by the circles of radii 2 and 5. In polar coordinates, this region is given by 2le rle 5 quad text{and} quad 0le thetale frac{pi}{2}. Therefore, the double integral becomes begin{split} int !!! int_D x^3y^2 ln (x^2+y^2) , dxdy &= int_0^{pi/2} int_2^5 r^6 cos^3theta sin^2theta ln (r^2) , dr,dtheta &= int_0^{pi/2} int_2^5 r^6 cos^3theta (1-cos^2theta) ln (r^2) , dr,dtheta &= int_0^{pi/2} int_2^5 r^6 (cos^3theta-cos^5theta) ln (r^2) , dr,dtheta &= int_0^{pi/2} left[ frac{r^7}{7} (cos^3theta-cos^5theta) ln (r^2) right]_2^5 , dtheta &= int_0^{pi/2} frac{125}{7} (cos^3theta-cos^5theta) ln 25 , dtheta &= frac{125}{7} ln 25 int_0^{pi/2} (cos^3theta-cos^5theta) , dtheta &= frac{125}{7} ln 25 left[ frac{sin 3theta}{3} - frac{sin 5theta}{5} right]_0^{pi/2} &= frac{125}{21} ln 25 end{split}

answer:The eigenvalues of the matrix are: [ lambda_1 = -2.923 - 1.786 i, quad lambda_2 = -2.923 + 1.786 i, quad lambda_3 = 3.846 ]

answer:To find x - y, we subtract the corresponding real parts and the imaginary parts of x and y: x - y = left(-frac{14}{pi} + frac{8i}{pi}right) - left(-frac{5}{pi} + frac{19i}{pi}right) Combine like terms: x - y = -frac{14}{pi} + frac{5}{pi} + frac{8i}{pi} - frac{19i}{pi} x - y = -frac{14 - 5}{pi} + frac{8 - 19}{pi}i x - y = -frac{9}{pi} - frac{11}{pi}i So, the difference is -frac{9 + 11i}{pi}.

answer:To find the acceleration, we can apply Newton's second law to each block: Block 1: ``` T - f_k = m_1a ``` where: * T is the tension in the string * f_k is the force of kinetic friction * m_1 is the mass of Block 1 (5.0 kg) * a is the acceleration of the system Block 2: ``` m_2g - T = m_2a ``` where: * m_2 is the mass of Block 2 (7.0 kg) * g is the acceleration due to gravity (9.8 m/s^2) The force of kinetic friction can be calculated as: ``` f_k = μ_k * N = μ_k * m_1 * g ``` where: * μ_k is the coefficient of kinetic friction (0.30) Substituting the expression for f_k into the equation for Block 1 and solving for T, we get: ``` T = m_1a + μ_k * m_1 * g ``` Substituting this expression for T into the equation for Block 2, we get: ``` m_2g - (m_1a + μ_k * m_1 * g) = m_2a ``` Solving for a, we get: ``` a = (m_2g - μ_k * m_1 * g) / (m_1 + m_2) ``` Plugging in the given values, we get: ``` a = (7.0 kg * 9.8 m/s^2 - 0.30 * 5.0 kg * 9.8 m/s^2) / (5.0 kg + 7.0 kg) ``` ``` a = 4.07 m/s^2 ``` Therefore, the acceleration of the system is 4.07 m/s^2.