answer:To multiply the polynomials, we can use the FOIL method (First, Outer, Inner, Last). First, we multiply the first terms of each polynomial: left(frac{33 x^2}{4}right)left(-13 x^2right) = -frac{429 x^4}{4} Next, we multiply the outer terms of each polynomial: left(frac{33 x^2}{4}right)(4 x) = frac{132 x^3}{4} Then, we multiply the inner terms of each polynomial: left(-frac{51 x}{4}right)(-13 x^2) = frac{663 x^3}{4} Finally, we multiply the last terms of each polynomial: left(-frac{51 x}{4}right)left(frac{29}{2}right) = -frac{1473 x}{8} Adding these products together, we get: -frac{429 x^4}{4}+frac{132 x^3}{4}+frac{663 x^3}{4}-frac{1473 x}{8} Combining like terms, we get: -frac{429 x^4}{4}+frac{795 x^3}{4}-frac{335 x^2}{8}-frac{1207 x}{8}+frac{493}{4} The answer is -frac{429 x^4}{4}+frac{795 x^3}{4}-frac{335 x^2}{8}-frac{1207 x}{8}+frac{493}{4}

answer:The given initial conditions allow us to set up a system of equations to solve for the unknowns c and a: 1. ( y(0) = 2 ) implies ( c = 2 ) since ( e^0 = 1 ). 2. ( y(1) = 1 ) gives us ( 1 = 2e^a ). Now, we solve for a using the second equation: ( 1 = 2e^a ) ( frac{1}{2} = e^a ) Taking the natural logarithm of both sides (assuming ( a neq 0 ) to avoid division by zero): ( lnleft(frac{1}{2}right) = a ) ( -ln(2) = a ) ( a = -ln(2) approx -0.693 ) Therefore, the specific form of the function is: ( y(x) = 2e^{-0.693x} ) To find ( y(2) ), we substitute ( x = 2 ) into the function: ( y(2) = 2e^{-0.693 cdot 2} ) ( y(2) = 2e^{-1.386} ) ( y(2) = 2 cdot frac{1}{e^{1.386}} ) ( y(2) approx frac{2}{3.94} ) ( y(2) approx 0.508 ) Hence, ( y(2) approx 0.508 ).

answer:The maximum value of the quadratic form on the unit sphere is equal to the largest eigenvalue of A, which is 4. This value is achieved by the corresponding eigenvector. The minimum value of the quadratic form on the unit sphere is equal to the smallest eigenvalue of A, which is -1. This value is achieved by the corresponding eigenvector.

answer:Using the formula derived in the previous answer, we have: int^frac{beta-1}{beta+1}_frac{alpha-1}{alpha+1}frac{{rm d}xi}{e^{2mu}-xi}=frac{pi}{2}lnleft(frac{e^{2mu}-frac{alpha-1}{alpha+1}}{e^{2mu}-frac{beta-1}{beta+1}}right) Substituting the given values of α, β, and μ, we obtain the following numerical values: | No. | Numerical Value | |---|---| | 1 | 0.037363172275853 | | 2 | 0.0050398726261960 | | 3 | 0.086272634856537 | | 4 | 0.097077534919905 | | 5 | 0.093245471867952 |