answer:Let the dimensions of the prism be x, y, and z. Given, xy = 18, yz = 12, and xz = 8. Assume z leq x leq y or any permutation where the condition y = 2z (longest side y is twice the shortest side z) holds. 1. From xy = 18 and yz = 12, solve for z: - Multiply the equations: (xy)(yz)(xz) = 18 cdot 12 cdot 8, - Thus, x^2y^2z^2 = 1728, - Taking the cube root gives xyz = 12. 2. Substitute y = 2z into yz = 12: - 2z^2 = 12, - z^2 = 6, - z = sqrt{6}. 3. Solve for other dimensions: - y = 2sqrt{6}, - Substitute z = sqrt{6} into xz = 8, - xsqrt{6} = 8, - x = frac{8}{sqrt{6}} = frac{8sqrt{6}}{6} = frac{4sqrt{6}}{3}. 4. Calculate the volume: - V = xyz = left(frac{4sqrt{6}}{3}right)(sqrt{6})(2sqrt{6}), - V = frac{4 cdot 6 cdot 2}{3} = frac{48}{3} = 16 cubic inches. Therefore, the volume of the prism is boxed{16} cubic inches.

answer:First, we simplify the denominator: [1 + 3^n + 3^{n+1} + 3^{2n+2} = (1 + 3^n) + 3^{n+1}(1 + 3^{n+1}) = (1 + 3^n)(1 + 3^{n+1}).] Next, express the numerator (3^n) as a difference: [ 3^n = (1 + 3^{n+1}) - (1 + 3^n), ] thus, the fraction becomes: [ frac{3^n}{1 + 3^n + 3^{n+1} + 3^{2n+2}} = frac{(1 + 3^{n+1}) - (1 + 3^n)}{(1 + 3^n)(1 + 3^{n+1})} = frac{1}{1 + 3^n} - frac{1}{1 + 3^{n+1}}. ] This setup allows us to use telescoping series: [ sum_{n=1}^infty left(frac{1}{1 + 3^n} - frac{1}{1 + 3^{n+1}}right), ] where all terms cancel except the first term of the series: [ frac{1}{1 + 3^1} = frac{1}{4}. ] Thus, the sum of the series is: [ boxed{frac{1}{4}}. ]

answer:# Problem Calculate the definite integral: [ int_{0}^{pi / 4} frac{2 tan^2 x - 11 tan x - 22}{4 - tan x} , dx ] Solution: To simplify the integration, we employ the substitution ( t = tan x ). Accordingly, the differential ( dx ) is transformed as [ dx = frac{dt}{1+t^2} ] Thus, the integral becomes: [ int_{0}^{1} frac{2t^2 - 11t - 22}{4 - t} cdot frac{1}{1 + t^2} , dt ] Next, let's simplify the integrand by decomposing it into partial fractions: [ frac{2 t^2 - 11 t - 22}{(4 - t)(t^2 + 1)} = frac{A}{4 - t} + frac{Bt + C}{t^2 + 1} ] To find (A), (B), and (C), we'll compare coefficients after clearing the denominators: [ 2 t^2 - 11 t - 22 = A(t^2 + 1) + (Bt + C)(4 - t) ] Let's substitute some values to solve for (A), (B), and (C): 1. **Substitute ( t = 4 ):** [ 2(4)^2 - 11(4) - 22 = A((4)^2 + 1) + (B(4) + C)(4 - 4) ] [ 2(16) - 44 - 22 = 17A ] [ -34 = 17A quad Rightarrow quad A = -2 ] 2. **Substitute ( t = 0 ):** [ 2(0)^2 - 11(0) - 22 = -2(0^2 + 1) + (B(0) + C)(4 - 0) ] [ -22 = -2 + 4C ] [ 4C = -20 quad Rightarrow quad C = -5 ] 3. **Substitute ( t = 1 ):** [ 2(1)^2 - 11(1) - 22 = -2(1^2 + 1) + (B(1) + (-5))(4 - 1) ] [ 2 - 11 - 22 = -2(2) + (B - 5)(3) ] [ -31 = -4 + 3B - 15 ] [ -31 = -19 + 3B quad Rightarrow quad 3B = -12 quad Rightarrow quad B = -4 ] Thus, the partial fraction decomposition is: [ frac{2t^2 - 11t - 22}{(4 - t)(t^2 + 1)} = frac{-2}{4 - t} - frac{4t + 5}{t^2 + 1} ] The integral then becomes: [ int_{0}^{1} left( frac{-2}{4 - t} - frac{4t + 5}{t^2 + 1} right) , dt ] Separate into two integrals and evaluate each one: [ int_{0}^{1} frac{-2}{4 - t} , dt - int_{0}^{1} frac{4t + 5}{t^2 + 1} , dt ] For the first integral: Perform the substitution ( u = 4 - t ), ( du = -dt ): [ int_{0}^{1} frac{-2}{4 - t} , dt = int_{4}^{3} frac{-2}{u} (-du) = 2 int_{3}^{4} frac{1}{u} , du = 2 left[ ln|u| right]_{3}^{4} = 2 left( ln 4 - ln 3 right) = 2 ln frac{4}{3} ] For the second integral: [ int_{0}^{1} frac{4t + 5}{t^2 + 1} , dt = 4 int_{0}^{1} frac{t}{t^2 + 1} , dt + 5 int_{0}^{1} frac{1}{t^2 + 1} , dt ] The first part: [ int_{0}^{1} frac{4t}{t^2 + 1} , dt quad text{using substitution } u = t^2 + 1, du = 2t , dt Rightarrow 2 int_{1}^{2} frac{1}{u} , du = 2 ln frac{2}{1} = 2 ln 2 ] The second part: [ 5 int_{0}^{1} frac{1}{t^2 + 1} quad dt = 5 left[ arctan t right]_{0}^{1} = 5 left( frac{pi}{4} - 0 right) = frac{5pi}{4} ] Combining these results: [ 2 ln frac{4}{3} - left(2 ln 2 + frac{5pi}{4}right) ] Finally, simplify: [ 2 ln frac{4}{3} - 2 ln 2 - frac{5pi}{4} = 2 ln frac{4/3}{2} - frac{5pi}{4} = 2 ln frac{4}{3 cdot 2} - frac{5pi}{4} = 2 ln frac{4}{6} - frac{5pi}{4} = 2 ln frac{2}{3} - frac{5pi}{4} ] Hence, the final answer is: [ boxed{2 ln frac{3}{8} - frac{5pi}{4}} ]

answer:Given (5x+4)^3 = a_0 + a_1x + a_2x^2 + a_3x^3, let x = -1, then (-1)^3 = a_0 - a_1 + a_2 - a_3 = (a_0 + a_2) - (a_1 + a_3) = -1 Therefore, the correct choice is boxed{text{A}}. By assigning a value to the variable according to the given equation, when x is -1, the desired value can be obtained. This problem tests the properties of the binomial theorem, focusing on the issue of assigning values to variables. Combined with the required result and observing the assigned values, it is considered a basic question.