answer:To find the minimum value of the function [ f(x, y)=6left(x^{2}+y^{2}right)(x+y)-4left(x^{2}+x y+y^{2}right)-3(x+y)+5 ] in the region ( D={(x, y) mid x>0, y>0} ), we proceed as follows: 1. **Introduce New Variable ( k ):** Given ( x > 0 ) and ( y > 0 ), we set ( x + y = k ) with ( k > 0 ). Then, we observe: [ x cdot y leq frac{1}{4}(x+y)^{2} = frac{1}{4}k^{2} ] and [ x^{2} + y^{2} geq frac{1}{2}(x+y)^{2} = frac{1}{2}k^{2}. ] 2. **Choosing ( k = 1 ):** When ( k = 1 ), i.e., ( x + y = 1 ), it follows that: [ xy leq frac{1}{4}(x+y)^{2} leq frac{1}{4}. ] Thus, we can rewrite the function: [ begin{aligned} f(x, y) &= 6left(x^{3}+y^{3}right) + 6xy(x+y) - 4xy - 4left(x^{2} + y^{2}right) - 3(x+y) + 5 &= 6left(xy - frac{1}{4}right)(x+y-1) + 6xleft(x-frac{1}{2}right)^{2} + 6yleft(y-frac{1}{2}right)^{2} + left(x-frac{1}{2}right)^{2} + left(y-frac{1}{2}right)^{2} + (x+y-1)^{2} + 2. end{aligned} ] 3. **Analyzing the Expression:** Now we simplify: [ begin{aligned} f(x, y) &= 6left(xy - frac{1}{4}right)(x + y - 1) + left(6x + 1right)left(x - frac{1}{2}right)^{2} + left(6y + 1right)left(y - frac{1}{2}right)^{2} + (x + y - 1)^{2} + 2. end{aligned} ] Since all squares are non-negative and the terms involving ((x+frac{1}{2})^2) and ((y+frac{1}{2})^2) are zero when (x = y = frac{1}{2}), we have: [ 6left(xy - frac{1}{4}right)(x+y-1) = 0. ] 4. **Determining the Minimum Value:** - When ( x = y = frac{1}{2} ): [ fleft(frac{1}{2}, frac{1}{2}right) = 2. ] - When ( x + y > 1 ): [ x^{2} + y^{2} geq frac{1}{2}(x+y)^{2} > frac{1}{2}. ] Therefore: [ begin{aligned} f(x, y) &= 6left(x^{2}+y^{2}-frac{1}{2}right)(x+y-1) + 2(x-y)^{2}+2 &> 2. end{aligned} ] 5. **Conclusion:** For any ((x, y) in D), ( f(x, y) geq 2 ). The minimum value of ( f(x, y) ) is achieved when ( x = y = frac{1}{2} ), and it is: [ boxed{2} ]

answer:The rectangle's length l = 10, width w, and perimeter P = 30. The formula for the perimeter of a rectangle is given by P = 2l + 2w. Substituting the given values: [ 2(10) + 2w = 30 ] [ 20 + 2w = 30 ] [ 2w = 30 - 20 ] [ 2w = 10 ] [ w = frac{10}{2} ] [ w = 5 ] Hence, the width, w, of the rectangle is 5. The ratio of the width to the length is then: [ frac{w}{l} = frac{5}{10} = boxed{1:2} ] Conclusion: The problem's hypothesis are valid and lead to a unique and coherent solution. The width of the rectangle as calculated is 5, which is consistent with the perimeter calculation.

answer:First, simplify the general term of the sequence: a_n = cos nπ = begin{cases} -1, & text{if } n text{ is odd} 1, & text{if } n text{ is even} end{cases} Next, analyze the sequence based on the value of n: 1. If n is odd, a_n = -1. 2. If n is even, a_n = 1. Since the sequence alternates between -1 and 1 as n increases, it is an oscillating sequence. Therefore, the answer is: boxed{D: text{Oscillating sequence}}

answer:We first expand the determinant: [ begin{vmatrix} x + 2 & x & x x & x + 2 & x x & x & x + 2 end{vmatrix} = (x + 2)begin{vmatrix} x + 2 & x x & x + 2 end{vmatrix} - x begin{vmatrix} x & x x & x + 2 end{vmatrix} + x begin{vmatrix} x & x + 2 x & x end{vmatrix} ] We calculate the 2x2 determinants: [ begin{vmatrix} x + 2 & x x & x + 2 end{vmatrix} = (x + 2)^2 - x^2 = x^2 + 4x + 4 - x^2 = 4x + 4 ] [ begin{vmatrix} x & x x & x + 2 end{vmatrix} = x(x + 2) - x^2 = 2x ] [ begin{vmatrix} x & x + 2 x & x end{vmatrix} = x^2 - x(x + 2) = -2x ] Substitute these values back into the determinant expansion: [ (x + 2)(4x + 4) - x(2x) + x(-2x) = 4x^2 + 8x + 4 - 2x^2 - 2x^2 = 4x^2 + 8x + 4 - 4x^2 = boxed{8x + 4} ]