answer:Proof: (1) Assume that l_1 and l_2 do not intersect, then l_1 and l_2 are either parallel or coincide, which implies k_1 = k_2. Substituting into k_1k_2 + 2 = 0, we get k^2 + 2 = 0. This contradicts the fact that k_1 is a real number, thus k_1 neq k_2, meaning l_1 and l_2 do intersect. (2) Solving the system of equations yields the coordinates of the intersection point P(x, y) as Thus, 2x^2 + y^2 = 2^2 + 2 = 4 + 2 = 6. This indicates that the intersection point P(x, y) lies on the ellipse 2x^2 + y^2 = 1. Therefore, the final answers are: (1) l_1 and l_2 intersect: boxed{text{True}} (2) The intersection point lies on the ellipse 2x^2 + y^2 = 1: boxed{text{True}}

answer:To determine the weight of a 6 m piece of the same rod, we need to find the volume of the original rod and then use the density to find the weight of the 6 m piece. First, let's find the volume of the original rod. The volume ( V ) of a cylinder (which is the shape of the rod) is given by the formula: [ V = pi r^2 h ] where ( r ) is the radius of the circular cross-section and ( h ) is the height (or length) of the cylinder. Given that the length of the rod is ( h = 11.25 ) m, we can express the volume of the rod as: [ V = pi r^2 times 11.25 ] Now, the weight ( W ) of the rod is related to its volume and density by the formula: [ W = V times rho ] Given that the weight of the rod is ( W = 42.75 ) kg, we can write: [ 42.75 = pi r^2 times 11.25 times rho ] Now, we need to find the weight of a 6 m piece of the same rod. The volume ( V' ) of the 6 m piece is: [ V' = pi r^2 times 6 ] The weight ( W' ) of the 6 m piece is then: [ W' = V' times rho ] [ W' = pi r^2 times 6 times rho ] We can find ( rho ) from the original equation: [ rho = frac{42.75}{pi r^2 times 11.25} ] Substituting ( rho ) into the equation for ( W' ), we get: [ W' = pi r^2 times 6 times frac{42.75}{pi r^2 times 11.25} ] [ W' = frac{6 times 42.75}{11.25} ] [ W' = frac{256.5}{11.25} ] [ W' = 22.8 text{ kg} ] So, the weight of a 6 m piece of the same rod is boxed{22.8} kg.

answer:Let's analyze each option step by step: **Option A**: {0}in {x|x^{2}-2x=0} - The set {x|x^{2}-2x=0} consists of elements that satisfy the equation x^{2}-2x=0. - Solving x^{2}-2x=0 gives x(x-2)=0, so x=0 or x=2. - Therefore, the set is {0, 2}. - The notation "in" is used to denote that an element is a member of a set. {0} is a set, not an element of the set {0, 2}. - Hence, option A is incorrect. **Option B**: The necessary and sufficient condition for "Acup B=B" is Acap B=A - Acup B=B implies that all elements of A are also elements of B, which means Asubseteq B. - If Asubseteq B, then Acap B will have all elements of A (since all elements of A are also in B), so Acap B=A. - Conversely, if Acap B=A, it means all elements of A are in B, so Asubseteq B, which implies Acup B=B. - Therefore, option B is correct. **Option C**: The solution set of the inequality x^{2}-7ax+12a^{2} lt 0left(ain Rright) is {xleft|right.3a lt x lt 4a} - Solving x^{2}-7ax+12a^{2}=0 gives x=3a or x=4a. - For a<0, 3a>4a, indicating that the solution set should be {x|4a<x<3a}, not {x|3a<x<4a}. - Therefore, option C is incorrect. **Option D**: If x gt 1, y gt 1, and satisfy x+y=xy, then the minimum value of frac{2x}{x-1}+frac{4y}{y-1} is 6+4sqrt{2} - We start by simplifying frac{2x}{x-1}+frac{4y}{y-1}: [ frac{2x}{x-1}+frac{4y}{y-1}=frac{2x(y-1)+4y(x-1)}{(x-1)(y-1)}=frac{6xy-2x-4y}{xy-(x+y)+1} ] - Since x+y=xy, we have frac{1}{x}+frac{1}{y}=1. - Substituting x+y=xy into the expression, we get: [ 4x+2y=(4x+2y)left(frac{1}{x}+frac{1}{y}right)=6+frac{2y}{x}+frac{4x}{y}geq6+2sqrt{8}=6+4sqrt{2} ] - The equality holds when frac{2y}{x}=frac{4x}{y} and x+y=xy, i.e., x=1+frac{sqrt{2}}{2}, y=1+sqrt{2}. - Therefore, option D is correct. Concluding, the correct options are boxed{BD}.

answer:First, compute the initial few powers of (mathbf{B}) to detect any patterns: [ mathbf{B}^2 = begin{pmatrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 end{pmatrix} begin{pmatrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 end{pmatrix} = begin{pmatrix} 0 & 0 & 1 1 & 0 & 0 0 & 1 & 0 end{pmatrix}, ] [ mathbf{B}^3 = mathbf{B} mathbf{B}^2 = begin{pmatrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 end{pmatrix} begin{pmatrix} 0 & 0 & 1 1 & 0 & 0 0 & 1 & 0 end{pmatrix} = begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix} = mathbf{I}. ] Thus, (mathbf{B}^3 = mathbf{I}), and this cycle repeats every three multiplications. Therefore, [ mathbf{B}^{150} = (mathbf{B}^3)^{50} = mathbf{I}^{50} = mathbf{I} = boxed{begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix}}. ]