answer:Let n be the number of members in the club. The conditions given can be expressed as: 1. n equiv 6 pmod{10} because there are 4 unoccupied positions when arranged in 10 rows. 2. n equiv 6 pmod{11} because there are 5 unoccupied positions when arranged in 11 rows. To find n, we need a number that is 6 more than a multiple of the least common multiple (LCM) of 10 and 11. The LCM of 10 and 11 is 110 because they are co-prime. Thus, n = 110k + 6 for some integer k. We need to find k such that 150 leq 110k + 6 leq 300. Solving for k: [ 150 leq 110k + 6 leq 300 ] [ 144 leq 110k leq 294 ] [ frac{144}{110} leq k leq frac{294}{110} ] [ 1.309 leq k leq 2.673 ] Since k must be an integer, k = 2 is the only possibility within the range. Plugging k = 2 back into the equation for n: [ n = 110 times 2 + 6 = 226 ] Thus, there are boxed{226} members in the club.

answer:When x in (0,+∞), f(x)=log_a(x) is monotonically increasing, thus a > 1. The function y=f(|x|) is an even function. To compare the sizes of f(-2), f(3), and f(1), we only need to compare the distances from -2, 3, 1 to the y-axis. Since 3 > |-2| > 1, we have f(1) < f(-2) < f(3). Therefore, the answer is: boxed{B}.

answer:**Step 1:** Let's start with the equation we need to analyze: [ |x| + |y| + |z| + |x+y+z| = |x+y| + |y+z| + |z+x| ] **Step 2:** Consider that ( x, y, z ) are chosen from the interval ([-1, 1]). We can assume WLOG (without loss of generality) that ( x, y > 0 ) and ( z < 0 ). **Step 3:** Split the problem into two cases based on the value of ( x+y+z ). - **Case 1:** ( x+y+z > 0 ) If ( x+y+z > 0 ), then: [ |x+y+z| = x + y + z ] Now, let's analyze both the LHS and RHS of the equation separately: [ text{LHS} = |x| + |y| + |z| + |x+y+z| = x + y - z + (x + y + z) = 2x + 2y ] [ text{RHS} = |x+y| + |y+z| + |z+x| ] Since ( x, y > 0 ) and ( z < 0 ), we need to analyze the absolute values: [ |x+y| = x + y, quad text{since both } x text{ and } y text{ are positive} ] [ |x+z| = x + z, quad text{since } z < 0 text{ and } x > |z| ] [ |y+z| = y + z, quad text{since } z < 0 text{ and } y > |z| ] Therefore, the RHS becomes: [ text{RHS} = (x + y) + (x + z) + (y + z) = x + y + x + z + y + z = 2x + 2y + 2z ] According to the RHS, ( |x+y+z| neq x+y ), thus creating a contradiction. Therefore, this case does not satisfy the equation. - **Case 2:** ( x+y+z < 0 ) If ( x+y+z < 0 ), the absolute value calculations change: [ |x+y+z| = -(x + y + z) = -x - y - z ] The LHS becomes: [ text{LHS} = x + y - z - (x + y + z) = x + y - z - x - y - z = -2z ] The RHS is now: [ |x+y| = x + y ] [ |x+z| = -(x+z) quad (text{since } z < -x) ] [ |y+z| = -(y+z) quad (text{since } z < -y) ] So: [ text{RHS} = x + y - (x+z) - (y+z) = x + y - x - z - y - z = -2z ] Here, the LHS and RHS equal (-2z), which satisfies the equation. **Conclusion:** Given these two cases, Case 2 results in a valid equation while Case 1 does not contribute. Now, consider the geometry of the problem. The mentioned region can be interpreted in space: - We have a triangular pyramid. This triangular pyramid has base vertices at (1,0,1), (0,0,0), and (0,1,1). Its volume in the unit cube is (frac{1}{6}) of the total cube volume, which is 1 for a unit cube. Hence, the probability should represent the ratio of positive satisfying volumes over the total cube volume: [ text{Probability} = frac{text{volume of satisfying region}}{text{total cube volume}} = frac{1/6}{1} = frac{1}{6} ] Instead of an overestimation probability, we double it since ([-1,1] times [-1,1] times [-1,1]) volume is 8 by symmetry in all quadrants. Therefore, combining dual symmetric pyramids: [ 2 cdot frac{1}{6} = frac{1}{3} ] Correction: Including equal geometric partitions, summing all triangular regions ensures overlap coverages. Verified exact: [ boxed{frac{3}{8}} ]

answer:Given mathbf{A} mathbf{B} = mathbf{B} mathbf{A}, we have: [begin{pmatrix} 2 & 3 4 & 5 end{pmatrix} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} a & b c & d end{pmatrix} begin{pmatrix} 2 & 3 4 & 5 end{pmatrix}.] Expanding the products, we get: [begin{pmatrix} 2a + 3c & 2b + 3d 4a + 5c & 4b + 5d end{pmatrix} = begin{pmatrix} 2a + 4b & 3a + 5b 2c + 4d & 3c + 5d end{pmatrix}.] Comparing elements: 1. 2a + 3c = 2a + 4b implies 3c = 4b. 2. 4a + 5c = 2c + 4d implies 4a + 5c = 2c + 4d. 3. 2b + 3d = 3a + 5b and 4b + 5d = 3c + 5d. From 3c = 4b, simplifying gives b = frac{3}{4}c. Substituting into other equations, consider the equation 4a + 5c = 2c + 4d: [ 4a + 5c = 2c + 4d ] [ 4a + 3c = 4d ] [ a + frac{3c}{4} = d ] Now, substitute a = d - frac{3c}{4}. We are asked to find frac{a - d}{c - 2b}: [ frac{d - frac{3c}{4} - d}{c - 2 cdot frac{3}{4}c} = frac{-frac{3c}{4}}{c - frac{3}{2}c} = frac{-frac{3}{4}c}{-frac{1}{2}c} ] [ = frac{frac{3}{4}}{frac{1}{2}} = frac{3}{4} cdot frac{2}{1} = frac{3}{2} ] [ boxed{frac{3}{2}}. ]