answer:Given the reflection matrix, [ begin{pmatrix} frac{5}{13} & frac{12}{13} frac{12}{13} & -frac{5}{13} end{pmatrix} ] we need to find begin{pmatrix} a b end{pmatrix} such that [ begin{pmatrix} frac{5}{13} & frac{12}{13} frac{12}{13} & -frac{5}{13} end{pmatrix} begin{pmatrix} a b end{pmatrix} = begin{pmatrix} a b end{pmatrix}. ] This leads to the equations: [ begin{pmatrix} frac{5}{13}a + frac{12}{13}b frac{12}{13}a - frac{5}{13}b end{pmatrix} = begin{pmatrix} a b end{pmatrix}. ] Hence, we have: [ frac{5}{13}a + frac{12}{13}b = a quad text{and} quad frac{12}{13}a - frac{5}{13}b = b. ] Solving these yields frac{8}{13}a = frac{12}{13}b and frac{18}{13}a = frac{18}{13}b, respectively. Thus, b = frac{2}{3}a. Choose a = 3 to satisfy a>0 and get b = 2. Thus, begin{pmatrix} 3 2 end{pmatrix}. Since gcd(3, 2) = 1, the direction vector that corresponds to the problem’s requirements is: [ boxed{begin{pmatrix} 3 2 end{pmatrix}} ]

answer:Let's denote the sides of the rectangular field as 3x and 4x, where x is a common multiplier. The area of the field (A) is then: A = (3x) * (4x) = 12x^2 The perimeter (P) of the field, which is the total length of fencing needed, is: P = 2 * (3x + 4x) = 2 * 7x = 14x The cost of fencing the field is given as 94.5, and the rate is 25 paise per meter. Since 1 rupee = 100 paise, the rate in rupees per meter is 25/100 = 0.25 rupees/meter. The total cost (C) for fencing is then: C = P * (cost per meter) 94.5 = 14x * 0.25 94.5 = 3.5x Now, solve for x: x = 94.5 / 3.5 x = 27 Now that we have the value of x, we can find the area of the field: A = 12x^2 A = 12 * (27)^2 A = 12 * 729 A = 8748 square meters Therefore, the area of the field is boxed{8748} square meters.

answer:Since there are 50 students and we want to draw a sample of size 4 using systematic sampling, we start by determining the sampling interval. The sampling interval (k) can be found by dividing the population size (N) by the sample size (n): k = frac{N}{n} = frac{50}{4} = 12.5 Since we can't have a fraction of a student, we round the interval down to the nearest whole number, which is 12: k = 12 Systematic sampling starts by selecting a random starting point between 1 and k. In this case, since we know student number 6 is in the sample, we can assume that 6 is the random starting point. Then, we add the interval k to this number to get the identification numbers of the other students in the sample: 1. Starting with student number 6: 6 + k = 6 + 12 = 18 2. Adding the interval k to get to the next known student: 18 + 12 = 30 Thus, the student number 18 is also correct, as that leads us to student number 30, which is in the sample. 3. Continuing with the same process: 30 + 12 = 42 Since students number 30 and 42 are included in the sample, the systematic sampling method has been applied correctly. Therefore, we can confirm that the identification number of the fourth student should be 18. Thus, the complete sample consists of student numbers 6, 18, 30, and 42. boxed{18}

answer:Given a_n=n^{2}-9n-100, we can rewrite the expression as a_n=(n- frac {9}{2})^{2}- frac {481}{4}. Notice that when n leqslant 4, the sequence {a_n} is strictly decreasing, and when n geqslant 5, the sequence {a_n} is strictly increasing. Therefore, the smallest term in the sequence {a_n} is either the 4th or the 5th term. Hence, the answer is boxed{D}. This solution utilized the monotonicity of quadratic functions. This problem tests understanding of the monotonicity of sequences and quadratic functions, as well as reasoning and computational skills. This is considered a foundational problem.