answer:First, calculate f(12): The divisors of 12 are 1, 2, 3, 4, 6, 12. The sum of all divisors except 12 itself is: [ f(12) = 1 + 2 + 3 + 4 + 6 = 16. ] Next, calculate f(16): The divisors of 16 are 1, 2, 4, 8, 16. The sum of all divisors except 16 itself is: [ f(16) = 1 + 2 + 4 + 8 = 15. ] Finally, calculate f(15): The divisors of 15 are 1, 3, 5, 15. The sum of all divisors except 15 itself is: [ f(15) = 1 + 3 + 5 = 9. ] Thus, the value of f(f(f(12))) is 9. 9 Conclusion: This solution confirms the calculation of f(f(f(12))) yields a final result of 9, with each step recalculating the sum of divisors (excluding the number itself) of the previous result. The final answer is boxed{C}.

answer:Given an exponential sequence {a_{n}}, where the first term and the common ratio are equal, we denote the first term as a_{1} and the common ratio as q. Thus, we have a_{1} = q neq 0. The general formula for the n-th term in an exponential sequence is a_{n} = a_{1}q^{n-1}. To find which term in the sequence {a_{n}} is equal to a_{3}a_{7}, we calculate a_{3} and a_{7} using the general formula: - For a_{3}, we have a_{3} = a_{1}q^{3-1} = a_{1}q^{2} = q^{2}, since a_{1} = q. - For a_{7}, we have a_{7} = a_{1}q^{7-1} = a_{1}q^{6} = q^{6}, since a_{1} = q. Multiplying a_{3} and a_{7}, we get: [a_{3}a_{7} = q^{2} cdot q^{6} = q^{2+6} = q^{8}] Next, we calculate the terms given in the options to see which one equals q^{8}: - For a_{5}, we have a_{5} = a_{1}q^{5-1} = a_{1}q^{4} = q^{4}. - For a_{7}, we have a_{7} = a_{1}q^{7-1} = a_{1}q^{6} = q^{6}. - For a_{9}, we have a_{9} = a_{1}q^{9-1} = a_{1}q^{8} = q^{8}. - For a_{10}, we have a_{10} = a_{1}q^{10-1} = a_{1}q^{9} = q^{9}. Upon reviewing the calculations, it appears there was a mistake in the initial solution provided. The correct calculation for a_{3}a_{7} should be q^{2+6} = q^{8}, not q^{10}. Therefore, the term that equals a_{3}a_{7} is a_{9}, not a_{10} as initially stated. Thus, the corrected answer is: [boxed{C: a_{9}}]

answer:The probability of each individual being selected is dfrac{180}{5400}= dfrac{1}{30}, Therefore, the number of junior college students to be drawn is dfrac{1}{30} times 1500 = 50, The number of undergraduate students to be drawn is dfrac{1}{30} times 3000 = 100, The number of graduate students to be drawn is dfrac{1}{30} times 900 = 30, Hence, the correct choice is boxed{D}. First, based on the total number of individuals and the sample drawn, calculate the probability of each individual being selected. Then, multiply the number of individuals in each stratum by the probability of each individual being selected to get the number of individuals to be drawn from each stratum. This question examines the definition and method of stratified sampling, where the number of individuals to be drawn from each stratum is obtained by multiplying the number of individuals in each stratum by the probability of each individual being selected. This is a basic question.

answer:1. **Select a generic point M on the desired line:** Let's denote this point as M(x, y). 2. **Find the vector overrightarrow{BM}:** Since B(-1, -4) is a point on the line, the vector overrightarrow{BM} from B to M is: [ overrightarrow{BM} = M - B = (x - (-1), y - (-4)) = (x + 1, y + 4) ] 3. **Find the vector overrightarrow{AC}:** Points A(5, 2) and C(-5, -3) define the vector overrightarrow{AC} as: [ overrightarrow{AC} = C - A = (-5 - 5, -3 - 2) = (-10, -5) ] 4. **Establish parallelism condition:** Since the desired line through B is parallel to AC, the vectors overrightarrow{BM} and overrightarrow{AC} must be collinear. That is, there exists a scalar k such that: [ overrightarrow{BM} = k cdot overrightarrow{AC} ] Hence, we can write: [ (x + 1, y + 4) = k cdot (-10, -5) ] 5. **Equate components and solve for k:** This yields two equations: [ x + 1 = -10k quad text{and} quad y + 4 = -5k ] 6. **Express k in terms of x and y:** Solving the equations above for k, we have: [ k = frac{x + 1}{-10} quad text{and} quad k = frac{y + 4}{-5} ] 7. **Set the expressions for k equal to each other:** [ frac{x + 1}{-10} = frac{y + 4}{-5} ] 8. **Simplify the equation:** Cross-multiplying gives: [ -5(x + 1) = -10(y + 4) ] 9. **Expand and simplify:** [ -5x - 5 = -10y - 40 ] [ -5x - 5 + 40 = -10y ] [ -5x + 35 = -10y ] 10. **Rearrange to standard linear form:** Divide both sides by -1 to get: [ 5x - 35 = 10y ] [ 5x - 10y - 35 = 0 ] 11. **Simplify the final equation:** Although this is already a linear equation in standard form, the coefficients can be simplified if desired (by dividing the entire equation by 5): [ x - 2y - 7 = 0 ] Thus, the equation of the line passing through point B and parallel to AC is: [ boxed{x - 2y - 7 = 0} ]