answer:Given the function f(x)=frac{e^{x}}{x}-ax, where xin (0,+infty), we are to find the range of the real number a for which the inequality frac{f(x_{1})}{x_{2}}-frac{f(x_{2})}{x_{1}} < 0 always holds when x_{2} > x_{1}. First, let's analyze the given inequality under the condition that x_{2} > x_{1} and xin (0,+infty). Given that x_{1}f(x_{1}) < x_{2}f(x_{2}), we can infer that the function g(x)=xf(x)=e^{x}-ax^{2} is monotonically increasing for xin (0,+infty). This means as x increases, g(x) also increases. To further understand the behavior of g(x), we find its derivative: {g'}(x)=e^{x}-2ax. For g(x) to be monotonically increasing, we require that {g'}(x)geqslant 0 for all xin (0,+infty). This gives us the inequality: e^{x}-2axgeqslant 0 Rightarrow 2aleqslant frac{e^{x}}{x}. Let's define m(x)=frac{e^x}{x} and find its derivative to understand its behavior: {m'}(x)=frac{(x-1)e^x}{x^2}. Analyzing {m'}(x), we find that for xin (0,1), {m'}(x) < 0, indicating that m(x) is monotonically decreasing in this interval. Conversely, for xin (1,+infty), {m'}(x) > 0, indicating that m(x) is monotonically increasing in this interval. Since m(x) decreases until x=1 and then increases, its minimum value occurs at x=1. Therefore, we have: 2aleqslant m(x)_{min}=m(1)=e Rightarrow aleqslant frac{e}{2}. Thus, the range of real number a for which the given inequality always holds is (-infty,frac{e}{2}]. Therefore, the correct answer is boxed{text{D}}.

answer:To determine how many 5-digit numbers contain at least one digit 6 and are divisible by 3, we will solve the problem using both inclusion-exclusion principle and direct counting approach. **Approach 1: Direct Counting by Subtraction** 1. **Total Number of 5-Digit Numbers:** - The range of 5-digit numbers is from (10000) to (99999). - The total number of such numbers is: [ 99999 - 10000 + 1 = 90000 ] 2. **Count of Numbers Divisible by 3:** - Every third number between (10000) and (99999) is divisible by 3. - Therefore, the count of numbers divisible by 3 can be calculated as: [ frac{90000}{3} = 30000 ] 3. **Count of Numbers Without a Digit 6:** - For a number to not contain the digit 6 and still be a 5-digit number: - The first digit (thousands place) can be (1, 2, 3, 4, 5, 7, 8, 9) (8 possibilities). - Each of the remaining four digits has 9 possibilities ((0, 1, 2, 3, 4, 5, 7, 8, 9)), because 6 is excluded. - Thus, the total count of 5-digit numbers without the digit 6 is: [ 8 times 9 times 9 times 9 times 9 = 8 times 9^4 ] - The exact number of such numbers is: [ 8 times 9^4 = 8 times 6561 = 52488 ] 4. **Count of Numbers Without a Digit 6 and Divisible by 3:** - Out of 5-digit numbers, those satisfying no digit 6 and divisible by 3: - First, combinations of the first four digits can be checked to ensure divisibility by 3. - If the sum of digits modulo 3 is 0, such combinations for the fifth digit are (0, 3, 9). - Calculating the numbers without digit 6 for each valid combination separately, we have: - First digit choices ((1, 2, 3, 4, 5, 7, 8, 9)): a. Sum of four digits divisible by 3: [ 8 times 9^3 times 3 = 17496 ] 5. **Final Count by Subtraction:** - Subtract the numbers without digit 6 and divisible by 3 from total numbers divisible by 3: [ 30000 - 17496 = 12504 ] Thus, the number of 5-digit numbers containing at least one digit 6 and divisible by 3 is: [ boxed{12504} ]

answer:1. **Identify relationships between roots and coefficients**: For the cubic equation x^3 - px^2 + qx - r = 0, Vieta's formulas give us: - The sum of the roots r+s+t = p. - The sum of the products of the roots taken two at a time rs + rt + st = q. - The product of the roots rst = r. 2. **Express r^2 + s^2 + t^2 using the sums and products of roots**: [ (r+s+t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st). ] Solving for r^2 + s^2 + t^2, we get: [ r^2 + s^2 + t^2 = (r+s+t)^2 - 2(rs + rt + st). ] 3. **Substitute the values from Vieta’s formulas**: [ r^2 + s^2 + t^2 = p^2 - 2q. ] 4. **Conclusion**: The expression for r^2 + s^2 + t^2 in terms of p and q is p^2 - 2q. Thus, [ p^2 - 2q ] The final answer is boxed{B}

answer:1. **Calculate additions in half an hour**: - There are 1800 seconds in half an hour (30 minutes). - Total additions = 12,000 text{ additions/second} times 1800 text{ seconds} = 21,600,000 text{ additions}. 2. **Calculate multiplications in half an hour**: - Total multiplications = 8,000 text{ multiplications/second} times 1800 text{ seconds} = 14,400,000 text{ multiplications}. 3. **Add both results to get total operations**: [ 21,600,000 text{ additions} + 14,400,000 text{ multiplications} = 36,000,000 text{ total operations} ] Therefore, the computer can perform thirty-six million total operations in half an hour. 4. **Conclusion with boxed answer**: The answer is [ 36,000,000 text{ total operations} ] The final answer is boxed{D. 36 million total operations}