answer:Identify numbers that must be excluded to avoid having one number be 4 times another. If n is in the set, then 4n must be excluded if 4n leq 150. 1. **Integers from 38 to 150** can be safely included since 4 times 38 = 152, which is out of the range, and these numbers do not exclude any others. This gives 150 - 38 + 1 = 113 numbers. 2. **Considering integers from 1 to 37**: We must exclude any number that, when multiplied by 4, results in another number ≤ 150. Specifically, including any number n leq 37 would require excluding 4n leq 150. This exclusion primarily affects numbers from 1 to 37, as the multiples of these (from 4 to 148) would cover most numbers. A detailed count: - Including numbers 1 to 9 (since 4 times 10 = 40 > 37): - Exclude 4, 8, 12, 16, 20, 24, 28, 32, 36 (by including 1, 2, 3, 4, 5, 6, 7, 8, 9) - This strategy excludes every fourth number starting from 4, resulting in 37 - 9 = 28 remaining numbers. Total numbers in the subset = 113 + 28 = boxed{141}.

answer:Given the inequality ax^{2}+bx+c < 0 has a solution set of (frac{1}{t},t) where t > 0, we aim to determine the correct statements among the provided options. 1. **Solution Set Analysis:** Since (frac{1}{t}, t) is the solution set, frac{1}{t} and t are the roots of the quadratic equation ax^{2}+bx+c=0. Given that the roots are real and positive (t > 0), we can infer that a > 0 to ensure the parabola opens upwards and the inequality ax^{2}+bx+c < 0 holds between the roots. Also, since t > 0, and we are dealing with real numbers, t > 1 to maintain the order frac{1}{t} < t. 2. **Sum and Product of Roots:** From the sum and product of roots of a quadratic equation, we have: - Sum of roots: frac{1}{t} + t = -frac{b}{a} - Product of roots: frac{1}{t} cdot t = frac{c}{a} These imply b = -a(frac{1}{t} + t) and c = a cdot t cdot frac{1}{t} = a, showing a = c > 0. 3. **Analysis of Inequalities:** - We see that frac{1}{t} + t > 2sqrt{frac{1}{t} cdot t} = 2, which leads to b < -2a < 0. Therefore, abc < 0 and 2a + b < 0, implying both A and B are correct. - For option C, (frac{1}{4}a+frac{1}{2}b+c)(4a+2b+c) = left[frac{1}{4}a-frac{1}{2}a(frac{1}{t}+t)+aright]cdot[4a-2a(frac{1}{t}+t)+a] = frac{1}{4}a^{2}[5-2(frac{1}{t}+t)]^{2} geq 0, showing C is incorrect because this expression is always non-negative. - For option D, considering the equation ax+bsqrt{x}+c=0 and substituting sqrt{x} = m (with m geq 0), we get am^{2} + bm + c = 0, which has solutions identical to our initial setup, i.e., frac{1}{t} and t. Thus, x_{1} + x_{2} = (frac{1}{t} + t)^{2} - 2. Since frac{1}{t} + t > 2, we have (frac{1}{t} + t)^{2} - 2 > frac{1}{t} + t, indicating D is correct. Therefore, the correct choices are encapsulated as boxed{ABD}.

answer:1. Determine the difference in distances: (600 - 540 = 60) miles. 2. Calculate the time for 60 miles at 60 miles per hour: Since the car travels at 60 miles per hour, it needs exactly 1 hour to cover 60 miles. 3. Convert the time difference from hours to minutes to answer the problem: (1text{ hour} = 60text{ minutes}). Since the 600-mile journey takes 60 minutes more than the 540-mile journey, the 540-mile journey takes (boxed{60}) minutes less than the 600-mile journey.

answer:1. **Labeling the Children**: We assign a number to each child based on their height in increasing order, from the shortest (1) to the tallest (10). 2. **Initial Analysis**: Consider the worst-case scenario where the children are initially arranged in the exact reverse order. In other words, they are standing in the sequence 10, 9, ..., 1. 3. **First Method - Counting Minimum Moves**: We need to determine the least number of swaps sufficient to arrange the children in the correct order. - Suppose less than 8 swaps are made. This means at least 3 children would remain in their initial positions. However, since their initial sequence is entirely reverse, retaining any 3 children in their initial positions will ensure that their order would still be wrong. Hence, you will need at least 8 moves. 4. **Second Method - Complexity Measure**: Define complexity as the number of inversions that need to be corrected by making swaps. - Initially, the complexity is maximal because the children are in completely reverse order. - Complexity = total number of places a child must move to reach its correct spot. - Let ( Delta ) be the change in complexity after each step: (Delta leq 1). 5. **Detailed Calculation**: - Initially, when all are in reverse order ( (10, 9, ..., 1) ), swap complexity is highest. - Each swap of positions reduces the complexity by exactly one. Hence, transitioning from a complexity of 9 (full reverse order) to 1 implies a minimum of 8 swaps to correct the entire sequence. 6. **Practical Example - 8 Required Moves**: Consider an example that achieves the optimal number of swaps: - Leave 1 and 2 in their original positions. - Place 3 after 2. - Place 4 after 3. - Continue in this manner. 7. **Conclusion**: Given this rigorous assessment, the minimum number of moves necessary to arrange the children correctly in a circle is exactly 8. [ boxed{8} ]