answer:1. Let's denote the distances traveled by the tips of the second hand, minute hand, and hour hand on the clock face in one unit time as follows: - Second hand: 1 unit - Minute hand: frac{1}{60} unit - Hour hand: frac{1}{720} unit 2. Consider the situation t = n cdot 60 + x seconds after 12:00 noon. At this time: - The distance the second hand has traveled from the 12 o'clock position is x units, - The distance the minute hand has traveled is n + frac{x}{60} units, - The distance the hour hand has traveled is frac{n}{12} + frac{x}{720} units. 3. The problem statement requires that the second hand divides the angle between the hour and minute hand in a ratio of frac{H O S}{S O M} = k, where (H) is the hour hand, (O) is the origin (center of the clock), (M) is the minute hand, and (S) is the second hand. 4. Therefore, we set up the equation for the given ratio: [ k = frac{x - left(frac{n}{12} + frac{x}{720}right)}{n + frac{x}{60} - x} ] 5. Simplify the numerator: [ x - left(frac{n}{12} + frac{x}{720}right) = x - frac{n}{12} - frac{x}{720} = frac{720x - x - 60n}{720} = frac{719x - 60n}{720} ] 6. Simplify the denominator: [ n + frac{x}{60} - x = n - x + frac{x}{60} = n - frac{60x}{60} + frac{x}{60} = n - frac{59x}{60} ] 7. Substitute these back into the equation: [ k = frac{frac{719x - 60n}{720}}{n - frac{59x}{60}} ] 8. Clear the denominators by multiplying both the numerator and the denominator by 720 and 60 respectively on both sides: [ k = frac{(719x - 60n) cdot 60}{720(n cdot 60 - 59x)} ] [ k cdot 720(n cdot 60 - 59x) = 719x cdot 60 - 60n cdot 60 ] [ k cdot 43200n - k cdot 708 cdot x = 43140x - 3600n ] 9. Group the x and n terms: [ 43200kn - 708kx = 43140x - 3600n ] [ k cdot 43200n + 3600n = 43140x + 708 cdot k cdot x ] 10. Collecting x term and factoring out n: [ n cdot (43200k + 3600) = frac{x cdot (43140 + 708k)}{x} ] 11. Solving for x, we get: [ x = frac{60n(1 + 12k)}{719 + 708k} ] 12. Finally, solving for (t): [ t = n cdot 60 + x = n times 60 + frac{60n (1+12k)}{719 + 708k} ] [ t = frac{43200n(1+k)}{719 + 708k} ] # Conclusion: [ boxed{t = frac{43200(1+k) n}{719 + 708k}} ]

answer:Let's denote the amount of sugar as S, the amount of flour as F, and the amount of baking soda as B. From the given information, we have the following ratios: 1) Sugar to flour: S/F = 5/2 2) Flour to baking soda: F/B = 10/1 From the second ratio, we can express B in terms of F: B = F/10 Now, we are given that if there were 60 more pounds of baking soda, the ratio of flour to baking soda would be 8 to 1. So we can write: F / (B + 60) = 8/1 Substituting B = F/10 into the equation, we get: F / (F/10 + 60) = 8/1 F = 8(F/10 + 60) F = 8F/10 + 480 10F = 8F + 4800 2F = 4800 F = 2400 Now that we have the amount of flour, we can find the amount of sugar using the first ratio: S/F = 5/2 S/2400 = 5/2 S = 2400 * (5/2) S = 2400 * 2.5 S = 6000 Therefore, there are boxed{6000} pounds of sugar stored in the room.

answer:To find the average salary of the associates, we need to calculate the total salary of all employees and then subtract the total salary of the managers. Let's denote the average salary of the associates as A. The total salary of the 9 managers is: 9 managers * 1300/manager = 11700 The total salary of the 18 associates is: 18 associates * A/associate The total salary of all employees (9 managers and 18 associates) is the sum of the total salaries of managers and associates: Total salary = 11700 + (18 * A) The average salary for the departmental store is given as 3989, and there are a total of 9 managers + 18 associates = 27 employees. So, the total salary of all employees is also: Total salary = 27 employees * 3989/employee = 107703 Now we can set up the equation: 11700 + (18 * A) = 107703 Solving for A: 18 * A = 107703 - 11700 18 * A = 96003 A = 96003 / 18 A = 5333.50 Therefore, the average salary of the associates is boxed{5333.50} .

answer:Method 1: When x=0, the inequality |x-5|+|x+3|=8geqslant 10 does not hold. Thus, options A and B can be eliminated. When x=-4, the inequality |x-5|+|x+3|=10geqslant 10 holds. Thus, option C can be eliminated. Therefore, the correct answer is option D. Method 2: When x < -3, the inequality |x-5|+|x+3|geqslant 10 can be simplified to: -(x-5)-(x+3)geqslant 10, which simplifies to xleqslant -4. When -3leqslant xleqslant 5, the inequality becomes -(x-5)+(x+3)=8geqslant 10, which never holds true. When x > 5, the inequality becomes (x-5)+(x+3)geqslant 10, which simplifies to xgeqslant 6. Thus, the set of solutions to the inequality |x-5|+|x+3|geqslant 10 is (-∞,-4]∪[6,+∞). Hence, the correct answer is boxed{text{D: }(-∞,-4]∪[6,+∞)}. Method 1 utilizes the special value method, which enables us to use the method of elimination to solve the problem. By taking x=0 and x=-4, we can make conclusions about the possible correct answers and the incorrect ones based on whether the inequality holds true or not. Method 2 involves the zero-point segmentation method, where we discuss the possible solutions to the inequality under three different cases. We then combine the ranges of x for each case to obtain the solution set. This method emphasizes understanding the key knowledge point of solving absolute value inequalities, which is transforming absolute value inequalities into polynomial inequalities and solving them accordingly.