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question:A company has 500 employees, each paid 12 per hour, working 10 hours a day, five days a week, 4 weeks a month. In August, the amount of work increased, and the company hired 200 more people. How much money does the company pay in total per month after the new hires?
answer:To calculate the total amount of money the company pays its employees per month after hiring 200 more people, we follow these steps: 1. **Calculate the daily pay per employee:** The pay per hour is 12, and each employee works 10 hours a day. Therefore, the daily pay per employee is: [ 12 , text{ per hour} times 10 , text{hours} = 120 , text{} ] 2. **Calculate the number of working days in a month:** Since an employee works 5 days a week for 4 weeks a month, the total number of working days in a month is: [ 5 , text{days} times 4 , text{weeks} = 20 , text{days} ] 3. **Calculate the monthly pay per employee:** With the daily pay and the number of working days, the monthly pay per employee is: [ 120 , text{ per day} times 20 , text{days} = 2400 , text{} ] 4. **Calculate the total number of employees after the new hires:** The company originally had 500 employees and hired 200 more, making the total: [ 500 + 200 = 700 , text{employees} ] 5. **Calculate the total monthly payment for all employees:** With the monthly pay per employee and the total number of employees, the total monthly payment is: [ 700 , text{employees} times 2400 , text{ per employee} = 1,680,000 , text{} ] Therefore, the total amount of money the company pays per month after the new hires is boxed{1,680,000 , }.
question:When x in (2, 1), the inequality (x+1)^2 < log_a |x| always holds. Then, the range of the real number a is ( ) A: [2, +infty) B: (1, 2] C: (1, 2) D: (0, 1)
answer:The correct answer is boxed{text{B: }(1, 2]}. Solution is omitted.
question:Find y if [2 arctan frac{1}{3} + arctan frac{1}{15} + arctan frac{1}{y} = frac{pi}{4}.]
answer:Using the connection between arctan and complex numbers, note that arctan frac{1}{3} is the argument of 3 + i, and arctan frac{1}{15} is the argument of 15 + i. Likewise, arctan frac{1}{y} is the argument of y + i. Thus, 2 arctan frac{1}{3} + arctan frac{1}{15} + arctan frac{1}{y} is the argument given by [ (3+i)^2 (15+i)(y+i) = (8+6i)(15+i)(y+i) = (120+138i)(y+i) = (120y - 138) + (120 + 138y)i. ] This should be equal to frac{pi}{4}, the argument of 1 + i. Therefore, equating real and imaginary parts we get: [120y - 138 = 120 + 138y.] Solving for y, we have: [120y - 138y = 120 + 138 quad Rightarrow quad -18y = 258 quad Rightarrow quad y = -frac{258}{18} = -frac{43}{3}.] Conclusion with boxed answer: [ boxed{y = -frac{43}{3}} ]
question:The range of the function f(x) = (x 1)^2 1, where x in {1, 0, 1, 2, 3}, is __________.
answer:We calculate the function values for each x in the domain: - f(-1) = f(3) = 3 - f(0) = f(2) = 0 - f(1) = -1 Therefore, the range of the function f(x) is boxed{{-1, 0, 3}}.