answer:1. **Calculate the total workday time in minutes**: Jerome's workday is 10 hours long. There are 60 minutes in an hour, so the total number of minutes in his workday is: [ 10 times 60 = 600 text{ minutes} ] 2. **Determine the duration of the second meeting**: The first meeting took 60 minutes and the second meeting took twice as long as the first. Therefore, the duration of the second meeting is: [ 2 times 60 = 120 text{ minutes} ] 3. **Calculate the total time spent in meetings**: The total time spent in meetings is the sum of the durations of the first and second meetings: [ 60 + 120 = 180 text{ minutes} ] 4. **Calculate the percentage of the workday spent in meetings**: To find the percentage of the workday spent in meetings, divide the total time spent in meetings by the total workday time, and then multiply by 100 to convert it to a percentage: [ frac{180}{600} times 100% = 30% ] Thus, Jerome spent 30% of his workday in meetings. Conclusion: [ 30% ] The final answer is boxed{30%} or boxed{(C)}

answer:Let's denote the distance they were even with each other at the beginning of the race as ( x ) feet. When Alex gets ahead by 300 feet, the distance covered by both is ( x + 300 ) feet. When Max gets ahead by 170 feet, the distance covered by both is ( x + 300 - 170 = x + 130 ) feet. When Alex gets ahead by 440 feet, the distance covered by both is ( x + 130 + 440 = x + 570 ) feet. Now, we are told that there are 3,890 feet left for Max to catch up to Alex. This means that Alex has covered ( 5000 - 3890 = 1110 ) feet. So, we have the equation: [ x + 570 = 1110 ] Solving for ( x ): [ x = 1110 - 570 ] [ x = 540 ] Therefore, Alex and Max were even with each other for boxed{540} feet at the beginning of the race.

answer:The condition for a number to be divisible by 11 is the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions to be a multiple of 11 (including zero). For the number 8n46325, the odd position digits are 8, 4, 3, 5 and the even position digits are n, 6, 2. 1. Odd sum: 8 + 4 + 3 + 5 = 20. 2. Even sum: n + 6 + 2 = n + 8. The expression for divisibility is 20 - (n + 8), or simplifying, 12 - n. This difference must be a multiple of 11. Therefore, [ 12 - n equiv 0 pmod{11} ] [ n equiv 12 pmod{11} ] [ n equiv 1 pmod{11} ] Since n is a digit, n = boxed{1} satisfies this condition.

answer:(I) The function f(x)=2cos x(sin x+cos x) can be rewritten as: f(x)=2sin xcos x+2cos^2 x f(x)=sin 2x+cos 2x+1 f(x)=sqrt{2}sin(2x+frac{pi}{4})+1 Therefore, the smallest positive period of the function f(x) is: T=frac{2pi}{omega}=pi. (II) The function f(x) is monotonically increasing when 2kpi-frac{pi}{2}leq 2x+frac{pi}{4}leq 2kpi+frac{pi}{2}. Solving for x, we get: kpi-frac{3pi}{8}leq xleq kpi+frac{pi}{8} Therefore, the interval where the function f(x) is monotonically increasing is: [kpi-frac{3pi}{8},kpi+frac{pi}{8}], where kin Z. (III) Given the interval [-frac{pi}{4},frac{pi}{4}], we have -frac{pi}{4}leq xleqfrac{pi}{4}, which leads to -frac{pi}{4}leq 2x+frac{pi}{4}leqfrac{3pi}{4}. Let 2x+frac{pi}{4}=-frac{pi}{4}, solving for x gives x=-frac{pi}{4}. Therefore, the minimum value of f(x) is: f(x)_{text{min}}=f(-frac{pi}{4})=sqrt{2}times(-frac{sqrt{2}}{2})+1=0 Let 2x+frac{pi}{4}=frac{pi}{2}, solving for x gives x=frac{pi}{8}. Therefore, the maximum value of f(x) is: f(x)_{text{max}}=f(frac{pi}{8})=sqrt{2}times1+1=boxed{sqrt{2}+1}