answer:1. **Identify the Cycles**: Chris's cycle is 6 days (4 work-days, 2 rest-days). Dana's cycle is 7 days (5 work-days, 2 rest-days). 2. **Calculate the Least Common Multiple (LCM)**: [ text{LCM}(6, 7) = 42 ] This means every 42 days, both Chris and Dana start their cycles anew simultaneously. 3. **Analyze the 42-Day Cycle**: We need to find days when both Chris and Dana rest. We check each day within the 42-day cycle: - Chris rests on days: 5, 6 (repeats every 6 days) - Dana rests on days: 6, 7 (repeats every 7 days) The only day that matches within the 42-day cycle is Day 6. 4. **Calculate Total Coinciding Rest-Days**: [ frac{1000}{42} approx 23.81 approx 23 text{ full cycles} ] Therefore, the total number of coinciding rest-days is: [ 1 times 23 = 23 ] 5. **Conclusion**: Over the course of the first 1000 days, Chris and Dana have rest-days on the same day 23 times. (23) The final answer is boxed{D}.

answer:First, we consider the scenario where the boxes are distinguishable. Each ball can be placed in any of the 3 boxes, independently of the others. Therefore, there are 3^5 = 243 ways to distribute 5 distinguishable balls into 3 distinguishable boxes. Next, we adjust this number to account for the indistinguishability of the boxes. The adjustment factor is the number of ways to arrange 3 indistinguishable boxes, which is 3! = 6. However, since distribution among indistinguishable boxes can be complex, we must be careful here. Not all configurations are equally likely when boxes are indistinguishable, so a simple division by 3! isn't accurate. Instead, we need to consider different cases of ball distribution: - All five balls in one box: 1 way. - Four balls in one box, one in another: binom{5}{4} = 5 ways to choose the four balls. - Three balls in one box, two in another: binom{5}{3} = 10 ways to choose the three balls. - Three balls in one box, one in each of the others: binom{5}{3} = 10 ways to choose the three balls. - Two balls in two boxes, one in the third: binom{5}{2} binom{3}{2} = 10 times 3 = 30 ways to choose the pairs. Adding these together gives 1 + 5 + 10 + 10 + 30 = 56 distinct distributions of balls into indistinguishable boxes. Conclusion: The number of ways to place 5 distinguishable balls into 3 indistinguishable boxes is boxed{56}.

answer:To find the greatest two-digit multiple of 17, we must first identify the largest two-digit number, which is 99. We then divide 99 by 17 to find the greatest integer quotient: [ frac{99}{17} approx 5.8235 ] Rounding down to the nearest whole number, we get 5. Now multiplying 5 by 17 to get the greatest two-digit multiple: [ 5 times 17 = 85 ] Thus, the greatest two-digit multiple of 17 is boxed{85}.

answer:1. Consider the function ( 2 arccosleft(frac{3}{5}right) ). 2. To simplify this function, we use the double angle identity for cosine: [ cos(2theta) = 2cos^2(theta) - 1 ] where (theta = arccosleft(frac{3}{5}right)). 3. Calculate (cosleft(2 arccosleft(frac{3}{5}right)right)) by substituting (theta): [ cosleft(2 arccosleft(frac{3}{5}right)right) = 2cos^2left(arccosleft(frac{3}{5}right)right) - 1 ] 4. Since (cosleft(arccosleft(frac{3}{5}right)right) = frac{3}{5}), we have: [ cos^2left(arccosleft(frac{3}{5}right)right) = left(frac{3}{5}right)^2 = frac{9}{25} ] 5. Now plug this value back into the double angle identity: [ cosleft(2 arccosleft(frac{3}{5}right)right) = 2 cdot frac{9}{25} - 1 = frac{18}{25} - 1 = frac{18}{25} - frac{25}{25} = frac{18 - 25}{25} = -frac{7}{25} ] 6. Hence, [ cosleft(2 arccosleft(frac{3}{5}right)right) = -frac{7}{25} ] 7. Since (0 < arccosleft(frac{3}{5}right) < frac{pi}{2}), it implies (0 < 2 arccosleft(frac{3}{5}right) < pi). 8. We know that for some angle (phi), [ cos(phi) = -frac{7}{25} ] must correspond to an angle in the first or second quadrant since (phi = 2 arccosleft(frac{3}{5}right)) is within ([0, pi]). 9. To find the corresponding (sinleft(2 arccosleft(frac{3}{5}right)right)), note: [ sin^2(x) + cos^2(x) = 1 ] where (cos(x) = -frac{7}{25}). 10. Thus, [ sin^2left(2 arccosleft(frac{3}{5}right)right) = 1 - left(-frac{7}{25}right)^2 = 1 - left(frac{49}{625}right) = frac{625}{625} - frac{49}{625} = frac{576}{625} ] 11. Then, [ sinleft(2 arccosleft(frac{3}{5}right)right) = sqrt{frac{576}{625}} = frac{24}{25} ] 12. Hence, [ 2 arccosleft(frac{3}{5}right) = arcsinleft(frac{24}{25}right) ] 13. Conclusion: [ boxed{text{A. } arcsin frac{24}{25}} ]