answer:To find the domain of the function g(x)=frac{f(x+1)}{sqrt{x-1}}, we need to consider the domain of f(x) and the restrictions imposed by the square root in the denominator. Given that the domain of f(x) is left(1,3right), we translate this condition to f(x+1), which means: [1 < x + 1 < 3] Simplifying this inequality, we subtract 1 from all parts of the inequality to keep the balance: [1 - 1 < x < 3 - 1] [0 < x < 2] For the denominator sqrt{x-1} to be defined, x-1 must be greater than 0: [x - 1 > 0] Adding 1 to both sides gives us: [x > 1] Combining both conditions, we need x to satisfy both 0 < x < 2 and x > 1. The intersection of these conditions is: [1 < x < 2] Therefore, the domain of the function g(x) is boxed{left(1,2right)}.

answer:Let ( a ) be the smallest number of cookies Mohan could have. We are given: [ a equiv 5 pmod{6} ] [ a equiv 6 pmod{8} ] [ a equiv 9 pmod{10} ] We also know ( a ) is a perfect square. 1. From ( a equiv 5 pmod{6} ), there are integers ( m ) for which ( a = 6m + 5 ). 2. Substituting into ( a equiv 6 pmod{8} ): [ 6m + 5 equiv 6 pmod{8} implies 6m equiv 1 pmod{8} implies m equiv 1 pmod{4} ] Let ( m = 4n + 1 ), then ( a = 6(4n + 1) + 5 = 24n + 11 ). 3. Substituting into ( a equiv 9 pmod{10} ): [ 24n + 11 equiv 9 pmod{10} implies 4n + 1 equiv 9 pmod{10} implies 4n equiv 8 pmod{10} implies n equiv 2 pmod{5} ] So, let ( n = 5k + 2 ), then ( a = 24(5k + 2) + 11 = 120k + 59 ). As ( a ) is supposed to be the minimum square, check the smallest square number for ( k ). ( k = 0 ) gives ( a = 59 ), not a square. By increasing ( k ), ( a = 179 ) for ( k = 1 ) also is not a square. ( k = 2 ) gives ( a = 299 ), also not a square. Continue until you find the smallest ( a ) that is a square. For ( k = 21 ) gives ( a = 2549 ), which is ( 51^2 ). Thus, the smallest number of cookies Mohan could have that satisfies all conditions is ( a = 2549 ), and ( 2549 = 51^2 ). Conclusion: [ boxed{2549} ]

answer:First, let's calculate the total study time required for each subject: Mathematics: - Chapters: 4 chapters * 2.5 hours/chapter = 10 hours - Worksheets: 7 worksheets * 1.5 hours/worksheet = 10.5 hours Total for Mathematics = 10 + 10.5 = 20.5 hours Physics: - Chapters: 5 chapters * 3 hours/chapter = 15 hours - Worksheets: 9 worksheets * 2 hours/worksheet = 18 hours Total for Physics = 15 + 18 = 33 hours Chemistry: - Chapters: 6 chapters * 3.5 hours/chapter = 21 hours - Worksheets: 8 worksheets * 1.75 hours/worksheet = 14 hours Total for Chemistry = 21 + 14 = 35 hours Now, let's add up the total study time for all subjects: Total study time = 20.5 (Math) + 33 (Physics) + 35 (Chemistry) = 88.5 hours Next, we need to calculate the effective study time per day, considering the breaks: - First 3 hours: 3 hours * 60 minutes/hour = 180 minutes - 3 breaks * 10 minutes/break = 30 minutes Effective time = 180 - 30 = 150 minutes - Next 3 hours: 3 hours * 60 minutes/hour = 180 minutes - 3 breaks * 15 minutes/break = 45 minutes Effective time = 180 - 45 = 135 minutes - Final hour: 1 hour * 60 minutes/hour = 60 minutes - 1 break * 20 minutes/break = 20 minutes Effective time = 60 - 20 = 40 minutes - Snack breaks: 2 breaks * 20 minutes/break = 40 minutes - Lunch break: 45 minutes Total break time = 30 (first 3 hours) + 45 (next 3 hours) + 20 (final hour) + 40 (snack breaks) + 45 (lunch) = 180 minutes or 3 hours So, the effective study time per day is 7 hours - 3 hours (breaks) = 4 hours. Now, let's calculate the number of days needed: Total study days = Total study time / Effective study time per day Total study days = 88.5 hours / 4 hours/day = 22.125 days Since they cannot study for a fraction of a day, they will need to round up to the nearest whole number. Therefore, Angelo and Melanie should plan to study for boxed{23} days in total over the next three weeks to cover all subjects.

answer:# Problem: Let ( f(x) ) be a function defined on the interval ([0,1]) as follows: [ f(x) = begin{cases} x + frac{1}{2} & text{if } 0 leqslant x leqslant frac{1}{2} 2(1-x) & text{if } frac{1}{2} < x leqslant 1 end{cases} ] Find five distinct points ( x_0, x_1, x_2, x_3, x_4 ) in ([0,1]) such that [ f(x_0) = x_1, quad f(x_1) = x_2, quad f(x_2) = x_3, quad f(x_3) = x_4, quad f(x_4) = x_0. ] 1. Evaluate (f) at key points (0), ( frac{1}{2} ), and (1): [ f(0) = frac{1}{2}, quad fleft(frac{1}{2}right) = 1, quad f(1) = 0. ] These values indicate a 3-cycle (left{0, frac{1}{2}, 1right}). 2. By Sharkovskii's theorem, which states that if a continuous function on a compact interval has a periodic point of period 3, then it has periodic points of all periods. Thus, ( f(x) ) must also have a periodic point of period 5. 3. Consider the intervals: - If ( x in left[0, frac{1}{2}right] ), then ( f(x) in left[frac{1}{2}, 1right] ). - If ( x in left[frac{1}{2}, 1right] ), ( f(x) ) could be anywhere in ([0, 1]). 4. Assume ( x_0 in left[0, frac{1}{2}right] ), and ( x_1, x_2, x_3, x_4 in left[frac{1}{2}, 1right] ). 5. Function values of ( f ) for iterates when ( x, f(x), dots, f^{[n-1]}(x) in left[frac{1}{2}, 1right] ): [ f^{[n]}(x) = (-2)^nleft(x - frac{2}{3}right) + frac{2}{3}. ] 6. Given requirements for points: [ begin{cases} x_0 + frac{1}{2} = x_1, (-2)^4 left(x_1 - frac{2}{3}right) + frac{2}{3} = x_0. end{cases} ] 7. Solve the system: - From the first equation: [ x_1 = x_0 + frac{1}{2}. ] - Substitute (x_1) in the second equation: [ (-2)^4 left(x_0 + frac{1}{2} - frac{2}{3}right) + frac{2}{3} = x_0. ] Simplify: [ 16left(x_0 + frac{1}{2} - frac{2}{3}right) + frac{2}{3} = x_0, ] [ 16left(x_0 - frac{1}{6}right) + 8 + frac{2}{3} = x_0, ] [ 16x_0 - frac{8}{3} + frac{26}{3} = x_0, ] [ 16x_0 - frac{2}{3} = x_0, ] [ 15x_0 = frac{2}{3}, ] [ x_0 = frac{2}{15}. ] Therefore: [ x_1 = frac{2}{15} + frac{1}{2} = frac{2}{15} + frac{7.5}{15} = frac{9.5}{15} = frac{19}{30}. ] 8. Calculate subsequent iterations: [ x_2 = fleft(x_1right) = 2(1 - frac{19}{30}) = frac{11}{15}, ] [ x_3 = fleft(x_2right) = 2(1 - frac{11}{15}) = frac{8}{15}, ] [ x_4 = fleft(x_3right) = 2(1 - frac{8}{15}) = frac{14}{15}. ] 9. Verify: [ fleft(x_4right) = fleft(frac{14}{15}right) = 2(1 - frac{14}{15}) = frac{2}{15} = x_0. ] Thus, the five distinct points (left{frac{2}{15}, frac{19}{30}, frac{11}{15}, frac{8}{15}, frac{14}{15}right}) form the desired cycle. [ boxed{left{frac{2}{15}, frac{19}{30}, frac{11}{15}, frac{8}{15}, frac{14}{15}right}} ]