answer：Okay, so I have this problem where I need to write the number 965 as 22q plus r, where both q and r are positive integers. Then, I have to find the greatest possible value of 2q minus r. Hmm, let me think about how to approach this.First, I know that when you divide a number by another, you get a quotient and a remainder. In this case, dividing 965 by 22 should give me q as the quotient and r as the remainder. But wait, the problem says both q and r are positive integers, so I need to make sure that r is also a positive integer, not zero. That makes sense because if r were zero, it would just be a multiple of 22, but since 965 isn't a multiple of 22, r will be some positive number less than 22.So, let me try dividing 965 by 22. Let me do this step by step. 22 times 40 is 880. Subtracting that from 965 gives me 85. Then, 22 times 3 is 66, which I can subtract from 85, leaving me with 19. So, that means 22 times 43 is 946, and 965 minus 946 is 19. So, q is 43 and r is 19.Now, the problem asks for the greatest possible value of 2q minus r. So, plugging in the values I just found, that would be 2 times 43 minus 19. Let me calculate that: 2 times 43 is 86, and 86 minus 19 is 67. So, the greatest possible value is 67.Wait a second, is that the only possible value? Or can I get a higher value by choosing different q and r? Let me think. If I choose a smaller q, then r would have to be larger because 22 times a smaller q would leave a larger remainder. But since r has to be less than 22, the maximum r can be is 21. So, if I take q as 42, then 22 times 42 is 924, and 965 minus 924 is 41. But wait, 41 is larger than 22, which isn't allowed because r has to be less than 22. So, that doesn't work.What if I take q as 44? Then, 22 times 44 is 968, which is larger than 965, so that's not possible because r would be negative, which isn't allowed since r has to be positive. So, q can't be 44.So, the only possible q is 43, which gives r as 19. Therefore, 2q minus r is 67, and that seems to be the maximum possible value.Wait, but let me double-check. Maybe there's another way to express 965 as 22q plus r with different q and r where 2q minus r is larger. Let's see. If I take q as 43, r is 19, giving 2q - r as 67. If I take q as 42, then r would be 965 - 22*42. Let me calculate that: 22*42 is 924, so 965 - 924 is 41. But 41 is more than 22, so that's not allowed because r has to be less than 22. So, I can't use q as 42.What if I take q as 41? Then, 22*41 is 902, and 965 - 902 is 63. Again, 63 is way larger than 22, so that's not allowed. So, it seems like q has to be 43 to keep r within the required range.Alternatively, maybe I can adjust r to be as small as possible to maximize q, which in turn would maximize 2q - r. Since r has to be at least 1, the smallest r can be is 1. Let me see what q would be in that case. If r is 1, then 22q = 965 - 1 = 964. Dividing 964 by 22, let's see: 22*43 is 946, and 964 - 946 is 18. So, 22*43 + 18 = 964, but we need 965, so that doesn't work. Wait, maybe I'm getting confused.Let me try a different approach. Since 965 divided by 22 is 43 with a remainder of 19, that's the standard division algorithm. So, q is 43 and r is 19. Any other representation would require r to be adjusted, but r has to be less than 22, so I can't have r as 20 or 21 without adjusting q accordingly, but that would decrease q, which might not be beneficial for maximizing 2q - r.Wait, let me test that. Suppose I take r as 20. Then, 22q = 965 - 20 = 945. Dividing 945 by 22, let's see: 22*42 is 924, and 945 - 924 is 21. So, 22*42 + 21 = 945, but we need 965, so that's not correct. Wait, I think I'm making a mistake here. If I set r to 20, then 22q = 965 - 20 = 945. But 945 divided by 22 is 42 with a remainder of 21, which means 22*42 + 21 = 945, but we need 22q + r = 965, so that doesn't work because 945 + 20 = 965, but 22*42 + 21 + 20 = 965, which would mean q is 42 and r is 41, but r can't be 41 because it's more than 22. So, that approach doesn't work.Alternatively, if I set r to 21, then 22q = 965 - 21 = 944. Dividing 944 by 22: 22*42 is 924, and 944 - 924 is 20. So, 22*42 + 20 = 944, but we need 965, so again, that doesn't work because 944 + 21 = 965, but that would mean r is 21 and q is 42, but then 22*42 + 21 = 965, which is correct. Wait, that works. So, q is 42 and r is 21. Then, 2q - r would be 2*42 - 21 = 84 - 21 = 63, which is less than 67. So, that's not better.Wait, so if I take q as 42 and r as 21, I get 2q - r as 63, which is less than 67. So, that's worse. What if I take q as 44? Then, 22*44 is 968, which is more than 965, so r would be negative, which isn't allowed. So, q can't be 44.What if I take q as 43 and r as 19, which gives me 2q - r as 67. If I take q as 43 and r as 19, that's the standard division. If I try to adjust r to be smaller, like 18, then 22q would be 965 - 18 = 947. Dividing 947 by 22: 22*43 is 946, so 947 - 946 is 1. So, q would be 43 and r would be 1, but then 22*43 + 1 = 947, which is less than 965. Wait, that doesn't add up. I think I'm getting confused again.Wait, no, if I set r to 18, then 22q = 965 - 18 = 947. But 947 divided by 22 is 43 with a remainder of 1, so 22*43 + 1 = 947, but we need 22q + r = 965, so that would mean q is 43 and r is 19, not 18. So, I can't just set r to 18; it has to be such that 22q + r = 965, with r < 22.So, it seems like the only valid way is q = 43 and r = 19, giving 2q - r = 67. Any other values of q and r either result in r being too large or q being too small, which doesn't help in maximizing 2q - r.Wait, let me try another angle. Maybe I can express 965 as 22q + r, and then express 2q - r in terms of 965. Let's see:We have 965 = 22q + r, so r = 965 - 22q.Then, 2q - r = 2q - (965 - 22q) = 2q - 965 + 22q = 24q - 965.So, 2q - r = 24q - 965.Now, to maximize 24q - 965, I need to maximize q, because 24 is positive. So, the larger q is, the larger the expression becomes.But q can't be larger than 43 because 22*44 = 968, which is more than 965, so r would be negative, which isn't allowed. So, the maximum q can be is 43, which gives r = 965 - 22*43 = 19.Therefore, 2q - r = 24*43 - 965. Let me calculate that: 24*43 is 1032, and 1032 - 965 is 67. So, that confirms the earlier result.So, the greatest possible value of 2q - r is 67.I think that's solid. I've approached it from different angles, checked different values of q and r, and each time, the maximum value comes out to be 67 when q is 43 and r is 19.

answer：Alright, so I have this integral to work on: the integral from 0 to π/6 of 1 over (1 + 3 sin²x) dx. The task is to estimate the upper and lower bounds of this integral. Hmm, okay. I remember that to find bounds for an integral, especially when the integrand is a function that's either increasing or decreasing over the interval, we can use the minimum and maximum values of the function on that interval to bound the integral.First, let me understand the integrand: 1/(1 + 3 sin²x). Since sin²x is always between 0 and 1, multiplying by 3 will make it between 0 and 3, so 1 + 3 sin²x will be between 1 and 4. Therefore, the integrand 1/(1 + 3 sin²x) will be between 1/4 and 1. But wait, that's over the entire domain of x. However, our interval is from 0 to π/6, so I need to see what sin²x does in this specific interval.At x = 0, sin(0) = 0, so sin²(0) = 0. Therefore, the integrand at x = 0 is 1/(1 + 0) = 1. At x = π/6, sin(π/6) = 1/2, so sin²(π/6) = (1/2)² = 1/4. Therefore, the integrand at x = π/6 is 1/(1 + 3*(1/4)) = 1/(1 + 3/4) = 1/(7/4) = 4/7.So, in the interval [0, π/6], sin²x increases from 0 to 1/4, which means that 1/(1 + 3 sin²x) decreases from 1 to 4/7. Therefore, the maximum value of the integrand is 1, and the minimum value is 4/7.Now, to find the upper and lower bounds of the integral, I can use these maximum and minimum values. The integral of a function over an interval is bounded by the minimum value times the length of the interval and the maximum value times the length of the interval.So, the length of the interval from 0 to π/6 is π/6. Therefore, the lower bound of the integral should be (4/7)*(π/6), and the upper bound should be 1*(π/6).Let me compute these:Lower bound: (4/7)*(π/6) = (4π)/42 = (2π)/21.Upper bound: 1*(π/6) = π/6.So, putting it all together, the integral is bounded below by 2π/21 and above by π/6.Wait, let me double-check my reasoning. I assumed that since the integrand is decreasing over [0, π/6], the maximum occurs at the left endpoint and the minimum at the right endpoint. Is that correct?Yes, because as x increases from 0 to π/6, sinx increases, so sin²x increases, making the denominator of the integrand increase, and thus the integrand decreases. So, the function is indeed decreasing on this interval, meaning the maximum is at x=0 and the minimum at x=π/6.Therefore, my bounds should be correct.Another way to think about it is to consider that for a decreasing function, the integral is less than the maximum value times the interval length and greater than the minimum value times the interval length.So, I think I'm confident with these bounds.**Final Answer**The integral is bounded by boxed{frac{2pi}{21}} and boxed{frac{pi}{6}}.

answer：First, I'll assign variables to represent the current ages of Alice and Bob. Let A be Alice's current age and B be Bob's current age.From the problem, I know two key pieces of information:1. Ten years ago, Alice was half of Bob's age. This gives me the equation: A - 10 = 1/2 * (B - 10)2. The ratio of their current ages is 3:4, which translates to: A/B = 3/4Using the second equation, I can express Alice's age in terms of Bob's age:A = 3B/4Next, I'll substitute this expression for A into the first equation to solve for Bob's current age:(3B/4) - 10 = 1/2 * (B - 10)Simplifying this equation:3B/4 - 10 = 1/2B - 53B/4 - 1/2B = 5(3B - 2B)/4 = 5B/4 = 5B = 20Now that I have Bob's current age, I can find Alice's current age:A = 3B/4 = 3*20/4 = 15Finally, I'll calculate Carol's age. Carol is 5 years older than the sum of Alice and Bob's current ages:C = A + B + 5 = 15 + 20 + 5 = 40Adding up the current ages of Alice, Bob, and Carol:Total = A + B + C = 15 + 20 + 40 = 75Therefore, the total of their present ages is 75 years.

answer：First, I need to define the variables involved in the problem. Let ( x ) represent the price of the sheep in coins, and let ( n ) represent the number of people contributing.From the first scenario, where each person contributes 5 coins and there is a shortage of 45 coins, I can set up the equation:[5n + 45 = x]This equation indicates that the total amount collected (( 5n )) plus the shortage (( 45 )) equals the price of the sheep (( x )).From the second scenario, where each person contributes 7 coins and there is a shortage of 3 coins, I can set up another equation:[7n + 3 = x]Similarly, this equation shows that the total amount collected (( 7n )) plus the shortage (( 3 )) equals the price of the sheep (( x )).Since both expressions equal ( x ), I can set them equal to each other to find a relationship between ( n ) and ( x ):[5n + 45 = 7n + 3]Simplifying this equation will help me solve for ( n ) and subsequently find ( x ).

answer：Alright, so I have this problem here: the function is f(x) = sin(ωx) - cos(ωx), and ω is greater than 2/3. The condition is that the x-coordinate of the intersection of any axis of symmetry of the graph of f(x) with the x-axis does not belong to the interval (2π, 3π). I need to determine the range of ω.Hmm, okay. First, I remember that functions like sin and cos can be combined into a single sine or cosine function using phase shifts. Maybe that will help simplify things. Let me try rewriting f(x).So, f(x) = sin(ωx) - cos(ωx). I think I can write this as a single sine function with a phase shift. The formula for combining sine and cosine is A sin(θ) + B cos(θ) = C sin(θ + φ), where C = sqrt(A² + B²) and tan(φ) = B/A. Wait, actually, it's A sin(θ) + B cos(θ) = C sin(θ + φ), right? Or maybe it's A sin(θ) + B cos(θ) = C cos(θ - φ). Let me double-check.I think it's A sin(θ) + B cos(θ) = C sin(θ + φ), where C = sqrt(A² + B²) and φ = arctan(B/A). So in this case, A is 1 and B is -1 because it's sin(ωx) - cos(ωx). So, C would be sqrt(1² + (-1)²) = sqrt(2). And φ would be arctan(-1/1) = arctan(-1) = -π/4. So, f(x) = sqrt(2) sin(ωx - π/4).Okay, so f(x) simplifies to sqrt(2) sin(ωx - π/4). That should make it easier to analyze.Now, the problem mentions the axes of symmetry of the graph of f(x). For a sine function, the axes of symmetry are the vertical lines passing through the maxima and minima of the function. So, for f(x) = sqrt(2) sin(ωx - π/4), the maxima occur where the argument of the sine function is π/2 + 2πk, and the minima occur where the argument is 3π/2 + 2πk, for integers k.So, let's find the x-coordinates of these maxima and minima. For maxima:ωx - π/4 = π/2 + 2πk=> ωx = π/2 + π/4 + 2πk=> ωx = 3π/4 + 2πk=> x = (3π/4 + 2πk)/ωSimilarly, for minima:ωx - π/4 = 3π/2 + 2πk=> ωx = 3π/2 + π/4 + 2πk=> ωx = 7π/4 + 2πk=> x = (7π/4 + 2πk)/ωSo, the axes of symmetry are at x = (3π/4 + 2πk)/ω and x = (7π/4 + 2πk)/ω for integers k.Now, the problem states that the x-coordinate of the intersection of any axis of symmetry with the x-axis does not belong to the interval (2π, 3π). So, none of these x-values should be in (2π, 3π).That means for all integers k, (3π/4 + 2πk)/ω ∉ (2π, 3π) and (7π/4 + 2πk)/ω ∉ (2π, 3π).So, we need to find ω such that for all integers k, (3π/4 + 2πk)/ω ≤ 2π or (3π/4 + 2πk)/ω ≥ 3π, and similarly for (7π/4 + 2πk)/ω.Let me handle the maxima first: x = (3π/4 + 2πk)/ω.We need (3π/4 + 2πk)/ω ≤ 2π or (3π/4 + 2πk)/ω ≥ 3π.Let's solve these inequalities:1. (3π/4 + 2πk)/ω ≤ 2π=> 3π/4 + 2πk ≤ 2πω=> 3/4 + 2k ≤ 2ω=> ω ≥ (3/4 + 2k)/2=> ω ≥ 3/8 + k2. (3π/4 + 2πk)/ω ≥ 3π=> 3π/4 + 2πk ≥ 3πω=> 3/4 + 2k ≥ 3ω=> ω ≤ (3/4 + 2k)/3=> ω ≤ 1/4 + (2k)/3Similarly, for the minima: x = (7π/4 + 2πk)/ω.Again, we need (7π/4 + 2πk)/ω ≤ 2π or (7π/4 + 2πk)/ω ≥ 3π.Solving these:1. (7π/4 + 2πk)/ω ≤ 2π=> 7π/4 + 2πk ≤ 2πω=> 7/4 + 2k ≤ 2ω=> ω ≥ 7/8 + k2. (7π/4 + 2πk)/ω ≥ 3π=> 7π/4 + 2πk ≥ 3πω=> 7/4 + 2k ≥ 3ω=> ω ≤ 7/12 + (2k)/3So, now we have four inequalities:From maxima:- ω ≥ 3/8 + k- ω ≤ 1/4 + (2k)/3From minima:- ω ≥ 7/8 + k- ω ≤ 7/12 + (2k)/3We need to find ω such that for all integers k, these inequalities hold. But since k can be any integer, positive or negative, we need to find ω that satisfies these inequalities for all k.But wait, actually, the axes of symmetry occur at specific x-values, so k is such that x is positive. Since ω > 2/3, and x must be positive, k must be such that (3π/4 + 2πk)/ω > 0, which is always true for k ≥ 0 because 3π/4 + 2πk is positive. For k negative, x could be negative, but since we're only concerned with the x-axis intersection in (2π, 3π), which is positive, we can focus on k ≥ 0.So, let's consider k ≥ 0.Now, we need to ensure that for all k ≥ 0, the x-values from maxima and minima do not fall into (2π, 3π). So, for each k, either the x-value is ≤ 2π or ≥ 3π.So, for each k, either:From maxima:(3π/4 + 2πk)/ω ≤ 2π or (3π/4 + 2πk)/ω ≥ 3πSimilarly for minima:(7π/4 + 2πk)/ω ≤ 2π or (7π/4 + 2πk)/ω ≥ 3πLet me analyze these inequalities for k=0, k=1, etc., and see what constraints they impose on ω.Starting with k=0:From maxima:(3π/4)/ω ≤ 2π => 3/4 ≤ 2ω => ω ≥ 3/8Or (3π/4)/ω ≥ 3π => 3/4 ≥ 3ω => ω ≤ 1/4But ω > 2/3, so ω ≤ 1/4 is impossible. Therefore, for k=0, we must have ω ≥ 3/8.But since ω > 2/3, which is approximately 0.666, and 3/8 is 0.375, so this condition is automatically satisfied.From minima:(7π/4)/ω ≤ 2π => 7/4 ≤ 2ω => ω ≥ 7/8Or (7π/4)/ω ≥ 3π => 7/4 ≥ 3ω => ω ≤ 7/12 ≈ 0.583But again, ω > 2/3 ≈ 0.666, so ω ≤ 7/12 is impossible. Therefore, for k=0, we must have ω ≥ 7/8.So, from k=0, we get ω ≥ 7/8.Now, let's check k=1:From maxima:(3π/4 + 2π)/ω ≤ 2π => (11π/4)/ω ≤ 2π => 11/4 ≤ 2ω => ω ≥ 11/8 ≈ 1.375Or (11π/4)/ω ≥ 3π => 11/4 ≥ 3ω => ω ≤ 11/12 ≈ 0.9167But ω > 2/3 ≈ 0.666, so ω must be ≤ 11/12 or ≥ 11/8. But since ω > 2/3, and 11/8 is about 1.375, which is greater than 1, but we don't know yet. However, we need to satisfy this for all k, so let's see.Similarly, from minima:(7π/4 + 2π)/ω ≤ 2π => (15π/4)/ω ≤ 2π => 15/4 ≤ 2ω => ω ≥ 15/8 ≈ 1.875Or (15π/4)/ω ≥ 3π => 15/4 ≥ 3ω => ω ≤ 15/12 = 5/4 = 1.25So, for k=1, we have two possibilities:From maxima: ω ≥ 11/8 or ω ≤ 11/12From minima: ω ≥ 15/8 or ω ≤ 5/4But since ω > 2/3, and we need to satisfy both, let's see:If we take ω ≤ 11/12 ≈ 0.9167, but from k=0, we already have ω ≥ 7/8 ≈ 0.875. So, combining these, ω must be in [7/8, 11/12].Alternatively, if we take ω ≥ 11/8 ≈ 1.375, but then from minima, we need ω ≥ 15/8 ≈ 1.875 or ω ≤ 5/4 = 1.25. But 11/8 ≈ 1.375 is less than 1.25, so this doesn't work. Therefore, the only feasible range from k=1 is ω ≤ 11/12 and ω ≥ 7/8.Wait, but 11/12 is approximately 0.9167, which is less than 1. So, combining with k=0, which requires ω ≥ 7/8 ≈ 0.875, we get ω ∈ [7/8, 11/12].Now, let's check k=2:From maxima:(3π/4 + 4π)/ω = (19π/4)/ω ≤ 2π => 19/4 ≤ 2ω => ω ≥ 19/8 ≈ 2.375Or (19π/4)/ω ≥ 3π => 19/4 ≥ 3ω => ω ≤ 19/12 ≈ 1.583From minima:(7π/4 + 4π)/ω = (23π/4)/ω ≤ 2π => 23/4 ≤ 2ω => ω ≥ 23/8 ≈ 2.875Or (23π/4)/ω ≥ 3π => 23/4 ≥ 3ω => ω ≤ 23/12 ≈ 1.9167So, for k=2, we have:From maxima: ω ≥ 19/8 or ω ≤ 19/12From minima: ω ≥ 23/8 or ω ≤ 23/12But our current range is [7/8, 11/12] ≈ [0.875, 0.9167]. So, ω is less than 1. So, for k=2, the conditions would require ω ≤ 19/12 ≈ 1.583 or ω ≤ 23/12 ≈ 1.9167, which are both greater than our current upper bound of 11/12 ≈ 0.9167. Therefore, these conditions are automatically satisfied because ω is already less than 1. So, no additional constraints from k=2.Similarly, for higher k, the required ω would be even larger, which is beyond our current range, so they don't impose any new constraints.Now, let's check k=-1, but since k is an integer, and we're considering k ≥ 0 because x must be positive, k=-1 would give x negative, which is outside our interval of interest (2π, 3π). So, we can ignore k negative.Therefore, combining all the constraints from k=0 and k=1, we get ω ∈ [7/8, 11/12].Wait, but let me double-check. For k=1, we had to satisfy both maxima and minima conditions. From maxima, ω ≤ 11/12 or ω ≥ 11/8. From minima, ω ≤ 5/4 or ω ≥ 15/8. Since we're in the range [7/8, 11/12], which is less than 1, we only need to satisfy ω ≤ 11/12 and ω ≥ 7/8.Yes, that seems correct.So, the range of ω is [7/8, 11/12].But let me check if this makes sense. If ω is in [7/8, 11/12], then the axes of symmetry for k=0 are at x = (3π/4)/ω and x = (7π/4)/ω.For ω = 7/8, x = (3π/4)/(7/8) = (3π/4)*(8/7) = 6π/7 ≈ 2.72, which is less than 2π ≈ 6.28, so it's fine.For ω = 11/12, x = (3π/4)/(11/12) = (3π/4)*(12/11) = 9π/11 ≈ 2.57, which is also less than 2π.Similarly, for k=1, x = (3π/4 + 2π)/ω = (11π/4)/ω.For ω = 7/8, x = (11π/4)/(7/8) = (11π/4)*(8/7) = 22π/7 ≈ 9.77, which is greater than 3π ≈ 9.42, so it's outside the interval.For ω = 11/12, x = (11π/4)/(11/12) = (11π/4)*(12/11) = 3π, which is exactly at 3π, so it's not in the open interval (2π, 3π).Similarly, for minima:For ω = 7/8, x = (7π/4)/ω = (7π/4)/(7/8) = 2π, which is at the boundary, not in the open interval.For ω = 11/12, x = (7π/4)/(11/12) = (7π/4)*(12/11) = 21π/11 ≈ 5.93, which is less than 2π ≈ 6.28, so it's fine.For k=1, minima x = (7π/4 + 2π)/ω = (15π/4)/ω.For ω = 7/8, x = (15π/4)/(7/8) = (15π/4)*(8/7) = 30π/7 ≈ 13.46, which is greater than 3π ≈ 9.42.For ω = 11/12, x = (15π/4)/(11/12) = (15π/4)*(12/11) = 45π/11 ≈ 12.73, which is also greater than 3π.So, it seems that with ω in [7/8, 11/12], the axes of symmetry do not intersect the x-axis within (2π, 3π).Therefore, the range of ω is [7/8, 11/12].

answer：First, I need to find the vector (overrightarrow{AB}) by subtracting the coordinates of point (A) from point (B).Next, I'll substitute the vectors into the given equation (overrightarrow{OP} = 2overrightarrow{OA} + moverrightarrow{AB}) to express (overrightarrow{OP}) in terms of (m).Since point (P) lies on the (y)-axis, its (x)-coordinate must be zero. I'll set the (x)-component of (overrightarrow{OP}) equal to zero and solve for (m).Finally, I'll verify the solution to ensure it satisfies the condition that (P) is on the (y)-axis.