answer：First, I recognize that the cafeteria ordered a total number of apples, which includes both red and green apples. They ordered 6 red apples and an unknown number of green apples, which I'll denote as ( G ).The total number of apples ordered is therefore ( 6 + G ).Next, I consider the number of apples that were actually used. Only 5 students wanted fruit, and assuming each student took one apple, that means 5 apples were consumed.The cafeteria ended up with 16 extra apples. This means that the total number of apples ordered minus the number of apples used equals the extra apples:[(6 + G) - 5 = 16]Simplifying this equation:[6 + G - 5 = 16][G + 1 = 16][G = 15]Therefore, the cafeteria ordered 15 green apples.

answer：Alright, so I've got this geometry problem here about lines and planes, and I need to figure out which of the statements A, B, C, or D is correct. Let's take it step by step.First, let's understand the problem. We have a line ( l ) and two different planes ( alpha ) and ( beta ). There are four statements, each making a claim about the relationships between ( l ), ( alpha ), and ( beta ). I need to determine which of these statements is correct.Let's start by recalling some basic concepts about lines and planes in three-dimensional space:1. **Parallel Lines and Planes**: If a line is parallel to a plane, it means the line doesn't intersect the plane and lies in a direction that's consistent with the plane's orientation.2. **Perpendicular Planes**: If two planes are perpendicular, their dihedral angle is 90 degrees, meaning they intersect along a line, and each plane is at a right angle to the other.3. **Perpendicular Line and Plane**: If a line is perpendicular to a plane, it means the line intersects the plane at a right angle and is orthogonal to every line in the plane that it intersects.Now, let's analyze each statement one by one.**Statement A**: If ( l ) is parallel to ( alpha ) and ( l ) is parallel to ( beta ), then ( alpha ) is parallel to ( beta ).Hmm. So, if a single line is parallel to two different planes, does that mean the two planes must be parallel to each other? Intuitively, I might think that if both planes are parallel to the same line, they could either be parallel to each other or they could intersect along some line. For example, imagine two different planes both parallel to the x-axis; they could still intersect along the x-axis. So, just because both planes are parallel to the same line doesn't necessarily mean they are parallel to each other. Therefore, Statement A might not always be true.**Statement B**: If ( alpha ) is perpendicular to ( beta ) and ( l ) is parallel to ( alpha ), then ( l ) is parallel to ( beta ).Okay, so if two planes are perpendicular, and a line is parallel to one of them, does that make the line parallel to the other plane? Let's visualize this. Imagine two perpendicular planes, like the floor and a wall in a room. If I have a line that's parallel to the floor, would it necessarily be parallel to the wall? Well, a line parallel to the floor could be running along the wall, but it could also be intersecting the wall at some angle. So, it's not guaranteed that the line is parallel to the other plane. Therefore, Statement B might not always hold true.**Statement C**: If ( alpha ) is perpendicular to ( beta ) and ( l ) is perpendicular to ( alpha ), then ( l ) is parallel to ( beta ).Alright, so if two planes are perpendicular, and a line is perpendicular to one of them, does that mean the line is parallel to the other plane? Let's think about this. If ( alpha ) and ( beta ) are perpendicular, and ( l ) is perpendicular to ( alpha ), then ( l ) would be parallel to ( beta ) because it's pointing in the direction that's orthogonal to ( alpha ), which aligns with ( beta ). Wait, is that always the case? Actually, if ( l ) is perpendicular to ( alpha ), it means ( l ) is orthogonal to every line in ( alpha ). Since ( alpha ) and ( beta ) are perpendicular, ( l ) would be parallel to ( beta ). So, this might be correct.**Statement D**: If ( l ) is parallel to ( alpha ) and ( l ) is perpendicular to ( beta ), then ( alpha ) is perpendicular to ( beta ).Okay, so if a line is parallel to one plane and perpendicular to another, does that mean the two planes are perpendicular? Let's see. If ( l ) is parallel to ( alpha ), it means ( l ) lies in a direction consistent with ( alpha ). If ( l ) is also perpendicular to ( beta ), then ( beta ) must be oriented such that it's at a right angle to ( l ). Since ( l ) is parallel to ( alpha ), this would imply that ( alpha ) is also at a right angle to ( beta ). So, yes, ( alpha ) and ( beta ) must be perpendicular. This seems correct.Now, let's recap:- **Statement A**: Not necessarily true, because two planes parallel to the same line can intersect.- **Statement B**: Not necessarily true, because a line parallel to one plane might not be parallel to the other.- **Statement C**: Seems correct, because a line perpendicular to one plane would be parallel to the other if they're perpendicular.- **Statement D**: Also seems correct, because if a line is parallel to one plane and perpendicular to another, the planes must be perpendicular.Wait, both C and D seem correct. Did I make a mistake? Let me double-check.For **Statement C**: If ( alpha ) is perpendicular to ( beta ), and ( l ) is perpendicular to ( alpha ), then ( l ) is parallel to ( beta ). Yes, because ( l ) being perpendicular to ( alpha ) means it's aligned with the normal vector of ( alpha ), which, since ( alpha ) and ( beta ) are perpendicular, would make ( l ) parallel to ( beta ).For **Statement D**: If ( l ) is parallel to ( alpha ) and perpendicular to ( beta ), then ( alpha ) is perpendicular to ( beta ). Yes, because ( l )'s direction is consistent with ( alpha ) and orthogonal to ( beta ), which implies ( alpha ) and ( beta ) are perpendicular.Hmm, so both C and D seem correct. But the question asks for which statement is correct, implying only one. Maybe I need to re-examine.Wait, in the original problem, it's stated that ( alpha ) and ( beta ) are two different planes. Does that affect anything? For Statement C, if ( l ) is perpendicular to ( alpha ) and ( alpha ) is perpendicular to ( beta ), then ( l ) could either be parallel to ( beta ) or lie within ( beta ). But since ( l ) is a line and ( beta ) is a plane, ( l ) being perpendicular to ( alpha ) and ( alpha ) being perpendicular to ( beta ) would mean ( l ) is parallel to ( beta ). So, C is correct.For Statement D, if ( l ) is parallel to ( alpha ) and perpendicular to ( beta ), then ( alpha ) must be perpendicular to ( beta ). This seems correct as well. So, both C and D are correct? But the options are A, B, C, D, and usually, such questions have only one correct answer. Maybe I need to check the logic again.Wait, perhaps Statement C isn't necessarily always true. If ( l ) is perpendicular to ( alpha ), and ( alpha ) is perpendicular to ( beta ), ( l ) could be parallel to ( beta ) or lie within ( beta ). But since ( l ) is a line and ( beta ) is a plane, ( l ) being perpendicular to ( alpha ) doesn't necessarily mean it's parallel to ( beta ); it could also lie within ( beta ). So, Statement C might not always be true.On the other hand, Statement D seems more definitive. If ( l ) is parallel to ( alpha ) and perpendicular to ( beta ), then ( alpha ) must be perpendicular to ( beta ). This seems to hold because the direction of ( l ) being parallel to ( alpha ) and perpendicular to ( beta ) enforces the perpendicularity between the planes.Therefore, after reconsidering, Statement D is the correct one.

answer：First, I need to solve the inequality ((x - 1)(x - 2) leqslant 0).I'll start by finding the critical points where the expression equals zero. Setting each factor equal to zero:[x - 1 = 0 quad Rightarrow quad x = 1][x - 2 = 0 quad Rightarrow quad x = 2]These critical points divide the number line into three intervals: ((-infty, 1)), ((1, 2)), and ((2, infty)).Next, I'll test each interval to determine where the product ((x - 1)(x - 2)) is less than or equal to zero.1. **Interval ((-infty, 1)):** Choose (x = 0): [ (0 - 1)(0 - 2) = (-1)(-2) = 2 quad (text{Positive}) ] Since the result is positive, this interval does not satisfy the inequality.2. **Interval ((1, 2)):** Choose (x = 1.5): [ (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 quad (text{Negative}) ] Since the result is negative, this interval satisfies the inequality.3. **Interval ((2, infty)):** Choose (x = 3): [ (3 - 1)(3 - 2) = (2)(1) = 2 quad (text{Positive}) ] Since the result is positive, this interval does not satisfy the inequality.Additionally, the critical points (x = 1) and (x = 2) make the expression equal to zero, which satisfies the inequality.Combining these results, the solution set is the interval ([1, 2]).

answer：Okay, so I've got these six math problems to solve. Let me take them one by one and think through each step carefully. I might make some mistakes along the way, but I'll try to catch them as I go.**Problem 1:** Given that sin(π/3 - α) = 1/6, find the value of cos(π/6 + α).Hmm, trigonometric identities. I remember that sine and cosine are related through co-function identities. Specifically, sin(θ) = cos(π/2 - θ). Maybe I can use that here.So, sin(π/3 - α) = 1/6. Let me rewrite the angle inside the sine function. π/3 is 60 degrees, so π/3 - α. If I think about the co-function identity, sin(θ) = cos(π/2 - θ). So, sin(π/3 - α) = cos(π/2 - (π/3 - α)).Let me compute π/2 - (π/3 - α). That would be π/2 - π/3 + α. π/2 is 3π/6 and π/3 is 2π/6, so 3π/6 - 2π/6 = π/6. So, it becomes cos(π/6 + α). Wait, that's exactly the expression we need to find! So, sin(π/3 - α) = cos(π/6 + α) = 1/6. So, the value is 1/6.**Problem 2:** Let vectors a and b be non-collinear. AB = 2a + k b, BC = a + b, CD = a - 2b. If points A, B, D are collinear, find k.Alright, vectors. So, points A, B, D are collinear, meaning vectors AB and AD are scalar multiples. So, AB = λ AD for some scalar λ.First, I need to find vector AD. Since AB + BC + CD = AD. Let's compute that.AB is given as 2a + k b. BC is a + b. CD is a - 2b. So, adding them up:AB + BC + CD = (2a + k b) + (a + b) + (a - 2b) = 2a + a + a + k b + b - 2b = 4a + (k + 1 - 2)b = 4a + (k - 1)b.So, AD = 4a + (k - 1)b.But AB is 2a + k b, and since AB and AD are collinear, AB must be a scalar multiple of AD. So, 2a + k b = λ (4a + (k - 1)b).Since a and b are non-collinear, their coefficients must be proportional. So, 2 = 4λ and k = λ(k - 1).From 2 = 4λ, we get λ = 1/2.Now, plug λ = 1/2 into the second equation: k = (1/2)(k - 1).Multiply both sides by 2: 2k = k - 1.Subtract k: k = -1.So, k is -1.**Problem 3:** Given α = -35π/6, find the value of [2 sin(π + α) cos(π - α) - sin(3π/2 + α)] / [1 + sin²α - cos(π/2 + α) - cos²(π + α)].This looks complicated, but maybe we can simplify using trigonometric identities.First, let's simplify each term in the numerator and denominator.Starting with the numerator:2 sin(π + α) cos(π - α) - sin(3π/2 + α).Using identities:sin(π + α) = -sin α.cos(π - α) = -cos α.sin(3π/2 + α) = -cos α.So, substituting these in:2*(-sin α)*(-cos α) - (-cos α) = 2 sin α cos α + cos α.So, numerator simplifies to 2 sin α cos α + cos α.Now, the denominator:1 + sin²α - cos(π/2 + α) - cos²(π + α).Simplify each term:cos(π/2 + α) = -sin α.cos(π + α) = -cos α, so cos²(π + α) = cos²α.So, substituting:1 + sin²α - (-sin α) - cos²α = 1 + sin²α + sin α - cos²α.We can simplify this further. Remember that sin²α + cos²α = 1, so sin²α - cos²α = -cos(2α). But let's see:1 + sin²α + sin α - cos²α = (1 - cos²α) + sin²α + sin α. Wait, 1 - cos²α is sin²α, so this becomes sin²α + sin²α + sin α = 2 sin²α + sin α.So, denominator is 2 sin²α + sin α.Putting it all together, the expression becomes:(2 sin α cos α + cos α) / (2 sin²α + sin α).Factor numerator and denominator:Numerator: cos α (2 sin α + 1).Denominator: sin α (2 sin α + 1).So, the expression simplifies to (cos α / sin α) = cot α.So, we need to find cot α where α = -35π/6.First, simplify α. Since angles are periodic with period 2π, let's find an equivalent angle between 0 and 2π.-35π/6 divided by 2π is -35/12, which is about -2.9167. So, adding 3*2π = 6π to get a positive angle:-35π/6 + 6π = -35π/6 + 36π/6 = π/6.So, α is equivalent to π/6, but since it's negative, it's in the fourth quadrant. Cotangent is cos/sin, which is positive in the fourth quadrant.cot(π/6) is √3. But since α is negative, cot(-π/6) is -√3. Wait, but earlier we found that the expression simplifies to cot α, which is cot(-35π/6). Since cotangent is periodic with period π, so cot(-35π/6) = cot(-35π/6 + 6π) = cot(π/6) = √3.Wait, maybe I messed up the sign. Let me double-check.α = -35π/6. Adding multiples of 2π:-35π/6 + 6π = -35π/6 + 36π/6 = π/6.So, cot(-35π/6) = cot(π/6) because cotangent has period π, but wait, actually, cotangent has period π, so cot(-35π/6) = cot(-35π/6 + 6π) = cot(π/6). But cotangent is an odd function, so cot(-θ) = -cot θ. So, cot(-35π/6) = -cot(35π/6). But 35π/6 is more than 2π, so 35π/6 - 6π = 35π/6 - 36π/6 = -π/6. So, cot(35π/6) = cot(-π/6) = -cot(π/6) = -√3. Therefore, cot(-35π/6) = -cot(35π/6) = -(-√3) = √3.Wait, this is getting confusing. Maybe it's simpler to note that cotangent is periodic with period π, so cot(-35π/6) = cot(-35π/6 + 6π) = cot(π/6) = √3.But since α is negative, does that affect the value? Wait, in our expression, we had cot α, which is cot(-35π/6). But cotangent is an odd function, so cot(-x) = -cot x. So, cot(-35π/6) = -cot(35π/6). But 35π/6 is equivalent to 35π/6 - 6π = 35π/6 - 36π/6 = -π/6. So, cot(35π/6) = cot(-π/6) = -cot(π/6) = -√3. Therefore, cot(-35π/6) = -cot(35π/6) = -(-√3) = √3.So, the value is √3.**Problem 4:** If sinθ + cosθ = 17/13 and θ ∈ (0, π/4), find tanθ.Alright, so we have sinθ + cosθ = 17/13. Let's square both sides to use the Pythagorean identity.(sinθ + cosθ)^2 = (17/13)^2.So, sin²θ + 2 sinθ cosθ + cos²θ = 289/169.We know sin²θ + cos²θ = 1, so 1 + 2 sinθ cosθ = 289/169.Subtract 1: 2 sinθ cosθ = 289/169 - 1 = (289 - 169)/169 = 120/169.So, sinθ cosθ = 60/169.Now, we can use the identity (sinθ - cosθ)^2 = sin²θ - 2 sinθ cosθ + cos²θ = 1 - 2*(60/169) = 1 - 120/169 = 49/169.So, sinθ - cosθ = ±7/13.But θ is in (0, π/4), so sinθ < cosθ, which means sinθ - cosθ is negative. Therefore, sinθ - cosθ = -7/13.Now, we have two equations:sinθ + cosθ = 17/13,sinθ - cosθ = -7/13.Adding them: 2 sinθ = 10/13 ⇒ sinθ = 5/13.Subtracting them: 2 cosθ = 24/13 ⇒ cosθ = 12/13.Therefore, tanθ = sinθ / cosθ = (5/13)/(12/13) = 5/12.**Problem 5:** If f(x) = sin(ωx + π/3) (ω > 0) attains its unique maximum 1 and minimum -1 on [0, 2π], find the range of ω.So, the function f(x) must complete at least a half-period to reach both max and min. But since it's unique, it must complete exactly one full period, but wait, the maximum and minimum are attained once each. So, the function must go from 1 to -1 and back to 1, but since it's on [0, 2π], we need to ensure that the function completes at least half a period but not more than a full period.Wait, actually, for f(x) to attain both maximum and minimum, the argument ωx + π/3 must cover at least π radians (half a period). But since it's on [0, 2π], the total change in the argument is ω*(2π) + π/3 - π/3 = 2πω.We need 2πω ≥ π ⇒ ω ≥ 1/2. But also, to have unique max and min, it shouldn't complete more than a full period, otherwise, it would have multiple maxima and minima.Wait, actually, if ω is too large, the function could oscillate multiple times, leading to multiple maxima and minima. But the problem states that the maximum and minimum are unique, meaning they occur exactly once each on [0, 2π].So, the function must go from some point, reach maximum once, then minimum once, and then not reach maximum again. So, the total phase shift should be such that the function doesn't complete a full period.The period of f(x) is 2π / ω. For the function to have only one maximum and one minimum on [0, 2π], the interval [0, 2π] must be less than a full period. So, 2π < 2π / ω ⇒ ω < 1.But wait, that contradicts the earlier thought. Wait, no. If ω is too small, the function would have a longer period, meaning it might not even reach the maximum and minimum once. So, we need to balance.Wait, let's think differently. The function f(x) = sin(ωx + π/3). Its maxima occur where ωx + π/3 = π/2 + 2πn, and minima where ωx + π/3 = 3π/2 + 2πn.We need exactly one maximum and one minimum in [0, 2π].So, the first maximum occurs at x = (π/2 - π/3)/ω = (π/6)/ω.The first minimum occurs at x = (3π/2 - π/3)/ω = (7π/6)/ω.We need both these x-values to be within [0, 2π], and the next maximum after the minimum should be outside [0, 2π].So, the next maximum after the first minimum would be at x = (π/2 - π/3 + 2π)/ω = (π/6 + 2π)/ω.We need this x to be greater than 2π.So, (π/6 + 2π)/ω > 2π ⇒ (13π/6)/ω > 2π ⇒ 13/6 > 2ω ⇒ ω < 13/12.Similarly, the first maximum should be within [0, 2π], so (π/6)/ω ≤ 2π ⇒ ω ≥ π/6 / 2π = 1/12. But wait, that's too small. Wait, actually, the first maximum is at π/6ω, which must be ≤ 2π. So, π/6ω ≤ 2π ⇒ ω ≥ 1/12. But that seems too low.Wait, perhaps a better approach is to consider the total change in the argument over [0, 2π]. The argument goes from π/3 to ω*2π + π/3. The sine function completes a full cycle every 2π, so the change in argument is 2πω.To have exactly one maximum and one minimum, the change in argument must be at least π (to reach from max to min) but less than 3π (to not reach the next max). Wait, no. Let me think.Actually, for the sine function to have exactly one maximum and one minimum, the argument must cover between π and 2π. Because from 0 to π, you get one maximum. From π to 2π, you get one minimum. So, to have both, the argument must cover at least π, but to not have another maximum, it must not reach 2π.Wait, no. Let me think again. The sine function has a maximum at π/2 and a minimum at 3π/2. So, to have both a maximum and a minimum, the argument must cover from some point to at least 3π/2. But to have only one maximum and one minimum, the argument must not reach the next maximum at 5π/2.So, the total change in argument is 2πω. We need 2πω ≥ π (to reach the first minimum) and 2πω < 2π (to not complete a full period and have another maximum). Wait, no.Wait, the argument starts at π/3. So, the first maximum is at π/2, which is π/6 after the start. The first minimum is at 3π/2, which is 4π/3 after the start. The next maximum is at 5π/2, which is 7π/3 after the start.So, the change in argument from π/3 to ω*2π + π/3 must include π/2 and 3π/2 but not 5π/2.So, ω*2π + π/3 ≥ 3π/2 (to reach the minimum) and ω*2π + π/3 < 5π/2 (to not reach the next maximum).So, solving for ω:First inequality: 2πω + π/3 ≥ 3π/2 ⇒ 2πω ≥ 3π/2 - π/3 = 9π/6 - 2π/6 = 7π/6 ⇒ ω ≥ 7/12.Second inequality: 2πω + π/3 < 5π/2 ⇒ 2πω < 5π/2 - π/3 = 15π/6 - 2π/6 = 13π/6 ⇒ ω < 13/12.So, the range of ω is 7/12 ≤ ω < 13/12.**Problem 6:** For f(x) = 3 sin(2x - π/3), determine which conclusions are correct:① Symmetric about x = 11π/12.② Symmetric about (2π/3, 0).③ Not monotonic on (-π/12, 5π/12).④ Translating y = 3 sin 2x right by π/3 gives C.Let me analyze each option.First, let's write f(x) = 3 sin(2x - π/3). We can rewrite this as 3 sin[2(x - π/6)]. So, it's a sine wave with amplitude 3, period π, shifted right by π/6.① Symmetric about x = 11π/12.For a sine function, the axis of symmetry is at the midpoint between a maximum and a minimum, or between two consecutive points of inflection. The general form is sin(Bx + C). The phase shift is -C/B, so here it's π/6.The function has maxima at 2x - π/3 = π/2 + 2πn ⇒ x = (π/2 + π/3)/2 + π n = (5π/6)/2 + π n = 5π/12 + π n.Similarly, minima at 2x - π/3 = 3π/2 + 2πn ⇒ x = (3π/2 + π/3)/2 + π n = (11π/6)/2 + π n = 11π/12 + π n.So, the first maximum is at 5π/12, the first minimum at 11π/12. The axis of symmetry between them is at (5π/12 + 11π/12)/2 = (16π/12)/2 = (4π/3)/2 = 2π/3.Wait, but 2π/3 is approximately 2.094, and 11π/12 is approximately 2.879. Wait, no, 11π/12 is about 2.879, and 5π/12 is about 1.308. So, the midpoint between 5π/12 and 11π/12 is (5π/12 + 11π/12)/2 = 16π/12 / 2 = 8π/12 = 2π/3. So, the axis of symmetry is at x = 2π/3, not 11π/12. So, ① is incorrect.Wait, but the question says symmetric about x = 11π/12. Hmm, maybe I made a mistake.Wait, the axis of symmetry is at the midpoint between a maximum and the next minimum. So, between 5π/12 and 11π/12, the midpoint is indeed 8π/12 = 2π/3. So, ① is incorrect.Wait, but maybe I'm misunderstanding. The function is symmetric about its midline, which is the horizontal line y=0, but the question is about vertical symmetry, i.e., reflection over a vertical line.So, for the function to be symmetric about x = a, f(a + t) = f(a - t) for all t.Let's test a = 11π/12.f(11π/12 + t) = 3 sin[2(11π/12 + t) - π/3] = 3 sin[11π/6 + 2t - π/3] = 3 sin[11π/6 - π/3 + 2t] = 3 sin[9π/6 + 2t] = 3 sin(3π/2 + 2t).Similarly, f(11π/12 - t) = 3 sin[2(11π/12 - t) - π/3] = 3 sin[11π/6 - 2t - π/3] = 3 sin[11π/6 - π/3 - 2t] = 3 sin[9π/6 - 2t] = 3 sin(3π/2 - 2t).Now, sin(3π/2 + 2t) = -cos(2t), and sin(3π/2 - 2t) = -cos(2t). So, f(11π/12 + t) = -cos(2t) and f(11π/12 - t) = -cos(2t). So, they are equal. Therefore, the function is symmetric about x = 11π/12. So, ① is correct.Wait, that contradicts my earlier conclusion. Hmm, maybe I was wrong earlier.Wait, let's double-check. The axis of symmetry is at x = 11π/12. Let me pick t = 0: f(11π/12) = 3 sin(2*(11π/12) - π/3) = 3 sin(11π/6 - π/3) = 3 sin(11π/6 - 2π/6) = 3 sin(9π/6) = 3 sin(3π/2) = -3. So, the point is (11π/12, -3).Now, let's pick t = π/12:f(11π/12 + π/12) = f(12π/12) = f(π) = 3 sin(2π - π/3) = 3 sin(5π/3) = 3*(-√3/2) = -3√3/2.f(11π/12 - π/12) = f(10π/12) = f(5π/6) = 3 sin(2*(5π/6) - π/3) = 3 sin(5π/3 - π/3) = 3 sin(4π/3) = 3*(-√3/2) = -3√3/2.So, they are equal. Similarly, for t = π/6:f(11π/12 + π/6) = f(13π/12) = 3 sin(2*(13π/12) - π/3) = 3 sin(13π/6 - π/3) = 3 sin(13π/6 - 2π/6) = 3 sin(11π/6) = 3*(-1/2) = -3/2.f(11π/12 - π/6) = f(9π/12) = f(3π/4) = 3 sin(2*(3π/4) - π/3) = 3 sin(3π/2 - π/3) = 3 sin(7π/6) = 3*(-1/2) = -3/2.They are equal. So, it seems that ① is correct.Wait, but earlier I thought the axis of symmetry was at 2π/3. Maybe both are correct? Or maybe I made a mistake.Wait, let's check symmetry about x = 2π/3.f(2π/3 + t) = 3 sin[2*(2π/3 + t) - π/3] = 3 sin(4π/3 + 2t - π/3) = 3 sin(π + 2t).f(2π/3 - t) = 3 sin[2*(2π/3 - t) - π/3] = 3 sin(4π/3 - 2t - π/3) = 3 sin(π - 2t).Now, sin(π + 2t) = -sin(2t), and sin(π - 2t) = sin(2t). So, f(2π/3 + t) = -sin(2t) and f(2π/3 - t) = sin(2t). These are not equal unless sin(2t) = 0, which is not generally true. So, ② is incorrect.Wait, but ② says symmetric about the point (2π/3, 0). That's a point symmetry, not a line symmetry. So, for point symmetry, f(2π/3 + t) = -f(2π/3 - t).Let's check:f(2π/3 + t) = 3 sin(π + 2t) = -3 sin(2t).f(2π/3 - t) = 3 sin(π - 2t) = 3 sin(2t).So, f(2π/3 + t) = -f(2π/3 - t), which means it's symmetric about the point (2π/3, 0). So, ② is correct.Wait, but earlier I thought ① was correct, but now ② is also correct. Let me double-check.Wait, for ①, we saw that f(11π/12 + t) = f(11π/12 - t), so it's symmetric about x = 11π/12. For ②, f(2π/3 + t) = -f(2π/3 - t), so it's symmetric about the point (2π/3, 0). So, both ① and ② are correct? But the options are ①, ②, ③, ④.Wait, let me check ③: Function f(x) is not monotonic on (-π/12, 5π/12).Monotonic means it's either entirely increasing or decreasing. Let's find the derivative: f'(x) = 6 cos(2x - π/3).We need to see if f'(x) changes sign in (-π/12, 5π/12).Compute 2x - π/3 at x = -π/12: 2*(-π/12) - π/3 = -π/6 - π/3 = -π/2.At x = 5π/12: 2*(5π/12) - π/3 = 5π/6 - π/3 = 5π/6 - 2π/6 = 3π/6 = π/2.So, the argument 2x - π/3 goes from -π/2 to π/2 as x goes from -π/12 to 5π/12.In this interval, cos(θ) is positive from -π/2 to π/2, except at the endpoints where it's zero. So, f'(x) is positive throughout (-π/12, 5π/12), meaning f(x) is increasing on this interval. Therefore, it is monotonic, so ③ is incorrect.Wait, but wait, at x = -π/12, 2x - π/3 = -π/6 - π/3 = -π/2. At x = 5π/12, it's π/2. So, cos(θ) is positive in (-π/2, π/2), so f'(x) is positive, so f(x) is increasing. Therefore, it's monotonic, so ③ is incorrect.④: Translating y = 3 sin 2x right by π/3 gives C.Original function: y = 3 sin 2x.Translating right by π/3: y = 3 sin[2(x - π/3)] = 3 sin(2x - 2π/3). But our function is 3 sin(2x - π/3). So, it's not the same. Therefore, ④ is incorrect.So, the correct conclusions are ① and ②.Wait, but earlier I thought ① was correct, but then I got confused. Let me confirm again.For ①, we saw that f(11π/12 + t) = f(11π/12 - t), so it's symmetric about x = 11π/12. So, ① is correct.For ②, f(2π/3 + t) = -f(2π/3 - t), so it's symmetric about the point (2π/3, 0). So, ② is correct.But wait, earlier I thought ② was correct, but now I'm confused because the function is symmetric about both a line and a point? That seems possible because the function is sinusoidal.Wait, actually, a sine function is symmetric about its midline (a horizontal line) and also has point symmetry about its inflection points. But in this case, the question is about vertical line symmetry and point symmetry.So, yes, ① and ② are correct. But wait, in my earlier analysis, I thought ① was correct, but then I got confused. Let me double-check.Yes, f(11π/12 + t) = f(11π/12 - t), so it's symmetric about x = 11π/12. And f(2π/3 + t) = -f(2π/3 - t), so it's symmetric about (2π/3, 0). So, both ① and ② are correct.But wait, the options are ①, ②, ③, ④. So, the correct conclusions are ①, ②, and ③? Wait, no, ③ is incorrect.Wait, no, ③ is incorrect because the function is monotonic on that interval. So, only ① and ② are correct.Wait, but earlier I thought ③ was incorrect. So, the correct conclusions are ① and ②.Wait, but in my initial analysis, I thought ① was incorrect, but after checking, it's correct. So, the correct conclusions are ① and ②.But wait, let me check ③ again. The function is 3 sin(2x - π/3). Its derivative is 6 cos(2x - π/3). On (-π/12, 5π/12), 2x - π/3 goes from -π/2 to π/2, where cosine is positive. So, derivative is positive, function is increasing. Therefore, it's monotonic, so ③ is incorrect.So, the correct conclusions are ① and ②.Wait, but in the options, ① is correct, ② is correct, ③ is incorrect, ④ is incorrect. So, the correct conclusions are ① and ②.But wait, in my earlier analysis, I thought ② was correct, but now I'm confused because the function is symmetric about both a line and a point. But in reality, a sine function is symmetric about its midline (a horizontal line) and has point symmetry about its inflection points. But in this case, the question is about vertical line symmetry and point symmetry.So, yes, ① and ② are correct.Wait, but in the options, ① is correct, ② is correct, ③ is incorrect, ④ is incorrect. So, the correct conclusions are ① and ②.But wait, the options are ①, ②, ③, ④. So, the correct ones are ① and ②.Wait, but in my initial analysis, I thought ② was correct, but then I got confused. Let me confirm again.Yes, f(2π/3 + t) = -f(2π/3 - t), so it's symmetric about (2π/3, 0). So, ② is correct.So, the correct conclusions are ① and ②.Wait, but in the options, ① is correct, ② is correct, ③ is incorrect, ④ is incorrect. So, the correct conclusions are ① and ②.But wait, in the initial problem statement, the options are ①, ②, ③, ④. So, the correct ones are ① and ②.Wait, but in the problem statement, it's written as:(6) If the graph of the function f(x)=3 sin(2x−π/3) is denoted as C, determine the correct conclusion(s) from the following options:(①) Graph C is symmetric with respect to the line x=11π/12.(②) Graph C is symmetric with respect to the point (2π/3,0).(③) Function f(x) is not a monotonic function on the interval (−π/12,5π/12).(④) The graph of y=3 sin 2x can be translated to the right by π/3 units to obtain graph C.So, based on my analysis, ① and ② are correct, ③ is incorrect, ④ is incorrect.But wait, earlier I thought ③ was incorrect because the function is monotonic on that interval. So, the correct conclusions are ① and ②.Wait, but in the problem statement, it's asking to determine the correct conclusions, so the answer is ① and ②.But in the initial problem, the user wrote the answers as:(1) boxed{frac{1}{6}}(2) boxed{-1}(3) boxed{sqrt{3}}(4) boxed{frac{5}{12}}(5) boxed{frac{7}{12}leqomega<frac{13}{12}}(6) The correct conclusions are (①) and (③).Wait, but according to my analysis, ① and ② are correct, but the user's answer says ① and ③. So, there's a discrepancy.Wait, let me double-check ③ again. The function is 3 sin(2x - π/3). Its derivative is 6 cos(2x - π/3). On the interval (-π/12, 5π/12), 2x - π/3 ranges from -π/2 to π/2. Cosine is positive in this interval, so the derivative is positive, meaning the function is increasing. Therefore, it is monotonic, so ③ is incorrect.Therefore, the correct conclusions are ① and ②, not ③.But in the user's answer, they concluded ① and ③. So, perhaps they made a mistake.Alternatively, maybe I made a mistake in analyzing ③. Let me check again.The interval is (-π/12, 5π/12). Let's compute the derivative at some points.At x = -π/12: 2x - π/3 = -π/6 - π/3 = -π/2. cos(-π/2) = 0.At x = 0: 2x - π/3 = -π/3. cos(-π/3) = 0.5.At x = π/12: 2x - π/3 = π/6 - π/3 = -π/6. cos(-π/6) = √3/2.At x = 5π/12: 2x - π/3 = 5π/6 - π/3 = 5π/6 - 2π/6 = 3π/6 = π/2. cos(π/2) = 0.So, the derivative starts at 0, increases to 6*(√3/2) = 3√3 at x = π/12, then decreases back to 0 at x = 5π/12.So, the derivative is always non-negative in this interval, meaning the function is non-decreasing. Therefore, it is monotonic on this interval. So, ③ is incorrect.Therefore, the correct conclusions are ① and ②.But in the user's answer, they concluded ① and ③. So, perhaps they made a mistake.Alternatively, maybe I made a mistake in analyzing ②.Wait, let's check ② again. The function is symmetric about the point (2π/3, 0). So, f(2π/3 + t) = -f(2π/3 - t).Let me pick t = π/12:f(2π/3 + π/12) = f(3π/4) = 3 sin(2*(3π/4) - π/3) = 3 sin(3π/2 - π/3) = 3 sin(7π/6) = 3*(-1/2) = -3/2.f(2π/3 - π/12) = f(7π/12) = 3 sin(2*(7π/12) - π/3) = 3 sin(7π/6 - π/3) = 3 sin(5π/6) = 3*(1/2) = 3/2.So, f(2π/3 + π/12) = -3/2 and f(2π/3 - π/12) = 3/2. So, f(2π/3 + t) = -f(2π/3 - t), which confirms the point symmetry. So, ② is correct.Therefore, the correct conclusions are ① and ②.But the user's answer says ① and ③. So, perhaps they made a mistake in analyzing ③.In conclusion, the correct conclusions are ① and ②.

answer：First, I need to understand the operation ( @ ) defined as ( X@Y = frac{2(X + Y)}{3} ). This means that for any two numbers ( X ) and ( Y ), I multiply their sum by 2 and then divide by 3.Next, I'll compute ( 4@7 ) using this definition. Adding 4 and 7 gives 11. Multiplying this sum by 2 results in 22. Dividing 22 by 3 gives ( frac{22}{3} ).Now, I'll use this result to compute ( (4@7)@5 ). Substituting ( frac{22}{3} ) for ( 4@7 ), I add ( frac{22}{3} ) and 5. Converting 5 to thirds, it becomes ( frac{15}{3} ). Adding ( frac{22}{3} ) and ( frac{15}{3} ) gives ( frac{37}{3} ). Multiplying this sum by 2 results in ( frac{74}{3} ). Finally, dividing by 3 gives ( frac{74}{9} ).Therefore, the value of ( (4@7)@5 ) is ( frac{74}{9} ).

answer：First, calculate the total length of rope Tanesha needs. She requires 10 pieces, each 6 inches long.10 pieces × 6 inches per piece = 60 inchesConvert inches to feet:60 inches ÷ 12 inches per foot = 5 feetTanesha needs 5 feet of rope. She can either buy a 6-foot rope for 5 or purchase five 1-foot ropes. Let the cost of each 1-foot rope be ( x ) dollars.Total cost for 5 one-foot ropes = 5x dollarsTo determine the least cost, compare the two options:- Cost of 6-foot rope: 5- Cost of 5 one-foot ropes: 5xIf ( 5x < 5 ), then buying the one-foot ropes is cheaper. If ( 5x geq 5 ), buying the 6-foot rope is more economical.Assuming the cost of each 1-foot rope is 1 or more, the minimum amount Tanesha has to spend is 5.