🧠:<think>Okay, so I have this problem where there are 10 numbers: x₁, x₂, up to x₁₀. And for each i from 1 through 9, the equation x_i + 2x_{i+1} = 1 holds. I need to find the value of x₁ + 512x_{10}. Hmm, that seems a bit tricky, but let me try to break it down step by step.First, let me write down the given equations to see if I can spot a pattern or find a way to express each x in terms of the next one. The equations are:1. x₁ + 2x₂ = 12. x₂ + 2x₃ = 13. x₃ + 2x₄ = 14. x₄ + 2x₅ = 15. x₅ + 2x₆ = 16. x₆ + 2x₇ = 17. x₇ + 2x₈ = 18. x₈ + 2x₉ = 19. x₉ + 2x_{10} = 1So, each equation relates a variable to the next one with a coefficient of 2. Maybe I can express each x in terms of the next variable and substitute them recursively.Starting with the first equation: x₁ + 2x₂ = 1. Let's solve for x₁:x₁ = 1 - 2x₂Similarly, from the second equation: x₂ = 1 - 2x₃If I substitute this into the expression for x₁:x₁ = 1 - 2(1 - 2x₃) = 1 - 2 + 4x₃ = -1 + 4x₃Okay, so x₁ is expressed in terms of x₃ now. Let's do the same for x₃ from the third equation: x₃ = 1 - 2x₄Substituting into x₁:x₁ = -1 + 4(1 - 2x₄) = -1 + 4 - 8x₄ = 3 - 8x₄Hmm, I see a pattern here. Each time I substitute, the coefficient of the next variable doubles and alternates in sign. Let me continue this process to see if I can express x₁ in terms of x_{10}.Continuing:From the fourth equation: x₄ = 1 - 2x₅Substitute into x₁:x₁ = 3 - 8(1 - 2x₅) = 3 - 8 + 16x₅ = -5 + 16x₅Fifth equation: x₅ = 1 - 2x₆Substitute:x₁ = -5 + 16(1 - 2x₆) = -5 + 16 - 32x₆ = 11 - 32x₆Sixth equation: x₆ = 1 - 2x₇Substitute:x₁ = 11 - 32(1 - 2x₇) = 11 - 32 + 64x₇ = -21 + 64x₇Seventh equation: x₇ = 1 - 2x₈Substitute:x₁ = -21 + 64(1 - 2x₈) = -21 + 64 - 128x₈ = 43 - 128x₈Eighth equation: x₈ = 1 - 2x₉Substitute:x₁ = 43 - 128(1 - 2x₉) = 43 - 128 + 256x₉ = -85 + 256x₉Ninth equation: x₉ = 1 - 2x_{10}Substitute:x₁ = -85 + 256(1 - 2x_{10}) = -85 + 256 - 512x_{10} = 171 - 512x_{10}So, now I have x₁ expressed in terms of x_{10}:x₁ = 171 - 512x_{10}But the problem asks for x₁ + 512x_{10}. Let me compute that:x₁ + 512x_{10} = (171 - 512x_{10}) + 512x_{10} = 171Wait, that's interesting. The terms involving x_{10} cancel out, leaving just 171. So, regardless of the value of x_{10}, the expression x₁ + 512x_{10} equals 171.Let me double-check my steps to make sure I didn't make a mistake. Starting from x₁ and substituting each subsequent x, the coefficients seem to follow a pattern of powers of 2 with alternating signs. Each substitution doubles the coefficient and changes the sign. After substituting all the way to x_{10}, the coefficient becomes 512 with a negative sign, which cancels out when adding 512x_{10}. The constants add up to 171, which seems correct.So, I think my conclusion is right. The value of x₁ + 512x_{10} is 171.</think>

🧠:<think>Okay, so I have these four points in space: P = (1, -2, 1), Q = (4, -7, 4), R = (2, -1, -1), and S = (5, -6, 2). I need to figure out if they form a parallelogram. If they do, I also have to calculate the area of that parallelogram. Hmm, let me think about how to approach this.First, I remember that in a parallelogram, opposite sides are equal and parallel. So, in terms of vectors, the vector from P to Q should be equal to the vector from R to S, and the vector from Q to R should be equal to the vector from P to S. Alternatively, another way is to check if the midpoints of the diagonals are the same because in a parallelogram, the diagonals bisect each other. Maybe I'll try both methods to be thorough.Let me start by finding the vectors between these points. So, vector PQ is Q - P, which is (4-1, -7-(-2), 4-1) = (3, -5, 3). Similarly, vector SR is R - S, which is (2-5, -1-(-6), -1-2) = (-3, 5, -3). Wait, that's just negative of PQ. So, PQ and SR are equal in magnitude but opposite in direction. That makes sense because in a parallelogram, opposite sides are equal and parallel, so their vectors should be equal. So, PQ equals SR, which is a good sign.Next, let me check vector QR and vector PS. Vector QR is R - Q, which is (2-4, -1-(-7), -1-4) = (-2, 6, -5). Vector PS is S - P, which is (5-1, -6-(-2), 2-1) = (4, -4, 1). Hmm, these don't look equal. Wait, did I do that right? Let me recalculate.Vector QR: R is (2, -1, -1), Q is (4, -7, 4). So, 2-4 = -2, -1 - (-7) = 6, -1 -4 = -5. That's correct. Vector PS: S is (5, -6, 2), P is (1, -2, 1). So, 5-1=4, -6 - (-2)= -4, 2-1=1. So, yes, PS is (4, -4, 1). But QR is (-2, 6, -5). These are not equal. Hmm, that seems problematic. Maybe I made a mistake in choosing the sides?Wait, perhaps I should check another pair of sides. Maybe vector PR and QS? Let me see. Vector PR is R - P, which is (2-1, -1 - (-2), -1 -1) = (1, 1, -2). Vector QS is S - Q, which is (5-4, -6 - (-7), 2 -4) = (1, 1, -2). Oh, okay, so PR equals QS. That's another pair of equal vectors. So, maybe the sides I checked earlier, QR and PS, aren't supposed to be equal? Wait, in a parallelogram, it's the opposite sides that are equal, not the adjacent ones. So, PQ and SR are opposite sides, and PR and QS are the diagonals? Wait, no, PR and QS are diagonals, not sides.Wait, I think I confused the sides. Let me clarify. In a quadrilateral, the sides are PQ, QR, RS, and SP. So, opposite sides are PQ and RS, and QR and SP. So, I should check if PQ equals RS and QR equals SP.So, vector PQ is (3, -5, 3), as before. Vector RS is S - R, which is (5-2, -6 - (-1), 2 - (-1)) = (3, -5, 3). So, PQ equals RS. Good.Vector QR is (-2, 6, -5), as before. Vector SP is P - S, which is (1-5, -2 - (-6), 1 -2) = (-4, 4, -1). Wait, that's not equal to QR. Hmm, that's confusing. Did I do something wrong?Wait, maybe I should think about the vectors in terms of edges rather than just subtracting points. So, in a parallelogram, the edges PQ and SR should be equal, and edges QR and PS should be equal. Let me check that.Vector PQ is (3, -5, 3). Vector SR is R - S, which is (2-5, -1 - (-6), -1 -2) = (-3, 5, -3). So, that's the negative of PQ, which is fine because they are in opposite directions but same magnitude.Vector QR is (-2, 6, -5). Vector PS is S - P, which is (5-1, -6 - (-2), 2 -1) = (4, -4, 1). Wait, that's not equal to QR. So, that seems like a problem. Maybe I made a mistake in defining the points.Wait, perhaps the order of the points is different. Maybe it's not PQRS in that order? Maybe it's a different order? Let me check.Alternatively, maybe it's better to check the midpoints of the diagonals. In a parallelogram, the midpoints of the diagonals should coincide. So, the midpoint of PR should be the same as the midpoint of QS.Midpoint of PR: P = (1, -2, 1), R = (2, -1, -1). So, midpoint is ((1+2)/2, (-2 + (-1))/2, (1 + (-1))/2) = (1.5, -1.5, 0).Midpoint of QS: Q = (4, -7, 4), S = (5, -6, 2). Midpoint is ((4+5)/2, (-7 + (-6))/2, (4 + 2)/2) = (4.5, -6.5, 3). Hmm, these midpoints are different. That means the diagonals don't bisect each other, so it's not a parallelogram? But earlier, PQ equals RS, and PR equals QS. Wait, that seems contradictory.Wait, maybe I messed up the points. Let me list the points again: P = (1, -2, 1), Q = (4, -7, 4), R = (2, -1, -1), S = (5, -6, 2). So, if I connect P to Q to R to S and back to P, is that the correct order? Maybe not. Maybe the order is different.Alternatively, perhaps the points form a parallelogram but not in the order PQRS. Maybe it's P, Q, S, R? Let me try that.So, if I consider the quadrilateral PQSR, then the sides would be PQ, QS, SR, and RP. Let me check the vectors.Vector PQ is (3, -5, 3). Vector SR is R - S, which is (2-5, -1 - (-6), -1 -2) = (-3, 5, -3). So, PQ is equal to SR, which is good.Vector QS is S - Q, which is (5-4, -6 - (-7), 2 -4) = (1, 1, -2). Vector RP is P - R, which is (1-2, -2 - (-1), 1 - (-1)) = (-1, -1, 2). So, QS is equal to RP, but in opposite direction. So, that's also good because in a parallelogram, opposite sides are equal and parallel.Wait, so if I consider the quadrilateral PQSR, then it is a parallelogram because both pairs of opposite sides are equal and parallel. So, maybe the original order PQRS is not a parallelogram, but PQSR is. So, perhaps the points form a parallelogram, but not in the order given.Alternatively, maybe I should check all possible pairs. Let me try another approach.Another way to check if four points form a parallelogram is to see if the vectors representing the sides satisfy the condition that the vector from P to Q plus the vector from P to R equals the vector from P to S. Wait, is that correct? Let me think.In a parallelogram, the position vectors of the points satisfy the relation: if P, Q, R, S are the vertices, then Q - P = S - R and R - P = Q - S. Wait, maybe I should use vector addition.Alternatively, in a parallelogram, the vector from P to Q should be equal to the vector from R to S, and the vector from Q to R should be equal to the vector from S to P. Let me check that.Vector PQ is (3, -5, 3). Vector RS is S - R, which is (5-2, -6 - (-1), 2 - (-1)) = (3, -5, 3). So, PQ equals RS.Vector QR is R - Q, which is (2-4, -1 - (-7), -1 -4) = (-2, 6, -5). Vector SP is P - S, which is (1-5, -2 - (-6), 1 -2) = (-4, 4, -1). Hmm, these are not equal. So, that's a problem.Wait, but earlier when I considered PQSR, the vectors worked out. So, maybe the order of the points is important. If I take the points in the order P, Q, S, R, then the vectors PQ and QS and SR and RP would form a parallelogram.Wait, let me try that. So, quadrilateral PQSR: P to Q is (3, -5, 3), Q to S is (1, 1, -2), S to R is (-3, 5, -3), and R to P is (-1, -1, 2). So, PQ equals SR, and QS equals RP. So, that works.But the original question is about points P, Q, R, S. So, unless specified otherwise, the order is important. So, if the points are given in order P, Q, R, S, then the quadrilateral PQRS may not be a parallelogram, but the points can form a parallelogram in a different order.Wait, the question says "form a parallelogram in space." It doesn't specify the order, so maybe it's just about whether any four points form a parallelogram regardless of order. So, in that case, since we can arrange them as PQSR to form a parallelogram, then yes, they form a parallelogram.Alternatively, maybe the original order is PQRS, and we have to check if that specific quadrilateral is a parallelogram. In that case, since the midpoints of the diagonals are different, it's not a parallelogram. But the question is a bit ambiguous.Wait, let me check the midpoints again. If the quadrilateral is PQRS, then the diagonals are PR and QS. Midpoint of PR is (1.5, -1.5, 0), and midpoint of QS is (4.5, -6.5, 3). These are different, so diagonals don't bisect each other, so PQRS is not a parallelogram.But if we arrange the points as PQSR, then the diagonals would be PS and QR. Let me check their midpoints.Midpoint of PS: P = (1, -2, 1), S = (5, -6, 2). Midpoint is ((1+5)/2, (-2 + (-6))/2, (1 + 2)/2) = (3, -4, 1.5).Midpoint of QR: Q = (4, -7, 4), R = (2, -1, -1). Midpoint is ((4+2)/2, (-7 + (-1))/2, (4 + (-1))/2) = (3, -4, 1.5). So, the midpoints coincide. Therefore, PQSR is a parallelogram.So, the answer is yes, the points form a parallelogram, but not in the order PQRS. So, the question is a bit ambiguous, but since it just asks if they form a parallelogram, regardless of order, then yes.Now, to calculate the area of the parallelogram. I remember that the area can be found by the magnitude of the cross product of two adjacent sides. So, if I take vectors PQ and PS, their cross product's magnitude will give the area.Wait, but in the parallelogram PQSR, the sides are PQ and PS. Wait, no, in PQSR, the sides are PQ and QS. Wait, no, let me think.In the parallelogram PQSR, the sides are PQ and QS. So, vectors PQ and QS. Let me compute those.Vector PQ is (3, -5, 3). Vector QS is S - Q, which is (5-4, -6 - (-7), 2 -4) = (1, 1, -2). So, cross product of PQ and QS is:|i   j   k||3  -5   3||1   1  -2|Calculating determinant:i * [(-5)(-2) - (3)(1)] - j * [(3)(-2) - (3)(1)] + k * [(3)(1) - (-5)(1)]= i * (10 - 3) - j * (-6 - 3) + k * (3 + 5)= 7i + 9j + 8kSo, the cross product vector is (7, 9, 8). The magnitude of this vector is sqrt(7^2 + 9^2 + 8^2) = sqrt(49 + 81 + 64) = sqrt(194). Therefore, the area is sqrt(194).Wait, but earlier when I thought of PQRS, I got the same cross product. Maybe because regardless of the order, the area remains the same. So, as long as the points form a parallelogram, the area can be calculated using the cross product of two adjacent sides.So, to summarize, the points do form a parallelogram, and the area is sqrt(194).</think>

🧠:<think>Alright, so I've got this problem here that I need to prove. It says that if ( a sin alpha + b sin beta + c sin gamma = 0 ) and ( a cos alpha + b cos beta + c cos gamma = 0 ), and all of ( a, b, c ) are nonzero, then I need to show that ( frac{sin (beta - gamma)}{a} = frac{sin (gamma - alpha)}{b} = frac{sin (alpha - beta)}{c} ).Okay, let's start by understanding what's given. We have two equations involving sine and cosine terms of three angles ( alpha, beta, gamma ), each multiplied by constants ( a, b, c ) respectively, and both sums equal zero. The goal is to relate the sines of the differences of these angles to the constants ( a, b, c ).First, I should probably write down the given equations:1. ( a sin alpha + b sin beta + c sin gamma = 0 )2. ( a cos alpha + b cos beta + c cos gamma = 0 )And we need to prove:( frac{sin (beta - gamma)}{a} = frac{sin (gamma - alpha)}{b} = frac{sin (alpha - beta)}{c} )Hmm. So, the idea is to manipulate the given equations to get expressions involving ( sin (beta - gamma) ), ( sin (gamma - alpha) ), and ( sin (alpha - beta) ), and then show that these are proportional to ( a, b, c ) respectively.One strategy that comes to mind is to eliminate one of the variables, say ( c ), from the equations, and then express the other terms in terms of the remaining variables. Alternatively, since both equations equal zero, perhaps I can square and add them to use the Pythagorean identity, but that might complicate things.Wait, another thought: if I consider these equations as components of a vector equation. Let me think about that. If I have a vector in the plane with components ( (a cos alpha, a sin alpha) ), another with ( (b cos beta, b sin beta) ), and another with ( (c cos gamma, c sin gamma) ), then the sum of these vectors is zero because both the x-components and y-components sum to zero. So, these three vectors form a closed triangle.That's an interesting geometric interpretation. So, the vectors ( a vec{v}_alpha ), ( b vec{v}_beta ), and ( c vec{v}_gamma ) add up to zero, where ( vec{v}_theta ) is the unit vector at angle ( theta ). This means that these vectors form a triangle when placed head-to-tail.From vector addition, I know that in such a case, the magnitudes and angles must satisfy certain relationships. Maybe I can use the law of sines or something similar here.But before jumping into that, let's see if I can manipulate the given equations algebraically. Perhaps I can express ( c sin gamma ) and ( c cos gamma ) from the first two equations and then relate them.From equation 1:( c sin gamma = -a sin alpha - b sin beta )From equation 2:( c cos gamma = -a cos alpha - b cos beta )So, if I square both equations and add them, I can use the identity ( sin^2 gamma + cos^2 gamma = 1 ):( c^2 (sin^2 gamma + cos^2 gamma) = (-a sin alpha - b sin beta)^2 + (-a cos alpha - b cos beta)^2 )Simplifying the left side:( c^2 = (a sin alpha + b sin beta)^2 + (a cos alpha + b cos beta)^2 )Expanding the right side:( c^2 = a^2 sin^2 alpha + 2ab sin alpha sin beta + b^2 sin^2 beta + a^2 cos^2 alpha + 2ab cos alpha cos beta + b^2 cos^2 beta )Combine like terms:( c^2 = a^2 (sin^2 alpha + cos^2 alpha) + b^2 (sin^2 beta + cos^2 beta) + 2ab (sin alpha sin beta + cos alpha cos beta) )Using the Pythagorean identity again:( c^2 = a^2 + b^2 + 2ab cos (alpha - beta) )Because ( sin alpha sin beta + cos alpha cos beta = cos (alpha - beta) ).So, we have:( c^2 = a^2 + b^2 + 2ab cos (alpha - beta) )Hmm, interesting. This relates ( c ) to ( a, b ) and the angle difference ( alpha - beta ).But we need to relate ( sin (beta - gamma) ), ( sin (gamma - alpha) ), and ( sin (alpha - beta) ) to ( a, b, c ). Maybe I need another approach.Let me think about expressing ( sin (beta - gamma) ) in terms of ( a, b, c ). Using the sine subtraction formula:( sin (beta - gamma) = sin beta cos gamma - cos beta sin gamma )From the given equations, we have expressions for ( sin gamma ) and ( cos gamma ):( sin gamma = frac{ -a sin alpha - b sin beta }{ c } )( cos gamma = frac{ -a cos alpha - b cos beta }{ c } )So, substituting these into ( sin (beta - gamma) ):( sin (beta - gamma) = sin beta left( frac{ -a cos alpha - b cos beta }{ c } right) - cos beta left( frac{ -a sin alpha - b sin beta }{ c } right) )Simplify:( sin (beta - gamma) = frac{ -a sin beta cos alpha - b sin beta cos beta + a cos beta sin alpha + b cos beta sin beta }{ c } )Notice that ( -b sin beta cos beta + b cos beta sin beta = 0 ), so those terms cancel out.So, we have:( sin (beta - gamma) = frac{ -a sin beta cos alpha + a cos beta sin alpha }{ c } )Factor out ( a ):( sin (beta - gamma) = frac{ a ( - sin beta cos alpha + cos beta sin alpha ) }{ c } )Recognize the sine subtraction formula in the numerator:( - sin beta cos alpha + cos beta sin alpha = sin (alpha - beta ) )So,( sin (beta - gamma) = frac{ a sin (alpha - beta ) }{ c } )Therefore,( frac{ sin (beta - gamma) }{ a } = frac{ sin (alpha - beta ) }{ c } )That's one equality down. Now, let's try to find another relation. Maybe ( sin (gamma - alpha ) ).Using the sine subtraction formula again:( sin (gamma - alpha ) = sin gamma cos alpha - cos gamma sin alpha )Again, substitute ( sin gamma ) and ( cos gamma ) from the given equations:( sin (gamma - alpha ) = left( frac{ -a sin alpha - b sin beta }{ c } right) cos alpha - left( frac{ -a cos alpha - b cos beta }{ c } right) sin alpha )Simplify:( sin (gamma - alpha ) = frac{ -a sin alpha cos alpha - b sin beta cos alpha + a cos alpha sin alpha + b cos beta sin alpha }{ c } )Again, ( -a sin alpha cos alpha + a cos alpha sin alpha = 0 ), so those terms cancel.So,( sin (gamma - alpha ) = frac{ -b sin beta cos alpha + b cos beta sin alpha }{ c } )Factor out ( b ):( sin (gamma - alpha ) = frac{ b ( - sin beta cos alpha + cos beta sin alpha ) }{ c } )Again, recognize the sine subtraction formula:( - sin beta cos alpha + cos beta sin alpha = sin (alpha - beta ) )So,( sin (gamma - alpha ) = frac{ b sin (alpha - beta ) }{ c } )Therefore,( frac{ sin (gamma - alpha ) }{ b } = frac{ sin (alpha - beta ) }{ c } )So now, we have:( frac{ sin (beta - gamma ) }{ a } = frac{ sin (alpha - beta ) }{ c } )and( frac{ sin (gamma - alpha ) }{ b } = frac{ sin (alpha - beta ) }{ c } )This implies that:( frac{ sin (beta - gamma ) }{ a } = frac{ sin (gamma - alpha ) }{ b } = frac{ sin (alpha - beta ) }{ c } )Which is exactly what we needed to prove.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the expressions for ( sin (beta - gamma) ) and ( sin (gamma - alpha) ), I substituted the expressions for ( sin gamma ) and ( cos gamma ) correctly. Then, I expanded and simplified, canceling out the terms that were zero. Then, I recognized the sine subtraction identities correctly. It seems solid.Also, the geometric interpretation as vectors forming a triangle might provide another way to see this, perhaps using the law of sines or something similar, but the algebraic approach worked out.So, I think this proves the required equality.Final Answerboxed{frac{sin (beta - gamma)}{a} = frac{sin (gamma - alpha)}{b} = frac{sin (alpha - beta)}{c}}</think>

🧠:<think>Okay, so I have this problem about a high-speed train's travel plotted on a graph where distance is shown against time. The question is asking me to determine during which hour interval the train's average speed was the highest. The options are from 0-1, 1-2, 3-4, 5-6, and 10-11 hours.First, I need to recall what average speed means. Average speed is basically the total distance traveled divided by the total time taken. So, in this case, since we're looking at hour intervals, it's the distance covered in each hour divided by one hour. That simplifies things a bit because the denominator is always one hour, so the average speed is just the distance covered in that hour.Now, the graph is distance versus time, which means it's a line graph where the x-axis is time in hours and the y-axis is distance in kilometers. The slope of this graph at any point gives the speed at that instant. But since we're dealing with average speed over an hour, we need to look at the change in distance over each hour interval.So, for each hour interval, I need to calculate the difference in distance between the end and the start of that hour. The interval with the largest difference in distance will have the highest average speed because average speed is directly proportional to the distance covered in that time.Let me think about how to approach this without the actual graph. Since I don't have the graph in front of me, I need to rely on the information given in the problem or any clues that might be implied. The options given are specific intervals: first (0-1), second (1-2), third (3-4), sixth (5-6), and last (10-11). I wonder if there's a pattern or if certain intervals are more likely to have higher speeds. High-speed trains usually accelerate at the beginning, maintain a constant speed, and then decelerate towards the end. So, maybe the initial intervals have increasing speeds, then a constant speed, and then decreasing speeds towards the end.But that's just a general thought. Without specific data points, it's hard to be certain. Maybe the steepest part of the graph, which would indicate the highest speed, is somewhere in the middle of the journey. Or perhaps it's at the beginning or the end.Wait, the problem mentions it's a route from city A to city B, so it's a one-way trip. High-speed trains often have a period where they reach their maximum speed after accelerating, and then they might maintain that speed for a while before slowing down to approach the destination.Given that, the highest average speed might be in the middle of the journey, not necessarily at the very beginning or the very end. So, looking at the options, the third interval (3-4) and the sixth interval (5-6) are in the middle of the journey, assuming the total journey is longer than 11 hours.But the last interval is 10-11, which is near the end. If the train is slowing down towards the end, that interval might have a lower average speed. Similarly, the first two intervals (0-1 and 1-2) are at the beginning, so the train might be accelerating, but maybe not yet at its maximum speed.So, perhaps the third interval (3-4) is when the train has reached its maximum speed and is maintaining it, leading to the highest average speed during that hour. Alternatively, the sixth interval (5-6) could also be a candidate if the journey is longer and the train maintains high speed for several hours.But without the actual graph, it's a bit tricky. Maybe I can think about how the graph would look. If the graph is a straight line, the speed is constant, but since it's a high-speed train, it's more likely to have varying speeds with acceleration and deceleration phases.So, the steepest part of the graph would be where the train is moving the fastest. If the steepest part is in the third hour, then that's when the average speed is the highest. If it's in the sixth hour, then that's the answer.Given the options, and considering that the train might have a period of constant high speed after initial acceleration, the third interval (3-4) seems plausible as the interval with the highest average speed.Alternatively, if the train's maximum speed is maintained for several hours, then multiple intervals could have high average speeds, but since the options only include specific intervals, I have to choose the one that is most likely.Considering all this, I think the third interval (3-4) is the most probable answer.</think>