answer：First, let's define the variables:- Let ( P ) represent the current population of Pirajussaraí.- Let ( T ) represent the current population of Tucupira.From the problem, we know:1. Three years ago, the population of Pirajussaraí was equal to the current population of Tucupira. Therefore, ( P = T_{text{three years ago}} ).2. The population of Pirajussaraí has remained unchanged, so ( P_{text{three years ago}} = P ).3. The population of Tucupira has grown by 50%, so ( T = T_{text{three years ago}} times 1.5 ).4. The combined current population of both cities is 9,000, so ( P + T = 9,000 ).Using these relationships, we can express ( T ) in terms of ( P ):[ T = 1.5P ]Substituting this into the combined population equation:[ P + 1.5P = 9,000 ][ 2.5P = 9,000 ][ P = frac{9,000}{2.5} = 3,600 ]Now, calculate ( T ):[ T = 1.5 times 3,600 = 5,400 ]Three years ago, the population of Pirajussaraí was ( P = 3,600 ), and the population of Tucupira was ( frac{T}{1.5} = 3,600 ).Therefore, the combined population three years ago was:[ 3,600 + 3,600 = 7,200 ]

answer：Okay, so I have this problem about a parallelogram ABCD. There are these vectors AA', BB', CC', and DD' that are all parallel and on the same side of the plane of ABCD. The lengths of these vectors are given: AA' is 10, BB' is 8, CC' is 18, and DD' is 22. Then M and N are the midpoints of A'C' and B'D', respectively. I need to find the length of MN, and the options are from 0 to 4. Hmm, okay.First, let me visualize this. ABCD is a parallelogram, so opposite sides are equal and parallel. Then, each vertex is moved along a parallel ray to form A', B', C', D'. Since all the rays are parallel, the movement is in the same direction for each point. So, A moves 10 units in that direction, B moves 8, C moves 18, D moves 22.Since ABCD is a parallelogram, I can assign coordinates to make this easier. Maybe I can place point A at the origin, so A is (0,0,0). Then, since it's a parallelogram, let me denote vector AB as vector b and vector AD as vector d. So, point B would be at vector b, point D at vector d, and point C would be at vector b + d.Now, the vectors AA', BB', CC', DD' are all parallel. Let's assume they are along the z-axis for simplicity, since they are all parallel and on the same side of the plane. So, the movement is only in the z-direction. Therefore, the coordinates of A' would be (0,0,10), B' would be (b_x, b_y, 8), C' would be (b_x + d_x, b_y + d_y, 18), and D' would be (d_x, d_y, 22).Wait, but since ABCD is a parallelogram, vectors AB and AD are in the plane, so their z-components are zero. So, the coordinates of A, B, C, D are all in the z=0 plane. Then, when we move them along the z-axis, their z-coordinates become the given lengths.So, A' is (0,0,10), B' is (b_x, b_y,8), C' is (b_x + d_x, b_y + d_y,18), and D' is (d_x, d_y,22).Now, M is the midpoint of A'C', and N is the midpoint of B'D'. So, let's find the coordinates of M and N.Midpoint formula: the average of the coordinates.Coordinates of A' are (0,0,10), and C' are (b_x + d_x, b_y + d_y,18). So, M is [(0 + b_x + d_x)/2, (0 + b_y + d_y)/2, (10 + 18)/2] = [(b_x + d_x)/2, (b_y + d_y)/2, 14].Similarly, coordinates of B' are (b_x, b_y,8), and D' are (d_x, d_y,22). So, N is [(b_x + d_x)/2, (b_y + d_y)/2, (8 + 22)/2] = [(b_x + d_x)/2, (b_y + d_y)/2, 15].Wait, so M has coordinates [(b_x + d_x)/2, (b_y + d_y)/2, 14], and N has coordinates [(b_x + d_x)/2, (b_y + d_y)/2, 15]. So, the x and y coordinates of M and N are the same, and the z-coordinates differ by 1. So, the distance between M and N is just the difference in the z-coordinates, which is 15 - 14 = 1. So, MN is 1 unit long.Wait, but let me double-check. Is that correct? Because in 3D space, if two points have the same x and y coordinates, then the distance between them is just the difference in z. So, yes, that should be correct.But hold on, let me think again. Is there something I missed? Because the answer is 1, which is one of the options, option B. But sometimes these problems can be tricky.Wait, another way to think about it: Since ABCD is a parallelogram, the midpoints of the diagonals AC and BD coincide. So, the midpoint of AC is the same as the midpoint of BD. Let's call this point O. Then, when we move A, B, C, D to A', B', C', D', the midpoints of A'C' and B'D' would be the points M and N. Since all the movements are along the same direction, the line connecting M and N is just the movement from O to O', where O' is the midpoint of A'C' and B'D', but wait, actually, M and N are midpoints of A'C' and B'D', so their coordinates are averages.But in my earlier calculation, I saw that M and N have the same x and y coordinates, only differing in z by 1. So, MN is 1. Hmm, that seems straightforward.Alternatively, maybe I can think in terms of vectors. Let me denote the translation vectors. Since all the rays are parallel, let's say the direction is along vector **v**. Then, AA' = 10**v**, BB' = 8**v**, CC' = 18**v**, DD' = 22**v**.Then, the coordinates of A' = A + 10**v**, B' = B + 8**v**, C' = C + 18**v**, D' = D + 22**v**.Then, midpoint M of A'C' is (A' + C')/2 = (A + 10**v** + C + 18**v**)/2 = (A + C)/2 + (28**v**)/2 = (A + C)/2 + 14**v**.Similarly, midpoint N of B'D' is (B' + D')/2 = (B + 8**v** + D + 22**v**)/2 = (B + D)/2 + (30**v**)/2 = (B + D)/2 + 15**v**.But in a parallelogram, (A + C)/2 is equal to (B + D)/2, because in a parallelogram, the diagonals bisect each other. So, both M and N have the same position vector except for the translation part. So, M = O + 14**v**, N = O + 15**v**, where O is the midpoint of the original parallelogram.Therefore, the vector MN is N - M = (O + 15**v**) - (O + 14**v**) = **v**. So, the length of MN is the magnitude of **v**. But wait, what is **v**?Wait, actually, in my earlier coordinate system, I assumed **v** was along the z-axis, so **v** would be a unit vector in the z-direction. But in reality, the length of **v** isn't given. Wait, no, actually, the translations are given as AA' = 10, BB' = 8, etc. So, if **v** is a unit vector, then AA' = 10**v**, so the magnitude is 10. But in my coordinate system, I took **v** as the z-axis, so the movement was purely in z. So, in that case, **v** is a unit vector, so the distance between M and N is 1, as I found earlier.Wait, but if **v** isn't a unit vector, then the length of MN would be |**v**|, but since the translations are given as lengths, then **v** must be a unit vector. Because AA' is 10, which is the magnitude of the translation vector. So, yeah, if **v** is a unit vector, then MN is 1.Alternatively, maybe I can think in terms of coordinates without assigning specific vectors. Let me denote the direction of the rays as some vector **k**, which is a unit vector. Then, the coordinates of A' = A + 10**k**, B' = B + 8**k**, C' = C + 18**k**, D' = D + 22**k**.Then, midpoint M = (A' + C')/2 = (A + C)/2 + (10 + 18)/2 **k** = O + 14**k**.Midpoint N = (B' + D')/2 = (B + D)/2 + (8 + 22)/2 **k** = O + 15**k**.So, vector MN = N - M = (O + 15**k**) - (O + 14**k**) = **k**.Therefore, the length of MN is |**k**|, which is 1, since **k** is a unit vector. So, MN = 1.Therefore, the answer is 1, which is option B.Wait, but let me think again. Is there a possibility that **k** is not a unit vector? Because the problem says the rays are parallel, but doesn't specify that they are unit vectors. Hmm.Wait, no, because the lengths AA', BB', etc., are given as 10, 8, 18, 22. So, if the direction vector is **k**, then the translation vectors are 10**k**, 8**k**, etc. So, the magnitude of AA' is |10**k**| = 10|**k**| = 10, so |**k**| must be 1. Therefore, **k** is a unit vector. So, MN is 1.Therefore, the answer is 1.**Final Answer**boxed{B}

answer：Okay, so I have this problem: I need to find the coefficient of ( x^3 ) in the expansion of ( (2x - 1)left( frac{1}{x} + x right)^6 ). Hmm, let's see. I remember that when dealing with expansions like this, especially with binomial expressions, the binomial theorem is really useful. The binomial theorem says that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ). So maybe I can apply that here.First, let me focus on expanding ( left( frac{1}{x} + x right)^6 ). Using the binomial theorem, each term in the expansion will be ( binom{6}{k} left( frac{1}{x} right)^{6 - k} x^k ). Let me write that out:( left( frac{1}{x} + x right)^6 = sum_{k=0}^{6} binom{6}{k} left( frac{1}{x} right)^{6 - k} x^k ).Simplifying each term, ( left( frac{1}{x} right)^{6 - k} x^k = x^{k - (6 - k)} = x^{2k - 6} ). So each term becomes ( binom{6}{k} x^{2k - 6} ).Therefore, the expansion is ( sum_{k=0}^{6} binom{6}{k} x^{2k - 6} ).Now, I need to find the coefficient of ( x^3 ) in the entire expression ( (2x - 1)left( frac{1}{x} + x right)^6 ). So, I can think of this as multiplying ( (2x - 1) ) with each term of the expansion of ( left( frac{1}{x} + x right)^6 ).Let me denote the expansion as ( sum_{k=0}^{6} binom{6}{k} x^{2k - 6} ). So, when I multiply by ( (2x - 1) ), each term becomes ( 2x times binom{6}{k} x^{2k - 6} ) and ( -1 times binom{6}{k} x^{2k - 6} ).So, the entire expression becomes:( sum_{k=0}^{6} binom{6}{k} (2x cdot x^{2k - 6} - 1 cdot x^{2k - 6}) )Simplifying each term inside the sum:First term: ( 2x cdot x^{2k - 6} = 2 x^{2k - 5} )Second term: ( -1 cdot x^{2k - 6} = -x^{2k - 6} )So, combining these, the expression is:( sum_{k=0}^{6} binom{6}{k} (2 x^{2k - 5} - x^{2k - 6}) )Now, I need to find the coefficient of ( x^3 ) in this sum. That means I need to find all the terms in the sum where the exponent of ( x ) is 3.So, let's set up equations for each part:For the first part, ( 2 x^{2k - 5} ), we set ( 2k - 5 = 3 ).Solving for ( k ):( 2k - 5 = 3 )Add 5 to both sides:( 2k = 8 )Divide by 2:( k = 4 )So, when ( k = 4 ), the term ( 2 x^{2k - 5} ) will contribute to ( x^3 ).Similarly, for the second part, ( -x^{2k - 6} ), we set ( 2k - 6 = 3 ).Solving for ( k ):( 2k - 6 = 3 )Add 6 to both sides:( 2k = 9 )Divide by 2:( k = 4.5 )Wait, ( k ) has to be an integer between 0 and 6 because it's a binomial coefficient. So, ( k = 4.5 ) isn't valid. That means there is no term in the second part that contributes to ( x^3 ).Therefore, only the first part, when ( k = 4 ), contributes to the ( x^3 ) term.So, the coefficient from the first part is ( 2 times binom{6}{4} ).Calculating ( binom{6}{4} ):( binom{6}{4} = frac{6!}{4!2!} = frac{720}{24 times 2} = frac{720}{48} = 15 ).So, the coefficient from the first part is ( 2 times 15 = 30 ).Since the second part doesn't contribute anything, the total coefficient of ( x^3 ) is 30.Wait, hold on. Let me double-check. Maybe I missed something.So, in the expansion, each term is multiplied by both ( 2x ) and ( -1 ). So, for each ( k ), we have two terms: one with exponent ( 2k - 5 ) and another with exponent ( 2k - 6 ). So, for each ( k ), we can check if either exponent equals 3.We found that for ( 2k - 5 = 3 ), ( k = 4 ), which is valid.For ( 2k - 6 = 3 ), ( k = 4.5 ), which is invalid because ( k ) must be an integer.Therefore, only ( k = 4 ) contributes, and the coefficient is 30.But wait, let me think again. Maybe I should consider the entire expansion and see if any other terms could contribute when multiplied by ( 2x ) or ( -1 ).Alternatively, perhaps I can approach this problem differently. Instead of expanding the entire expression, maybe I can find the coefficients of specific terms in ( left( frac{1}{x} + x right)^6 ) that, when multiplied by ( 2x ) or ( -1 ), result in an ( x^3 ) term.So, let's denote ( left( frac{1}{x} + x right)^6 = sum_{k=0}^{6} binom{6}{k} x^{2k - 6} ). So, each term is ( binom{6}{k} x^{2k - 6} ).When we multiply by ( 2x ), each term becomes ( 2 binom{6}{k} x^{2k - 5} ).When we multiply by ( -1 ), each term becomes ( -binom{6}{k} x^{2k - 6} ).So, to get the coefficient of ( x^3 ), we need to find all ( k ) such that either ( 2k - 5 = 3 ) or ( 2k - 6 = 3 ).We already saw that ( 2k - 5 = 3 ) gives ( k = 4 ), which is valid.( 2k - 6 = 3 ) gives ( k = 4.5 ), which is not an integer, so no contribution.Therefore, only ( k = 4 ) contributes, and the coefficient is ( 2 times binom{6}{4} = 2 times 15 = 30 ).Wait, but let me make sure that I'm not missing any other terms. For example, maybe ( k ) can be such that ( 2k - 5 = 3 ) or ( 2k - 6 = 3 ). Since ( k ) must be an integer between 0 and 6, let's list all possible exponents:For ( k = 0 ): exponent is ( 2(0) - 6 = -6 )For ( k = 1 ): exponent is ( 2(1) - 6 = -4 )For ( k = 2 ): exponent is ( 2(2) - 6 = -2 )For ( k = 3 ): exponent is ( 2(3) - 6 = 0 )For ( k = 4 ): exponent is ( 2(4) - 6 = 2 )For ( k = 5 ): exponent is ( 2(5) - 6 = 4 )For ( k = 6 ): exponent is ( 2(6) - 6 = 6 )So, in the original expansion, the exponents are -6, -4, -2, 0, 2, 4, 6.When multiplied by ( 2x ), the exponents become:For each term, exponent increases by 1:-6 +1 = -5-4 +1 = -3-2 +1 = -10 +1 = 12 +1 = 34 +1 = 56 +1 = 7Similarly, when multiplied by ( -1 ), the exponents remain the same:-6, -4, -2, 0, 2, 4, 6.So, in the entire expression ( (2x - 1)left( frac{1}{x} + x right)^6 ), the exponents present are:From ( 2x times ) expansion: -5, -3, -1, 1, 3, 5, 7From ( -1 times ) expansion: -6, -4, -2, 0, 2, 4, 6So, combining these, the exponents are from -6 to 7.Therefore, the exponent 3 is present only in the ( 2x times ) expansion, specifically when the exponent was 2 in the original expansion, which corresponds to ( k = 4 ).So, the coefficient is ( 2 times binom{6}{4} = 30 ).Wait, but hold on. Let me think again. When I multiplied by ( 2x ), the exponent 3 comes from the term with exponent 2 in the original expansion. So, the coefficient for exponent 2 in the original expansion is ( binom{6}{4} ), which is 15. Then, multiplying by 2x gives 2 * 15 = 30 for exponent 3.Similarly, if I look at the term from the ( -1 times ) expansion, exponent 3 is not present because the original expansion doesn't have exponent 3. So, only the 2x term contributes.Therefore, the coefficient is 30.But wait, let me just compute the entire expansion to make sure.Alternatively, maybe I can compute ( left( frac{1}{x} + x right)^6 ) and then multiply by ( (2x - 1) ) and see.But that might be time-consuming, but perhaps for verification.First, let's compute ( left( frac{1}{x} + x right)^6 ).Using binomial expansion:( sum_{k=0}^{6} binom{6}{k} left( frac{1}{x} right)^{6 - k} x^k = sum_{k=0}^{6} binom{6}{k} x^{2k - 6} ).So, let's list each term:For k=0: ( binom{6}{0} x^{-6} = 1 times x^{-6} )k=1: ( binom{6}{1} x^{-4} = 6 x^{-4} )k=2: ( binom{6}{2} x^{-2} = 15 x^{-2} )k=3: ( binom{6}{3} x^{0} = 20 times 1 = 20 )k=4: ( binom{6}{4} x^{2} = 15 x^{2} )k=5: ( binom{6}{5} x^{4} = 6 x^{4} )k=6: ( binom{6}{6} x^{6} = 1 x^{6} )So, the expansion is:( x^{-6} + 6x^{-4} + 15x^{-2} + 20 + 15x^{2} + 6x^{4} + x^{6} )Now, multiply this by ( (2x - 1) ):First, multiply each term by ( 2x ):( 2x times x^{-6} = 2x^{-5} )( 2x times 6x^{-4} = 12x^{-3} )( 2x times 15x^{-2} = 30x^{-1} )( 2x times 20 = 40x )( 2x times 15x^{2} = 30x^{3} )( 2x times 6x^{4} = 12x^{5} )( 2x times x^{6} = 2x^{7} )Then, multiply each term by ( -1 ):( -1 times x^{-6} = -x^{-6} )( -1 times 6x^{-4} = -6x^{-4} )( -1 times 15x^{-2} = -15x^{-2} )( -1 times 20 = -20 )( -1 times 15x^{2} = -15x^{2} )( -1 times 6x^{4} = -6x^{4} )( -1 times x^{6} = -x^{6} )Now, combine all these terms:From ( 2x times ) expansion:( 2x^{-5} + 12x^{-3} + 30x^{-1} + 40x + 30x^{3} + 12x^{5} + 2x^{7} )From ( -1 times ) expansion:( -x^{-6} -6x^{-4} -15x^{-2} -20 -15x^{2} -6x^{4} -x^{6} )Now, combine like terms:Let's list all the terms:- ( x^{-6} ): -1- ( x^{-5} ): +2- ( x^{-4} ): -6- ( x^{-3} ): +12- ( x^{-2} ): -15- ( x^{-1} ): +30- ( x^{0} ): -20- ( x^{1} ): +40- ( x^{2} ): -15- ( x^{3} ): +30- ( x^{4} ): -6- ( x^{5} ): +12- ( x^{6} ): -1- ( x^{7} ): +2So, looking at the ( x^3 ) term: it's +30.Therefore, the coefficient of ( x^3 ) is 30.Wait, that's consistent with what I found earlier. So, that confirms it.Alternatively, another way to think about this is to realize that when multiplying ( (2x - 1) ) by the expansion, the ( x^3 ) term can only come from two places:1. Multiplying ( 2x ) by the ( x^2 ) term in the expansion.2. Multiplying ( -1 ) by the ( x^3 ) term in the expansion.But in the expansion of ( left( frac{1}{x} + x right)^6 ), is there an ( x^3 ) term? Let's check.Looking back at the expansion:( x^{-6} + 6x^{-4} + 15x^{-2} + 20 + 15x^{2} + 6x^{4} + x^{6} )No, there is no ( x^3 ) term. So, the second contribution is zero. Therefore, the only contribution is from multiplying ( 2x ) by the ( x^2 ) term, which is ( 15x^2 ). So, ( 2x times 15x^2 = 30x^3 ). Therefore, the coefficient is 30.Yep, that's another way to see it.So, in summary, the coefficient of ( x^3 ) is 30.**Final Answer**The coefficient of ( x^3 ) is boxed{30}.

answer：Okay, so I need to figure out how to maximize the number of primary school students served by arranging the number of students from the east and west schools, given some constraints. Let me try to break this down step by step.First, let's define some variables. Let me call the number of students from the east school "x" and the number from the west school "y". The problem says that at least one student from each school must participate, so x ≥ 1 and y ≥ 1.Next, the problem mentions that the number of students from the west school must be at least one more than those from the east school. So that translates to y ≥ x + 1. I should note that down as a constraint.Now, the bus fare: each east student costs 3 yuan round trip, and each west student costs 5 yuan. The total fare shouldn't exceed 37 yuan. So, the total cost is 3x + 5y ≤ 37. That's another constraint.Our goal is to maximize the number of primary school students served. Each east student can serve 5 primary students, and each west student can serve 3. So, the total number served is 5x + 3y. We need to maximize this.So, summarizing the constraints:1. x ≥ 12. y ≥ 13. y ≥ x + 14. 3x + 5y ≤ 37And the objective is to maximize 5x + 3y.Hmm, this seems like a linear programming problem. I can try to graph the feasible region defined by these constraints and find the vertices, then evaluate the objective function at each vertex to find the maximum.Let me list all the constraints again:1. x ≥ 12. y ≥ 13. y ≥ x + 14. 3x + 5y ≤ 37I can also note that x and y must be integers because you can't have a fraction of a student. So, this is actually an integer linear programming problem.Let me try to visualize the feasible region.First, plot the y ≥ x + 1 line. This is a straight line with a slope of 1 and y-intercept at 1. The region above this line is feasible.Next, the 3x + 5y ≤ 37 line. Let me find its intercepts.If x = 0, then 5y = 37 => y = 7.4.If y = 0, then 3x = 37 => x ≈ 12.33.But since x and y must be at least 1, and y must be at least x + 1, the feasible region is a polygon bounded by these lines and the axes, but only in the first quadrant with x and y ≥ 1.But since y must be at least x + 1, the feasible region is actually a polygon with vertices at the intersection points of these constraints.Let me find the intersection points.First, find where y = x + 1 intersects with 3x + 5y = 37.Substitute y = x + 1 into 3x + 5y = 37:3x + 5(x + 1) = 373x + 5x + 5 = 378x + 5 = 378x = 32x = 4Then, y = 4 + 1 = 5.So, one vertex is at (4,5).Next, find where y = x + 1 intersects with y = 1.Set y = 1 in y = x + 1:1 = x + 1 => x = 0But x must be at least 1, so this point (0,1) is not in the feasible region.Similarly, if we set x =1, then y must be at least 2.So, another vertex is at (1,2). Let me check if this satisfies 3x + 5y ≤ 37.3*1 + 5*2 = 3 + 10 = 13 ≤ 37. Yes, it does.Now, find where 3x + 5y = 37 intersects with y = 1.Set y =1:3x + 5*1 = 37 => 3x = 32 => x ≈ 10.666. But x must be integer, so x=10 would give y=1, but let's check if y=1 satisfies y ≥ x +1? If x=10, y=1, then 1 ≥ 10 +1? No, 1 ≥11 is false. So, this point is not feasible.Similarly, if x=1, y=2 is feasible, but when x increases, y needs to be at least x+1. So, the feasible region is bounded by (1,2), (4,5), and another point where 3x +5y=37 intersects with y as high as possible.Wait, maybe I need to check if 3x +5y=37 intersects with another constraint.Wait, when x is maximum, y is minimum, but y must be at least x +1.Let me see, if I set y as x +1, and then find x such that 3x +5(x+1) ≤37.Which is 8x +5 ≤37 => 8x ≤32 => x ≤4. So, x can be up to 4, which gives y=5.So, the feasible region is a polygon with vertices at (1,2), (4,5), and another point where 3x +5y=37 intersects with y-axis or x-axis, but since y must be at least x +1, which is above y=1, the other intersection points may not be in the feasible region.Wait, let me check if when x=1, y=2 is a vertex, and when x=4, y=5 is another vertex. What about when y is maximum?Wait, if I fix x=1, y can be up to (37 -3*1)/5 = 34/5=6.8, so y=6. But y must be at least x +1=2, so y can be 2,3,4,5,6.But we need to see if y=6 is feasible with x=1.Check 3*1 +5*6=3+30=33 ≤37, yes. So, (1,6) is a feasible point.But wait, is (1,6) on the constraint y ≥x +1? Yes, 6 ≥1 +1=2, which is true.Similarly, if x=2, then y must be at least 3.Compute 3*2 +5*3=6+15=21 ≤37, so (2,3) is feasible.But also, if x=2, y can be higher. Let me see, maximum y when x=2 is (37 -6)/5=31/5=6.2, so y=6.So, (2,6) is feasible.Similarly, for x=3, y must be at least 4.Compute 3*3 +5*4=9+20=29 ≤37, so (3,4) is feasible.And maximum y when x=3 is (37 -9)/5=28/5=5.6, so y=5.So, (3,5) is also feasible.For x=4, y must be at least 5.Compute 3*4 +5*5=12+25=37, which is exactly the budget. So, (4,5) is feasible.For x=5, y must be at least 6.Compute 3*5 +5*6=15+30=45 >37, which is over budget. So, x cannot be 5.So, the feasible integer points are:(1,2), (1,3), (1,4), (1,5), (1,6),(2,3), (2,4), (2,5), (2,6),(3,4), (3,5),(4,5).Wait, that's a lot of points. Maybe I can list all of them and compute 5x +3y for each to find the maximum.Alternatively, since the objective function is 5x +3y, which is linear, the maximum will occur at one of the vertices of the feasible region.But since we're dealing with integers, the vertices might not all be integers, but the maximum will be at one of the integer points near the vertices.But perhaps it's easier to list all feasible integer points and compute the total served.Let me try that.First, starting with x=1:x=1:y must be ≥2, and 3*1 +5y ≤37 => 5y ≤34 => y ≤6.8, so y=2,3,4,5,6.Compute total served:(1,2): 5*1 +3*2=5+6=11(1,3):5+9=14(1,4):5+12=17(1,5):5+15=20(1,6):5+18=23x=2:y must be ≥3, and 3*2 +5y ≤37 =>5y ≤31 => y ≤6.2, so y=3,4,5,6.Compute:(2,3):10+9=19(2,4):10+12=22(2,5):10+15=25(2,6):10+18=28x=3:y must be ≥4, and 3*3 +5y ≤37 =>5y ≤28 => y ≤5.6, so y=4,5.Compute:(3,4):15+12=27(3,5):15+15=30x=4:y must be ≥5, and 3*4 +5y ≤37 =>5y ≤25 => y ≤5.So y=5.Compute:(4,5):20+15=35x=5:Not feasible as above.So, compiling all the totals:(1,2):11(1,3):14(1,4):17(1,5):20(1,6):23(2,3):19(2,4):22(2,5):25(2,6):28(3,4):27(3,5):30(4,5):35Looking at these totals, the maximum is 35 at (4,5).Wait, but let me double-check if there are any other points I might have missed.For example, when x=3, y=5 gives 30, and x=4, y=5 gives 35.Is there a point beyond x=4? x=5 is not feasible.What about when x=2, y=6 gives 28, which is less than 35.Similarly, x=3, y=5 is 30, which is less than 35.So, the maximum seems to be at (4,5) with 35 primary students served.But wait, let me confirm if (4,5) satisfies all constraints.x=4, y=5.Check y ≥x +1: 5 ≥4 +1=5, which is true.Total fare:3*4 +5*5=12+25=37, which is exactly the budget.So, yes, it's feasible.Therefore, the optimal arrangement is 4 students from the east school and 5 from the west school, serving a total of 35 primary school students.**Final Answer**The maximum number of primary school students that can be served is boxed{35}.

answer：Okay, so I have this problem about a sector of a circle. The perimeter is 8 cm, and the area is 4 cm². I need to find the central angle in radians. Hmm, let's see. I remember that a sector has two radii and an arc, so the perimeter would be the sum of the two radii and the arc length. The area of a sector is given by (1/2)*r²*θ, where θ is the central angle in radians. Let me write down the formulas:Perimeter (P) of the sector = 2r + arc length (L)Area (A) of the sector = (1/2)*r²*θI know that the arc length L is equal to rθ, right? So substituting that into the perimeter equation, I get:P = 2r + rθGiven that the perimeter is 8 cm, so:2r + rθ = 8 ...(1)And the area is 4 cm², so:(1/2)*r²*θ = 4 ...(2)Okay, so I have two equations here with two variables, r and θ. I need to solve for θ. Maybe I can express θ from equation (1) and substitute into equation (2). Let's try that.From equation (1):2r + rθ = 8Let's factor out r:r(2 + θ) = 8So,r = 8 / (2 + θ) ...(3)Now, plug this expression for r into equation (2):(1/2)*(8 / (2 + θ))² * θ = 4Let me simplify this step by step.First, square the 8/(2 + θ):(8)^2 = 64(2 + θ)^2 = (2 + θ)^2So,(1/2)*(64 / (2 + θ)^2) * θ = 4Simplify (1/2)*64:(32 / (2 + θ)^2) * θ = 4So,32θ / (2 + θ)^2 = 4Let me write that as:32θ = 4*(2 + θ)^2Divide both sides by 4:8θ = (2 + θ)^2Now, expand the right side:(2 + θ)^2 = 4 + 4θ + θ²So,8θ = 4 + 4θ + θ²Bring all terms to one side:θ² + 4θ + 4 - 8θ = 0Simplify:θ² - 4θ + 4 = 0Hmm, that's a quadratic equation in θ. Let me write it as:θ² - 4θ + 4 = 0I can try to factor this. Let's see, does it factor into (θ - a)(θ - b) = 0?Looking for two numbers that multiply to 4 and add to -4. Hmm, -2 and -2. So,(θ - 2)(θ - 2) = 0Which is (θ - 2)^2 = 0So, θ = 2Wait, so θ is 2 radians? Let me check if that makes sense.Let me substitute θ = 2 back into equation (3) to find r.r = 8 / (2 + 2) = 8 / 4 = 2 cmSo, radius is 2 cm. Let's check the perimeter:2r + rθ = 2*2 + 2*2 = 4 + 4 = 8 cm. That's correct.Now, check the area:(1/2)*r²*θ = (1/2)*(2)^2*2 = (1/2)*4*2 = (1/2)*8 = 4 cm². Perfect, that's correct.Wait, so θ is 2 radians. That seems straightforward. But let me make sure I didn't make any mistakes in my algebra.Starting from 32θ / (2 + θ)^2 = 4, then 32θ = 4*(2 + θ)^2, which simplifies to 8θ = (2 + θ)^2. Expanding gives 8θ = 4 + 4θ + θ², so θ² - 4θ + 4 = 0, which factors to (θ - 2)^2 = 0, so θ = 2. Yep, that seems solid.I don't think I made any calculation errors. So, the central angle is 2 radians.**Final Answer**The central angle of the sector is boxed{2} radians.

answer：Okay, so I have this problem about an arithmetic progression. Let me try to understand what it's asking. An arithmetic progression, or arithmetic sequence, is a sequence of numbers where each term after the first is obtained by adding a constant difference. So, if I denote the first term as ( a ) and the common difference as ( d ), then the terms go like ( a, a + d, a + 2d, a + 3d, ) and so on.The problem says two things: first, the sum of the first and fifth terms is ( frac{5}{3} ). Second, the product of the third and fourth terms is ( frac{65}{72} ). I need to find the sum of the first 17 terms of this progression.Let me write down what each term is in terms of ( a ) and ( d ). The first term is ( a ), the second is ( a + d ), the third is ( a + 2d ), the fourth is ( a + 3d ), and the fifth is ( a + 4d ). So, the sum of the first and fifth terms is ( a + (a + 4d) = 2a + 4d ). According to the problem, this equals ( frac{5}{3} ). So, I can write the equation:( 2a + 4d = frac{5}{3} ) ...(1)Next, the product of the third and fourth terms is ( (a + 2d)(a + 3d) = frac{65}{72} ). So, that gives me another equation:( (a + 2d)(a + 3d) = frac{65}{72} ) ...(2)Now, I have two equations with two variables, ( a ) and ( d ). I need to solve these equations to find ( a ) and ( d ), and then use them to find the sum of the first 17 terms.Let me see. Equation (1) can be simplified. If I factor out a 2, I get:( 2(a + 2d) = frac{5}{3} )Divide both sides by 2:( a + 2d = frac{5}{6} ) ...(1a)So, ( a + 2d = frac{5}{6} ). Let me denote this as equation (1a).Looking at equation (2), it's ( (a + 2d)(a + 3d) = frac{65}{72} ). But from equation (1a), I know that ( a + 2d = frac{5}{6} ). So, I can substitute this into equation (2):( left( frac{5}{6} right)(a + 3d) = frac{65}{72} )Let me solve for ( a + 3d ). Multiply both sides by ( frac{6}{5} ):( a + 3d = frac{65}{72} times frac{6}{5} )Calculating the right-hand side:( frac{65}{72} times frac{6}{5} = frac{65 times 6}{72 times 5} )Simplify numerator and denominator:65 divided by 5 is 13, and 6 divided by 72 is ( frac{1}{12} ). So,( frac{13 times 1}{12} = frac{13}{12} )Therefore, ( a + 3d = frac{13}{12} ) ...(2a)Now, I have two equations:1. ( a + 2d = frac{5}{6} ) (from 1a)2. ( a + 3d = frac{13}{12} ) (from 2a)I can subtract the first equation from the second to find ( d ):( (a + 3d) - (a + 2d) = frac{13}{12} - frac{5}{6} )Simplify left side:( a + 3d - a - 2d = d )Right side:( frac{13}{12} - frac{5}{6} ). Let me convert ( frac{5}{6} ) to twelfths: ( frac{5}{6} = frac{10}{12} ). So,( frac{13}{12} - frac{10}{12} = frac{3}{12} = frac{1}{4} )Therefore, ( d = frac{1}{4} )Now that I have ( d ), I can substitute back into equation (1a) to find ( a ):( a + 2d = frac{5}{6} )Substitute ( d = frac{1}{4} ):( a + 2 times frac{1}{4} = frac{5}{6} )Simplify:( a + frac{1}{2} = frac{5}{6} )Subtract ( frac{1}{2} ) from both sides:( a = frac{5}{6} - frac{1}{2} )Convert ( frac{1}{2} ) to sixths: ( frac{3}{6} )So,( a = frac{5}{6} - frac{3}{6} = frac{2}{6} = frac{1}{3} )Therefore, ( a = frac{1}{3} ) and ( d = frac{1}{4} )Now, I need to find the sum of the first 17 terms of this arithmetic progression.The formula for the sum of the first ( n ) terms of an arithmetic progression is:( S_n = frac{n}{2} times [2a + (n - 1)d] )So, for ( n = 17 ):( S_{17} = frac{17}{2} times [2a + 16d] )Let me compute ( 2a + 16d ):( 2a = 2 times frac{1}{3} = frac{2}{3} )( 16d = 16 times frac{1}{4} = 4 )So, ( 2a + 16d = frac{2}{3} + 4 )Convert 4 to thirds: ( 4 = frac{12}{3} ), so:( frac{2}{3} + frac{12}{3} = frac{14}{3} )Therefore, ( S_{17} = frac{17}{2} times frac{14}{3} )Multiply the fractions:( frac{17 times 14}{2 times 3} = frac{238}{6} )Simplify ( frac{238}{6} ). Let's divide numerator and denominator by 2:( frac{119}{3} )So, ( S_{17} = frac{119}{3} )Let me just double-check my calculations to make sure I didn't make any mistakes.First, ( a = frac{1}{3} ), ( d = frac{1}{4} ). So, the first term is ( frac{1}{3} ), second is ( frac{1}{3} + frac{1}{4} = frac{4}{12} + frac{3}{12} = frac{7}{12} ), third is ( frac{7}{12} + frac{1}{4} = frac{7}{12} + frac{3}{12} = frac{10}{12} = frac{5}{6} ), fourth is ( frac{5}{6} + frac{1}{4} = frac{10}{12} + frac{3}{12} = frac{13}{12} ), fifth is ( frac{13}{12} + frac{1}{4} = frac{13}{12} + frac{3}{12} = frac{16}{12} = frac{4}{3} ).Sum of first and fifth terms: ( frac{1}{3} + frac{4}{3} = frac{5}{3} ). That's correct.Product of third and fourth terms: ( frac{5}{6} times frac{13}{12} = frac{65}{72} ). That's also correct.So, the values of ( a ) and ( d ) are correct.Now, computing the sum of the first 17 terms:( S_{17} = frac{17}{2} times [2a + 16d] )We found ( 2a + 16d = frac{14}{3} ), so:( S_{17} = frac{17}{2} times frac{14}{3} = frac{17 times 14}{6} = frac{238}{6} = frac{119}{3} ). That's approximately 39.666..., but as a fraction, it's ( frac{119}{3} ).So, I think that's the correct answer.**Final Answer**The sum of the first 17 terms is boxed{dfrac{119}{3}}.