answer：First, I need to recall the formula for the nth term of an arithmetic sequence, which is (a_n = a + (n-1)d), where (a) is the first term and (d) is the common difference.Given that the first term (a = frac{2}{3}) and the seventeenth term (a_{17} = frac{3}{2}), I can set up the equation for the seventeenth term:[a_{17} = a + 16d]Substituting the known values:[frac{3}{2} = frac{2}{3} + 16d]To find the common difference (d), I'll subtract (frac{2}{3}) from both sides:[frac{3}{2} - frac{2}{3} = 16d]Finding a common denominator to subtract the fractions:[frac{9}{6} - frac{4}{6} = 16d][frac{5}{6} = 16d]Now, solve for (d):[d = frac{5}{6 times 16} = frac{5}{96}]With the common difference determined, I can now find the ninth term (a_9) using the nth term formula:[a_9 = a + 8d]Substituting the known values:[a_9 = frac{2}{3} + 8 times frac{5}{96}]Simplify the multiplication:[a_9 = frac{2}{3} + frac{40}{96}]Convert (frac{2}{3}) to a fraction with a denominator of 96:[frac{2}{3} = frac{64}{96}]Now, add the two fractions:[a_9 = frac{64}{96} + frac{40}{96} = frac{104}{96}]Simplify the fraction by dividing both numerator and denominator by 8:[a_9 = frac{13}{12}]

answer：Okay, so I have this problem about a regular hexagon with area 1. The hexagon is labeled ABCDEF, and M is the midpoint of DE. Then there are points X, Y, Z defined as intersections of certain lines: X is where AC and BM intersect, Y is where BF and AM intersect, and Z is where AC and BF intersect. I need to find the value of [BXC] + [AYF] + [ABZ] - [MXZY], where [P] denotes the area of polygon P.First, I should probably draw a diagram to visualize everything. Since it's a regular hexagon, all sides are equal, and all internal angles are 120 degrees. Let me recall that the area of a regular hexagon can be divided into six equilateral triangles, each with area 1/6. So, each of those small triangles has an area of 1/6.Now, let me label the hexagon: A, B, C, D, E, F in order. So, A is connected to B, B to C, and so on until F connects back to A. The diagonals AC and BF are mentioned, so I should note those. Also, M is the midpoint of DE, so that's halfway between D and E.Next, I need to find the points X, Y, Z. X is the intersection of AC and BM. So, BM is the line from B to M, which is the midpoint of DE. Similarly, Y is the intersection of BF and AM, so AM is the line from A to M. Z is the intersection of AC and BF.I think using coordinate geometry might help here. Let me assign coordinates to the hexagon. Let's place the hexagon with its center at the origin (0,0) and one vertex at (1,0). Since it's regular, the coordinates of the vertices can be determined using trigonometric functions.The coordinates for a regular hexagon with side length 's' can be given by:- A: (s, 0)- B: (s/2, (s√3)/2)- C: (-s/2, (s√3)/2)- D: (-s, 0)- E: (-s/2, -(s√3)/2)- F: (s/2, -(s√3)/2)But since the area is 1, I need to find the appropriate side length. The area of a regular hexagon is given by (3√3/2) * s². So, setting that equal to 1:(3√3/2) * s² = 1s² = 2/(3√3)s = √(2/(3√3)) = (√6)/(3√3) = √(2)/√(3√3) Hmm, maybe it's better to keep it symbolic for now.Alternatively, maybe I can use a unit circle approach, setting the distance from the center to each vertex as 1. Then the area would be (3√3)/2, but since we need area 1, we can scale accordingly. Let me see.If the distance from the center to each vertex is 'r', then the area is (3√3/2) * r². So, setting that equal to 1:(3√3/2) * r² = 1r² = 2/(3√3)r = √(2/(3√3)) = same as before.Alternatively, maybe it's simpler to assign coordinates with the center at (0,0) and one vertex at (1,0), then scale the entire figure so that the area becomes 1. That might complicate the coordinates, but perhaps manageable.Alternatively, maybe using vectors or barycentric coordinates could help, but I think coordinate geometry is more straightforward for me.Let me proceed step by step.1. Assign coordinates to the hexagon with center at (0,0) and vertex A at (1,0). Then, the coordinates of the vertices can be determined using angles.The angle between each vertex is 60 degrees. So, starting from A at (1,0), the next vertex B is at (cos(60°), sin(60°)) = (0.5, √3/2). Then C is at (-0.5, √3/2), D is at (-1,0), E is at (-0.5, -√3/2), F is at (0.5, -√3/2).But wait, the area of this hexagon is (3√3)/2, which is approximately 2.598. But we need the area to be 1. So, we need to scale down the coordinates.Let me denote the scaling factor as k. So, if the original area is (3√3)/2, then scaling by k will make the area (3√3)/2 * k² = 1. Therefore, k² = 2/(3√3), so k = √(2/(3√3)).Simplify √(2/(3√3)):Multiply numerator and denominator by √3:√(2√3 / (3*3)) = √(2√3 / 9) = (√(2√3))/3.But this might complicate the coordinates. Alternatively, maybe I can just keep the coordinates as is and adjust the areas proportionally.Alternatively, maybe I can work with a unit hexagon and then scale the areas accordingly. Since the total area is 1, each small equilateral triangle has area 1/6.Wait, in a regular hexagon, it's composed of six equilateral triangles, each with area 1/6. So, each triangle has area 1/6.So, if I can express the areas [BXC], [AYF], [ABZ], and [MXZY] in terms of these small triangles, I can sum them up.But I'm not sure if that's straightforward. Maybe I need to find the coordinates of points X, Y, Z.Let me try to find the coordinates of X, Y, Z.First, let's assign coordinates to the hexagon with center at (0,0) and vertex A at (1,0). Then, as above:A: (1, 0)B: (0.5, √3/2)C: (-0.5, √3/2)D: (-1, 0)E: (-0.5, -√3/2)F: (0.5, -√3/2)But since the area is 1, we need to scale these coordinates. Let me compute the scaling factor.The area of the regular hexagon with side length s is (3√3/2) s². So, if we set (3√3/2) s² = 1, then s² = 2/(3√3), so s = √(2/(3√3)).Therefore, the coordinates need to be scaled by s. So, the coordinates become:A: (s, 0)B: (s/2, (s√3)/2)C: (-s/2, (s√3)/2)D: (-s, 0)E: (-s/2, -(s√3)/2)F: (s/2, -(s√3)/2)But this might complicate the calculations. Alternatively, maybe I can work with the unit hexagon and then scale the areas accordingly.Alternatively, perhaps it's better to use barycentric coordinates or vector methods.Alternatively, maybe using coordinate geometry without scaling, and then scaling the areas at the end.Let me proceed with the unit hexagon (side length 1, area (3√3)/2), compute the areas, and then scale them to make the total area 1.So, let's proceed with the unit hexagon.First, find the coordinates:A: (1, 0)B: (0.5, √3/2)C: (-0.5, √3/2)D: (-1, 0)E: (-0.5, -√3/2)F: (0.5, -√3/2)M is the midpoint of DE. So, D is (-1, 0) and E is (-0.5, -√3/2). So, midpoint M is average of coordinates:M_x = (-1 + (-0.5))/2 = (-1.5)/2 = -0.75M_y = (0 + (-√3/2))/2 = (-√3/2)/2 = -√3/4So, M is at (-0.75, -√3/4)Now, find X, Y, Z.X is the intersection of AC and BM.First, find equations of lines AC and BM.Line AC connects A(1,0) and C(-0.5, √3/2).Compute the slope of AC:m_AC = (√3/2 - 0)/(-0.5 - 1) = (√3/2)/(-1.5) = -√3/3Equation of AC: y - 0 = m_AC (x - 1)So, y = (-√3/3)(x - 1)Similarly, line BM connects B(0.5, √3/2) and M(-0.75, -√3/4).Compute the slope of BM:m_BM = (-√3/4 - √3/2)/( -0.75 - 0.5) = (-3√3/4)/(-1.25) = (-3√3/4)/(-5/4) = (3√3)/5Equation of BM: y - √3/2 = (3√3/5)(x - 0.5)Now, find intersection X of AC and BM.Set the two equations equal:(-√3/3)(x - 1) = (3√3/5)(x - 0.5) + √3/2Let me solve for x.Multiply both sides by 15 to eliminate denominators:-5√3 (x - 1) = 9√3 (x - 0.5) + (15√3)/2Divide both sides by √3:-5(x - 1) = 9(x - 0.5) + 15/2Expand:-5x + 5 = 9x - 4.5 + 7.5Simplify right side:9x - 4.5 + 7.5 = 9x + 3So:-5x + 5 = 9x + 3Bring variables to left and constants to right:-5x - 9x = 3 - 5-14x = -2x = (-2)/(-14) = 1/7Now, plug x = 1/7 into equation of AC:y = (-√3/3)(1/7 - 1) = (-√3/3)(-6/7) = (6√3)/21 = (2√3)/7So, X is at (1/7, 2√3/7)Next, find Y, which is the intersection of BF and AM.First, find equations of BF and AM.Line BF connects B(0.5, √3/2) and F(0.5, -√3/2). Wait, both points have x-coordinate 0.5, so BF is a vertical line at x = 0.5.Line AM connects A(1,0) and M(-0.75, -√3/4).Compute the slope of AM:m_AM = (-√3/4 - 0)/(-0.75 - 1) = (-√3/4)/(-1.75) = (√3/4)/(7/4) = √3/7Equation of AM: y - 0 = (√3/7)(x - 1)So, y = (√3/7)(x - 1)Intersection Y is where x = 0.5 (from BF) and y = (√3/7)(0.5 - 1) = (√3/7)(-0.5) = -√3/14So, Y is at (0.5, -√3/14)Now, find Z, which is the intersection of AC and BF.But BF is x = 0.5, so plug x = 0.5 into equation of AC:y = (-√3/3)(0.5 - 1) = (-√3/3)(-0.5) = (√3)/6So, Z is at (0.5, √3/6)Now, we have coordinates for X, Y, Z:X: (1/7, 2√3/7)Y: (0.5, -√3/14)Z: (0.5, √3/6)Now, we need to find the areas [BXC], [AYF], [ABZ], and [MXZY].Let's compute each area one by one.1. [BXC]: Triangle with vertices B(0.5, √3/2), X(1/7, 2√3/7), C(-0.5, √3/2)We can use the shoelace formula for the area of a triangle given coordinates.Formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in:x1 = 0.5, y1 = √3/2x2 = 1/7, y2 = 2√3/7x3 = -0.5, y3 = √3/2Compute:0.5*(2√3/7 - √3/2) + (1/7)*(√3/2 - √3/2) + (-0.5)*(√3/2 - 2√3/7)Simplify term by term:First term: 0.5*(2√3/7 - √3/2) = 0.5*( (4√3 - 7√3)/14 ) = 0.5*(-3√3/14) = -3√3/28Second term: (1/7)*(0) = 0Third term: -0.5*(√3/2 - 2√3/7) = -0.5*( (7√3 - 4√3)/14 ) = -0.5*(3√3/14) = -3√3/28Sum: -3√3/28 + 0 - 3√3/28 = -6√3/28 = -3√3/14Take absolute value and divide by 2:Wait, no, the formula already includes division by 2. So, the area is |sum| / 2, but in this case, the sum is -3√3/14, so area is | -3√3/14 | = 3√3/14Wait, no, let me double-check.Wait, the formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|So, plugging in:= |0.5*(2√3/7 - √3/2) + (1/7)*(√3/2 - √3/2) + (-0.5)*(√3/2 - 2√3/7)| / 2Wait, no, the formula is:= |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| / 2So, compute:0.5*(2√3/7 - √3/2) + (1/7)*(√3/2 - √3/2) + (-0.5)*(√3/2 - 2√3/7)Compute each term:First term: 0.5*(2√3/7 - √3/2) = 0.5*( (4√3 - 7√3)/14 ) = 0.5*(-3√3/14) = -3√3/28Second term: (1/7)*(0) = 0Third term: -0.5*(√3/2 - 2√3/7) = -0.5*( (7√3 - 4√3)/14 ) = -0.5*(3√3/14) = -3√3/28Sum: -3√3/28 + 0 - 3√3/28 = -6√3/28 = -3√3/14Take absolute value: 3√3/14Divide by 2? Wait, no, the formula already includes division by 2. Wait, no, the formula is:Area = |sum| / 2Wait, no, the formula is:Area = |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| / 2So, in this case, the sum is -3√3/14, so area is | -3√3/14 | / 2 = (3√3/14)/2 = 3√3/28Wait, that contradicts my earlier thought. Let me recast.Wait, no, actually, the formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))| / 2So, the sum inside is -3√3/14, so area is | -3√3/14 | / 2 = (3√3/14)/2 = 3√3/28Wait, but that seems too small. Let me check with another method.Alternatively, since points B, X, C are all in the upper half of the hexagon, and X is closer to A, the area should be a small triangle.Alternatively, maybe using vectors.But perhaps I made a mistake in the calculation.Let me recompute:Compute x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)x1 = 0.5, y1 = √3/2x2 = 1/7, y2 = 2√3/7x3 = -0.5, y3 = √3/2Compute:0.5*(2√3/7 - √3/2) + (1/7)*(√3/2 - √3/2) + (-0.5)*(√3/2 - 2√3/7)First term: 0.5*(2√3/7 - √3/2) = 0.5*( (4√3 - 7√3)/14 ) = 0.5*(-3√3/14) = -3√3/28Second term: (1/7)*(0) = 0Third term: -0.5*(√3/2 - 2√3/7) = -0.5*( (7√3 - 4√3)/14 ) = -0.5*(3√3/14) = -3√3/28Total sum: -3√3/28 + 0 -3√3/28 = -6√3/28 = -3√3/14So, area is | -3√3/14 | / 2 = (3√3/14)/2 = 3√3/28So, [BXC] = 3√3/282. [AYF]: Triangle with vertices A(1,0), Y(0.5, -√3/14), F(0.5, -√3/2)Again, using shoelace formula.Compute:x1 = 1, y1 = 0x2 = 0.5, y2 = -√3/14x3 = 0.5, y3 = -√3/2Compute:1*(-√3/14 - (-√3/2)) + 0.5*(-√3/2 - 0) + 0.5*(0 - (-√3/14))Simplify term by term:First term: 1*(-√3/14 + √3/2) = 1*( ( -√3 + 7√3 ) /14 ) = 1*(6√3/14) = 6√3/14Second term: 0.5*(-√3/2) = -√3/4Third term: 0.5*(√3/14) = √3/28Sum: 6√3/14 - √3/4 + √3/28Convert to common denominator, which is 28:6√3/14 = 12√3/28-√3/4 = -7√3/28√3/28 = √3/28Total sum: 12√3/28 -7√3/28 + √3/28 = (12 -7 +1)√3/28 = 6√3/28So, area is |6√3/28| / 2 = (6√3/28)/2 = 6√3/56 = 3√3/28Wait, no, the formula is:Area = |sum| / 2But in this case, the sum is 6√3/28, so area is |6√3/28| / 2 = 6√3/56 = 3√3/28So, [AYF] = 3√3/283. [ABZ]: Triangle with vertices A(1,0), B(0.5, √3/2), Z(0.5, √3/6)Using shoelace formula.x1 = 1, y1 = 0x2 = 0.5, y2 = √3/2x3 = 0.5, y3 = √3/6Compute:1*(√3/2 - √3/6) + 0.5*(√3/6 - 0) + 0.5*(0 - √3/2)Simplify term by term:First term: 1*( (3√3 - √3)/6 ) = 1*(2√3/6) = √3/3Second term: 0.5*(√3/6) = √3/12Third term: 0.5*(-√3/2) = -√3/4Sum: √3/3 + √3/12 - √3/4Convert to common denominator 12:√3/3 = 4√3/12√3/12 = √3/12-√3/4 = -3√3/12Total sum: 4√3/12 + √3/12 -3√3/12 = (4 +1 -3)√3/12 = 2√3/12 = √3/6So, area is |√3/6| / 2 = √3/12Wait, no, the formula is:Area = |sum| / 2But in this case, the sum is √3/6, so area is |√3/6| / 2 = √3/12Wait, but that seems small. Let me check.Alternatively, since ABZ is a triangle with base AB and height from Z.But AB is a side of the hexagon, length 1. The height from Z is the vertical distance from Z to AB.But AB is from A(1,0) to B(0.5, √3/2). The equation of AB is y = √3(x -1)Wait, no, let me compute the equation of AB.Points A(1,0) and B(0.5, √3/2). Slope m = (√3/2 - 0)/(0.5 -1) = (√3/2)/(-0.5) = -√3Equation: y -0 = -√3(x -1), so y = -√3 x + √3Point Z is at (0.5, √3/6). The distance from Z to AB is | -√3*(0.5) + √3 - √3/6 | / sqrt( (√3)^2 +1 )Wait, no, the distance from a point (x0,y0) to line ax + by + c =0 is |ax0 + by0 + c| / sqrt(a² + b²)First, write AB in ax + by + c =0 form:y = -√3 x + √3 => √3 x + y - √3 =0So, a=√3, b=1, c=-√3Point Z: (0.5, √3/6)Distance = |√3*0.5 + 1*(√3/6) - √3| / sqrt( (√3)^2 +1^2 )Simplify numerator:√3/2 + √3/6 - √3 = (3√3/6 + √3/6 -6√3/6) = (-2√3/6) = -√3/3Absolute value: √3/3Denominator: sqrt(3 +1 )=2So, distance = (√3/3)/2 = √3/6So, area of triangle ABZ is (base * height)/2 = (1 * √3/6)/2 = √3/12Which matches the shoelace result.So, [ABZ] = √3/124. [MXZY]: Quadrilateral with vertices M(-0.75, -√3/4), X(1/7, 2√3/7), Z(0.5, √3/6), Y(0.5, -√3/14)This is a quadrilateral, so we can divide it into two triangles or use the shoelace formula for quadrilaterals.Let me use the shoelace formula.Order of points: M, X, Z, YCoordinates:M: (-0.75, -√3/4)X: (1/7, 2√3/7)Z: (0.5, √3/6)Y: (0.5, -√3/14)Back to M: (-0.75, -√3/4)Compute shoelace sum:Sum1 = (-0.75)*(2√3/7) + (1/7)*(√3/6) + 0.5*(-√3/14) + 0.5*(-√3/4)Sum2 = (-√3/4)*(1/7) + (2√3/7)*(0.5) + (√3/6)*(0.5) + (-√3/14)*(-0.75)Compute Sum1:First term: (-0.75)*(2√3/7) = (-3/4)*(2√3/7) = (-6√3)/28 = (-3√3)/14Second term: (1/7)*(√3/6) = √3/42Third term: 0.5*(-√3/14) = (-√3)/28Fourth term: 0.5*(-√3/4) = (-√3)/8Sum1: (-3√3)/14 + √3/42 - √3/28 - √3/8Convert to common denominator, which is 168:-3√3/14 = -36√3/168√3/42 = 4√3/168-√3/28 = -6√3/168-√3/8 = -21√3/168Total Sum1: (-36 +4 -6 -21)√3/168 = (-59√3)/168Compute Sum2:First term: (-√3/4)*(1/7) = (-√3)/28Second term: (2√3/7)*(0.5) = (√3)/7Third term: (√3/6)*(0.5) = √3/12Fourth term: (-√3/14)*(-0.75) = (√3/14)*(3/4) = 3√3/56Sum2: (-√3)/28 + √3/7 + √3/12 + 3√3/56Convert to common denominator 168:-√3/28 = -6√3/168√3/7 = 24√3/168√3/12 =14√3/1683√3/56 =9√3/168Total Sum2: (-6 +24 +14 +9)√3/168 = (41√3)/168Now, shoelace area is |Sum1 - Sum2| / 2Compute Sum1 - Sum2:(-59√3/168) - (41√3/168) = (-100√3)/168Take absolute value: 100√3/168Divide by 2: 50√3/168 = 25√3/84So, [MXZY] =25√3/84Now, we have all areas:[BXC] =3√3/28[AYF] =3√3/28[ABZ] =√3/12[MXZY] =25√3/84Now, compute the expression:[BXC] + [AYF] + [ABZ] - [MXZY] = (3√3/28 +3√3/28 +√3/12) -25√3/84First, sum the first three terms:3√3/28 +3√3/28 =6√3/28 =3√3/14Add √3/12:3√3/14 +√3/12Convert to common denominator 84:3√3/14 =18√3/84√3/12 =7√3/84Total:18√3/84 +7√3/84=25√3/84Now, subtract [MXZY] =25√3/84:25√3/84 -25√3/84=0Wait, that can't be right. The expression evaluates to zero?But the problem states to evaluate [BXC] + [AYF] + [ABZ] - [MXZY], which according to my calculations is 0.But the total area of the hexagon is 1, so getting zero seems odd.Wait, let me check my calculations again.First, [BXC] =3√3/28[AYF] =3√3/28[ABZ] =√3/12Sum:3√3/28 +3√3/28 +√3/12 =6√3/28 +√3/12 =3√3/14 +√3/12Convert to common denominator 84:3√3/14 =18√3/84√3/12=7√3/84Total:25√3/84[MXZY] =25√3/84So, 25√3/84 -25√3/84=0Hmm, that suggests the expression evaluates to zero.But let me think geometrically. The areas [BXC], [AYF], [ABZ] are parts of the hexagon, and [MXZY] is another area. It might be that these areas overlap or cancel out in such a way that their combination results in zero.Alternatively, perhaps I made a mistake in the shoelace calculations.Let me recheck [MXZY].Quadrilateral MXZY with points M(-0.75, -√3/4), X(1/7, 2√3/7), Z(0.5, √3/6), Y(0.5, -√3/14)Using shoelace formula:List the coordinates in order:M(-0.75, -√3/4)X(1/7, 2√3/7)Z(0.5, √3/6)Y(0.5, -√3/14)M(-0.75, -√3/4)Compute Sum1 = x1y2 +x2y3 +x3y4 +x4y1Wait, no, shoelace formula is:Sum1 = x1y2 +x2y3 +x3y4 +x4y1Sum2 = y1x2 + y2x3 + y3x4 + y4x1Area = |Sum1 - Sum2| /2So, let's compute Sum1:x1y2 = (-0.75)*(2√3/7) = (-3/4)*(2√3/7) = (-6√3)/28 = (-3√3)/14x2y3 = (1/7)*(√3/6) = √3/42x3y4 =0.5*(-√3/14) = -√3/28x4y1 =0.5*(-√3/4) = -√3/8Sum1: (-3√3)/14 + √3/42 -√3/28 -√3/8Convert to denominator 168:-3√3/14 = -36√3/168√3/42 =4√3/168-√3/28= -6√3/168-√3/8= -21√3/168Total Sum1: (-36 +4 -6 -21)√3/168 = (-59√3)/168Sum2:y1x2 = (-√3/4)*(1/7) = -√3/28y2x3 = (2√3/7)*(0.5) = √3/7y3x4 = (√3/6)*(0.5) = √3/12y4x1 = (-√3/14)*(-0.75) = (3√3)/56Sum2: -√3/28 +√3/7 +√3/12 +3√3/56Convert to denominator 168:-√3/28 = -6√3/168√3/7=24√3/168√3/12=14√3/1683√3/56=9√3/168Total Sum2: (-6 +24 +14 +9)√3/168=41√3/168So, Sum1 - Sum2 = (-59√3/168) - (41√3/168)= -100√3/168Absolute value:100√3/168Divide by 2:50√3/168=25√3/84So, [MXZY]=25√3/84Thus, the calculation seems correct.Therefore, the expression [BXC] + [AYF] + [ABZ] - [MXZY] =25√3/84 -25√3/84=0But the total area of the hexagon is 1, so getting zero seems counterintuitive. Maybe the areas overlap in such a way that their combination cancels out.Alternatively, perhaps I made a mistake in interpreting the problem. Let me re-read it."Let M be the midpoint of DE. Let X be the intersection of AC and BM, let Y be the intersection of BF and AM, and let Z be the intersection of AC and BF. If [P] denotes the area of polygon P for any polygon P in the plane, evaluate [BXC] + [AYF] + [ABZ] - [MXZY]."Wait, perhaps the areas are not all positive? Or maybe the orientation affects the sign.But in the shoelace formula, the area is always positive, so the subtraction is just numerical.Alternatively, maybe the areas [BXC], [AYF], [ABZ] are parts of the hexagon, and [MXZY] is another part, and their combination results in zero.But that seems odd.Alternatively, perhaps the areas are being considered with signs based on orientation, but in the problem statement, [P] denotes the area, which is always positive.Wait, but the problem says "evaluate [BXC] + [AYF] + [ABZ] - [MXZY]". So, it's a combination of areas with a subtraction.Given that all areas are positive, and their combination results in zero, it suggests that the sum of [BXC] + [AYF] + [ABZ] equals [MXZY].But in the hexagon, these areas might be overlapping or covering the same regions.Alternatively, perhaps the expression is designed to result in zero.But let me think differently. Maybe the areas [BXC], [AYF], [ABZ] are each 1/7 of the hexagon, and [MXZY] is 3/7, so 1/7 +1/7 +1/7 -3/7=0.But in my calculations, each [BXC] and [AYF] are 3√3/28, which is approximately 0.18, and [ABZ] is √3/12≈0.144, and [MXZY]≈25√3/84≈0.517.But 3√3/28 +3√3/28 +√3/12≈0.18+0.18+0.144≈0.504Subtract [MXZY]≈0.517, so≈-0.013, which is close to zero, but not exactly.But in exact terms, it's zero.Wait, in exact terms, [BXC] + [AYF] + [ABZ] =25√3/84And [MXZY]=25√3/84So, their difference is zero.Therefore, the expression evaluates to zero.But the problem states the hexagon has area 1, so 25√3/84 is a portion of it.But in the unit hexagon, the total area is (3√3)/2≈2.598, so 25√3/84≈0.517, which is roughly 1/5 of the total area.But in our scaled hexagon, the total area is 1, so 25√3/84 corresponds to 25√3/84 divided by (3√3)/2)=25√3/84 * 2/(3√3)=50/(252)=25/126≈0.198Wait, that complicates things.Wait, actually, in our initial calculations, we used the unit hexagon with area (3√3)/2, but the problem states the hexagon has area 1. So, all areas need to be scaled accordingly.So, scaling factor is 1 / ( (3√3)/2 )=2/(3√3)Therefore, each area computed needs to be multiplied by 2/(3√3)So, [BXC] =3√3/28 *2/(3√3)= (6√3)/(84√3)=6/84=1/14Similarly, [AYF]=3√3/28 *2/(3√3)=1/14[ABZ]=√3/12 *2/(3√3)= (2√3)/(36√3)=2/36=1/18[MXZY]=25√3/84 *2/(3√3)=50√3/(252√3)=50/252=25/126Now, compute the expression:[BXC] + [AYF] + [ABZ] - [MXZY] =1/14 +1/14 +1/18 -25/126Convert to common denominator 126:1/14=9/1261/14=9/1261/18=7/12625/126=25/126So, total=9+9+7 -25=25-25=0Therefore, the expression evaluates to zero.So, the final answer is 0.But the problem asks to evaluate [BXC] + [AYF] + [ABZ] - [MXZY], which is zero.But in the initial problem statement, the hexagon has area 1, so scaling is necessary.Therefore, the answer is 0.**Final Answer**boxed{0}

answer：Okay, so I have this problem here about a set T of points (x, y) in the coordinate plane. The condition is that two of the three quantities 5, x - 3, and y + 2 are equal, and the third quantity is no greater than this common value. I need to figure out what T looks like and choose the best description from the options given.Let me break it down. There are three quantities: 5, x - 3, and y + 2. The problem says that two of these are equal, and the third is less than or equal to that common value. So, I think I need to consider all possible cases where two of them are equal and then see what the third one does.First, let me list the three possible cases:1. 5 equals x - 3.2. 5 equals y + 2.3. x - 3 equals y + 2.For each case, I need to find the corresponding relationship between x and y and then determine the condition on the third quantity.Starting with the first case: 5 equals x - 3.If 5 = x - 3, then solving for x gives x = 8. So, in this case, x is fixed at 8. Now, the third quantity is y + 2, which must be less than or equal to 5. So, y + 2 ≤ 5, which simplifies to y ≤ 3. Therefore, in this case, the set of points is all points where x = 8 and y ≤ 3. That sounds like a vertical ray starting at (8, 3) and going downward.Moving on to the second case: 5 equals y + 2.If 5 = y + 2, then solving for y gives y = 3. So, y is fixed at 3. The third quantity here is x - 3, which must be less than or equal to 5. So, x - 3 ≤ 5, which simplifies to x ≤ 8. Therefore, in this case, the set of points is all points where y = 3 and x ≤ 8. That sounds like a horizontal ray starting at (8, 3) and going to the left.Now, the third case: x - 3 equals y + 2.If x - 3 = y + 2, then solving for y gives y = x - 5. So, this is a line with a slope of 1 and y-intercept at -5. Now, the third quantity here is 5, which must be less than or equal to the common value. Wait, actually, the third quantity is 5, and it must be no greater than the common value. So, since x - 3 = y + 2, the common value is x - 3 (or y + 2). Therefore, 5 must be less than or equal to x - 3. So, 5 ≤ x - 3, which simplifies to x ≥ 8. Similarly, since 5 ≤ y + 2, that gives y ≥ 3. But wait, if x ≥ 8, then y = x - 5 would mean y ≥ 3 as well. So, in this case, the set of points is the line y = x - 5, but only where x ≥ 8 and y ≥ 3. That would be a ray starting at (8, 3) and going upwards and to the right.Wait, hold on. Let me double-check that. If x - 3 = y + 2, then y = x - 5. The third quantity is 5, which must be ≤ the common value, which is x - 3. So, 5 ≤ x - 3 implies x ≥ 8. Similarly, since y + 2 = x - 3, and x ≥ 8, then y + 2 ≥ 5, so y ≥ 3. So, yes, the ray starts at (8, 3) and goes upwards and to the right.So, summarizing the three cases:1. When 5 = x - 3, we get a vertical ray starting at (8, 3) going downward.2. When 5 = y + 2, we get a horizontal ray starting at (8, 3) going left.3. When x - 3 = y + 2, we get a ray starting at (8, 3) going upwards and to the right.So, all three rays start at the same point (8, 3) and go in different directions. Therefore, the set T is composed of three rays with a common endpoint at (8, 3).Looking back at the options:A. A single point – No, because there are infinitely many points along the rays.B. Two intersecting lines – No, because we have three rays, not just two lines.C. Three lines whose pairwise intersections are three distinct points – No, because all three rays intersect at the same point (8, 3), not three distinct points.D. A triangle – No, because a triangle would require three line segments forming a closed figure, but here we have rays extending infinitely.E. Three rays with a common endpoint – Yes, this matches our conclusion.So, the correct answer should be E.

answer：First, I need to determine what percentage of Rs. 500 is equal to Rs. 125.I will use the percentage formula: Percentage = (Part / Whole) * 100.Here, the "Part" is Rs. 125 and the "Whole" is Rs. 500.Plugging in the values, the calculation becomes: (125 / 500) * 100.Performing the division, 125 divided by 500 equals 0.25.Multiplying 0.25 by 100 gives the final percentage: 25%.Therefore, Rs. 125 is 25% of Rs. 500.

answer：Okay, so I have this problem here: I need to find the maximum value of the function y = sin²x - 4cosx + 2. The options are A: 8, B: 7, C: 6, D: 5. Hmm, let me think about how to approach this.First, I remember that sin²x and cosx are related through the Pythagorean identity, which is sin²x + cos²x = 1. Maybe I can use that to rewrite the equation in terms of cosx only, which might make it easier to handle.So, if I substitute sin²x with 1 - cos²x, the equation becomes y = (1 - cos²x) - 4cosx + 2. Let me simplify that: 1 - cos²x - 4cosx + 2. Combining the constants, 1 + 2 is 3, so now it's y = -cos²x - 4cosx + 3.Hmm, that looks like a quadratic in terms of cosx. Maybe I can treat cosx as a variable, say t, and rewrite the equation as y = -t² - 4t + 3, where t = cosx. Since cosx has a range of [-1, 1], t can only be between -1 and 1.Now, I have a quadratic function y = -t² - 4t + 3. I need to find its maximum value within the interval t ∈ [-1, 1]. Quadratic functions have their maximum or minimum at the vertex. Since the coefficient of t² is negative (-1), the parabola opens downward, meaning the vertex is the maximum point.The formula for the vertex of a quadratic y = at² + bt + c is at t = -b/(2a). Plugging in the values, a = -1 and b = -4, so t = -(-4)/(2*(-1)) = 4/(-2) = -2. So, the vertex is at t = -2. But wait, t = cosx, and cosx can only be between -1 and 1. So, t = -2 is outside the domain of our function. That means the maximum must occur at one of the endpoints of the interval, either t = -1 or t = 1.Let me evaluate y at both endpoints.First, at t = -1:y = -(-1)² - 4*(-1) + 3 = -1 + 4 + 3 = 6.Next, at t = 1:y = -(1)² - 4*(1) + 3 = -1 - 4 + 3 = -2.So, y is 6 when t = -1 and y is -2 when t = 1. Since we're looking for the maximum value, 6 is larger than -2. Therefore, the maximum value of y is 6.Wait, but let me double-check. Maybe I made a mistake in the calculations. Let me go through it again.Starting with y = sin²x - 4cosx + 2. Using sin²x = 1 - cos²x, so y = 1 - cos²x - 4cosx + 2. Combine constants: 1 + 2 = 3, so y = -cos²x - 4cosx + 3. That's correct.Then, treating cosx as t, y = -t² - 4t + 3. The vertex is at t = -b/(2a) = -(-4)/(2*(-1)) = 4/(-2) = -2. Right, that's outside the domain. So, evaluate at t = -1 and t = 1.At t = -1: y = -(-1)^2 -4*(-1) +3 = -1 +4 +3 = 6. Correct.At t = 1: y = -(1)^2 -4*(1) +3 = -1 -4 +3 = -2. Correct.So, yes, the maximum value is 6, which is option C.But wait, just to be thorough, maybe there's another way to approach this problem. Let me think.Alternatively, I could express the function in terms of sinx and then take the derivative to find the maximum. But since it's a trigonometric function, maybe calculus is overkill, but let's try it.Let me write y = sin²x - 4cosx + 2. To find the maximum, take the derivative dy/dx and set it to zero.dy/dx = 2sinx cosx + 4sinx. Wait, is that right? Let's see:Derivative of sin²x is 2sinx cosx (using chain rule). Derivative of -4cosx is 4sinx. Derivative of 2 is 0. So, dy/dx = 2sinx cosx + 4sinx.Set dy/dx = 0: 2sinx cosx + 4sinx = 0. Factor out 2sinx: 2sinx (cosx + 2) = 0.So, either 2sinx = 0 or cosx + 2 = 0.Case 1: 2sinx = 0 => sinx = 0 => x = 0, π, 2π, etc.Case 2: cosx + 2 = 0 => cosx = -2. But cosx can't be less than -1, so this case has no solution.So, critical points are where sinx = 0, i.e., x = 0, π, 2π, etc.Now, evaluate y at these points.At x = 0:sin0 = 0, cos0 = 1y = 0 - 4*1 + 2 = -4 + 2 = -2.At x = π:sinπ = 0, cosπ = -1y = 0 - 4*(-1) + 2 = 0 + 4 + 2 = 6.At x = 2π:Same as x = 0, y = -2.So, again, the maximum value is 6 at x = π.Therefore, both methods confirm that the maximum value is 6, which is option C.I think I'm confident now that the answer is C: 6.

answer：First, I need to determine the original purchase price of the product. Given that Bill sold the product for 549.9999999999995 with a 10% profit, I can calculate the original purchase price by dividing the selling price by 1.10.Next, I'll calculate the new purchase price if Bill had bought the product for 10% less. This involves multiplying the original purchase price by 0.90.Then, I'll find the new selling price if Bill sold the product at a 30% profit based on the reduced purchase price. This is done by multiplying the new purchase price by 1.30.Finally, to find out how much more Bill would have received, I'll subtract the original selling price from the new selling price.