💡Okay, so I have this problem about quadrilaterals and their areas. Let me try to understand it step by step. We have quadrilateral ( WXYZ ) with some given side lengths and the area of ( WXYZ ) is 15. The question is asking for the area of another quadrilateral ( W'X'Y'Z' ). The given side lengths are:- ( WZ = ZW' = 10 )- ( WX = XX' = 6 )- ( XY = YY' = 7 )- ( YZ = Z'W = 12 )Hmm, okay. So, each side of quadrilateral ( WXYZ ) is equal to the corresponding side of another quadrilateral ( W'X'Y'Z' ). Wait, no, not exactly. It seems like each side is extended to form another point. For example, ( WZ = ZW' = 10 ) suggests that point ( W' ) is such that ( ZW' = WZ = 10 ). Similarly, ( WX = XX' = 6 ) implies that ( X' ) is a point such that ( XX' = WX = 6 ). The same goes for the other sides.So, essentially, each side of quadrilateral ( WXYZ ) is being extended beyond each vertex to form a new quadrilateral ( W'X'Y'Z' ). Each extension is equal in length to the original side. So, for example, side ( WZ ) is extended beyond ( Z ) to ( W' ) such that ( ZW' = WZ = 10 ). Similarly, side ( WX ) is extended beyond ( X ) to ( X' ) such that ( XX' = WX = 6 ), and so on.Now, the area of ( WXYZ ) is given as 15. We need to find the area of ( W'X'Y'Z' ).I think the key here is to recognize that quadrilateral ( W'X'Y'Z' ) is formed by extending each side of ( WXYZ ) and connecting these new points. This likely creates a larger quadrilateral that encompasses ( WXYZ ) and some surrounding triangles.Let me visualize this. If I have quadrilateral ( WXYZ ), and I extend each side beyond each vertex by the length of that side, then connect these new points, I'll get a larger quadrilateral ( W'X'Y'Z' ). The area of ( W'X'Y'Z' ) should then be the area of ( WXYZ ) plus the areas of the four surrounding triangles.Wait, but how many triangles are we talking about here? Each extension of a side creates a triangle. So, for each side of ( WXYZ ), there is a corresponding triangle outside ( WXYZ ) that is part of ( W'X'Y'Z' ).Let me think about each triangle:1. Extending ( WX ) beyond ( X ) to ( X' ) creates triangle ( WX'X ).2. Extending ( XY ) beyond ( Y ) to ( Y' ) creates triangle ( XY'Y ).3. Extending ( YZ ) beyond ( Z ) to ( Z' ) creates triangle ( YZ'Z ).4. Extending ( ZW ) beyond ( W ) to ( W' ) creates triangle ( ZW'W ).So, there are four triangles outside ( WXYZ ) that are part of ( W'X'Y'Z' ). Therefore, the area of ( W'X'Y'Z' ) should be the area of ( WXYZ ) plus the areas of these four triangles.But how do I find the areas of these triangles? I know the lengths of the sides, but I don't know the heights or any angles. Hmm, maybe there's a relationship between the areas of these triangles and the area of ( WXYZ ).Wait a minute, each triangle is similar to a triangle inside ( WXYZ ) but scaled by a factor. Since each extension is equal in length to the original side, maybe each triangle has twice the area of a corresponding triangle inside ( WXYZ ).Let me explain. For example, triangle ( WX'X ) is formed by extending ( WX ) beyond ( X ) to ( X' ) such that ( XX' = WX = 6 ). So, the length from ( W ) to ( X' ) is ( WX + XX' = 6 + 6 = 12 ). Similarly, the other sides are extended in the same manner.But does this mean that each triangle outside has twice the area of the corresponding triangle inside? Maybe not exactly, because the height might not necessarily double. Hmm, perhaps I need to think differently.Alternatively, maybe the entire figure ( W'X'Y'Z' ) is a scaled version of ( WXYZ ). If each side is extended by its own length, the scaling factor might be 2, which would make the area four times as large. But wait, that might not be accurate because scaling all sides by 2 would scale the area by 4, but in this case, we're only extending each side by its own length, not scaling the entire figure.Wait, perhaps it's better to think in terms of vectors or coordinates. If I assign coordinates to the points of ( WXYZ ), I could then find the coordinates of ( W'X'Y'Z' ) and compute the area using the shoelace formula or something similar. But that might be complicated without more information.Alternatively, maybe the area of ( W'X'Y'Z' ) is related to the area of ( WXYZ ) through some proportional relationship. Since each side is extended by its own length, the new quadrilateral ( W'X'Y'Z' ) might be similar to ( WXYZ ) but scaled by a factor.Wait, if each side is extended by its own length, the distance from each original vertex to the new vertex is equal to the original side length. So, for example, the distance from ( W ) to ( W' ) is ( WZ = 10 ), but ( WZ ) is a side of ( WXYZ ), not a diagonal. Hmm, maybe I need to consider the vectors from each vertex.Let me try to think about this differently. If I consider each triangle formed by extending a side, such as triangle ( WX'X ), the base is ( XX' = 6 ) and the height would be the same as the height of triangle ( WX ) in ( WXYZ ). Therefore, the area of triangle ( WX'X ) would be twice the area of triangle ( WX ) in ( WXYZ ).Wait, is that true? If the base is doubled and the height remains the same, then yes, the area would double. So, each of these four triangles outside ( WXYZ ) would have twice the area of the corresponding triangles inside ( WXYZ ).But hold on, ( WXYZ ) is a quadrilateral, not necessarily composed of four triangles. However, any quadrilateral can be divided into two triangles by a diagonal. So, if I divide ( WXYZ ) into two triangles, say ( WXY ) and ( WYZ ), each of these triangles would have some area. Then, the triangles outside, such as ( WX'X ) and ( XY'Y ), would each have twice the area of ( WXY ) and ( WYZ ), respectively.Wait, no, that might not be accurate. Each triangle outside corresponds to a side, not a triangle inside. Maybe I need to think of each triangle outside as being congruent to the adjacent triangle inside.Alternatively, perhaps the area of each surrounding triangle is equal to the area of the corresponding triangle in ( WXYZ ). But I'm not sure.Wait, let's think about the triangles more carefully. For example, triangle ( WX'X ) is formed by extending side ( WX ) to ( X' ). The length ( XX' = WX = 6 ). So, the base of triangle ( WX'X ) is ( XX' = 6 ), and the height would be the same as the height from ( Y ) to ( WX ) in quadrilateral ( WXYZ ). Therefore, the area of triangle ( WX'X ) would be equal to the area of triangle ( WXY ) in ( WXYZ ).Similarly, triangle ( XY'Y ) would have the same area as triangle ( WXY ) because it's formed by extending ( XY ) to ( Y' ) with ( YY' = XY = 7 ). Wait, no, that might not be correct because the height could be different.Hmm, maybe I need to consider that each triangle outside is congruent to the triangle inside. If that's the case, then each surrounding triangle would have the same area as the corresponding triangle inside ( WXYZ ). Therefore, the total area of the four surrounding triangles would be equal to the area of ( WXYZ ), which is 15. So, the area of ( W'X'Y'Z' ) would be the area of ( WXYZ ) plus the areas of the four surrounding triangles, which would be ( 15 + 15 = 30 ). But 30 is one of the options, option A.But wait, I'm not sure if the surrounding triangles each have the same area as the triangles inside. Maybe they have twice the area. Let me reconsider.If each surrounding triangle has a base that's double the base of the corresponding triangle inside ( WXYZ ), and the same height, then the area would be double. For example, triangle ( WX'X ) has base ( XX' = 6 ), which is the same as ( WX = 6 ). Wait, no, ( XX' = WX = 6 ), so the base is the same, but the height might be different.Wait, actually, the height depends on the angle between the sides. If the sides are extended, the height might not necessarily be the same. This is getting complicated.Maybe a better approach is to consider that quadrilateral ( W'X'Y'Z' ) is formed by translating each side of ( WXYZ ) outward by the length of the side. This might create a larger quadrilateral whose area is related to the original by a factor.Alternatively, perhaps the area of ( W'X'Y'Z' ) is twice the area of ( WXYZ ). But 15 times 2 is 30, which is option A. However, I'm not sure if it's exactly twice.Wait, let me think about the figure. If each side is extended beyond each vertex by its own length, the resulting quadrilateral ( W'X'Y'Z' ) might enclose ( WXYZ ) and four surrounding triangles, each of which has the same area as the corresponding triangle in ( WXYZ ). Therefore, the total area would be the area of ( WXYZ ) plus four times the area of each triangle.But ( WXYZ ) can be divided into two triangles, say ( WXY ) and ( WYZ ). Let's denote the area of ( WXY ) as ( A ) and the area of ( WYZ ) as ( B ). So, ( A + B = 15 ).Now, each surrounding triangle would correspond to these two triangles. For example, triangle ( WX'X ) would correspond to triangle ( WXY ), and triangle ( XY'Y ) would also correspond to triangle ( WXY ). Similarly, triangle ( YZ'Z ) would correspond to triangle ( WYZ ), and triangle ( ZW'W ) would also correspond to triangle ( WYZ ).Therefore, each surrounding triangle has the same area as the corresponding triangle inside ( WXYZ ). So, the total area of the surrounding triangles would be ( 2A + 2B = 2(A + B) = 2 times 15 = 30 ).Therefore, the area of ( W'X'Y'Z' ) would be the area of ( WXYZ ) plus the area of the surrounding triangles, which is ( 15 + 30 = 45 ). So, the answer would be 45, which is option C.Wait, that makes sense. Each surrounding triangle has the same area as the corresponding triangle inside, and since there are four surrounding triangles (two for each original triangle), the total surrounding area is twice the area of ( WXYZ ). Therefore, the total area of ( W'X'Y'Z' ) is ( 15 + 30 = 45 ).I think that's the correct approach. So, the area of ( W'X'Y'Z' ) is 45.

💡Alright, so I have this problem here where I need to figure out which of the options A, B, C, or D is correct given that a and b are non-zero real numbers and that a is greater than b. Let me try to break this down step by step.First, let's look at option A: 1/a < 1/b. Hmm, okay. So, if a is greater than b, does that mean 1/a is less than 1/b? Well, I remember that when dealing with inequalities and reciprocals, the direction of the inequality can change depending on the signs of a and b. For example, if both a and b are positive, then yes, if a > b, 1/a < 1/b because as the denominator gets larger, the fraction gets smaller. But what if one of them is negative?Suppose a is positive and b is negative. Then, since a > b, a is positive and b is negative. But 1/a would be positive, and 1/b would be negative. So, in this case, 1/a is actually greater than 1/b because a positive number is always greater than a negative number. That means option A isn't always true. So, A might not be the correct answer.Moving on to option B: a² > ab. Let's see. If I factor out an a, this becomes a(a) > a(b), which simplifies to a > b, which is given. But wait, that seems too straightforward. Is there something I'm missing? Maybe the sign of a?If a is positive, then multiplying both sides of the inequality a > b by a (which is positive) preserves the inequality, so a² > ab would hold. But what if a is negative? If a is negative, then multiplying both sides by a (a negative number) would reverse the inequality. So, a² < ab in that case. Therefore, option B isn't always true either because it depends on the sign of a. So, B might not be the correct answer.Now, let's check option C: 1/(ab²) > 1/(a²b). Hmm, this looks a bit more complicated. Let me try to manipulate this inequality. If I take the reciprocal of both sides, the inequality would reverse, so ab² < a²b. Then, I can divide both sides by ab (assuming ab ≠ 0, which is given since a and b are non-zero). Dividing both sides by ab gives b < a. But wait, we know that a > b, so this inequality holds. Therefore, the original inequality 1/(ab²) > 1/(a²b) must also hold. So, option C seems to be correct.But let me double-check. Starting from a > b, and wanting to get to 1/(ab²) > 1/(a²b). Let's subtract 1/(a²b) from both sides: 1/(ab²) - 1/(a²b) > 0. Combining these fractions, we get (a - b)/(a²b²) > 0. Since a > b, a - b is positive. Also, a²b² is always positive because squares are positive and a and b are non-zero. So, the entire expression is positive, which means the inequality holds. Okay, so C is definitely correct.Finally, let's look at option D: a² > b². Hmm, is this always true? Well, if both a and b are positive, then yes, since a > b implies a² > b². But what if a is positive and b is negative? Let's say a = 1 and b = -1. Then, a > b because 1 > -1, but a² = 1 and b² = 1, so a² is not greater than b². They are equal. Therefore, D isn't always true. So, D might not be the correct answer.Wait, but in this case, a² equals b², so it's not greater. So, D is not necessarily correct. Therefore, the only option that is always correct is option C.I think I've considered all the cases and checked each option thoroughly. So, I'm pretty confident that option C is the correct answer.**Final Answer**boxed{C}

💡Alright, so I have this problem about generating functions. It's divided into three parts: (Ⅰ), (Ⅱ), and (Ⅲ). Let me try to tackle each part step by step. I'll start with part (Ⅰ).**Part (Ⅰ):** There are two sets of functions given. I need to determine if the given function ( h(x) ) is a generating function of ( f_1(x) ) and ( f_2(x) ). That means I have to check if there exist real numbers ( a ) and ( b ) such that ( h(x) = a cdot f_1(x) + b cdot f_2(x) ).**Set 1:**- ( f_1(x) = sin x )- ( f_2(x) = cos x )- ( h(x) = sinleft(x + frac{pi}{3}right) )Hmm, okay. I remember that ( sin(x + phi) ) can be expanded using the sine addition formula: ( sin(x + phi) = sin x cos phi + cos x sin phi ). So, let me apply that here.( h(x) = sinleft(x + frac{pi}{3}right) = sin x cosleft(frac{pi}{3}right) + cos x sinleft(frac{pi}{3}right) )I know that ( cosleft(frac{pi}{3}right) = frac{1}{2} ) and ( sinleft(frac{pi}{3}right) = frac{sqrt{3}}{2} ). So substituting these values in:( h(x) = sin x cdot frac{1}{2} + cos x cdot frac{sqrt{3}}{2} )Which simplifies to:( h(x) = frac{1}{2} sin x + frac{sqrt{3}}{2} cos x )Comparing this with ( a cdot f_1(x) + b cdot f_2(x) ), it's clear that ( a = frac{1}{2} ) and ( b = frac{sqrt{3}}{2} ). So yes, ( h(x) ) is a generating function for ( f_1(x) ) and ( f_2(x) ) in Set 1.**Set 2:**- ( f_1(x) = x^2 - x )- ( f_2(x) = x^2 + x + 1 )- ( h(x) = x^2 - x + 1 )Alright, let's see if ( h(x) ) can be expressed as a linear combination of ( f_1(x) ) and ( f_2(x) ). So, we need to find ( a ) and ( b ) such that:( a(x^2 - x) + b(x^2 + x + 1) = x^2 - x + 1 )Let me expand the left side:( a x^2 - a x + b x^2 + b x + b = (a + b)x^2 + (-a + b)x + b )Now, this should equal ( x^2 - x + 1 ). So, we can set up equations by equating the coefficients:1. Coefficient of ( x^2 ): ( a + b = 1 )2. Coefficient of ( x ): ( -a + b = -1 )3. Constant term: ( b = 1 )Let me solve these equations step by step.From equation 3: ( b = 1 )Plugging ( b = 1 ) into equation 1: ( a + 1 = 1 ) => ( a = 0 )Now, plug ( a = 0 ) and ( b = 1 ) into equation 2: ( -0 + 1 = -1 ) => ( 1 = -1 ). Wait, that's not true. So, this system of equations doesn't have a solution. Therefore, there are no real numbers ( a ) and ( b ) that satisfy the equation. Hence, ( h(x) ) is not a generating function for ( f_1(x) ) and ( f_2(x) ) in Set 2.**Conclusion for Part (Ⅰ):**- Set 1: Yes, ( h(x) ) is a generating function.- Set 2: No, ( h(x) ) is not a generating function.**Part (Ⅱ):**Given:- ( f_1(x) = log_{2}x )- ( f_2(x) = log_{frac{1}{2}}x )- ( a = 2 )- ( b = 1 )First, I need to find the generating function ( h(x) ). So, ( h(x) = 2f_1(x) + 1f_2(x) ).Let me compute ( h(x) ):( h(x) = 2log_{2}x + log_{frac{1}{2}}x )I remember that ( log_{frac{1}{2}}x ) can be rewritten using the change of base formula. Since ( log_{frac{1}{2}}x = frac{log_{2}x}{log_{2}frac{1}{2}} ). And ( log_{2}frac{1}{2} = -1 ), so:( log_{frac{1}{2}}x = -log_{2}x )Therefore, substituting back into ( h(x) ):( h(x) = 2log_{2}x - log_{2}x = log_{2}x )So, ( h(x) = log_{2}x ).Now, the inequality given is ( 3h^2(x) + 2h(x) + t < 0 ). We need to find the range of ( t ) such that this inequality has a solution for ( x in [2, 4] ).Let me substitute ( h(x) = log_{2}x ) into the inequality:( 3(log_{2}x)^2 + 2log_{2}x + t < 0 )Let me denote ( s = log_{2}x ). Since ( x in [2, 4] ), ( s ) will range from ( log_{2}2 = 1 ) to ( log_{2}4 = 2 ). So, ( s in [1, 2] ).Substituting ( s ) into the inequality:( 3s^2 + 2s + t < 0 )We need this quadratic inequality to have at least one solution in ( s in [1, 2] ). Let me rearrange the inequality:( t < -3s^2 - 2s )So, ( t ) must be less than ( -3s^2 - 2s ) for some ( s ) in [1, 2]. To find the range of ( t ), I need to find the maximum value of ( -3s^2 - 2s ) over ( s in [1, 2] ). Because ( t ) has to be less than this maximum to satisfy the inequality for some ( s ).Let me define ( y = -3s^2 - 2s ). This is a quadratic function in terms of ( s ), opening downward (since the coefficient of ( s^2 ) is negative). The vertex of this parabola will give the maximum value.The vertex of a parabola ( y = as^2 + bs + c ) is at ( s = -frac{b}{2a} ).Here, ( a = -3 ), ( b = -2 ). So,( s = -frac{-2}{2 cdot -3} = frac{2}{-6} = -frac{1}{3} )But ( s = -frac{1}{3} ) is not in our interval [1, 2]. Therefore, the maximum of ( y ) on [1, 2] will occur at one of the endpoints.Let me compute ( y ) at ( s = 1 ) and ( s = 2 ):- At ( s = 1 ): ( y = -3(1)^2 - 2(1) = -3 - 2 = -5 )- At ( s = 2 ): ( y = -3(2)^2 - 2(2) = -12 - 4 = -16 )So, the maximum value of ( y ) on [1, 2] is ( -5 ) at ( s = 1 ). Therefore, for the inequality ( t < -3s^2 - 2s ) to hold for some ( s ) in [1, 2], ( t ) must be less than ( -5 ).**Conclusion for Part (Ⅱ):** The range of ( t ) is ( t < -5 ).**Part (Ⅲ):**Given:- ( f_1(x) = x )- ( f_2(x) = frac{1}{x} ) for ( 1 leq x leq 10 )- ( a = 1 )- ( b > 0 )We need to find the generating function ( h(x) ) such that ( h(x) geq b ) always holds, and then find the range of ( b ).So, ( h(x) = a f_1(x) + b f_2(x) = 1 cdot x + b cdot frac{1}{x} = x + frac{b}{x} )We need ( h(x) = x + frac{b}{x} geq b ) for all ( x ) in [1, 10].So, the inequality is:( x + frac{b}{x} geq b )Let me rearrange this inequality:( x + frac{b}{x} - b geq 0 )Factor out ( b ):( x + bleft(frac{1}{x} - 1right) geq 0 )Hmm, not sure if that helps. Maybe another approach. Let's consider the function ( h(x) = x + frac{b}{x} ). We need this function to be greater than or equal to ( b ) for all ( x ) in [1, 10].Let me write the inequality:( x + frac{b}{x} geq b )Subtract ( b ) from both sides:( x + frac{b}{x} - b geq 0 )Factor ( b ):( x + bleft(frac{1}{x} - 1right) geq 0 )Alternatively, maybe it's better to consider the function ( h(x) = x + frac{b}{x} ) and find its minimum value on [1, 10], then set that minimum to be greater than or equal to ( b ).Yes, that seems like a solid approach. So, to find the minimum of ( h(x) ) on [1, 10], we can take the derivative and find critical points.Let me compute the derivative ( h'(x) ):( h'(x) = 1 - frac{b}{x^2} )Set ( h'(x) = 0 ) to find critical points:( 1 - frac{b}{x^2} = 0 )( frac{b}{x^2} = 1 )( x^2 = b )( x = sqrt{b} ) (since ( x > 0 ))So, the critical point is at ( x = sqrt{b} ). Now, we need to check if this critical point lies within our interval [1, 10].Case 1: ( sqrt{b} in [1, 10] ) => ( b in [1, 100] )In this case, the minimum of ( h(x) ) occurs at ( x = sqrt{b} ). Let's compute ( h(sqrt{b}) ):( h(sqrt{b}) = sqrt{b} + frac{b}{sqrt{b}} = sqrt{b} + sqrt{b} = 2sqrt{b} )We need ( 2sqrt{b} geq b ). Let's solve this inequality:( 2sqrt{b} geq b )Divide both sides by ( sqrt{b} ) (since ( b > 0 )):( 2 geq sqrt{b} )Square both sides:( 4 geq b )So, ( b leq 4 ). But in this case, ( b in [1, 100] ). Therefore, combining these, ( b in [1, 4] ).Case 2: ( sqrt{b} < 1 ) => ( b < 1 )In this case, the critical point ( x = sqrt{b} ) is less than 1, so the function ( h(x) ) is increasing on [1, 10]. Therefore, the minimum occurs at ( x = 1 ).Compute ( h(1) = 1 + frac{b}{1} = 1 + b )We need ( 1 + b geq b ). Simplify:( 1 + b geq b ) => ( 1 geq 0 )Which is always true. So, for ( b < 1 ), the inequality ( h(x) geq b ) holds for all ( x in [1, 10] ).Case 3: ( sqrt{b} > 10 ) => ( b > 100 )In this case, the critical point ( x = sqrt{b} ) is greater than 10, so the function ( h(x) ) is decreasing on [1, 10]. Therefore, the minimum occurs at ( x = 10 ).Compute ( h(10) = 10 + frac{b}{10} )We need ( 10 + frac{b}{10} geq b )Multiply both sides by 10:( 100 + b geq 10b )( 100 geq 9b )( b leq frac{100}{9} approx 11.11 )But in this case, ( b > 100 ), which contradicts ( b leq frac{100}{9} ). Therefore, there are no solutions in this case.**Combining all cases:**- Case 1: ( b in [1, 4] )- Case 2: ( b in (0, 1) )- Case 3: No solutionsTherefore, the range of ( b ) is ( 0 < b leq 4 ).**Conclusion for Part (Ⅲ):** The range of ( b ) is ( 0 < b leq 4 ).**Final Answer**Ⅰ. Set 1: (boxed{a = frac{1}{2}, b = frac{sqrt{3}}{2}}); Set 2: (boxed{text{No solution}})Ⅱ. (boxed{t < -5})Ⅲ. (boxed{0 < b leq 4})

💡Okay, so I have this equation to solve: (frac{x+11}{x-4} = frac{x-1}{x+6}). Hmm, I remember that when you have fractions equal to each other, you can cross-multiply to get rid of the denominators. That should make things simpler.Alright, let me try cross-multiplying. So, I'll multiply the numerator of the left side by the denominator of the right side and set it equal to the numerator of the right side times the denominator of the left side. That gives me:[(x + 11)(x + 6) = (x - 1)(x - 4)]Now, I need to expand both sides. Let's start with the left side. Multiplying (x + 11) by (x + 6):First, (x times x = x^2).Then, (x times 6 = 6x).Next, (11 times x = 11x).Finally, (11 times 6 = 66).So, combining all those terms, the left side becomes:[x^2 + 6x + 11x + 66]Which simplifies to:[x^2 + 17x + 66]Okay, now let's expand the right side: ((x - 1)(x - 4)).First, (x times x = x^2).Then, (x times (-4) = -4x).Next, (-1 times x = -x).Finally, (-1 times (-4) = 4).So, combining those terms, the right side becomes:[x^2 - 4x - x + 4]Which simplifies to:[x^2 - 5x + 4]Now, I have both sides expanded:Left side: (x^2 + 17x + 66)Right side: (x^2 - 5x + 4)Hmm, I can subtract (x^2) from both sides to eliminate the quadratic terms. Let's do that:[x^2 + 17x + 66 - x^2 = x^2 - 5x + 4 - x^2]Which simplifies to:[17x + 66 = -5x + 4]Now, I need to get all the x terms on one side and the constants on the other. Let me add (5x) to both sides:[17x + 5x + 66 = -5x + 5x + 4]Which simplifies to:[22x + 66 = 4]Next, I'll subtract 66 from both sides to isolate the term with x:[22x + 66 - 66 = 4 - 66]Simplifying that:[22x = -62]Now, to solve for x, I'll divide both sides by 22:[x = frac{-62}{22}]I can simplify this fraction by dividing numerator and denominator by 2:[x = frac{-31}{11}]Wait, let me check if this solution makes sense. I should plug it back into the original equation to make sure it doesn't make any denominators zero.Original equation denominators are (x - 4) and (x + 6). Plugging (x = -frac{31}{11}) into (x - 4):[-frac{31}{11} - 4 = -frac{31}{11} - frac{44}{11} = -frac{75}{11}], which is not zero.And plugging into (x + 6):[-frac{31}{11} + 6 = -frac{31}{11} + frac{66}{11} = frac{35}{11}], which is also not zero.So, the solution is valid. Therefore, (x = -frac{31}{11}) is the correct answer.

💡Okay, so I need to find the variance of the daily reading time for all students in the school. The data is given for three different grades: 10, 11, and 12. Each grade has its own mean reading time, variance, and number of students. First, I remember that variance measures how spread out the data is from the mean. Since there are three different grades with different numbers of students, I think I need to calculate a combined variance that accounts for all students together. I recall that when combining variances from different groups, especially when the groups have different sizes, I need to consider the weighted average of the variances and also account for the differences between each group's mean and the overall mean. So, maybe the first step is to find the overall mean reading time for all students combined. To do that, I can use the formula for the weighted mean. That is, multiply each grade's mean by the number of students in that grade, sum those products, and then divide by the total number of students.Let me write that down:Total mean (μ) = (Σ (mean_i * n_i)) / NWhere:- mean_i is the mean reading time for each grade- n_i is the number of students in each grade- N is the total number of studentsSo, plugging in the numbers:Grade 10: mean = 2.7 hours, n = 800Grade 11: mean = 3.1 hours, n = 600Grade 12: mean = 3.3 hours, n = 600Total number of students, N = 800 + 600 + 600 = 2000Calculating the total mean:μ = (2.7 * 800 + 3.1 * 600 + 3.3 * 600) / 2000Let me compute each term:2.7 * 800 = 21603.1 * 600 = 18603.3 * 600 = 1980Adding these up: 2160 + 1860 + 1980 = 6000So, μ = 6000 / 2000 = 3 hoursOkay, so the overall mean reading time is 3 hours.Now, to find the combined variance, I think I need to use the formula for the weighted variance. The formula is a bit more complex because it involves not just the variances of each group but also the squared differences between each group's mean and the overall mean.The formula for the combined variance (σ²) is:σ² = Σ [ (n_i / N) * (s_i² + (μ_i - μ)²) ]Where:- n_i is the number of students in grade i- N is the total number of students- s_i² is the variance of grade i- μ_i is the mean of grade i- μ is the overall meanSo, applying this formula to our data:For Grade 10:n_i = 800s_i² = 1μ_i = 2.7μ = 3For Grade 11:n_i = 600s_i² = 2μ_i = 3.1μ = 3For Grade 12:n_i = 600s_i² = 3μ_i = 3.3μ = 3Now, let's compute each term:For Grade 10:(n_i / N) = 800 / 2000 = 0.4(μ_i - μ)² = (2.7 - 3)² = (-0.3)² = 0.09So, the term is 0.4 * (1 + 0.09) = 0.4 * 1.09 = 0.436For Grade 11:(n_i / N) = 600 / 2000 = 0.3(μ_i - μ)² = (3.1 - 3)² = (0.1)² = 0.01So, the term is 0.3 * (2 + 0.01) = 0.3 * 2.01 = 0.603For Grade 12:(n_i / N) = 600 / 2000 = 0.3(μ_i - μ)² = (3.3 - 3)² = (0.3)² = 0.09So, the term is 0.3 * (3 + 0.09) = 0.3 * 3.09 = 0.927Now, summing up these terms:0.436 (Grade 10) + 0.603 (Grade 11) + 0.927 (Grade 12) = 1.966So, the combined variance is 1.966.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the overall mean was 3 hours, which seems correct.For Grade 10:0.4 * (1 + 0.09) = 0.4 * 1.09 = 0.436Grade 11:0.3 * (2 + 0.01) = 0.3 * 2.01 = 0.603Grade 12:0.3 * (3 + 0.09) = 0.3 * 3.09 = 0.927Adding them up: 0.436 + 0.603 = 1.039; 1.039 + 0.927 = 1.966Yes, that seems correct.I think that's the right approach. I accounted for both the variance within each grade and the variance between the grades by considering the squared differences from the overall mean. So, the combined variance should be 1.966.

💡Okay, so I have this problem about a class periodic geometric sequence. Hmm, I need to figure out the sum of the first 25 terms. Let me try to understand what a class periodic geometric sequence is first. The problem says that for any positive integer n, the sequence satisfies a_{n+m} = a_n * q, where m is the period and q is the common ratio. So, every m terms, the sequence multiplies by q. That means the sequence repeats its pattern every m terms but each time scaled by q.Given that, the sequence {b_n} has a period of 4 and a common ratio of 3. The first four terms are 1, 1, 2, 3. So, let me write that down:b1 = 1b2 = 1b3 = 2b4 = 3Since the period is 4, the next four terms (b5 to b8) should be 3 times the first four terms. Let me check that:b5 = b1 * 3 = 1 * 3 = 3b6 = b2 * 3 = 1 * 3 = 3b7 = b3 * 3 = 2 * 3 = 6b8 = b4 * 3 = 3 * 3 = 9So, the next four terms are 3, 3, 6, 9. That seems right. Then, the following four terms (b9 to b12) should be 3 times the previous four terms, which were 3, 3, 6, 9. So:b9 = b5 * 3 = 3 * 3 = 9b10 = b6 * 3 = 3 * 3 = 9b11 = b7 * 3 = 6 * 3 = 18b12 = b8 * 3 = 9 * 3 = 27So, b9 to b12 are 9, 9, 18, 27. Continuing this pattern, each block of four terms is 3 times the previous block.I need to find the sum of the first 25 terms. Let's see how many complete blocks of four terms are there in 25 terms. 25 divided by 4 is 6 with a remainder of 1. So, there are 6 complete blocks of four terms and one extra term.Each block of four terms is 3 times the previous block. The first block (b1 to b4) sums to 1 + 1 + 2 + 3 = 7. The second block (b5 to b8) sums to 3 + 3 + 6 + 9 = 21, which is 3 times 7. The third block (b9 to b12) sums to 9 + 9 + 18 + 27 = 63, which is 3 times 21. So, each block's sum is 3 times the previous block's sum.So, the sum of each block forms a geometric sequence where the first term is 7 and the common ratio is 3. The number of blocks is 6, so the sum of the first 6 blocks is:Sum = 7 * (3^6 - 1) / (3 - 1) = 7 * (729 - 1) / 2 = 7 * 728 / 2 = 7 * 364 = 2548Wait, that's the sum of the first 24 terms. Then, the 25th term is the first term of the 7th block, which would be b25. Since each block is 3 times the previous, the 7th block's first term is 3^6 * b1 = 729 * 1 = 729.So, the total sum of the first 25 terms is 2548 + 729 = 3277.Let me double-check my calculations. The sum of the first block is 7, the second is 21, third is 63, fourth is 189, fifth is 567, sixth is 1701. Adding these up: 7 + 21 = 28, 28 + 63 = 91, 91 + 189 = 280, 280 + 567 = 847, 847 + 1701 = 2548. Then, adding the 25th term, 729, gives 2548 + 729 = 3277. That seems correct.So, the answer should be 3277, which is option A.