answer：First, let's denote the initial number of people abducted as ( P ).The alien returns 80% of the people abducted, which means 20% of the people are not returned.The alien takes 10 people to another planet and 30 people to his home planet, totaling ( 10 + 30 = 40 ) people who are not returned.These 40 people represent the 20% of the people abducted who are not returned. Therefore, we can set up the equation:[0.20 times P = 40]To find the initial number of people abducted (( P )), we divide both sides of the equation by 0.20:[P = frac{40}{0.20} = 200]Thus, the alien initially abducted 200 people.

answer：Alright, so I have this geometry problem here, and I need to show that a certain quadrilateral is cyclic. Let me try to unpack the problem step by step.First, there are two circles, ω₁ and ω₂. There are two tangents, t and t', to these circles. The points where t touches ω₁ and ω₂ are T₁ and T₂, respectively. Similarly, the points where t' touches ω₁ and ω₂ are T₁' and T₂'. So, we have two tangent lines, each touching both circles at specific points.Next, M is defined as the midpoint of the segment [T₁T₂]. That seems straightforward enough. Then, P₁ and P₂ are the intersections of lines MT₁' and MT₂' with ω₁ and ω₂, respectively. So, if I draw a line from M to T₁', it will intersect ω₁ again at P₁, and similarly, a line from M to T₂' will intersect ω₂ again at P₂.The goal is to show that the quadrilateral P₁P₂T₁'T₂' is cyclic. That means all four points lie on a single circle. To prove this, I need to show that the opposite angles of the quadrilateral sum up to 180 degrees or that the power of a point with respect to the circle is equal for certain points.Let me try to visualize this. I imagine two circles, ω₁ and ω₂, with two common tangents t and t'. The points T₁, T₂, T₁', and T₂' are the points of tangency on each circle. M is the midpoint between T₁ and T₂. Then, lines from M to T₁' and T₂' intersect the circles again at P₁ and P₂.I think I should consider the power of point M with respect to both circles. The power of a point with respect to a circle is defined as the square of the length of the tangent from that point to the circle. Since M is the midpoint of [T₁T₂], the lengths MT₁ and MT₂ are equal. Therefore, the power of M with respect to both circles should be equal because the tangents from M to both circles are equal in length.Wait, let me write that down more formally. The power of M with respect to ω₁ is MT₁², and the power with respect to ω₂ is MT₂². Since M is the midpoint, MT₁ = MT₂, so MT₁² = MT₂². Therefore, the power of M with respect to both circles is the same.Now, considering the lines MT₁' and MT₂', these lines intersect ω₁ and ω₂ again at P₁ and P₂, respectively. By the power of a point theorem, the power of M with respect to ω₁ is equal to MP₁ * MT₁', and similarly, the power with respect to ω₂ is equal to MP₂ * MT₂'. Since the power of M is the same for both circles, we have:MP₁ * MT₁' = MP₂ * MT₂'This seems important. It tells us that the products of the lengths from M to P₁ and M to T₁' are equal to the products from M to P₂ and M to T₂'.I wonder if this implies that P₁, P₂, T₁', and T₂' lie on a circle. Maybe I can use the converse of the power of a point theorem or some cyclic quadrilateral properties.Alternatively, perhaps I can consider the angles involved. If I can show that the angles at P₁ and P₂ subtended by the segment T₁'T₂' are supplementary, that would prove the quadrilateral is cyclic.Let me think about the angles. Since P₁ is on ω₁ and T₁' is also on ω₁, the angle ∠P₁T₁'T₂' should be equal to the angle ∠P₁T₁T₂ because they both subtend the same arc in ω₁. Similarly, in ω₂, the angle ∠P₂T₂'T₁' should be equal to the angle ∠P₂T₂T₁.Wait, I'm not sure if that's correct. Maybe I need to consider the angles at P₁ and P₂ with respect to the lines MT₁' and MT₂'. Since P₁ and P₂ are intersections of these lines with the circles, perhaps there's some similarity or congruence I can exploit.Another approach could be to use homothety or inversion, but that might be more advanced than needed here. Let me stick to basic circle theorems and power of a point.Given that MP₁ * MT₁' = MP₂ * MT₂', maybe I can set up a ratio or proportion that would help me show that the quadrilateral is cyclic. If I can express the power of another point with respect to the potential circle through P₁, P₂, T₁', T₂', that might help.Alternatively, perhaps I can consider triangle similarities. If I can show that triangles MP₁T₁' and MP₂T₂' are similar, that might lead me somewhere.Wait, let's think about the power of point M again. Since M has equal power with respect to both circles, it lies on the radical axis of ω₁ and ω₂. The radical axis is the set of points with equal power with respect to both circles, so M is on the radical axis.But how does that help with showing P₁P₂T₁'T₂' is cyclic? Maybe I need to consider another radical axis or some other property.Let me try to consider the power of point P₁ with respect to ω₂. Since P₁ is on ω₁, its power with respect to ω₂ is equal to the square of the tangent from P₁ to ω₂. Similarly, the power of P₂ with respect to ω₁ is the square of the tangent from P₂ to ω₁.But I'm not sure if that's directly helpful. Maybe I need to look at the angles again.Alternatively, perhaps I can use the fact that the angles formed by the tangents and the lines MT₁' and MT₂' have some relationship.Wait, since t and t' are both tangents, the angles at T₁ and T₁' with respect to the centers of the circles are right angles. Maybe that can help me establish some perpendicularity or right triangles that can be used in the proof.Let me try to draw a diagram in my mind. There are two circles, two common tangents, points of tangency, midpoint M, and lines from M to the other points of tangency intersecting the circles again at P₁ and P₂.I think the key is the equal power of M with respect to both circles, leading to the equality MP₁ * MT₁' = MP₂ * MT₂'. This suggests that the points P₁, P₂, T₁', T₂' lie on a circle where M has equal power, but I need to formalize this.Wait, maybe I can use the power of point M with respect to the circle passing through P₁, P₂, T₁', T₂'. If I can show that the power of M with respect to this circle is the same as its power with respect to ω₁ and ω₂, then that would imply that P₁P₂T₁'T₂' is cyclic.But I'm not sure how to directly compute the power of M with respect to the new circle. Maybe I need another approach.Alternatively, perhaps I can consider the cyclic quadrilateral condition. For four points to be cyclic, the product of the lengths of the diagonals can be related to the sum of the products of opposite sides, but that might be too involved.Wait, another idea: if I can show that the angles ∠P₁T₁'T₂' and ∠P₂T₂'T₁' are equal or supplementary, that would help. Since P₁ and P₂ are intersections of lines from M, maybe there's a way to relate these angles through the properties of the circles.Alternatively, maybe I can use the fact that the angles subtended by the same chord are equal. If I can find such chords in the quadrilateral, that might work.Hmm, I'm going in circles a bit. Let me try to summarize what I have so far:1. M is the midpoint of [T₁T₂], so MT₁ = MT₂.2. Therefore, power of M with respect to ω₁ and ω₂ is equal: MT₁² = MT₂².3. By power of a point, MP₁ * MT₁' = MP₂ * MT₂'.4. I need to show that P₁P₂T₁'T₂' is cyclic.Maybe I can consider the circle passing through P₁, T₁', P₂, and T₂' and show that the power of M with respect to this circle is equal to MP₁ * MT₁' and MP₂ * MT₂', which are equal. Therefore, M has equal power with respect to this circle, implying that the points lie on a circle.Wait, that might be the way to go. If I can show that the power of M with respect to the circle through P₁, P₂, T₁', T₂' is equal to both MP₁ * MT₁' and MP₂ * MT₂', which are equal, then the points must lie on a circle.So, let me formalize that:Let’s assume that P₁, P₂, T₁', T₂' lie on a circle Γ. Then, the power of M with respect to Γ is equal to MP₁ * MT₁' and also equal to MP₂ * MT₂'. Since we already have MP₁ * MT₁' = MP₂ * MT₂', this would confirm that M has equal power with respect to Γ, which is consistent.But wait, I need to show that such a circle Γ exists. Perhaps I can construct it or use the radical axis theorem.Alternatively, since the power of M with respect to both ω₁ and ω₂ is equal, and the power with respect to Γ is also equal, M lies on the radical axis of ω₁ and ω₂, and also on the radical axis of Γ and ω₁ (or Γ and ω₂). Therefore, the radical axes must coincide, implying that Γ is the same as the radical axis, but that might not directly help.Wait, maybe I'm overcomplicating it. Since MP₁ * MT₁' = MP₂ * MT₂', and these products are equal, it suggests that P₁, P₂, T₁', T₂' lie on a circle for which M has equal power. Therefore, by the converse of the power of a point theorem, these four points must lie on a circle.Yes, that makes sense. Because if the power of M with respect to the circle through P₁, P₂, T₁', T₂' is equal to both MP₁ * MT₁' and MP₂ * MT₂', and these are equal, then the points must lie on a circle.So, putting it all together:1. M is the midpoint of [T₁T₂], so MT₁ = MT₂.2. Therefore, power of M with respect to ω₁ and ω₂ is equal: MT₁² = MT₂².3. By power of a point, MP₁ * MT₁' = MP₂ * MT₂'.4. Since these products are equal, the points P₁, P₂, T₁', T₂' lie on a circle where M has equal power, hence they are concyclic.Therefore, quadrilateral P₁P₂T₁'T₂' is cyclic.I think that's a solid line of reasoning. I started by understanding the given points and their properties, then used the power of a point theorem to establish an equality, and finally connected that to the cyclic quadrilateral condition. It makes sense, and I don't see any gaps in the logic.

answer：Okay, so I have this problem where I'm given a function ( f(x) = log_{2}(x + 1) ). Then, it says that as the point ((x, y)) moves along the graph of ( f(x) ), the point (left(frac{x}{3}, frac{y}{2}right)) moves along the graph of ( y = g(x) ). I need to find the maximum value of the function ( h(x) = g(x) - f(x) ).Alright, let's break this down step by step. First, I know that ( f(x) = log_{2}(x + 1) ). So, for any point ((x, y)) on the graph of ( f(x) ), ( y = log_{2}(x + 1) ). Now, the problem says that when we take the point (left(frac{x}{3}, frac{y}{2}right)), it lies on the graph of ( y = g(x) ). So, if I denote the new point as ((u, v)), then ( u = frac{x}{3} ) and ( v = frac{y}{2} ). Since ( v = g(u) ), this means ( frac{y}{2} = gleft(frac{x}{3}right) ).But since ( y = log_{2}(x + 1) ), substituting that in, we get ( frac{1}{2} log_{2}(x + 1) = gleft(frac{x}{3}right) ). Hmm, so to find ( g(x) ), I need to express this in terms of ( x ). Let me set ( u = frac{x}{3} ), which implies ( x = 3u ). Substituting back into the equation, we get ( g(u) = frac{1}{2} log_{2}(3u + 1) ). So, replacing ( u ) with ( x ), we have ( g(x) = frac{1}{2} log_{2}(3x + 1) ). That makes sense because we're scaling the input ( x ) by 3 and the output ( y ) by 2.Now, the function ( h(x) = g(x) - f(x) ) becomes ( h(x) = frac{1}{2} log_{2}(3x + 1) - log_{2}(x + 1) ). I need to find the maximum value of this function. To do that, I should probably simplify ( h(x) ) first. Let's use logarithm properties to combine these terms. Recall that ( a log_b c = log_b (c^a) ) and ( log_b c - log_b d = log_b left( frac{c}{d} right) ). Applying these properties:( h(x) = frac{1}{2} log_{2}(3x + 1) - log_{2}(x + 1) = log_{2}(3x + 1)^{1/2} - log_{2}(x + 1) = log_{2} left( frac{sqrt{3x + 1}}{x + 1} right) ).Hmm, that's a bit simpler. So, ( h(x) = log_{2} left( frac{sqrt{3x + 1}}{x + 1} right) ).To find the maximum value of ( h(x) ), I need to maximize the argument of the logarithm because the logarithm function is monotonically increasing. That is, the maximum of ( h(x) ) occurs at the maximum of ( frac{sqrt{3x + 1}}{x + 1} ).Let me denote ( k(x) = frac{sqrt{3x + 1}}{x + 1} ). So, I need to find the maximum of ( k(x) ).To find the maximum of ( k(x) ), I can take its derivative with respect to ( x ) and set it equal to zero. First, let's write ( k(x) = (3x + 1)^{1/2} cdot (x + 1)^{-1} ).Taking the derivative using the product rule:( k'(x) = frac{1}{2}(3x + 1)^{-1/2} cdot 3 cdot (x + 1)^{-1} + (3x + 1)^{1/2} cdot (-1)(x + 1)^{-2} ).Simplify each term:First term: ( frac{3}{2}(3x + 1)^{-1/2}(x + 1)^{-1} ).Second term: ( - (3x + 1)^{1/2}(x + 1)^{-2} ).So, ( k'(x) = frac{3}{2(3x + 1)^{1/2}(x + 1)} - frac{(3x + 1)^{1/2}}{(x + 1)^2} ).To combine these terms, let's find a common denominator. The common denominator would be ( 2(3x + 1)^{1/2}(x + 1)^2 ).So, rewriting each term:First term: ( frac{3(x + 1)}{2(3x + 1)^{1/2}(x + 1)^2} ).Second term: ( frac{2(3x + 1)}{2(3x + 1)^{1/2}(x + 1)^2} ).So, combining them:( k'(x) = frac{3(x + 1) - 2(3x + 1)}{2(3x + 1)^{1/2}(x + 1)^2} ).Simplify the numerator:( 3(x + 1) - 2(3x + 1) = 3x + 3 - 6x - 2 = -3x + 1 ).So, ( k'(x) = frac{-3x + 1}{2(3x + 1)^{1/2}(x + 1)^2} ).To find critical points, set ( k'(x) = 0 ):( frac{-3x + 1}{2(3x + 1)^{1/2}(x + 1)^2} = 0 ).The denominator is always positive for ( x > -1/3 ) (since ( 3x + 1 > 0 ) and ( x + 1 > 0 )), so the numerator must be zero:( -3x + 1 = 0 ) → ( x = frac{1}{3} ).So, the critical point is at ( x = frac{1}{3} ).Now, we should check if this is a maximum. Since ( k'(x) ) changes from positive to negative as ( x ) increases through ( frac{1}{3} ), it is indeed a maximum.Now, let's compute ( kleft( frac{1}{3} right) ):( kleft( frac{1}{3} right) = frac{sqrt{3 cdot frac{1}{3} + 1}}{frac{1}{3} + 1} = frac{sqrt{1 + 1}}{frac{4}{3}} = frac{sqrt{2}}{frac{4}{3}} = frac{3sqrt{2}}{4} ).So, the maximum value of ( k(x) ) is ( frac{3sqrt{2}}{4} ).Therefore, the maximum value of ( h(x) = log_{2} left( frac{sqrt{3x + 1}}{x + 1} right) ) is ( log_{2} left( frac{3sqrt{2}}{4} right) ).Let me simplify that:( log_{2} left( frac{3sqrt{2}}{4} right) = log_{2} left( frac{3}{4} cdot sqrt{2} right) = log_{2} left( frac{3}{4} right) + log_{2} left( sqrt{2} right) ).We know that ( log_{2} left( sqrt{2} right) = frac{1}{2} ) because ( sqrt{2} = 2^{1/2} ).Also, ( log_{2} left( frac{3}{4} right) = log_{2} 3 - log_{2} 4 = log_{2} 3 - 2 ).So, putting it all together:( log_{2} left( frac{3sqrt{2}}{4} right) = (log_{2} 3 - 2) + frac{1}{2} = log_{2} 3 - frac{3}{2} ).But wait, is this the maximum value? Let me double-check.Alternatively, perhaps I made a miscalculation earlier. Let's go back.Wait, when I found ( k(x) = frac{sqrt{3x + 1}}{x + 1} ), and its maximum at ( x = frac{1}{3} ) is ( frac{3sqrt{2}}{4} ). So, ( h(x) = log_{2} left( frac{3sqrt{2}}{4} right) ).Alternatively, maybe it's better to express ( frac{3sqrt{2}}{4} ) as ( frac{9}{8} ) raised to some power? Wait, let's see:( frac{3sqrt{2}}{4} = frac{3}{4} cdot 2^{1/2} = frac{3}{4} cdot 2^{1/2} = frac{3}{2^{2}} cdot 2^{1/2} = 3 cdot 2^{-2 + 1/2} = 3 cdot 2^{-3/2} ).Hmm, not sure if that helps. Alternatively, perhaps express it as ( frac{9}{8} ) inside the log.Wait, let me compute ( left( frac{3sqrt{2}}{4} right)^2 = frac{9 cdot 2}{16} = frac{18}{16} = frac{9}{8} ).Ah! So, ( frac{3sqrt{2}}{4} = sqrt{frac{9}{8}} ). Therefore, ( log_{2} left( sqrt{frac{9}{8}} right) = frac{1}{2} log_{2} left( frac{9}{8} right) ).Yes, that's a cleaner way to express it. So, ( h(x) ) reaches its maximum value at ( frac{1}{2} log_{2} left( frac{9}{8} right) ).Therefore, the maximum value of ( h(x) ) is ( frac{1}{2} log_{2} left( frac{9}{8} right) ).I think that's the answer. Let me just recap:1. Expressed ( g(x) ) in terms of ( f(x) ).2. Formulated ( h(x) = g(x) - f(x) ).3. Simplified ( h(x) ) using logarithm properties.4. Set up ( k(x) ) as the argument of the logarithm.5. Took the derivative of ( k(x) ) to find critical points.6. Found that the maximum occurs at ( x = frac{1}{3} ).7. Calculated the maximum value of ( k(x) ) and then transformed it back into the logarithmic form for ( h(x) ).Everything seems to check out. I don't see any mistakes in the calculations, so I'm confident this is the correct maximum value.**Final Answer**The maximum value of ( h(x) ) is boxed{frac{1}{2} log_{2} left( frac{9}{8} right)}.

answer：First, I need to determine the speed of the boat in still water given its downstream and upstream speeds. I know that when the boat is moving downstream, its effective speed is the sum of its speed in still water and the speed of the current. Conversely, when moving upstream, its effective speed is the difference between its speed in still water and the speed of the current.I'll set up two equations based on this information:1. Speed downstream: ( V_b + V_c = 16 ) km/hr2. Speed upstream: ( V_b - V_c = 9 ) km/hrTo find the speed of the boat in still water (( V_b )), I can add these two equations to eliminate the speed of the current (( V_c )):[ (V_b + V_c) + (V_b - V_c) = 16 + 9 ][ 2V_b = 25 ]Now, I'll solve for ( V_b ):[ V_b = frac{25}{2} ][ V_b = 12.5 ]Therefore, the speed of the boat in still water is 12.5 km/hr.

answer：Okay, so I have this problem about two lines, l₁ and l₂, and I need to figure out whether m=2 is a sufficient, necessary, both, or neither condition for the lines to be parallel. The options are A, B, C, D, which correspond to these possibilities.First, let me recall what it means for two lines to be parallel. I think two lines are parallel if their slopes are equal. So, I need to find the slopes of l₁ and l₂ and set them equal to each other, then solve for m. That should give me the condition for them to be parallel.The equations of the lines are given as:- l₁: 2x - my = 1- l₂: (m - 1)x - y = 1I remember that to find the slope of a line in the form Ax + By = C, the slope is -A/B. So, let me apply that here.For l₁: A = 2, B = -m. So, the slope of l₁ is -A/B = -2/(-m) = 2/m.For l₂: A = (m - 1), B = -1. So, the slope of l₂ is -A/B = -(m - 1)/(-1) = (m - 1)/1 = m - 1.Now, setting the slopes equal for parallelism:2/m = m - 1Hmm, okay, that gives me an equation to solve for m. Let me write that down:2/m = m - 1To solve for m, I can multiply both sides by m to eliminate the denominator:2 = m(m - 1)Expanding the right side:2 = m² - mBring all terms to one side to form a quadratic equation:m² - m - 2 = 0Now, let's solve this quadratic equation. I can use the quadratic formula, where a = 1, b = -1, c = -2.The quadratic formula is m = [-b ± sqrt(b² - 4ac)] / (2a)Plugging in the values:m = [1 ± sqrt(1 + 8)] / 2m = [1 ± sqrt(9)] / 2m = [1 ± 3] / 2So, the solutions are:m = (1 + 3)/2 = 4/2 = 2m = (1 - 3)/2 = (-2)/2 = -1Therefore, the solutions are m = 2 and m = -1.Wait, so m can be either 2 or -1 for the lines to be parallel. That means m=2 is one of the possible values, but not the only one. So, if m=2, the lines are definitely parallel, but there's another value, m=-1, that also makes them parallel.So, m=2 is sufficient because it guarantees the lines are parallel, but it's not necessary because there's another value, m=-1, that also works. Therefore, m=2 is a sufficient but not necessary condition.Let me double-check this. If m=2, then l₁ becomes 2x - 2y = 1, which simplifies to x - y = 0.5, and l₂ becomes (2 - 1)x - y = 1, which is x - y = 1. These two lines have the same slope, 1, so they are parallel.If m=-1, then l₁ becomes 2x - (-1)y = 1, which is 2x + y = 1, and l₂ becomes (-1 - 1)x - y = 1, which is -2x - y = 1. Let me find their slopes.For l₁: slope is -A/B = -2/1 = -2.For l₂: slope is -A/B = -(-2)/(-1) = 2/-1 = -2. So, both slopes are -2, meaning they are parallel.If m is neither 2 nor -1, say m=0, then l₁ becomes 2x = 1, which is a vertical line, and l₂ becomes (-1)x - y = 1, which is y = -x -1, a line with slope -1. These are not parallel because one is vertical and the other has a slope of -1.Similarly, if m=1, l₁ becomes 2x - y = 1, slope is 2, and l₂ becomes (0)x - y = 1, which is y = -1, a horizontal line. These are not parallel either.So, only when m=2 or m=-1 are the lines parallel. Therefore, m=2 is one way to make them parallel, but it's not the only way, so it's sufficient but not necessary.I think that makes sense. So, the answer should be A: A sufficient but not necessary condition.**Final Answer**boxed{A}

answer：First, I need to identify the relationships between the dimensions of the wall. The height is 6 times the width, and the length is 7 times the height. This means I can express all dimensions in terms of the width.Next, I'll use the formula for the volume of a rectangular prism, which is the product of its width, height, and length. By substituting the relationships into the volume formula, I can create an equation that includes only the width.Once the equation is set up, I'll simplify it to solve for the width. This involves dividing the volume by the product of the coefficients obtained from the relationships. Finally, by taking the cube root of the result, I'll find the value of the width.