answer：To find ( a_2 ) in the sequence, I start with the given initial term ( a_1 = 3 ).The recursive formula provided is ( 2a_{n+1} = a_n + 1 ). To find ( a_2 ), I substitute ( n = 1 ) into the formula:[2a_2 = a_1 + 1]Plugging in the value of ( a_1 ):[2a_2 = 3 + 1 = 4]Finally, I solve for ( a_2 ):[a_2 = frac{4}{2} = 2]

answer：Okay, so I need to find the limit as x approaches 0 of the function (1 - sqrt(cos x)) divided by (1 - cos sqrt(x)). Hmm, that looks a bit tricky, but I think I can figure it out step by step.First, let me write down the function again to make sure I have it right:lim _{x rightarrow 0} frac{1-sqrt{cos x}}{1-cos sqrt{x}}Alright, so both the numerator and the denominator are approaching 0 as x approaches 0 because cos(0) is 1, and sqrt(1) is also 1. So, we have a 0/0 indeterminate form. That means I can probably use L'Hospital's Rule here, which involves taking derivatives of the numerator and the denominator.But before I jump into taking derivatives, maybe I can simplify the expression a bit. Sometimes, multiplying numerator and denominator by the conjugate can help. Let me try that with the numerator first.The numerator is 1 - sqrt(cos x). If I multiply numerator and denominator by the conjugate, which is 1 + sqrt(cos x), that might help eliminate the square root. Let me do that:Numerator becomes: (1 - sqrt(cos x))(1 + sqrt(cos x)) = 1 - (sqrt(cos x))^2 = 1 - cos x.So, the numerator simplifies to 1 - cos x. Now, the denominator is still 1 - cos sqrt(x). So, the entire expression becomes:frac{1 - cos x}{(1 - cos sqrt{x})(1 + sqrt{cos x})}Hmm, okay, so now I have (1 - cos x) in the numerator and (1 - cos sqrt(x)) times (1 + sqrt(cos x)) in the denominator. Maybe I can also do something similar with the denominator? Let's see.The denominator has 1 - cos sqrt(x). If I multiply numerator and denominator by the conjugate of the denominator, which is 1 + cos sqrt(x), that might help. Let me try that:Denominator becomes: (1 - cos sqrt(x))(1 + cos sqrt(x)) = 1 - (cos sqrt(x))^2.So, the denominator simplifies to 1 - cos^2 sqrt(x). Now, the entire expression becomes:frac{(1 - cos x)(1 + cos sqrt{x})}{(1 - cos^2 sqrt{x})(1 + sqrt{cos x})}Wait, but 1 - cos^2 sqrt(x) is equal to sin^2 sqrt(x), right? Because 1 - cos^2 θ = sin^2 θ. So, that simplifies the denominator to sin^2 sqrt(x). So, now the expression is:frac{(1 - cos x)(1 + cos sqrt{x})}{sin^2 sqrt{x} (1 + sqrt{cos x})}Hmm, okay. So, now I have (1 - cos x) in the numerator and sin^2 sqrt(x) in the denominator. Maybe I can express 1 - cos x in terms of sin^2 (x/2). Because I remember that 1 - cos θ = 2 sin^2 (θ/2). Let me apply that identity here.So, 1 - cos x = 2 sin^2 (x/2). Similarly, 1 - cos sqrt(x) = 2 sin^2 (sqrt(x)/2). Wait, but I already used that identity earlier. Anyway, let's substitute that in:Numerator becomes: 2 sin^2 (x/2) * (1 + cos sqrt(x))Denominator becomes: sin^2 sqrt(x) * (1 + sqrt(cos x))So, putting it all together:frac{2 sin^2 (x/2) (1 + cos sqrt{x})}{sin^2 sqrt{x} (1 + sqrt{cos x})}Hmm, okay. So, now I have this expression. Maybe I can split the fraction into two parts:2 cdot frac{sin^2 (x/2)}{sin^2 sqrt{x}} cdot frac{1 + cos sqrt{x}}{1 + sqrt{cos x}}So, now I can consider the limit as x approaches 0 of each part separately.First, let's look at the term 2. That's just a constant, so it stays as is.Next, let's consider the term (sin^2 (x/2)) / (sin^2 sqrt(x)). Maybe I can write this as [sin(x/2) / sin(sqrt(x))]^2. So, the square of the ratio of sines.Similarly, the last term is (1 + cos sqrt(x)) / (1 + sqrt(cos x)). Hmm, not sure about that one yet.Let me tackle the first part: [sin(x/2) / sin(sqrt(x))]^2.I know that as x approaches 0, sin(theta) is approximately equal to theta. So, maybe I can use the small-angle approximation here.So, sin(x/2) ≈ x/2, and sin(sqrt(x)) ≈ sqrt(x). So, substituting these approximations in:sin(x/2) / sin(sqrt(x)) ≈ (x/2) / sqrt(x) = (x)/(2 sqrt(x)) = sqrt(x)/2.So, [sin(x/2) / sin(sqrt(x))]^2 ≈ (sqrt(x)/2)^2 = x/4.Therefore, the first part [sin(x/2) / sin(sqrt(x))]^2 approaches 0 as x approaches 0, because x is going to 0. But wait, that would make the entire expression go to 0, but I have a 2 multiplied by that. Hmm, but let me check if that's accurate.Wait, but maybe I should be more precise with the limits rather than just using approximations. Let me try applying L'Hospital's Rule to the term [sin(x/2) / sin(sqrt(x))]^2.First, let's consider the limit as x approaches 0 of sin(x/2) / sin(sqrt(x)).Let me set this up as a limit:lim_{x to 0} frac{sin(x/2)}{sin(sqrt{x})}Since both numerator and denominator approach 0 as x approaches 0, I can apply L'Hospital's Rule. So, take derivatives of numerator and denominator.Derivative of numerator: (1/2) cos(x/2)Derivative of denominator: (1/(2 sqrt(x))) cos(sqrt(x))So, applying L'Hospital's Rule, the limit becomes:lim_{x to 0} frac{(1/2) cos(x/2)}{(1/(2 sqrt{x})) cos(sqrt{x})}Simplify this expression:The 1/2 cancels out, so we have:lim_{x to 0} frac{cos(x/2)}{(1/sqrt{x}) cos(sqrt{x})} = lim_{x to 0} sqrt{x} cdot frac{cos(x/2)}{cos(sqrt{x})}Now, as x approaches 0, cos(x/2) approaches cos(0) = 1, and cos(sqrt(x)) approaches cos(0) = 1. So, the expression simplifies to:lim_{x to 0} sqrt{x} cdot frac{1}{1} = lim_{x to 0} sqrt{x} = 0So, the limit of sin(x/2)/sin(sqrt(x)) as x approaches 0 is 0. Therefore, the square of that limit is also 0.Wait, but hold on, that would mean that the first part of our expression is approaching 0, but let's not forget that we have another term multiplied by it: 2 * [sin(x/2)/sin(sqrt(x))]^2 * [ (1 + cos sqrt(x)) / (1 + sqrt(cos x)) ]So, if the first part is approaching 0, but the other term might be approaching some finite value, so the entire limit could be 0. But I need to check the other term as well.Let me look at the term (1 + cos sqrt(x)) / (1 + sqrt(cos x)).As x approaches 0, sqrt(x) approaches 0, so cos sqrt(x) approaches cos 0 = 1. Similarly, sqrt(cos x) approaches sqrt(1) = 1. So, both numerator and denominator approach 1 + 1 = 2. So, the ratio (1 + cos sqrt(x)) / (1 + sqrt(cos x)) approaches 2/2 = 1.Therefore, the entire expression becomes:2 * 0 * 1 = 0.So, is the limit 0? Hmm, but wait, I feel like I might have made a mistake here because sometimes when dealing with limits, especially with products, even if one part goes to 0 and the other goes to infinity, the overall limit might not be straightforward. But in this case, the first part is going to 0, and the other part is going to 1, so the product should indeed go to 0.But let me double-check my steps because sometimes when using approximations, we might miss something.Alternatively, maybe I can use series expansions for the functions involved. Let's try that approach.First, let's expand cos x and sqrt(cos x) around x = 0.We know that cos x can be expanded as:cos x = 1 - x^2/2 + x^4/24 - ... Similarly, sqrt(cos x) can be expanded. Let me recall that sqrt(1 + y) ≈ 1 + y/2 - y^2/8 + ... for small y.So, let me write cos x as 1 - x^2/2 + higher order terms, so y = -x^2/2 + ... Therefore, sqrt(cos x) ≈ sqrt(1 - x^2/2) ≈ 1 - (x^2)/4 - (x^4)/96 + ...Wait, let me compute that more carefully.Let me set y = -x^2/2, so sqrt(1 + y) ≈ 1 + y/2 - y^2/8 + ... So, substituting y = -x^2/2:sqrt(1 - x^2/2) ≈ 1 + (-x^2/2)/2 - ( (-x^2/2)^2 ) /8 + ... = 1 - x^2/4 - (x^4)/32 + ...So, sqrt(cos x) ≈ 1 - x^2/4 - x^4/32 + ...Therefore, 1 - sqrt(cos x) ≈ x^2/4 + x^4/32 + ...Similarly, let's expand the denominator: 1 - cos sqrt(x).First, compute cos sqrt(x). Let me set z = sqrt(x), so cos z ≈ 1 - z^2/2 + z^4/24 - ... So, substituting z = sqrt(x):cos sqrt(x) ≈ 1 - (sqrt(x))^2 / 2 + (sqrt(x))^4 / 24 - ... = 1 - x/2 + x^2 / 24 - ...Therefore, 1 - cos sqrt(x) ≈ x/2 - x^2 / 24 + ...So, putting it all together, the numerator is approximately x^2/4, and the denominator is approximately x/2.So, the ratio (1 - sqrt(cos x)) / (1 - cos sqrt(x)) ≈ (x^2/4) / (x/2) = (x^2/4) * (2/x) = x/2.Therefore, as x approaches 0, this ratio approaches 0.So, according to this series expansion, the limit is 0.Wait, but earlier when I used L'Hospital's Rule, I also got 0, so that seems consistent.But just to make sure, let me try another approach. Maybe using substitution.Let me set t = sqrt(x). Then, as x approaches 0, t approaches 0 as well. So, x = t^2.Substituting into the original limit:lim_{t to 0} frac{1 - sqrt{cos(t^2)}}{1 - cos t}So, now the limit is as t approaches 0 of [1 - sqrt(cos(t^2))] / [1 - cos t].Hmm, maybe this substitution makes it easier to handle.Again, both numerator and denominator approach 0 as t approaches 0, so we can apply L'Hospital's Rule here.Taking derivatives of numerator and denominator with respect to t.First, numerator: d/dt [1 - sqrt(cos(t^2))] = - (1/(2 sqrt(cos(t^2)))) * (-sin(t^2)) * 2t = (sin(t^2) * t) / sqrt(cos(t^2))Denominator: d/dt [1 - cos t] = sin tSo, applying L'Hospital's Rule, the limit becomes:lim_{t to 0} frac{sin(t^2) cdot t / sqrt{cos(t^2)}}{sin t}Simplify this expression:= lim_{t→0} [ t sin(t^2) / (sqrt(cos(t^2)) sin t) ]Again, as t approaches 0, sin(t^2) ≈ t^2, sin t ≈ t, and sqrt(cos(t^2)) ≈ 1 - (t^4)/8, but since t is approaching 0, sqrt(cos(t^2)) ≈ 1.So, substituting these approximations:≈ lim_{t→0} [ t * t^2 / (1 * t) ] = lim_{t→0} [ t^3 / t ] = lim_{t→0} t^2 = 0Therefore, the limit is 0.Wait, so both methods give me 0. Hmm, but earlier when I used the series expansion, I also got 0. So, seems consistent.But just to make sure, let me compute the limit numerically for some small x.Let me pick x = 0.0001.Compute numerator: 1 - sqrt(cos(0.0001)).cos(0.0001) ≈ 1 - (0.0001)^2 / 2 = 1 - 0.000000005 = 0.999999995sqrt(0.999999995) ≈ 0.9999999975So, numerator ≈ 1 - 0.9999999975 ≈ 0.0000000025Denominator: 1 - cos(sqrt(0.0001)) = 1 - cos(0.01)cos(0.01) ≈ 1 - (0.01)^2 / 2 = 1 - 0.00005 = 0.99995So, denominator ≈ 1 - 0.99995 = 0.00005So, the ratio ≈ 0.0000000025 / 0.00005 ≈ 0.00005Which is 5e-5, which is approaching 0 as x approaches 0.Similarly, if I take x = 0.000001.Numerator: 1 - sqrt(cos(0.000001)).cos(0.000001) ≈ 1 - (0.000001)^2 / 2 = 1 - 0.0000000000005 ≈ 0.9999999999995sqrt(0.9999999999995) ≈ 0.99999999999975So, numerator ≈ 1 - 0.99999999999975 ≈ 0.00000000000025Denominator: 1 - cos(sqrt(0.000001)) = 1 - cos(0.001)cos(0.001) ≈ 1 - (0.001)^2 / 2 = 1 - 0.0000005 = 0.9999995So, denominator ≈ 1 - 0.9999995 = 0.0000005Therefore, the ratio ≈ 0.00000000000025 / 0.0000005 ≈ 0.0000005Which is 5e-7, again approaching 0.So, numerically, it seems the limit is indeed 0.Wait, but just to make sure, let me try x = 0.01.Numerator: 1 - sqrt(cos(0.01)).cos(0.01) ≈ 0.99995sqrt(0.99995) ≈ 0.999975So, numerator ≈ 1 - 0.999975 ≈ 0.000025Denominator: 1 - cos(sqrt(0.01)) = 1 - cos(0.1)cos(0.1) ≈ 0.995004So, denominator ≈ 1 - 0.995004 ≈ 0.004996Therefore, ratio ≈ 0.000025 / 0.004996 ≈ 0.005Which is 5e-3, which is still approaching 0 as x approaches 0.So, all these numerical checks support the conclusion that the limit is 0.But just to make sure, let me think about the behavior of the functions.The numerator is 1 - sqrt(cos x). As x approaches 0, cos x approaches 1, so sqrt(cos x) approaches 1. Therefore, 1 - sqrt(cos x) approaches 0. Similarly, the denominator is 1 - cos sqrt(x), which also approaches 0 as x approaches 0.But how fast does each approach 0?From the series expansion, we saw that numerator ~ x^2/4 and denominator ~ x/2, so their ratio ~ x/2, which approaches 0.Alternatively, using substitution t = sqrt(x), we saw that the limit becomes [t sin(t^2)] / [sqrt(cos(t^2)) sin t], which simplifies to [t sin(t^2)] / [sqrt(cos(t^2)) sin t]. As t approaches 0, sin(t^2) ~ t^2, sin t ~ t, so the expression becomes [t * t^2] / [1 * t] = t^2, which approaches 0.So, all methods point to the limit being 0.Therefore, I think the limit is 0.**Final Answer**The limit is boxed{0}.

answer：Okay, so I have this problem where there are rectangles arranged in a specific pattern. Each rectangle is 4 units by 2 units. The way they're positioned is such that the top left vertex of each subsequent rectangle is placed at the midpoint of the bottom edge of the rectangle above it. I need to figure out the perimeter of the entire figure when ten rectangles are arranged this way. The answer choices are 48, 64, 90, 84, and 100. Hmm, let me try to visualize this.First, let me sketch a rough diagram in my mind. The first rectangle is straightforward, 4 units wide and 2 units tall. The next rectangle is placed such that its top left corner is at the midpoint of the bottom edge of the first rectangle. Since the bottom edge of the first rectangle is 4 units long, the midpoint would be at 2 units from the left. So, the top left corner of the second rectangle is at (2, 0) if the first rectangle is placed with its bottom left corner at (0,0). Wait, actually, the first rectangle's bottom edge goes from (0,0) to (4,0). The midpoint is at (2,0). So, the second rectangle is placed such that its top left corner is at (2,0). Since the rectangle is 4 units wide and 2 units tall, its bottom edge would extend from (2,0) to (6,0), but wait, that can't be right because the top left is at (2,0), so the rectangle would extend to the right and downward. Hmm, maybe I need to adjust my coordinate system.Let me think again. If the first rectangle is placed with its top left corner at (0,0), then it would extend to (4,0) at the bottom right. The midpoint of the bottom edge is at (2,0). So, the next rectangle's top left corner is at (2,0). Since the rectangle is 4 units wide, it would extend from (2,0) to (6,0) on the bottom edge, but it's also 2 units tall, so it would go down to (2,-2) and (6,-2). Wait, so the second rectangle is placed below and to the right of the first one.But then, how does this affect the overall figure? Each subsequent rectangle is placed such that its top left is at the midpoint of the bottom edge of the one above. So, each time, the horizontal position shifts by 2 units to the right, and the vertical position shifts down by 2 units, since the height of each rectangle is 2 units.Wait, no. The vertical shift isn't necessarily 2 units because the rectangle is placed such that its top is at the midpoint of the bottom of the previous rectangle. So, the vertical shift is actually 2 units downward because the height of each rectangle is 2 units. So, each new rectangle is shifted 2 units to the right and 2 units down relative to the previous one.But if we have ten rectangles arranged this way, the figure will form some sort of stepped shape, maybe like a zig-zag or a spiral? Hmm, not sure. Maybe it's a straight line? Wait, each rectangle is shifted 2 units right and 2 units down, so the horizontal and vertical shifts are equal. So, the figure would form a diagonal line, but with each step being a rectangle.Wait, actually, each rectangle is placed such that the top left is at the midpoint of the bottom edge of the previous one. So, the horizontal shift is half the width of the rectangle, which is 2 units, and the vertical shift is equal to the height of the rectangle, which is 2 units. So, each subsequent rectangle is shifted 2 units to the right and 2 units down.So, the figure is built by placing each rectangle diagonally, each time moving 2 units right and 2 units down. So, after ten rectangles, how does the overall figure look?Let me think about the coordinates. The first rectangle is from (0,0) to (4,2). The second rectangle is placed with its top left at (2,0), so it goes from (2,0) to (6,2). Wait, no, because if the top left is at (2,0), and it's 4 units wide and 2 units tall, then it would extend to (6,0) on the bottom edge and down to (2,-2) and (6,-2). Hmm, that seems like it's overlapping with the first rectangle.Wait, maybe I'm getting confused with the coordinate system. Maybe I should consider the first rectangle as being placed with its top left corner at (0,0), so it goes from (0,0) to (4,2). Then, the midpoint of its bottom edge is at (2,2). So, the next rectangle is placed with its top left corner at (2,2). Since it's 4 units wide and 2 units tall, it would extend from (2,2) to (6,4). Wait, that's going upwards, but the problem says the top left of each subsequent rectangle is placed at the midpoint of the bottom edge of the rectangle above it. So, actually, the second rectangle is placed below the first one.Wait, maybe I need to adjust my initial placement. Let me try again.Let's say the first rectangle is placed with its bottom left corner at (0,0). So, it extends to (4,0) on the bottom edge and up to (0,2) on the top left. The midpoint of the bottom edge is at (2,0). So, the second rectangle is placed with its top left corner at (2,0). Since it's 4 units wide and 2 units tall, its bottom edge would be from (2,0) to (6,0), and its top edge would be from (2,2) to (6,2). Wait, so the second rectangle is placed to the right and overlapping with the first rectangle? Because the first rectangle is from (0,0) to (4,2), and the second is from (2,0) to (6,2). So, they overlap between (2,0) to (4,2). Hmm, that seems like they are overlapping.But in the problem, it says "the top left vertex of each rectangle (after the top one) is placed at the midpoint of the bottom edge of the rectangle above it." So, maybe the first rectangle is the top one, and each subsequent rectangle is placed below it.Wait, perhaps the figure is built by stacking rectangles in a way where each new rectangle is partially overlapping the previous one, creating a kind of spiral or zig-zag pattern.Alternatively, maybe it's better to think in terms of the overall figure's dimensions. Let me try to figure out how the figure expands with each added rectangle.Each rectangle is 4 units wide and 2 units tall. When placing the second rectangle, its top left is at the midpoint of the first rectangle's bottom edge. So, the horizontal position shifts by 2 units to the right, and the vertical position shifts down by 2 units.So, for each rectangle added, the figure extends 2 units to the right and 2 units downward. So, after n rectangles, the figure would have extended 2*(n-1) units to the right and 2*(n-1) units downward.Wait, but that might not account for the entire width and height. Let me think.The first rectangle is 4 units wide and 2 units tall. The second rectangle is placed 2 units to the right and 2 units down, so its top left is at (2,0). But its width is 4 units, so it extends from (2,0) to (6,0) on the bottom edge, and up to (2,2) on the top left. So, the total width after two rectangles is 6 units, and the total height is still 2 units because the second rectangle is only 2 units tall, but placed below the first one.Wait, no, the height would be 4 units because the first rectangle is 2 units tall, and the second rectangle is placed 2 units below it, so the total height is 2 + 2 = 4 units.Wait, maybe not. Let me clarify.If the first rectangle is from (0,0) to (4,2). The second rectangle is placed with its top left at (2,0), so it extends from (2,0) to (6,2). So, the figure now spans from (0,0) to (6,2). But wait, the second rectangle is placed such that its top left is at (2,0), so it's actually overlapping with the first rectangle.Wait, maybe I need to think of it differently. Maybe each subsequent rectangle is placed such that it's half overlapping with the previous one. So, the figure's width increases by 2 units each time, and the height increases by 2 units each time.Wait, let's consider the overall figure after n rectangles. The first rectangle is 4x2. The second rectangle is placed 2 units to the right and 2 units down, so the figure's width becomes 4 + 2 = 6 units, and the height becomes 2 + 2 = 4 units. The third rectangle would be placed 2 units to the right of the second one's midpoint, so 2 + 2 = 4 units from the first rectangle's left edge, and 2 units down from the second rectangle's bottom edge, so 4 units down from the first rectangle's top edge.Wait, no, each subsequent rectangle is placed 2 units to the right and 2 units down relative to the previous one. So, after n rectangles, the total width would be 4 + 2*(n-1) units, and the total height would be 2 + 2*(n-1) units.Wait, let me test this with n=2. For n=2, width would be 4 + 2*(1) = 6, height would be 2 + 2*(1) = 4. That seems correct because the second rectangle extends the figure to 6 units wide and 4 units tall.Similarly, for n=3, width would be 4 + 2*2 = 8, height would be 2 + 2*2 = 6. So, each rectangle adds 2 units to both width and height.Therefore, for n=10, the total width would be 4 + 2*(10-1) = 4 + 18 = 22 units, and the total height would be 2 + 2*(10-1) = 2 + 18 = 20 units.Wait, but is that accurate? Because each rectangle is placed such that it's only overlapping partially, so the overall figure might not just be a simple rectangle of 22x20. It might have a more complex shape, with some parts sticking out.Alternatively, maybe the figure is a larger rectangle with these dimensions, but with some internal lines that don't contribute to the perimeter. Hmm, I'm not sure.Wait, perhaps I should think about the perimeter in terms of the outer edges. Each time a rectangle is added, it covers some part of the previous figure but also extends the perimeter.Alternatively, maybe the figure is a kind of spiral, where each rectangle adds to the perimeter on two sides.Wait, let me try to think step by step.First rectangle: 4x2. Perimeter is 2*(4+2) = 12.Second rectangle: placed such that its top left is at the midpoint of the first rectangle's bottom edge. So, the second rectangle is placed 2 units to the right and 2 units down from the first rectangle. So, the figure now has a kind of L-shape, but connected.Wait, actually, when you place the second rectangle, it overlaps the first one partially. So, the total figure's perimeter would not just be the sum of the perimeters of both rectangles minus the overlapping parts.But this might get complicated with ten rectangles. Maybe there's a pattern or formula.Alternatively, perhaps the figure after n rectangles has a width of 4 + 2*(n-1) and a height of 2 + 2*(n-1). So, for n=10, width=22, height=20. Then, the perimeter would be 2*(22+20)=84. Hmm, 84 is one of the answer choices, option D.But wait, is that correct? Because the figure isn't a perfect rectangle; it's more like a stepped figure. So, the perimeter might not just be that of a 22x20 rectangle.Wait, let me think again. If each rectangle is placed such that it's shifted 2 units right and 2 units down, then the overall figure would have a kind of diagonal edge, but with each step being a rectangle.So, the figure would have a sort of zig-zag perimeter on the right and bottom sides. So, the total perimeter would be more than that of a 22x20 rectangle.Wait, maybe I need to calculate the horizontal and vertical components separately.Each rectangle is 4 units wide and 2 units tall. When placed in this pattern, each subsequent rectangle adds 2 units to the width and 2 units to the height, but also creates a kind of step.Wait, perhaps the total width is 4 + 2*(n-1) and the total height is 2 + 2*(n-1). So, for n=10, width=22, height=20.But the perimeter isn't just 2*(22+20)=84 because the figure has internal edges that are not on the perimeter.Wait, maybe it's better to think about the figure as a combination of horizontal and vertical lines.Each time a rectangle is added, it contributes some new edges to the perimeter.Let me try to think about how the perimeter grows with each rectangle.First rectangle: 4x2. Perimeter is 12.Second rectangle: placed 2 units right and 2 units down from the first. So, the second rectangle's top edge is from (2,0) to (6,0), but the first rectangle's bottom edge is from (0,0) to (4,0). So, the overlapping part is from (2,0) to (4,0). Therefore, the new perimeter contributed by the second rectangle is the right side from (4,0) to (4,2), then up to (6,2), and then left to (2,2), but wait, that might not be accurate.Wait, maybe it's better to think in terms of the overall figure.After the first rectangle, the figure is 4x2.After the second rectangle, the figure extends to the right by 2 units and down by 2 units, so the figure is now 6 units wide and 4 units tall, but with a kind of notch where they overlap.Wait, actually, the perimeter might increase by 4 units each time a new rectangle is added.Wait, let me think about the first rectangle: perimeter 12.Adding the second rectangle: the overlapping part is 2 units in width and 2 units in height. So, the new perimeter added would be the right side of the second rectangle (4 units) and the bottom side (2 units), but subtracting the overlapping parts.Wait, this is getting confusing. Maybe I should look for a pattern.Alternatively, perhaps the figure is a kind of stepped figure where each step is 2 units wide and 2 units tall.So, for n rectangles, the figure would have n steps, each 2 units wide and 2 units tall.Wait, but each rectangle is 4 units wide and 2 units tall, so each step is actually 2 units in both directions.Wait, maybe the figure is a right triangle with legs of length 2*(n-1). So, for n=10, the legs would be 18 units each. Then, the perimeter would be 2*(18) + 18*sqrt(2). But that doesn't make sense because the figure is made of rectangles, not triangles.Wait, perhaps the figure is a rectangle with some extensions.Wait, maybe I should think about the figure as a combination of horizontal and vertical lines.Each rectangle added contributes 4 units to the width and 2 units to the height, but shifted by 2 units each time.Wait, no, each rectangle is 4 units wide and 2 units tall, but placed such that each subsequent one is shifted 2 units right and 2 units down.So, the figure's width increases by 2 units each time, and the height increases by 2 units each time.Therefore, after n rectangles, the width is 4 + 2*(n-1) and the height is 2 + 2*(n-1).So, for n=10, width=4 + 18=22, height=2 + 18=20.Then, the perimeter would be 2*(22 + 20)=84, which is option D.But wait, is that correct? Because the figure isn't a perfect rectangle; it's more like a stepped figure with each step being 2 units. So, the perimeter might actually be longer because of the steps.Wait, maybe the perimeter is calculated as if it's a rectangle, but with some additional edges.Alternatively, perhaps the figure's perimeter is the same as a rectangle of 22x20 because the steps are internal and don't contribute to the outer perimeter.Wait, let me think about how the figure looks. Each rectangle is placed such that it's overlapping the previous one by 2 units in width and 2 units in height. So, the figure's outer edges are formed by the outermost edges of the rectangles.So, the total width is 4 + 2*(n-1) and the total height is 2 + 2*(n-1). Therefore, the perimeter is 2*(width + height).So, for n=10, width=22, height=20, perimeter=2*(22+20)=84.Therefore, the answer is D) 84.But wait, let me confirm this with a smaller number of rectangles to see if the pattern holds.For n=1: width=4, height=2, perimeter=12. Correct.For n=2: width=6, height=4, perimeter=20. Let's calculate manually.First rectangle: 4x2. Second rectangle placed 2 units right and 2 units down. So, the figure spans from (0,0) to (6,4). But the actual figure is a combination of the two rectangles overlapping. The perimeter would be the outer edges.Top edge: from (0,2) to (4,2), then to (6,2). Wait, no, the second rectangle is from (2,0) to (6,2). So, the top edge is from (0,2) to (6,2). The bottom edge is from (0,0) to (6,0). The left side is from (0,0) to (0,2). The right side is from (6,0) to (6,2). But wait, that would make the figure a 6x2 rectangle, but that's not correct because the second rectangle is placed below the first one.Wait, maybe I'm overcomplicating it. If n=2, the figure is a combination of two rectangles overlapping. The total width is 6 units, and the total height is 4 units. So, the perimeter would be 2*(6+4)=20, which matches the formula. So, for n=2, the perimeter is 20.Similarly, for n=3, the width would be 8, height=6, perimeter=28. Let's see: three rectangles, each placed 2 units right and down. The figure would span 8 units wide and 6 units tall, so perimeter=28.Yes, that seems to hold. So, for n=10, the perimeter would be 2*(22+20)=84.Therefore, the answer is D) 84.

answer：Okay, so I have this function here, f(x) = x³ - 2x² + 3, and I need to find the number of extreme points it has. Hmm, extreme points... I think those are the points where the function reaches a local maximum or minimum. Right, so to find those, I remember from my calculus class that I need to take the derivative of the function and then find where that derivative is zero or undefined. Those points are called critical points, and they're where the function could have maxima or minima.Alright, let me start by finding the derivative of f(x). The function is a polynomial, so its derivative should be straightforward. The derivative of x³ is 3x², the derivative of -2x² is -4x, and the derivative of 3 is 0. So putting that all together, f'(x) = 3x² - 4x. Okay, that seems right.Now, I need to find where this derivative is zero because that's where the function could have extreme points. So I set f'(x) equal to zero and solve for x:3x² - 4x = 0.Hmm, factoring out an x, I get x(3x - 4) = 0. So, the solutions are x = 0 and 3x - 4 = 0. Solving 3x - 4 = 0 gives x = 4/3. So, the critical points are at x = 0 and x = 4/3. That means there are two critical points. But wait, does that mean there are two extreme points? Not necessarily, because sometimes critical points can be points of inflection instead of maxima or minima. So I need to check whether these critical points are maxima or minima.How do I do that? I think there are two methods: the first derivative test and the second derivative test. Maybe I'll try the second derivative test because it seems quicker. Let me find the second derivative of f(x). The first derivative was 3x² - 4x, so the derivative of that is 6x - 4. So f''(x) = 6x - 4.Now, I can plug in the critical points into the second derivative to see if they're maxima or minima. If f''(x) is positive, it's a local minimum; if it's negative, it's a local maximum.First, let's check x = 0. Plugging into f''(x): 6(0) - 4 = -4. That's negative, so at x = 0, the function has a local maximum.Next, let's check x = 4/3. Plugging into f''(x): 6*(4/3) - 4. Let me compute that: 6*(4/3) is 8, and 8 - 4 is 4. That's positive, so at x = 4/3, the function has a local minimum.So, both critical points are extreme points—one maximum and one minimum. Therefore, the function has two extreme points.Wait, but just to make sure I didn't make a mistake, let me double-check my calculations. First derivative: 3x² - 4x. Setting that to zero, factoring gives x = 0 and x = 4/3. Second derivative: 6x - 4. Plugging in x = 0 gives -4, which is concave down, so maximum. Plugging in x = 4/3 gives 6*(4/3) - 4 = 8 - 4 = 4, which is concave up, so minimum. Yeah, that seems correct.Alternatively, I could use the first derivative test to confirm. For x = 0, let's pick points just less than 0 and just greater than 0 to see the sign of the derivative. Let's say x = -1: f'(-1) = 3*(-1)² - 4*(-1) = 3 + 4 = 7, which is positive. Then x = 1: f'(1) = 3*(1)² - 4*(1) = 3 - 4 = -1, which is negative. So the derivative goes from positive to negative as x increases through 0, indicating a local maximum at x = 0.For x = 4/3, let's pick points just less than 4/3 and just greater than 4/3. Let's choose x = 1: f'(1) = 3 - 4 = -1, which is negative. Then x = 2: f'(2) = 3*(4) - 4*(2) = 12 - 8 = 4, which is positive. So the derivative goes from negative to positive as x increases through 4/3, indicating a local minimum at x = 4/3.So both methods agree. Therefore, the function has two extreme points: one local maximum at x = 0 and one local minimum at x = 4/3.Just to visualize, the function is a cubic polynomial. Since the leading term is x³, as x approaches positive infinity, f(x) goes to positive infinity, and as x approaches negative infinity, f(x) goes to negative infinity. So the graph should have a "hill" at x = 0 and a "valley" at x = 4/3. That makes sense with the two extreme points we found.I don't think I made any mistakes here. The critical points are correctly found, and both tests confirm their nature. So yeah, the number of extreme points is two.**Final Answer**The number of extreme points of the function is boxed{2}.

answer：To find the opposite of -2023, I need to determine the number that is the same distance from zero on the number line but in the opposite direction.The opposite of a negative number is a positive number with the same absolute value.Therefore, the opposite of -2023 is 2023.

answer：To solve (6! - 4!), I'll start by calculating each factorial separately.First, (6! = 6 times 5 times 4 times 3 times 2 times 1 = 720).Next, (4! = 4 times 3 times 2 times 1 = 24).Finally, subtracting the two results: (720 - 24 = 696).So, the answer is 696.