answer：Alright, so I've got these two problems about sets A and B. Let me try to figure them out step by step.First, let's understand what sets A and B are. Set A is defined as all real numbers x such that 0 ≤ x ≤ 2. So, A is the interval [0, 2]. Set B is defined as all real numbers x such that a ≤ x ≤ 3 - 2a. So, B is the interval [a, 3 - 2a]. Now, for part (1), the problem states that the union of the complement of A and B equals the universal set U. So, (complement of A) ∪ B = U. I need to find the range of real number a that satisfies this condition.Okay, let's break this down. The complement of A, denoted as ∁ₐ, would be all the elements not in A. Since A is [0, 2], the complement would be (-∞, 0) ∪ (2, ∞). So, ∁ₐ = (-∞, 0) ∪ (2, ∞).Now, the union of ∁ₐ and B should be U. That means every element in U is either in ∁ₐ or in B. Since ∁ₐ already covers (-∞, 0) and (2, ∞), B needs to cover the remaining part of U, which is [0, 2]. So, B must cover [0, 2] to make sure that the union is U.But wait, B is [a, 3 - 2a]. So, for B to cover [0, 2], the interval [a, 3 - 2a] must include [0, 2]. That means a must be less than or equal to 0, and 3 - 2a must be greater than or equal to 2.Let me write that down:1. a ≤ 0 (so that the lower bound of B is at most 0)2. 3 - 2a ≥ 2 (so that the upper bound of B is at least 2)Let's solve the second inequality:3 - 2a ≥ 2 Subtract 3 from both sides: -2a ≥ -1 Divide both sides by -2 (remembering to reverse the inequality sign): a ≤ 1/2So, from the first condition, a ≤ 0, and from the second condition, a ≤ 1/2. The more restrictive condition is a ≤ 0. Therefore, the range of a is (-∞, 0].Wait, but I should also make sure that B is a valid interval. For B to be valid, the lower bound a must be less than or equal to the upper bound 3 - 2a. So, a ≤ 3 - 2a Adding 2a to both sides: 3a ≤ 3 Divide both sides by 3: a ≤ 1But since we already have a ≤ 0, this condition is automatically satisfied. So, no additional constraints here.Therefore, for part (1), the range of a is (-∞, 0].Now, moving on to part (2). The problem states that A ∩ B ≠ B. I need to find the range of real number a that satisfies this condition.First, let's recall that A ∩ B is the set of elements common to both A and B. If A ∩ B ≠ B, that means B is not entirely contained within A. In other words, there are elements in B that are not in A.Alternatively, if A ∩ B = B, then B is a subset of A. So, to have A ∩ B ≠ B, B should not be a subset of A.So, let's first find the conditions under which B is a subset of A, and then take the complement of that.For B to be a subset of A, the interval [a, 3 - 2a] must be entirely within [0, 2]. That requires:1. a ≥ 0 (so that the lower bound of B is at least 0)2. 3 - 2a ≤ 2 (so that the upper bound of B is at most 2)Let's solve the second inequality:3 - 2a ≤ 2 Subtract 3 from both sides: -2a ≤ -1 Divide both sides by -2 (remembering to reverse the inequality sign): a ≥ 1/2So, for B to be a subset of A, we need a ≥ 1/2.But we also need to ensure that B is a valid interval, meaning a ≤ 3 - 2a Adding 2a to both sides: 3a ≤ 3 Divide both sides by 3: a ≤ 1So, combining these, for B to be a subset of A, a must satisfy 1/2 ≤ a ≤ 1.Therefore, if a is in [1/2, 1], then A ∩ B = B. Hence, to have A ∩ B ≠ B, a must not be in [1/2, 1]. So, the range of a is (-∞, 1/2) ∪ (1, ∞).Wait, but let me think again. If a > 1, then the upper bound of B, which is 3 - 2a, becomes less than 1. So, for a > 1, 3 - 2a < 1. But since a > 1, the lower bound of B is greater than 1, and the upper bound is less than 1. That would make B an empty set because the lower bound is greater than the upper bound.An empty set is a subset of every set, including A. So, if B is empty, then A ∩ B = B (since both are empty sets). Therefore, in this case, A ∩ B = B, which we don't want. So, a > 1 would actually satisfy A ∩ B = B, which is the opposite of what we want.Wait, that complicates things. So, if a > 1, B is empty, and thus A ∩ B = B. Therefore, to have A ∩ B ≠ B, we need B to not be a subset of A and also B not to be empty.So, let's reconsider. For A ∩ B ≠ B, either B is not a subset of A, or B is empty. But if B is empty, A ∩ B = B, which is the empty set. So, in that case, A ∩ B = B, which we don't want. Therefore, to have A ∩ B ≠ B, B must not be a subset of A and B must not be empty.So, B must be non-empty and not entirely contained within A.So, first, B must be non-empty. That requires a ≤ 3 - 2a Which simplifies to a ≤ 1.So, a ≤ 1.Second, B is not a subset of A. For B to not be a subset of A, either the lower bound of B is less than 0 or the upper bound of B is greater than 2.So, either a < 0 or 3 - 2a > 2.Let's solve these:1. a < 02. 3 - 2a > 2 Subtract 3: -2a > -1 Divide by -2 (reverse inequality): a < 1/2So, combining these, for B to not be a subset of A, a < 0 or a < 1/2.But since we already have a ≤ 1, the conditions are:- If a < 0, then B is not a subset of A.- If 0 ≤ a < 1/2, then B is not a subset of A because the upper bound of B is greater than 2.Wait, let me check that. If a is between 0 and 1/2, then 3 - 2a is greater than 2 because:If a = 0, 3 - 0 = 3 > 2 If a = 1/2, 3 - 2*(1/2) = 3 - 1 = 2So, for 0 ≤ a < 1/2, 3 - 2a > 2, meaning the upper bound of B is greater than 2, so B is not entirely contained within A.Therefore, combining these, the range of a where B is non-empty and not a subset of A is a < 1/2.But wait, when a < 0, B is [a, 3 - 2a]. Since a < 0, the lower bound is negative, so B extends into negative numbers, which are not in A. Therefore, B is not a subset of A.When 0 ≤ a < 1/2, B's upper bound is greater than 2, so B extends beyond A, hence not a subset.When a = 1/2, B is [1/2, 2], which is a subset of A.When 1/2 < a ≤ 1, B is [a, 3 - 2a]. Let's see:If a = 3/4, then 3 - 2*(3/4) = 3 - 1.5 = 1.5. So, B is [0.75, 1.5], which is within [0, 2], so B is a subset of A.If a = 1, B is [1, 1], which is just {1}, still a subset of A.So, for a in [1/2, 1], B is a subset of A.For a > 1, B is empty, which is a subset of A.Therefore, to have A ∩ B ≠ B, we need B to not be a subset of A and B to be non-empty. So, B is non-empty when a ≤ 1, and B is not a subset of A when a < 1/2.Therefore, the range of a is a < 1/2.But wait, when a < 0, B is non-empty and not a subset of A. When 0 ≤ a < 1/2, B is non-empty and not a subset of A. When a = 1/2, B is a subset of A. When 1/2 < a ≤ 1, B is a subset of A. When a > 1, B is empty, which is a subset of A.Therefore, the range of a where A ∩ B ≠ B is a < 1/2.So, the range of a is (-∞, 1/2).Wait, but earlier I thought it might be (-∞, 1/2) ∪ (1, ∞), but now I realize that for a > 1, B is empty, and thus A ∩ B = B, which is the empty set. So, in that case, A ∩ B = B, which we don't want. Therefore, a > 1 is not part of the solution.Hence, the correct range is a < 1/2.So, summarizing:For part (1), the range of a is (-∞, 0].For part (2), the range of a is (-∞, 1/2).I think that's it. Let me just double-check.For part (1), we needed (complement of A) ∪ B = U. Since complement of A is (-∞, 0) ∪ (2, ∞), and B needs to cover [0, 2]. So, B must include [0, 2], which requires a ≤ 0 and 3 - 2a ≥ 2. Solving gives a ≤ 0.For part (2), we needed A ∩ B ≠ B, meaning B is not a subset of A. B is a subset of A when a ≥ 1/2 and a ≤ 1. So, to have B not a subset of A, a < 1/2. Also, B must be non-empty, which requires a ≤ 1. So, combining, a < 1/2.Yes, that makes sense.

answer：First, I need to determine the dimensions of the photograph by subtracting the margins from the total dimensions of the paper.The paper has a total length of 10 inches and a total width of 8.5 inches. There is a 1.5-inch margin on all sides. This means I need to subtract twice the margin from both the length and the width to account for both sides.For the length:10 inches (total length) - 1.5 inches (top margin) - 1.5 inches (bottom margin) = 10 - 1.5 - 1.5 = 7 inchesFor the width:8.5 inches (total width) - 1.5 inches (left margin) - 1.5 inches (right margin) = 8.5 - 1.5 - 1.5 = 5.5 inchesNow that I have the dimensions of the photograph, I can calculate the area it covers by multiplying the length by the width.Area = length × widthArea = 7 inches × 5.5 inchesArea = 38.5 square inchesSo, the photograph covers an area of 38.5 square inches.

answer：Okay, so I have this problem about lines and areas. Let me try to figure it out step by step. First, part (1) says that line l passes through the intersection point P of lines l' and another line, which is given by 2x + y + 2 = 0. Also, line l is perpendicular to the line x - 2y - 1 = 0. I need to find the equation of line l.Alright, so I think the first thing I should do is find the intersection point P of lines l' and the other line. Line l' is given by 3x + 4y - 2 = 0, and the other line is 2x + y + 2 = 0. To find their intersection, I need to solve these two equations simultaneously.Let me write them down:1. 3x + 4y - 2 = 02. 2x + y + 2 = 0Hmm, maybe I can solve the second equation for y and substitute into the first equation. Let's try that.From equation 2: 2x + y + 2 = 0, so y = -2x - 2.Now, substitute this into equation 1:3x + 4(-2x - 2) - 2 = 0Let me compute that:3x - 8x - 8 - 2 = 0Combine like terms:(3x - 8x) + (-8 - 2) = 0-5x - 10 = 0So, -5x = 10Divide both sides by -5:x = -2Now, plug x = -2 back into equation 2 to find y:2(-2) + y + 2 = 0-4 + y + 2 = 0Simplify:y - 2 = 0So, y = 2Therefore, the intersection point P is (-2, 2). Got that.Now, line l passes through P and is perpendicular to the line x - 2y - 1 = 0. I need to find the equation of line l.First, let me find the slope of the line x - 2y - 1 = 0. To do that, I can rewrite it in slope-intercept form (y = mx + b).Starting with x - 2y - 1 = 0:x - 1 = 2yDivide both sides by 2:y = (1/2)x - 1/2So, the slope (m) of this line is 1/2.Since line l is perpendicular to this line, its slope will be the negative reciprocal of 1/2, which is -2.So, the slope of line l is -2.Now, I have a point (-2, 2) that line l passes through, and its slope is -2. I can use the point-slope form to write the equation of line l.Point-slope form is: y - y1 = m(x - x1)Plugging in the values:y - 2 = -2(x - (-2)) => y - 2 = -2(x + 2)Let me simplify this:y - 2 = -2x - 4Add 2 to both sides:y = -2x - 2So, the equation of line l is y = -2x - 2. Alternatively, I can write it in standard form:2x + y + 2 = 0Wait, let me check that. If I rearrange y = -2x - 2:2x + y + 2 = 0Yes, that seems correct.Okay, so part (1) is done. The equation of line l is 2x + y + 2 = 0.Now, moving on to part (2). I need to find the area S of the triangle formed by lines l, l', and the x-axis.Hmm, so the triangle is formed by three lines: line l, line l', and the x-axis. I need to find the points where these lines intersect each other and then calculate the area.First, let me find the x-intercepts of lines l and l' because the triangle is bounded by the x-axis.Starting with line l: 2x + y + 2 = 0To find the x-intercept, set y = 0:2x + 0 + 2 = 0 => 2x + 2 = 0 => 2x = -2 => x = -1So, the x-intercept of line l is (-1, 0).Now, for line l': 3x + 4y - 2 = 0Again, set y = 0 to find the x-intercept:3x + 0 - 2 = 0 => 3x - 2 = 0 => 3x = 2 => x = 2/3So, the x-intercept of line l' is (2/3, 0).Now, the triangle is formed by the points where these lines intersect each other and the x-axis. So, the three vertices of the triangle are:1. The intersection point P of lines l and l', which we already found as (-2, 2).2. The x-intercept of line l: (-1, 0)3. The x-intercept of line l': (2/3, 0)Wait, but hold on. The triangle is formed by lines l, l', and the x-axis. So, actually, the three lines intersect each other at three points, which are the vertices of the triangle.But in this case, line l and line l' intersect at P (-2, 2). Line l intersects the x-axis at (-1, 0), and line l' intersects the x-axis at (2/3, 0). So, the three vertices of the triangle are (-2, 2), (-1, 0), and (2/3, 0).Wait, but actually, the triangle is formed by the three lines: l, l', and the x-axis. So, the three intersection points are:1. Intersection of l and l': P (-2, 2)2. Intersection of l and x-axis: (-1, 0)3. Intersection of l' and x-axis: (2/3, 0)So, yes, those three points form the triangle.Now, to find the area of the triangle, I can use the coordinates of these three points.The formula for the area of a triangle given three vertices (x1, y1), (x2, y2), (x3, y3) is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Let me plug in the points:Point A: (-2, 2) => x1 = -2, y1 = 2Point B: (-1, 0) => x2 = -1, y2 = 0Point C: (2/3, 0) => x3 = 2/3, y3 = 0Plugging into the formula:Area = |(-2*(0 - 0) + (-1)*(0 - 2) + (2/3)*(2 - 0))/2|Simplify each term:First term: -2*(0 - 0) = 0Second term: (-1)*(0 - 2) = (-1)*(-2) = 2Third term: (2/3)*(2 - 0) = (2/3)*2 = 4/3So, adding them up:0 + 2 + 4/3 = 2 + 4/3 = 10/3Now, take the absolute value (which is still 10/3) and divide by 2:Area = |10/3| / 2 = (10/3)/2 = 10/6 = 5/3So, the area S is 5/3.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, the points are correct: (-2, 2), (-1, 0), and (2/3, 0). Using the area formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in:|(-2*(0 - 0) + (-1)*(0 - 2) + (2/3)*(2 - 0))/2|= |(0 + 2 + 4/3)/2|= |(10/3)/2|= 5/3Yes, that seems correct.Alternatively, I can think of the base of the triangle as the distance between the two x-intercepts: from (-1, 0) to (2/3, 0). The length of the base is |2/3 - (-1)| = |2/3 + 1| = |5/3| = 5/3.The height of the triangle is the y-coordinate of the point P, which is 2.So, area = (base * height)/2 = (5/3 * 2)/2 = (10/3)/2 = 5/3.Yes, same result. So, the area is indeed 5/3.Okay, so I think I've got both parts figured out. The equation of line l is 2x + y + 2 = 0, and the area of the triangle is 5/3.**Final Answer**(1) The equation of line ( l ) is boxed{2x + y + 2 = 0}.(2) The area ( S ) of the triangle is boxed{dfrac{5}{3}}.

answer：Okay, so I have this math problem here: "Given that ( a ) is an integer and ( a^2 ) is an even number, prove that ( a ) is also an even number." Hmm, interesting. I remember learning something about even and odd numbers in algebra, but I need to recall exactly how that works.Alright, let's start by understanding what it means for a number to be even or odd. An even number is any integer that can be written as ( 2k ) where ( k ) is also an integer. So, for example, 2, 4, 6, etc., are even because they can be expressed as ( 2 times 1 ), ( 2 times 2 ), ( 2 times 3 ), and so on. On the other hand, an odd number is any integer that can be written as ( 2k + 1 ). So, numbers like 1, 3, 5, etc., are odd because they are one more than an even number.Now, the problem says that ( a ) is an integer and ( a^2 ) is even. I need to prove that ( a ) must also be even. Hmm, so if ( a^2 ) is even, does that necessarily mean ( a ) is even? Intuitively, I think yes, but I need to prove it formally.Maybe I can approach this by contradiction. That is, assume the opposite of what I want to prove and show that it leads to a contradiction. So, let's assume that ( a ) is not even, which means ( a ) is odd. If ( a ) is odd, then by definition, ( a = 2k + 1 ) for some integer ( k ).Let me compute ( a^2 ) under this assumption. If ( a = 2k + 1 ), then ( a^2 = (2k + 1)^2 ). Expanding this, I get ( a^2 = 4k^2 + 4k + 1 ). Hmm, let's see. I can factor out a 4 from the first two terms: ( a^2 = 4(k^2 + k) + 1 ).Now, ( 4(k^2 + k) ) is clearly an even number because it's a multiple of 4, which is even. Adding 1 to an even number gives an odd number. So, ( a^2 = 4(k^2 + k) + 1 ) is odd. But wait, the problem states that ( a^2 ) is even. That's a contradiction because we assumed ( a ) is odd and ended up with ( a^2 ) being odd, which contradicts the given that ( a^2 ) is even.Therefore, our initial assumption that ( a ) is odd must be false. So, ( a ) cannot be odd, which means ( a ) must be even. That seems to make sense. Let me double-check my reasoning to make sure I didn't make any mistakes.I started by assuming ( a ) is odd, expressed it as ( 2k + 1 ), squared it, and found that ( a^2 ) is odd. But the problem says ( a^2 ) is even, so my assumption that ( a ) is odd leads to a contradiction. Therefore, ( a ) must be even. Yeah, that seems solid.Another way to think about it is by considering the properties of even and odd numbers. If ( a ) is even, then ( a = 2k ), and squaring it gives ( a^2 = 4k^2 ), which is clearly even. If ( a ) is odd, as I showed earlier, ( a^2 ) is odd. So, the square of an even number is even, and the square of an odd number is odd. Therefore, if the square is even, the original number must have been even.I think that covers it. I can't think of any exceptions or special cases where this wouldn't hold because we're dealing with integers, and the definitions of even and odd are pretty straightforward. So, I feel confident that this proof by contradiction works.**Final Answer**boxed{text{a is an even number}}

answer：Alright, so I have this problem here where I need to determine how many different triangles can be formed using the logarithms of three numbers: lg 12, lg 75, and lg(n² - 16n + 947), where n is a positive integer. Hmm, okay, let me try to break this down step by step.First, I remember that to form a triangle, the sum of any two sides must be greater than the third side. But in this case, the sides are given as logarithms. I wonder if I can apply the triangle inequality directly to the logarithms or if I need to convert them back to their original numbers. Hmm, since logarithm is a monotonically increasing function, maybe I can work with the logarithms directly. Let me think about that.So, if I have three sides a, b, and c, then the triangle inequalities are:1. a + b > c2. a + c > b3. b + c > aIn this problem, the sides are lg 12, lg 75, and lg(n² - 16n + 947). Let me denote these as A = lg 12, B = lg 75, and C = lg(n² - 16n + 947). So, I need to satisfy the following inequalities:1. A + B > C2. A + C > B3. B + C > ABut since A and B are constants, maybe I can find the range of C that satisfies these inequalities.Let me compute A and B first. A is lg 12, which is log base 10 of 12. Similarly, B is lg 75, which is log base 10 of 75. I can calculate these approximately:- lg 12 ≈ 1.07918- lg 75 ≈ 1.87506So, A ≈ 1.07918 and B ≈ 1.87506.Now, let's write down the inequalities with these approximate values.1. A + B > C: 1.07918 + 1.87506 > C So, 2.95424 > C2. A + C > B: 1.07918 + C > 1.87506 So, C > 1.87506 - 1.07918 C > 0.795883. B + C > A: 1.87506 + C > 1.07918 This will always be true since 1.87506 is already greater than 1.07918, and C is positive.So, combining the first two inequalities, we have:0.79588 < C < 2.95424But C is lg(n² - 16n + 947). So, let's write that:0.79588 < lg(n² - 16n + 947) < 2.95424Since logarithm is a monotonically increasing function, I can convert these inequalities into their exponential forms to solve for n.First, let's convert the lower bound:lg(n² - 16n + 947) > 0.79588Which means:n² - 16n + 947 > 10^0.79588Calculating 10^0.79588:10^0.79588 ≈ 6.25Similarly, the upper bound:lg(n² - 16n + 947) < 2.95424Which means:n² - 16n + 947 < 10^2.95424Calculating 10^2.95424:10^2.95424 ≈ 900So now, we have:6.25 < n² - 16n + 947 < 900Since n is a positive integer, let's write this as:7 ≤ n² - 16n + 947 ≤ 899Now, I need to find all positive integers n such that n² - 16n + 947 is between 7 and 899.Let me write the quadratic expression:n² - 16n + 947I can complete the square to find the minimum value of this quadratic.n² - 16n + 947= (n² - 16n + 64) + (947 - 64)= (n - 8)² + 883So, the expression (n - 8)² + 883 is always greater than or equal to 883, since (n - 8)² is non-negative.Wait, but earlier we had n² - 16n + 947 must be greater than 6.25, but since the minimum value is 883, which is way larger than 6.25, that inequality is automatically satisfied. So, the lower bound doesn't really restrict n. The real restriction comes from the upper bound.So, n² - 16n + 947 < 900Which simplifies to:n² - 16n + 947 - 900 < 0n² - 16n + 47 < 0Now, solve the quadratic inequality:n² - 16n + 47 < 0First, find the roots of the equation n² - 16n + 47 = 0.Using the quadratic formula:n = [16 ± sqrt(256 - 188)] / 2= [16 ± sqrt(68)] / 2= [16 ± 2*sqrt(17)] / 2= 8 ± sqrt(17)Approximating sqrt(17) ≈ 4.123So, the roots are approximately:8 + 4.123 ≈ 12.1238 - 4.123 ≈ 3.877So, the quadratic expression n² - 16n + 47 is less than zero between the roots, i.e., for n between approximately 3.877 and 12.123.Since n is a positive integer, n can be 4, 5, 6, 7, 8, 9, 10, 11, or 12.So, n can take values from 4 to 12 inclusive.Now, let's list the possible values of n: 4,5,6,7,8,9,10,11,12.But wait, the problem says "different triangles". So, even if different values of n give the same value of C, they would not form different triangles. So, I need to check if different n's give different values of C.Given that C = lg(n² - 16n + 947), and n² - 16n + 947 is a quadratic function, it's symmetric around its vertex. The vertex is at n = 8, as we saw earlier when completing the square.So, for n = 8 + k and n = 8 - k, the value of n² - 16n + 947 will be the same.Let me verify that.Take n = 4 and n = 12:For n=4:4² -16*4 +947 = 16 -64 +947 = 899For n=12:12² -16*12 +947 = 144 -192 +947 = 899Same result.Similarly, n=5 and n=11:n=5: 25 -80 +947 = 892n=11: 121 -176 +947 = 892Same result.n=6 and n=10:n=6: 36 -96 +947 = 887n=10: 100 -160 +947 = 887Same result.n=7 and n=9:n=7: 49 -112 +947 = 884n=9: 81 -144 +947 = 884Same result.n=8:n=8: 64 -128 +947 = 883So, n=8 gives a unique value.Therefore, the values of n² -16n +947 for n=4 to 12 are:899, 892, 887, 884, 883, 884, 887, 892, 899.So, the distinct values are 883, 884, 887, 892, 899.That's 5 distinct values.Therefore, there are 5 distinct values of C, which correspond to 5 different triangles.Wait, but let me make sure. Each distinct value of C gives a distinct triangle because the sides would be different.Since A and B are fixed, if C is different, the triangle is different.Yes, because the side lengths would be different, so the triangles are different.Therefore, the number of different triangles is 5.**Final Answer**The number of different triangles that can be formed is boxed{5}.

answer：I need to convert the binary number (1001_2) into its decimal equivalent.First, I'll identify the place value of each digit in the binary number. Starting from the right, the digits represent (2^0), (2^1), (2^2), and (2^3).Next, I'll multiply each binary digit by its corresponding place value:- The rightmost digit is (1 times 2^0 = 1)- The next digit is (0 times 2^1 = 0)- The following digit is (0 times 2^2 = 0)- The leftmost digit is (1 times 2^3 = 8)Finally, I'll add up all these values to get the decimal number:(8 + 0 + 0 + 1 = 9)