answer：Okay, so I have this proposition: "Given (a, b, c, d in mathbb{R}), if (a = b) and (c = d), then (a + c = b + d)." I need to find its contrapositive. Hmm, contrapositive... I remember that involves switching the hypothesis and the conclusion and negating both. Let me try to break this down step by step.First, let's identify the hypothesis and the conclusion in the original statement. The hypothesis is "if (a = b) and (c = d)", and the conclusion is "then (a + c = b + d)." So, in logical terms, the proposition can be written as: If (P) and (Q), then (R), where (P) is (a = b), (Q) is (c = d), and (R) is (a + c = b + d).Now, the contrapositive of a statement "If (P) and (Q), then (R)" is "If not (R), then not (P) or not (Q)." Wait, is that right? I think it's "If not (R), then not (P) or not (Q)." Yeah, that sounds familiar. So, applying that to our problem, the contrapositive should be: If (a + c neq b + d), then (a neq b) or (c neq d).Let me double-check that. The original statement says that if both (a = b) and (c = d) are true, then (a + c = b + d) must be true. The contrapositive flips this around and negates both parts. So, if (a + c) is not equal to (b + d), then at least one of the initial equalities must be false. That makes sense because if both (a = b) and (c = d) were true, then (a + c) would have to equal (b + d). Therefore, if (a + c) doesn't equal (b + d), it must be that either (a) isn't equal to (b) or (c) isn't equal to (d).I think I got it right. The contrapositive is logically equivalent to the original statement, so they should have the same truth value. If I can't find a counterexample where the contrapositive doesn't hold when the original does, then it's likely correct. Let me test it with some numbers.Suppose (a = 2), (b = 2), (c = 3), and (d = 3). Then (a + c = 5) and (b + d = 5), so (a + c = b + d). Now, if I change (a) to 3, making (a = 3), (b = 2), (c = 3), (d = 3). Then (a + c = 6) and (b + d = 5), so (a + c neq b + d). According to the contrapositive, this should mean that either (a neq b) or (c neq d). In this case, (a neq b) is true, so the contrapositive holds.Another example: Let (a = 2), (b = 2), (c = 4), (d = 5). Then (a + c = 6) and (b + d = 7), so (a + c neq b + d). The contrapositive says that either (a neq b) or (c neq d). Here, (c neq d) is true, so again, the contrapositive holds.What if both (a neq b) and (c neq d)? Let's say (a = 1), (b = 2), (c = 3), (d = 4). Then (a + c = 4) and (b + d = 6), so (a + c neq b + d). The contrapositive still applies because at least one of the inequalities holds.Conversely, if (a + c = b + d), does it mean that both (a = b) and (c = d)? Not necessarily. For example, (a = 1), (b = 2), (c = 3), (d = 2). Here, (a + c = 4) and (b + d = 4), but (a neq b) and (c neq d). So, the original statement only claims that if both equalities hold, then the sum equality holds, not the other way around. That's why the contrapositive is about the sum inequality implying at least one of the initial equalities doesn't hold.I think I've covered different scenarios and the logic holds. The contrapositive correctly flips and negates the original statement, ensuring logical equivalence. Therefore, I'm confident that the contrapositive is: "Given (a, b, c, d in mathbb{R}), if (a + c neq b + d), then (a neq b) or (c neq d)."**Final Answer**The contrapositive of the proposition is boxed{text{Given } a, b, c, d in mathbb{R}, text{ if } a+c neq b+d, text{ then } a neq b text{ or } c neq d.}

answer：Okay, so James got a cable program, and I need to figure out how much he paid. Let's see, the problem says the first 100 channels cost 100, and the next 100 channels cost half that much. Then he splits it evenly with his roommate. Hmm, okay, let's break this down.First, the first 100 channels cost 100. That seems straightforward. Now, the next 100 channels cost half of that. So half of 100 is 50. So, the next 100 channels are 50. So, in total, for 200 channels, it's 100 plus 50, which is 150.Now, James splits this cost evenly with his roommate. So, if the total is 150, and they split it equally, each person pays half of that. Half of 150 is 75. So, James pays 75.Wait, let me make sure I didn't miss anything. The first 100 channels are 100, and the next 100 are half of that, which is 50. So, total is 150. Splitting that evenly means each pays 75. Yeah, that seems right.I don't think there's anything else to consider here. The problem doesn't mention any additional fees or taxes, so it's just a straightforward calculation. So, James pays 75.

answer：Okay, so I need to find where the line 2y + 5x = 10 intersects the x-axis and y-axis. Hmm, I remember that when a line crosses the x-axis, the y-coordinate is zero, and when it crosses the y-axis, the x-coordinate is zero. So, I think I can find these points by plugging in zero for one of the variables and solving for the other.First, let me try finding the x-intercept. That's where the line crosses the x-axis, so y should be zero there. I'll substitute y = 0 into the equation:2(0) + 5x = 10Simplifying that, I get:0 + 5x = 10So, 5x = 10. To solve for x, I'll divide both sides by 5:x = 10 / 5x = 2Okay, so the x-intercept is at the point (2, 0). That makes sense because when y is zero, x is 2.Now, I'll find the y-intercept. This is where the line crosses the y-axis, so x should be zero here. I'll substitute x = 0 into the equation:2y + 5(0) = 10Simplifying that, I get:2y + 0 = 10So, 2y = 10. To solve for y, I'll divide both sides by 2:y = 10 / 2y = 5Alright, so the y-intercept is at the point (0, 5). That seems right because when x is zero, y is 5.Let me double-check my work. For the x-intercept, plugging y = 0 into the equation gives 5x = 10, which simplifies to x = 2. That seems correct. For the y-intercept, plugging x = 0 into the equation gives 2y = 10, which simplifies to y = 5. That also looks correct.I think I've got it. The line intersects the x-axis at (2, 0) and the y-axis at (0, 5). To visualize this, if I were to graph the line, it would cross the x-axis at 2 and the y-axis at 5, forming a straight line between those two points.Just to be thorough, maybe I can plot these points on a graph to make sure the line makes sense. Starting at (0, 5), which is on the y-axis, and then going to (2, 0), which is on the x-axis. The slope between these two points would be (0 - 5)/(2 - 0) = -5/2, which is a negative slope, so the line would be going downward from left to right. That seems consistent with the equation 2y + 5x = 10, which I can rewrite in slope-intercept form to check.Let me rewrite the equation:2y + 5x = 10Subtract 5x from both sides:2y = -5x + 10Divide both sides by 2:y = (-5/2)x + 5Yes, the slope is -5/2 and the y-intercept is 5, which matches what I found earlier. So, everything checks out.I think I'm confident with these intercepts. They make sense both algebraically and graphically. So, the line intersects the x-axis at (2, 0) and the y-axis at (0, 5).

answer：Alright, so I'm trying to solve this problem about a hyperbola. Let me read it again carefully:"Given the hyperbola (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) where (a > 0) and (b > 0), with foci (F_1) and (F_2) (left and right respectively), and a point (P) on the left branch of the hyperbola. We're told that the minimum value of (frac{|PF_2|^2}{|PF_1|}) is (8a). We need to find the range of the eccentricity (e) of the hyperbola."Okay, so first, I remember that for hyperbolas, the distance between the foci is (2c), where (c^2 = a^2 + b^2). The eccentricity (e) is defined as (e = frac{c}{a}), which is always greater than 1 for hyperbolas.Now, the point (P) is on the left branch, so it's closer to (F_1) and farther from (F_2). I recall that one of the key properties of hyperbolas is that the difference of distances from any point on the hyperbola to the two foci is constant, specifically (|PF_2| - |PF_1| = 2a). That seems important here.So, let me write that down:[|PF_2| - |PF_1| = 2a]Which means:[|PF_2| = |PF_1| + 2a]Alright, so we can express (|PF_2|) in terms of (|PF_1|). The expression we need to minimize is (frac{|PF_2|^2}{|PF_1|}). Let's substitute (|PF_2|) with (|PF_1| + 2a):[frac{(|PF_1| + 2a)^2}{|PF_1|}]Expanding the numerator:[frac{|PF_1|^2 + 4a|PF_1| + 4a^2}{|PF_1|}]Which simplifies to:[|PF_1| + 4a + frac{4a^2}{|PF_1|}]So, now we have:[|PF_1| + 4a + frac{4a^2}{|PF_1|}]Hmm, this looks like an expression that might be minimized using the AM-GM inequality. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, if I consider the terms (|PF_1|) and (frac{4a^2}{|PF_1|}), their sum is greater than or equal to twice the square root of their product.Let me write that:[|PF_1| + frac{4a^2}{|PF_1|} geq 2 sqrt{|PF_1| cdot frac{4a^2}{|PF_1|}} = 2 cdot 2a = 4a]So, adding the (4a) from the earlier expression:[|PF_1| + 4a + frac{4a^2}{|PF_1|} geq 4a + 4a = 8a]Therefore, the minimum value of (frac{|PF_2|^2}{|PF_1|}) is (8a), which matches the given condition. Now, equality in AM-GM holds when the two terms are equal, so:[|PF_1| = frac{4a^2}{|PF_1|}]Multiplying both sides by (|PF_1|):[|PF_1|^2 = 4a^2]Taking square roots:[|PF_1| = 2a]So, the minimum occurs when (|PF_1| = 2a). Now, I need to relate this back to the hyperbola's properties. Since (P) is on the left branch, the minimum distance from (P) to (F_1) should be (c - a), where (c) is the distance from the center to each focus. Wait, is that correct?Let me think. For hyperbolas, the distance from a focus to a vertex is (c), and the vertices are at ((pm a, 0)). So, the distance from (F_1) (which is at ((-c, 0))) to the left vertex ((-a, 0)) is (c - a). But (P) is any point on the left branch, so the distance (|PF_1|) can vary. The minimum distance occurs when (P) is at the vertex, right? So, the minimum (|PF_1|) is (c - a).But wait, we just found that (|PF_1| = 2a) at the minimum of the expression. So, does that mean (c - a = 2a)? That would imply (c = 3a). Therefore, the eccentricity (e = frac{c}{a} = 3).But the problem says the minimum value is (8a), and we've found that when (|PF_1| = 2a), the expression equals (8a). So, does this mean that (c - a leq 2a)? Because (|PF_1|) can't be less than (c - a), right?Wait, no. If (|PF_1|) is minimized at (c - a), but in our case, the minimum of the expression (frac{|PF_2|^2}{|PF_1|}) occurs at (|PF_1| = 2a), which must be greater than or equal to (c - a). So, (2a geq c - a), which simplifies to (3a geq c). Therefore, (c leq 3a), so (e = frac{c}{a} leq 3).But since (e > 1) for hyperbolas, the range of (e) is (1 < e leq 3).Wait, let me double-check. If (c leq 3a), then (e = frac{c}{a} leq 3). And since (e > 1), the range is (1 < e leq 3).Yes, that seems correct. So, the eccentricity (e) must be greater than 1 and at most 3.**Final Answer**The range of eccentricity (e) is (boxed{(1, 3]}).

answer：First, I need to find the average weight of the 24 boys in the class. I'll start by calculating the total weight of all the boys in the class.There are a total of 32 boys (24 + 8). The average weight of all the boys is 48.975 kg. So, the total weight of all the boys is:32 boys * 48.975 kg = 1567.2 kgNext, I'll calculate the total weight of the 8 boys whose average weight is 45.15 kg:8 boys * 45.15 kg = 361.2 kgNow, I'll subtract the total weight of the 8 boys from the total weight of all the boys to find the total weight of the 24 boys:1567.2 kg - 361.2 kg = 1206 kgFinally, to find the average weight of the 24 boys, I'll divide the total weight of the 24 boys by the number of boys:1206 kg / 24 boys = 50.25 kgSo, the average weight of the 24 boys is 50.25 kg.

answer：Alright, so I'm trying to figure out which numbers in the set {11, 13, 15, 17, 19} are "good numbers." From the problem statement, a "good number" is defined as a number ( n ) for which there exists a permutation ( a_1, a_2, ldots, a_n ) of the numbers 1 through ( n ) such that each ( k + a_k ) is a perfect square. First, I need to understand what exactly is being asked. Essentially, for each number ( n ) in the set, I need to check if I can rearrange the numbers 1 to ( n ) in such a way that when I add each number to its position index, the result is a perfect square. Let me start by breaking down the problem for each number in the set.**1. Checking if 11 is a "good number":**For ( n = 11 ), I need to find a permutation ( a_1, a_2, ldots, a_{11} ) such that ( k + a_k ) is a perfect square for all ( k ) from 1 to 11.I'll start by listing the possible perfect squares that could result from ( k + a_k ). The smallest possible sum is ( 1 + 1 = 2 ), and the largest is ( 11 + 11 = 22 ). So, the perfect squares in this range are 4, 9, 16, and 25. Wait, 25 is 5 squared, but 25 is larger than 22, so the relevant perfect squares are 4, 9, 16.Now, for each ( k ), I need to find an ( a_k ) such that ( k + a_k ) is one of these squares. Let's list the possible ( a_k ) values for each ( k ):- ( k = 1 ): ( a_1 ) can be 3 (since 1 + 3 = 4) or 8 (1 + 8 = 9) or 15 (1 + 15 = 16). But since ( a_1 ) must be between 1 and 11, 15 is out. So possible ( a_1 ) are 3 or 8.- ( k = 2 ): ( a_2 ) can be 2 (2 + 2 = 4) or 7 (2 + 7 = 9) or 14 (2 + 14 = 16). Again, 14 is out, so possible ( a_2 ) are 2 or 7.- ( k = 3 ): ( a_3 ) can be 1 (3 + 1 = 4) or 6 (3 + 6 = 9) or 13 (3 + 13 = 16). 13 is out, so possible ( a_3 ) are 1 or 6.- ( k = 4 ): ( a_4 ) can be 5 (4 + 5 = 9) or 12 (4 + 12 = 16). 12 is out, so ( a_4 = 5 ).- ( k = 5 ): ( a_5 ) can be 4 (5 + 4 = 9) or 11 (5 + 11 = 16). So possible ( a_5 ) are 4 or 11.- ( k = 6 ): ( a_6 ) can be 3 (6 + 3 = 9) or 10 (6 + 10 = 16). So possible ( a_6 ) are 3 or 10.- ( k = 7 ): ( a_7 ) can be 2 (7 + 2 = 9) or 9 (7 + 9 = 16). So possible ( a_7 ) are 2 or 9.- ( k = 8 ): ( a_8 ) can be 1 (8 + 1 = 9) or 8 (8 + 8 = 16). So possible ( a_8 ) are 1 or 8.- ( k = 9 ): ( a_9 ) can be 7 (9 + 7 = 16). So ( a_9 = 7 ).- ( k = 10 ): ( a_{10} ) can be 6 (10 + 6 = 16). So ( a_{10} = 6 ).- ( k = 11 ): ( a_{11} ) can be 5 (11 + 5 = 16). So ( a_{11} = 5 ).Now, let's try to assign these values without repeating any numbers.Starting with ( k = 4 ), ( a_4 = 5 ). So 5 is taken.For ( k = 11 ), ( a_{11} = 5 ). But 5 is already taken by ( a_4 ). That's a problem. Maybe I need to adjust.Wait, for ( k = 5 ), ( a_5 ) can be 4 or 11. If I set ( a_5 = 11 ), then 11 is taken, which might help.Let me try:- ( a_4 = 5 )- ( a_5 = 11 )- ( a_{11} = 5 ) is still a problem because 5 is taken.Alternatively, maybe ( a_5 = 4 ). Then ( a_4 = 5 ), ( a_5 = 4 ). That works.Now, ( a_4 = 5 ), ( a_5 = 4 ).Next, ( k = 1 ): possible ( a_1 = 3 ) or 8.Let's try ( a_1 = 3 ). Then 3 is taken.( k = 6 ): possible ( a_6 = 3 ) or 10. 3 is taken, so ( a_6 = 10 ).( k = 10 ): ( a_{10} = 6 ). So 6 is taken.( k = 3 ): possible ( a_3 = 1 ) or 6. 6 is taken, so ( a_3 = 1 ).( k = 8 ): possible ( a_8 = 1 ) or 8. 1 is taken, so ( a_8 = 8 ).( k = 1 ): ( a_1 = 3 ), which is okay.Now, ( k = 2 ): possible ( a_2 = 2 ) or 7.Let's try ( a_2 = 2 ).( k = 7 ): possible ( a_7 = 2 ) or 9. 2 is taken, so ( a_7 = 9 ).( k = 9 ): ( a_9 = 7 ).( k = 7 ): ( a_7 = 9 ).Now, let's check the assignments:- ( a_1 = 3 )- ( a_2 = 2 )- ( a_3 = 1 )- ( a_4 = 5 )- ( a_5 = 4 )- ( a_6 = 10 )- ( a_7 = 9 )- ( a_8 = 8 )- ( a_9 = 7 )- ( a_{10} = 6 )- ( a_{11} = 5 ) → Wait, ( a_{11} = 5 ) is already taken by ( a_4 ). That's a conflict.So, this assignment doesn't work. Maybe I need to try a different approach.Let's try ( a_1 = 8 ) instead.- ( a_1 = 8 )- ( k = 8 ): possible ( a_8 = 1 ) or 8. 8 is taken, so ( a_8 = 1 ).- ( k = 3 ): possible ( a_3 = 1 ) or 6. 1 is taken, so ( a_3 = 6 ).- ( k = 10 ): ( a_{10} = 6 ). 6 is taken by ( a_3 ), so conflict.Hmm, this is tricky. Maybe 11 isn't a "good number."**2. Checking if 13 is a "good number":**For ( n = 13 ), the perfect squares range from ( 1 + 1 = 2 ) to ( 13 + 13 = 26 ). So, the relevant perfect squares are 4, 9, 16, and 25.I'll try to find a permutation similar to the approach above, but it might be time-consuming. Alternatively, I recall that 13 is known to be a "good number" because such a permutation exists. For example, the permutation [8, 2, 13, 12, 11, 10, 9, 1, 7, 6, 5, 4, 3] works.**3. Checking if 15 is a "good number":**For ( n = 15 ), the perfect squares range from 2 to 30, so squares are 4, 9, 16, 25.Again, I might need to construct a permutation, but I remember that 15 is a "good number." A simple permutation that works is the reverse order: [15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]. Each ( k + a_k ) becomes ( k + (16 - k) = 16 ), which is a perfect square.**4. Checking if 17 is a "good number":**For ( n = 17 ), perfect squares up to 34 are 4, 9, 16, 25.I think 17 is also a "good number." There exists a permutation where each ( k + a_k ) is a perfect square. For example, one such permutation is [3, 7, 6, 5, 4, 10, 2, 17, 16, 15, 14, 13, 12, 11, 1, 9, 8].**5. Checking if 19 is a "good number":**For ( n = 19 ), perfect squares up to 38 are 4, 9, 16, 25.I believe 19 is a "good number" as well. A possible permutation is [8, 7, 6, 5, 4, 3, 2, 1, 16, 15, 14, 13, 12, 11, 10, 9, 19, 18, 17].**Conclusion:**From the above, it seems that 13, 15, 17, and 19 are "good numbers," while 11 is not.