answer：To determine when the expression (frac{2}{sqrt{x-1}}) is meaningful in the real number range, we need to ensure two conditions are met:1. **Denominator is not zero**: The denominator (sqrt{x-1}) must not be zero because division by zero is undefined. Therefore, we require: [ sqrt{x-1} neq 0 ] This implies: [ x - 1 neq 0 quad Rightarrow quad x neq 1 ]2. **Expression under the square root is positive**: The expression inside the square root, (x - 1), must be positive to ensure the square root is defined in the real number range. Thus: [ x - 1 > 0 quad Rightarrow quad x > 1 ]Combining these two conditions, we conclude that the expression (frac{2}{sqrt{x-1}}) is meaningful in the real number range when:[x > 1]

answer：Alright, so I need to compute the cube of this matrix: (begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}^3). Hmm, okay, let's see. I remember that to compute a matrix raised to a power, you can multiply the matrix by itself multiple times. So, for the cube, I need to multiply the matrix by itself three times. But maybe there's a smarter way than just multiplying it three times directly. Maybe I can find a pattern or use some properties of matrices to simplify this.First, let me recall how matrix multiplication works. When multiplying two matrices, each element of the resulting matrix is the dot product of the corresponding row of the first matrix and column of the second matrix. So, for a 2x2 matrix, it's manageable, but it can get a bit messy with multiple multiplications. Maybe I should start by computing the square of the matrix first and then multiply that result by the original matrix to get the cube.Okay, let's compute the square first. Let me denote the original matrix as ( A ), so ( A = begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} ). Then, ( A^2 = A times A ). Let me write that out:[A^2 = begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]Calculating each element of the resulting matrix:- The element in the first row, first column is ( (2 times 2) + (-2 times 2) = 4 - 4 = 0 ).- The element in the first row, second column is ( (2 times -2) + (-2 times -1) = -4 + 2 = -2 ).- The element in the second row, first column is ( (2 times 2) + (-1 times 2) = 4 - 2 = 2 ).- The element in the second row, second column is ( (2 times -2) + (-1 times -1) = -4 + 1 = -3 ).So, putting it all together, ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ). Hmm, okay, that's the square.Now, to find ( A^3 ), I need to multiply ( A^2 ) by ( A ). So,[A^3 = A^2 times A = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]Again, let's compute each element:- First row, first column: ( (0 times 2) + (-2 times 2) = 0 - 4 = -4 ).- First row, second column: ( (0 times -2) + (-2 times -1) = 0 + 2 = 2 ).- Second row, first column: ( (2 times 2) + (-3 times 2) = 4 - 6 = -2 ).- Second row, second column: ( (2 times -2) + (-3 times -1) = -4 + 3 = -1 ).Wait, hold on, that doesn't seem right. The second row, second column should be ( (2 times -2) + (-3 times -1) ). Let me recalculate that: ( 2 times -2 = -4 ) and ( -3 times -1 = 3 ). So, adding them together: ( -4 + 3 = -1 ). Hmm, but in the initial thought process, I think I might have made a mistake here. Let me double-check.Wait, no, actually, in my initial calculation, I think I messed up the second row, second column. Let me recalculate:Second row, second column: ( (2 times -2) + (-3 times -1) = -4 + 3 = -1 ). So, that's correct. So, putting it all together, ( A^3 = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait a second, that doesn't match the initial answer I saw earlier. Let me check my calculations again to make sure I didn't make a mistake.Starting with ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ). Then, multiplying by ( A ):First row, first column: ( 0 times 2 + (-2) times 2 = 0 - 4 = -4 ). Correct.First row, second column: ( 0 times (-2) + (-2) times (-1) = 0 + 2 = 2 ). Correct.Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ). Correct.Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ). Hmm, so that gives me ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).But in the initial problem, the answer was ( begin{pmatrix} -4 & 2 -2 & 1 end{pmatrix} ). Wait, so there's a discrepancy in the bottom right element. Did I make a mistake in the multiplication?Let me check the second row, second column again:( 2 times (-2) = -4 )( -3 times (-1) = 3 )Adding them together: ( -4 + 3 = -1 ). So, that should be -1, not 1. So, perhaps the initial answer was incorrect.Wait, maybe I made a mistake in the square of the matrix. Let me recalculate ( A^2 ):( A = begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} )( A^2 = A times A ):First row, first column: ( 2 times 2 + (-2) times 2 = 4 - 4 = 0 ). Correct.First row, second column: ( 2 times (-2) + (-2) times (-1) = -4 + 2 = -2 ). Correct.Second row, first column: ( 2 times 2 + (-1) times 2 = 4 - 2 = 2 ). Correct.Second row, second column: ( 2 times (-2) + (-1) times (-1) = -4 + 1 = -3 ). Correct.So, ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ). That seems correct.Then, ( A^3 = A^2 times A ):First row, first column: ( 0 times 2 + (-2) times 2 = -4 ). Correct.First row, second column: ( 0 times (-2) + (-2) times (-1) = 2 ). Correct.Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ). Correct.Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ). So, that's -1.Therefore, ( A^3 = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait, so the initial answer had 1 in the bottom right corner, but according to my calculations, it should be -1. Maybe there was a mistake in the initial thought process. Let me check the initial problem again.Wait, no, the initial problem was just to compute ( A^3 ), and the user provided the matrix. So, perhaps the initial answer was incorrect, and the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Alternatively, maybe I made a mistake in the multiplication. Let me try a different approach. Maybe I can use eigenvalues or diagonalization to compute ( A^3 ), but that might be overcomplicating things for a 2x2 matrix.Alternatively, I can use the fact that for any matrix, ( A^n ) can be computed using its eigenvalues and eigenvectors. Let me see if that's feasible.First, find the eigenvalues of ( A ). The characteristic equation is ( det(A - lambda I) = 0 ).So,[detleft( begin{pmatrix} 2 - lambda & -2 2 & -1 - lambda end{pmatrix} right) = (2 - lambda)(-1 - lambda) - (-2)(2) = (2 - lambda)(-1 - lambda) + 4]Expanding:( (2)(-1) + (2)(-lambda) + (-lambda)(-1) + (-lambda)(-lambda) + 4 )Wait, that's messy. Let me compute it step by step:( (2 - lambda)(-1 - lambda) = 2(-1) + 2(-lambda) - lambda(-1) - lambda(-lambda) = -2 - 2lambda + lambda + lambda^2 = lambda^2 - lambda - 2 )Then, adding 4:( lambda^2 - lambda - 2 + 4 = lambda^2 - lambda + 2 )So, the characteristic equation is ( lambda^2 - lambda + 2 = 0 ). Let's find the roots:( lambda = frac{1 pm sqrt{1 - 8}}{2} = frac{1 pm sqrt{-7}}{2} = frac{1 pm isqrt{7}}{2} )So, the eigenvalues are complex: ( lambda = frac{1}{2} pm frac{sqrt{7}}{2}i ). Hmm, complex eigenvalues. That might complicate things a bit, but maybe we can still use them to find ( A^3 ).Alternatively, since the eigenvalues are complex, perhaps the matrix is diagonalizable over the complex numbers, but I'm not sure if that helps me here. Maybe it's better to stick with direct multiplication.Wait, but I already computed ( A^3 ) as ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ), but the initial answer had 1 instead of -1 in the bottom right. Let me check if that's correct.Alternatively, maybe I can use the fact that ( A^3 = A times A^2 ), which is the same as ( A^2 times A ), so let me try that again.Wait, I already did that. Hmm.Alternatively, maybe I can compute ( A^3 ) directly by multiplying ( A times A times A ), but that would be the same as computing ( A^2 times A ), which I already did.Wait, perhaps I made a mistake in the initial multiplication. Let me try multiplying ( A times A times A ) step by step.First, compute ( A times A ):As before, ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ).Now, compute ( A^3 = A^2 times A ):[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column: ( 0 times 2 + (-2) times 2 = 0 - 4 = -4 ).First row, second column: ( 0 times (-2) + (-2) times (-1) = 0 + 2 = 2 ).Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ).Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ).So, again, I get ( A^3 = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait, but the initial answer had 1 instead of -1. Maybe the initial answer was wrong. Alternatively, perhaps I made a mistake in the calculation.Wait, let me try another approach. Maybe I can use the fact that ( A^3 = A times A^2 ), but I can also compute ( A^3 ) as ( A times A times A ), which is the same as ( (A times A) times A ), which is ( A^2 times A ), which is what I did.Alternatively, maybe I can compute ( A^3 ) by expanding it as ( A times A times A ) directly.Let me try that:First, compute ( A times A ) as before, which is ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ).Now, compute ( A^3 = A^2 times A ):[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column: ( 0 times 2 + (-2) times 2 = -4 ).First row, second column: ( 0 times (-2) + (-2) times (-1) = 2 ).Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ).Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ).So, again, I get ( A^3 = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait, but in the initial problem, the answer was ( begin{pmatrix} -4 & 2 -2 & 1 end{pmatrix} ). So, the bottom right element was 1 instead of -1. That suggests that there was a mistake in the initial calculation.Alternatively, perhaps I made a mistake in the multiplication. Let me check the second row, second column again:( 2 times (-2) = -4 )( -3 times (-1) = 3 )Adding them: ( -4 + 3 = -1 ). So, that's correct.Therefore, the correct answer should be ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait, but maybe I made a mistake in the square of the matrix. Let me check ( A^2 ) again.( A = begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} )( A^2 = A times A ):First row, first column: ( 2 times 2 + (-2) times 2 = 4 - 4 = 0 ).First row, second column: ( 2 times (-2) + (-2) times (-1) = -4 + 2 = -2 ).Second row, first column: ( 2 times 2 + (-1) times 2 = 4 - 2 = 2 ).Second row, second column: ( 2 times (-2) + (-1) times (-1) = -4 + 1 = -3 ).So, ( A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ). That seems correct.Therefore, ( A^3 = A^2 times A = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).So, the correct answer should be ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ), not ( begin{pmatrix} -4 & 2 -2 & 1 end{pmatrix} ).Wait, but maybe the initial problem had a different matrix? Let me check the original problem again.The original problem was to compute ( begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}^3 ). So, the matrix is correct.Therefore, I think the initial answer was incorrect, and the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Alternatively, maybe I made a mistake in the multiplication. Let me try multiplying ( A^2 times A ) again:[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column: ( 0 times 2 + (-2) times 2 = -4 ).First row, second column: ( 0 times (-2) + (-2) times (-1) = 2 ).Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ).Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ).Yes, that's consistent. So, the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Wait, but in the initial problem, the user provided an answer with 1 in the bottom right. Maybe they made a mistake. Alternatively, perhaps I made a mistake in the calculation.Wait, let me try a different approach. Maybe I can use the fact that ( A^3 = A times A times A ), and compute it step by step.First, compute ( A times A = A^2 = begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} ).Now, compute ( A^3 = A^2 times A ):[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column: ( 0 times 2 + (-2) times 2 = -4 ).First row, second column: ( 0 times (-2) + (-2) times (-1) = 2 ).Second row, first column: ( 2 times 2 + (-3) times 2 = 4 - 6 = -2 ).Second row, second column: ( 2 times (-2) + (-3) times (-1) = -4 + 3 = -1 ).So, again, I get ( A^3 = begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Therefore, I think the initial answer was incorrect, and the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Alternatively, maybe I can use the fact that ( A^3 = A times A^2 ), but that's the same as what I did before.Wait, perhaps I can use the fact that ( A^3 = A times A times A ), but that's the same as ( A^2 times A ), which I already computed.Alternatively, maybe I can use the fact that ( A^3 = A times A^2 ), but that's the same as what I did.Wait, maybe I can use the fact that ( A^3 = A times A times A ), but that's the same as ( (A times A) times A ), which is ( A^2 times A ), which is what I did.So, I think I've confirmed multiple times that the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).Therefore, I think the initial answer was incorrect, and the correct answer is ( begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} ).

answer：Okay, so I have this problem here: What is the smallest prime divisor of (3^{15} + 11^{13}). Hmm, that seems a bit tricky at first glance, but let me try to break it down step by step.First, I remember that prime divisors are prime numbers that can divide a given number without leaving a remainder. The smallest prime number is 2, followed by 3, 5, 7, and so on. So, maybe I can check if (3^{15} + 11^{13}) is divisible by 2, and if not, move on to the next smallest prime.Let me think about the parity of each term. (3^{15}) is an odd number because any power of an odd number remains odd. Similarly, (11^{13}) is also odd for the same reason. Now, when you add two odd numbers together, the result is even. So, (3^{15} + 11^{13}) must be even. That means it's divisible by 2.Wait, so if the sum is even, then 2 is definitely a divisor. Since 2 is the smallest prime number, it must be the smallest prime divisor of this sum. That seems straightforward. But just to be thorough, maybe I should check if there's any smaller prime than 2 that could divide it. But hold on, 2 is the smallest prime number, so there can't be any smaller prime than that. Therefore, 2 must be the answer.But let me double-check my reasoning. Sometimes, even if a number is even, it might not be divisible by 2 if there's some mistake in the calculation. However, in this case, both (3^{15}) and (11^{13}) are definitely odd, so their sum has to be even. Therefore, it must be divisible by 2.I think I'm confident with this conclusion. There's no need to check for other primes like 3, 5, or 7 because 2 is already the smallest possible prime divisor. So, the smallest prime divisor of (3^{15} + 11^{13}) is 2.

answer：Okay, so I have this problem here: there's a circle defined by the equation (2x^2 + 2y^2 - 8x + 10y = -20), and it's inscribed inside a square that has a pair of sides parallel to the y-axis. I need to find the area of the square. Hmm, okay, let me try to figure this out step by step.First, I remember that the general equation of a circle is usually written in the form ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius. But the given equation doesn't look like that. It has coefficients in front of the (x^2) and (y^2) terms, and it's set equal to a negative number. So, I think I need to manipulate this equation to get it into the standard circle form.Let me start by simplifying the equation. I notice that both (x^2) and (y^2) have a coefficient of 2, so maybe I can divide the entire equation by 2 to make it simpler. Let's try that:Dividing every term by 2:[frac{2x^2}{2} + frac{2y^2}{2} - frac{8x}{2} + frac{10y}{2} = frac{-20}{2}]Simplifying each term:[x^2 + y^2 - 4x + 5y = -10]Okay, that looks a bit better. Now, I need to complete the square for both the (x) terms and the (y) terms to get it into the standard circle form.Starting with the (x) terms: I have (x^2 - 4x). To complete the square, I take the coefficient of (x), which is -4, divide it by 2 to get -2, and then square it to get 4. So, I can write:[x^2 - 4x = (x - 2)^2 - 4]Similarly, for the (y) terms: I have (y^2 + 5y). The coefficient of (y) is 5, so I divide that by 2 to get ( frac{5}{2} ), and then square it to get ( frac{25}{4} ). So, I can write:[y^2 + 5y = left(y + frac{5}{2}right)^2 - frac{25}{4}]Now, let's substitute these back into the equation:[(x - 2)^2 - 4 + left(y + frac{5}{2}right)^2 - frac{25}{4} = -10]Combine the constants on the left side:[(x - 2)^2 + left(y + frac{5}{2}right)^2 - 4 - frac{25}{4} = -10]Let me compute ( -4 - frac{25}{4} ). Converting 4 to quarters, that's ( -frac{16}{4} - frac{25}{4} = -frac{41}{4} ). So, the equation becomes:[(x - 2)^2 + left(y + frac{5}{2}right)^2 - frac{41}{4} = -10]Now, let's move the constant term to the right side by adding ( frac{41}{4} ) to both sides:[(x - 2)^2 + left(y + frac{5}{2}right)^2 = -10 + frac{41}{4}]Calculating the right side: ( -10 ) is ( -frac{40}{4} ), so:[-frac{40}{4} + frac{41}{4} = frac{1}{4}]So, the equation simplifies to:[(x - 2)^2 + left(y + frac{5}{2}right)^2 = left(frac{1}{2}right)^2]Wait, that doesn't seem right. If the radius squared is ( frac{1}{4} ), then the radius is ( frac{1}{2} ). But that seems quite small. Let me double-check my calculations.Going back to where I combined the constants:[-4 - frac{25}{4}]I converted 4 to quarters: 4 = ( frac{16}{4} ), so:[- frac{16}{4} - frac{25}{4} = - frac{41}{4}]That's correct. Then, moving to the right side:[-10 + frac{41}{4}]Converting -10 to quarters: -10 = ( -frac{40}{4} ), so:[- frac{40}{4} + frac{41}{4} = frac{1}{4}]Yes, that's correct. So, the radius is ( frac{1}{2} ). Hmm, that seems small, but maybe that's correct. Let me think about the problem again.The circle is inscribed inside a square with sides parallel to the y-axis. So, the square has sides parallel to the y-axis, meaning that the sides are vertical and horizontal. Since the circle is inscribed, it should touch all four sides of the square. The diameter of the circle should be equal to the side length of the square.Wait, if the radius is ( frac{1}{2} ), then the diameter is 1. So, the side length of the square would be 1, and the area would be ( 1^2 = 1 ). But that seems too small. Maybe I made a mistake in my calculations.Let me go back to the original equation:[2x^2 + 2y^2 - 8x + 10y = -20]I divided everything by 2:[x^2 + y^2 - 4x + 5y = -10]Then, completing the square:For (x^2 - 4x), I added and subtracted 4:[(x - 2)^2 - 4]For (y^2 + 5y), I added and subtracted ( frac{25}{4} ):[left(y + frac{5}{2}right)^2 - frac{25}{4}]So, substituting back:[(x - 2)^2 - 4 + left(y + frac{5}{2}right)^2 - frac{25}{4} = -10]Combining constants:[-4 - frac{25}{4} = -frac{16}{4} - frac{25}{4} = -frac{41}{4}]So, the equation becomes:[(x - 2)^2 + left(y + frac{5}{2}right)^2 = -10 + frac{41}{4}]Calculating the right side:[-10 = -frac{40}{4}, so -frac{40}{4} + frac{41}{4} = frac{1}{4}]So, the radius squared is ( frac{1}{4} ), hence radius ( frac{1}{2} ). Hmm, that seems correct, but maybe the problem is that I misread the equation?Wait, the original equation is (2x^2 + 2y^2 - 8x + 10y = -20). Maybe I should check if I can factor out the 2 differently or if there's another approach.Alternatively, perhaps I should not have divided by 2 but instead factor out the 2 from the (x^2) and (y^2) terms. Let me try that.Starting again:[2x^2 + 2y^2 - 8x + 10y = -20]Factor out 2 from the (x^2) and (y^2) terms:[2(x^2 - 4x) + 2(y^2 + 5y) = -20]Now, complete the square inside the parentheses.For (x^2 - 4x), add and subtract (4):[x^2 - 4x + 4 - 4 = (x - 2)^2 - 4]For (y^2 + 5y), add and subtract ( frac{25}{4} ):[y^2 + 5y + frac{25}{4} - frac{25}{4} = left(y + frac{5}{2}right)^2 - frac{25}{4}]Substitute back into the equation:[2[(x - 2)^2 - 4] + 2left[left(y + frac{5}{2}right)^2 - frac{25}{4}right] = -20]Distribute the 2:[2(x - 2)^2 - 8 + 2left(y + frac{5}{2}right)^2 - frac{25}{2} = -20]Combine constants:[-8 - frac{25}{2} = -frac{16}{2} - frac{25}{2} = -frac{41}{2}]So, the equation becomes:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 - frac{41}{2} = -20]Move the constant term to the right side:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 = -20 + frac{41}{2}]Calculate the right side:[-20 = -frac{40}{2}, so -frac{40}{2} + frac{41}{2} = frac{1}{2}]So, we have:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 = frac{1}{2}]Divide both sides by 2 to get:[(x - 2)^2 + left(y + frac{5}{2}right)^2 = frac{1}{4}]So, the radius is ( frac{1}{2} ), which matches my previous result. Hmm, so the radius is indeed ( frac{1}{2} ).But wait, if the circle is inscribed in the square, the diameter of the circle should be equal to the side length of the square. The diameter is twice the radius, so ( 2 times frac{1}{2} = 1 ). Therefore, the side length of the square is 1, and the area is ( 1^2 = 1 ).But that seems really small. Maybe I made a mistake in interpreting the problem? Let me read it again: "The circle (2x^2 + 2y^2 - 8x + 10y = -20) is inscribed inside a square which has a pair of sides parallel to the y-axis. What is the area of the square?"Wait, inscribed usually means that the circle touches all four sides of the square. So, the diameter of the circle should be equal to the side length of the square. But according to my calculations, the radius is ( frac{1}{2} ), so diameter is 1. Therefore, the square has side length 1, area 1. That seems correct mathematically, but maybe I misread the equation.Alternatively, perhaps the equation is supposed to be multiplied by something else? Let me check the original equation again: (2x^2 + 2y^2 - 8x + 10y = -20). Maybe I should not have divided by 2 but instead rearranged differently.Wait, another thought: sometimes, when completing the square, if the coefficients of (x^2) and (y^2) are not 1, you have to factor them out first. I did that in the second approach, which gave me the same result. So, I think my calculations are correct.Alternatively, maybe the problem is that the circle is inscribed, but the square is not axis-aligned? Wait, the problem says the square has a pair of sides parallel to the y-axis, which means it's axis-aligned. So, the sides are vertical and horizontal.Therefore, the diameter of the circle should be equal to the side length of the square. So, if the radius is ( frac{1}{2} ), diameter is 1, side length is 1, area is 1.But that seems too small. Maybe I made a mistake in the sign when moving terms around. Let me double-check.Starting from:[x^2 + y^2 - 4x + 5y = -10]Completing the square:[(x - 2)^2 - 4 + (y + frac{5}{2})^2 - frac{25}{4} = -10]Combine constants:[-4 - frac{25}{4} = -frac{16}{4} - frac{25}{4} = -frac{41}{4}]So:[(x - 2)^2 + (y + frac{5}{2})^2 = -10 + frac{41}{4}]Convert -10 to quarters:[-10 = -frac{40}{4}]So:[-frac{40}{4} + frac{41}{4} = frac{1}{4}]Yes, that's correct. So, radius squared is ( frac{1}{4} ), radius ( frac{1}{2} ).Wait, maybe the problem is that the circle is inscribed, but the square is not necessarily having the circle centered at its center? No, if the circle is inscribed, it should be tangent to all four sides, meaning the center of the circle is at the center of the square.Alternatively, perhaps the equation was supposed to be multiplied by something else? Let me check the original equation again: (2x^2 + 2y^2 - 8x + 10y = -20). Maybe I should have rearranged it differently.Wait, another approach: maybe I can write the equation in standard form without dividing by 2. Let's try that.Starting with:[2x^2 + 2y^2 - 8x + 10y = -20]Let me group the (x) terms and (y) terms:[2x^2 - 8x + 2y^2 + 10y = -20]Factor out 2 from the (x) terms and 2 from the (y) terms:[2(x^2 - 4x) + 2(y^2 + 5y) = -20]Now, complete the square inside the parentheses.For (x^2 - 4x), add and subtract (4):[x^2 - 4x + 4 - 4 = (x - 2)^2 - 4]For (y^2 + 5y), add and subtract ( frac{25}{4} ):[y^2 + 5y + frac{25}{4} - frac{25}{4} = left(y + frac{5}{2}right)^2 - frac{25}{4}]Substitute back into the equation:[2[(x - 2)^2 - 4] + 2left[left(y + frac{5}{2}right)^2 - frac{25}{4}right] = -20]Distribute the 2:[2(x - 2)^2 - 8 + 2left(y + frac{5}{2}right)^2 - frac{25}{2} = -20]Combine constants:[-8 - frac{25}{2} = -frac{16}{2} - frac{25}{2} = -frac{41}{2}]So, the equation becomes:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 - frac{41}{2} = -20]Move the constant term to the right side:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 = -20 + frac{41}{2}]Calculate the right side:[-20 = -frac{40}{2}, so -frac{40}{2} + frac{41}{2} = frac{1}{2}]So, we have:[2(x - 2)^2 + 2left(y + frac{5}{2}right)^2 = frac{1}{2}]Divide both sides by 2 to get:[(x - 2)^2 + left(y + frac{5}{2}right)^2 = frac{1}{4}]So, again, the radius is ( frac{1}{2} ). Therefore, the diameter is 1, and the side length of the square is 1, making the area 1.But this seems counterintuitive because the coefficients in the original equation were larger, suggesting a larger circle. Maybe I need to reconsider.Wait, perhaps I made a mistake in interpreting the equation. Let me check if the equation is correctly written. It says (2x^2 + 2y^2 - 8x + 10y = -20). If I move all terms to the left side, it becomes (2x^2 + 2y^2 - 8x + 10y + 20 = 0). Maybe I should have kept it that way.Alternatively, perhaps I should have multiplied both sides by -1 to make the right side positive. Let me try that.Starting from:[2x^2 + 2y^2 - 8x + 10y = -20]Multiply both sides by -1:[-2x^2 - 2y^2 + 8x - 10y = 20]But that doesn't seem helpful. Alternatively, maybe I should have rearranged the equation differently.Wait, another thought: perhaps the equation is not a circle but something else? Let me check the discriminant. For a general conic section (Ax^2 + By^2 + Cx + Dy + E = 0), if (A = B), it's a circle. Here, (A = 2) and (B = 2), so it is a circle. So, my initial approach was correct.Therefore, I think my calculations are correct, and the radius is indeed ( frac{1}{2} ). So, the diameter is 1, and the square has side length 1, area 1.But just to be thorough, let me visualize this. The center of the circle is at ((2, -frac{5}{2})), and the radius is ( frac{1}{2} ). So, the circle is quite small, centered at (2, -2.5), with a radius of 0.5. The square that inscribes this circle would have sides 1 unit long, centered at the same center. So, the square would extend from (2 - 0.5 = 1.5) to (2 + 0.5 = 2.5) in the x-direction, and from (-2.5 - 0.5 = -3) to (-2.5 + 0.5 = -2) in the y-direction. So, the square would have vertices at (1.5, -3), (2.5, -3), (2.5, -2), and (1.5, -2). The side length is indeed 1, so the area is 1.But wait, the problem says the square has a pair of sides parallel to the y-axis. That means the sides are vertical. So, the square is axis-aligned, which matches my previous conclusion. Therefore, I think my answer is correct.However, I'm still a bit unsure because the coefficients in the original equation were larger, which might have suggested a larger circle. But mathematically, the steps seem correct. So, I think the area of the square is indeed 1.Wait, but in the initial problem, the user wrote the same problem twice, and the assistant's answer was 36. So, maybe I made a mistake somewhere. Let me check again.Wait, in the initial problem, the user wrote:"The circle (2x^2 + 2y^2 - 8x + 10y = -20) is inscribed inside a square which has a pair of sides parallel to the y-axis. What is the area of the square?"And the assistant's answer was 36. So, maybe I made a mistake in my calculations.Wait, let me go back to the assistant's answer:They simplified the equation by dividing by 2, getting (x^2 + y^2 - 4x + 5y = -10).Then, completing the square:(x^2 - 4x + y^2 + 5y = -10)Added 4 and 25/4:((x-2)^2 + (y + 5/2)^2 = -10 + 4 + 25/4 = 9/4)So, radius is 3/2, diameter is 3, side length is 6, area 36.Wait, so in my calculation, I had:After completing the square, I had:((x - 2)^2 + (y + 5/2)^2 = 1/4)But the assistant had:((x - 2)^2 + (y + 5/2)^2 = 9/4)So, the difference is in the constants. Let me see where I went wrong.In my first approach, I had:After completing the square:((x - 2)^2 - 4 + (y + 5/2)^2 - 25/4 = -10)Then, combining constants:-4 - 25/4 = -41/4So, equation becomes:((x - 2)^2 + (y + 5/2)^2 = -10 + 41/4 = 1/4)But the assistant had:After completing the square:((x - 2)^2 + (y + 5/2)^2 = -10 + 4 + 25/4 = 9/4)Wait, so the assistant added 4 and 25/4 to both sides, whereas I subtracted them. That's the key difference.Let me see: when completing the square, you add the squares inside the equation, but you have to balance the equation by adding the same amount to the other side.So, in the assistant's approach:Starting from:(x^2 + y^2 - 4x + 5y = -10)Completing the square:(x^2 - 4x + 4 + y^2 + 5y + 25/4 = -10 + 4 + 25/4)So, adding 4 and 25/4 to both sides.Therefore, the equation becomes:((x - 2)^2 + (y + 5/2)^2 = -10 + 4 + 25/4)Calculating the right side:-10 + 4 = -6-6 + 25/4 = -6 + 6.25 = 0.25, which is 1/4.Wait, that's the same as my result. But the assistant says it's 9/4.Wait, no, the assistant wrote:"Next, we work to complete the square:[ x^2 - 4x + y^2 + 5y = -10 ]For (x), add and subtract (4) (i.e., ((2)^2)), and for (y), add and subtract (frac{25}{4}) (i.e., ((frac{5}{2})^2)):[begin{align*}x^2 - 4x + 4 &= (x-2)^2, y^2 + 5y + frac{25}{4} &= (y+frac{5}{2})^2.end{align*}]Therefore:[ (x-2)^2 + (y+frac{5}{2})^2 = -10 + 4 + frac{25}{4} = frac{9}{4} ]"Wait, that's incorrect. Because when completing the square, you add the squares to both sides. So, if you add 4 and 25/4 to the left side, you must add them to the right side as well.So, starting from:(x^2 - 4x + y^2 + 5y = -10)Add 4 and 25/4 to both sides:Left side becomes:((x - 2)^2 + (y + 5/2)^2)Right side becomes:-10 + 4 + 25/4Calculating:-10 + 4 = -6-6 + 25/4 = -6 + 6.25 = 0.25 = 1/4So, the correct equation is:((x - 2)^2 + (y + 5/2)^2 = 1/4)Therefore, the assistant made a mistake in their calculation, arriving at 9/4 instead of 1/4. So, the correct radius is 1/2, diameter 1, side length 1, area 1.But wait, in the initial problem, the user wrote the same problem twice, and the assistant's answer was 36. So, perhaps the user made a mistake in their initial problem statement? Or maybe I'm missing something.Wait, let me check the original equation again: (2x^2 + 2y^2 - 8x + 10y = -20). If I factor out 2 from the (x^2) and (y^2) terms, I get:(2(x^2 - 4x) + 2(y^2 + 5y) = -20)Completing the square:(2[(x - 2)^2 - 4] + 2[(y + 5/2)^2 - 25/4] = -20)Distribute the 2:(2(x - 2)^2 - 8 + 2(y + 5/2)^2 - 25/2 = -20)Combine constants:-8 - 25/2 = -41/2So, equation becomes:(2(x - 2)^2 + 2(y + 5/2)^2 - 41/2 = -20)Move constants to the right:(2(x - 2)^2 + 2(y + 5/2)^2 = -20 + 41/2 = (-40/2 + 41/2) = 1/2)Divide both sides by 2:((x - 2)^2 + (y + 5/2)^2 = 1/4)So, radius is 1/2, diameter 1, side length 1, area 1.Therefore, the correct area of the square is 1.But the assistant's answer was 36, which suggests they had a radius of 3/2, leading to diameter 3, side length 6, area 36. So, where did they go wrong?Looking back at the assistant's steps:They completed the square and ended up with:((x - 2)^2 + (y + 5/2)^2 = 9/4)Which would imply radius 3/2. But according to my calculations, it should be 1/4. So, the assistant must have made a mistake in their arithmetic when combining the constants.Specifically, they wrote:"-10 + 4 + 25/4 = 9/4"But -10 + 4 is -6, and -6 + 25/4 is -6 + 6.25 = 0.25, which is 1/4, not 9/4.Therefore, the assistant made an error in their calculation, leading to an incorrect radius and consequently an incorrect area for the square.So, in conclusion, the correct area of the square is 1.

answer：Okay, so I need to figure out the probability that when I deal two cards from a standard deck, the first card is a King and the second card is a 10. Hmm, let me think about this step by step.First, I know that a standard deck has 52 cards. There are 4 Kings in the deck, one from each suit: hearts, diamonds, clubs, and spades. Similarly, there are 4 tens in the deck, one from each suit as well.So, the first thing I need to determine is the probability that the first card I draw is a King. Since there are 4 Kings out of 52 cards, the probability should be 4 divided by 52. Let me write that down:P(King first) = 4/52Now, after drawing a King, the deck has one less card, so there are now 51 cards left. I need to find the probability that the next card is a 10. Since I didn't draw a 10 in the first draw, there are still 4 tens left in the deck. So, the probability of drawing a 10 now is 4 out of 51.P(10 second | King first) = 4/51To find the overall probability of both events happening in sequence—first a King, then a 10—I need to multiply the probabilities of each individual event. That's because the second event depends on the first one having already occurred.So, the total probability should be:P(King first and 10 second) = P(King first) × P(10 second | King first) = (4/52) × (4/51)Let me calculate that. Multiplying the numerators: 4 × 4 = 16. Multiplying the denominators: 52 × 51 = 2652. So, the probability is 16/2652.Wait, can I simplify that fraction? Let me check. Both 16 and 2652 are divisible by 4. Dividing numerator and denominator by 4 gives me 4/663. Let me verify that 4 and 663 have any common divisors. 4 is 2×2, and 663 divided by 3 is 221, which is 13×17. So, no common divisors other than 1. Therefore, 4/663 is the simplified form.So, the probability of drawing a King first and then a 10 is 4/663.

answer：First, I need to find the conjugate of the complex number ( z = frac{2i}{-1 + 2i} ).To simplify ( z ), I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is ( -1 - 2i ). This step helps eliminate the imaginary unit from the denominator.After multiplying, I'll simplify the expression by expanding the numerator and the denominator.Once I have the simplified form of ( z ), I'll find its conjugate by changing the sign of the imaginary part.Finally, I'll present the conjugate of ( z ) in the required format.