answer：First, I need to determine how many of each item Alex plans to use in her collage. She intends to use 3 fabric swatches, 4 magazine cutouts, 2 buttons, 5 ribbons, and 1 sketch.Next, I'll calculate the total number of items she will use by adding up each of these quantities:3 (fabric swatches) + 4 (magazine cutouts) + 2 (buttons) + 5 (ribbons) + 1 (sketch) = 15 items in total.Then, I'll determine how many of each item Alex will have left after using them in the collage by subtracting the number used from the number she originally has:- Fabric swatches: 12 - 3 = 9 left- Magazine cutouts: 18 - 4 = 14 left- Buttons: 9 - 2 = 7 left- Ribbons: 15 - 5 = 10 left- Sketches: 6 - 1 = 5 leftFinally, I'll add up the remaining items to find out how many items she has left over:9 (fabric swatches) + 14 (magazine cutouts) + 7 (buttons) + 10 (ribbons) + 5 (sketches) = 45 items left over.

answer：First, I recognize that the player's scoring pattern forms an arithmetic sequence because each game he scores 2 more points than the previous one.Given:- Total points scored in the season: 1,200- Number of games played: 60- Points scored in the first game: 10- Common difference between consecutive games: 2I'll use the formula for the sum of an arithmetic series to find the total points:Sum = (Number of terms / 2) × (First term + Last term)Plugging in the known values:1,200 = (60 / 2) × (10 + Last term)Simplifying the equation:1,200 = 30 × (10 + Last term)1,200 = 300 + 30 × Last termSubtracting 300 from both sides:900 = 30 × Last termDividing both sides by 30:Last term = 30Therefore, the player scored 30 points in the last game.

answer：First, I need to determine the profit Jamie makes per pastry. The selling price is 3.00 and the cost to make each pastry is 1.50, so the profit per pastry is 1.50.Jamie wants to make a daily profit of 90 from the pastries. To find out how many pastries need to be sold, I divide the desired profit by the profit per pastry: 90 ÷ 1.50 = 60.Therefore, Jamie needs to sell 60 pastries each day to achieve the desired profit.

answer：First, I'll start with the number of visitors in the first year, which is 8,000.In the second year, there's a 25% increase due to the new conservation program. To find the number of visitors in the second year, I'll calculate 25% of 8,000 and add it to the original number.Next, in the third year, the number of visitors decreases by 15% from the second year's total because of new restrictions. I'll calculate 15% of the second year's visitor count and subtract that from the second year's total to find the number of visitors in the third year.Finally, I'll present the calculated number of visitors for the third year as the answer.

answer：First, I need to determine the total number of shows attended over 3 weeks. Since they attend 4 shows each week, the total number of shows is 4 multiplied by 3, which equals 12 shows.Next, I'll calculate the total cost for the entry fees. Each show costs 15, so for 12 shows, the entry fees amount to 12 multiplied by 15, totaling 180.Then, I'll calculate the total cost for snacks. They spend 10 on snacks after each show, so for 12 shows, the snack expenses are 12 multiplied by 10, totaling 120.Finally, I'll add the total entry fees and the total snack costs to find the overall expenditure. Adding 180 for entry fees and 120 for snacks gives a total of 300.

answer：Alright, so I have these two problems to solve. Let me start with the first one about the integral using the method of steepest descent. Hmm, I remember that the method of steepest descent is a technique used to approximate integrals of the form ( int e^{M f(x)} dx ) when M is large. But in this case, the integral is ( int_{-infty}^{infty} e^{-x^4 + 2x^2} dx ). Wait, so the exponent is ( -x^4 + 2x^2 ). I need to see if I can apply the method of steepest descent here.First, let me recall that the method of steepest descent involves finding the saddle points of the function in the exponent. So, I need to consider the function ( f(x) = -x^4 + 2x^2 ). To find the saddle points, I should take the derivative of f(x) and set it equal to zero.So, ( f'(x) = -4x^3 + 4x ). Setting this equal to zero: ( -4x^3 + 4x = 0 ). Factoring out 4x: ( 4x(-x^2 + 1) = 0 ). So, the critical points are at x = 0 and x = ±1.Now, I need to determine which of these are maxima or minima. Let's compute the second derivative: ( f''(x) = -12x^2 + 4 ). Evaluating at x = 0: ( f''(0) = 4 > 0 ), so it's a local minimum. At x = 1: ( f''(1) = -12 + 4 = -8 < 0 ), so it's a local maximum. Similarly, at x = -1: ( f''(-1) = -12 + 4 = -8 < 0 ), another local maximum.So, the function ( f(x) ) has a local minimum at x = 0 and local maxima at x = ±1. Since we're dealing with an integral over the entire real line, the main contributions to the integral will come from the regions around the maxima because the exponential function will be largest there.Wait, but in the method of steepest descent, usually, we consider the dominant contributions from the saddle points. However, in this case, our function is real, so the saddle points are just the critical points. Since the integral is over the real line, and the exponent is real, we can use the method of Laplace's method or steepest descent.But since the exponent is a polynomial, maybe we can expand around the maxima. Let me think. The function ( e^{-x^4 + 2x^2} ) is symmetric because it's even in x, so the integral from -infty to infty is twice the integral from 0 to infty.So, maybe I can approximate the integral by expanding around x = 1 and x = -1. Let me try expanding around x = 1.Let me set x = 1 + t, where t is small. Then, let's expand f(x) = -x^4 + 2x^2 around x = 1.First, compute f(1) = -1 + 2 = 1.Now, compute f'(1) = -4(1)^3 + 4(1) = -4 + 4 = 0, which makes sense since it's a critical point.Compute f''(1) = -12(1)^2 + 4 = -12 + 4 = -8.So, the expansion around x = 1 is f(x) ≈ f(1) + (1/2)f''(1)(t)^2 = 1 - 4t^2.Similarly, for x near -1, the expansion would be similar because of symmetry.Therefore, the integral near x = 1 is approximately ( e^{1} int e^{-4t^2} dt ). Similarly, near x = -1, it's the same.So, the total contribution from both sides is 2 * e^{1} * (1/2) sqrt(π / 4) )? Wait, let me think.Wait, the integral of ( e^{-at^2} dt ) from -infty to infty is sqrt(π/a). So, if I have an integral around t near 0, the integral is approximately sqrt(π / |f''(1)|) times e^{f(1)}.But since we're expanding around x = 1, and the original integral is over x, which we've shifted to t, so the integral near x = 1 is approximately e^{1} * sqrt(π / 8). Similarly, near x = -1, it's the same.Therefore, the total integral is approximately 2 * e * sqrt(π / 8). Simplifying sqrt(π / 8) is sqrt(π)/(2 sqrt(2)). So, 2 * e * sqrt(π)/(2 sqrt(2)) = e * sqrt(π)/sqrt(2).But wait, is that the whole story? Because the function also has a minimum at x = 0. Does that contribute significantly?At x = 0, f(x) = 0, so e^{0} = 1. But the second derivative is positive, so it's a minimum. So, the function is decreasing away from x = 0 towards x = ±1. But since the exponent is negative beyond x = ±1, the tails decay rapidly.Wait, actually, the exponent is ( -x^4 + 2x^2 ). So, as x becomes large, the dominant term is -x^4, which goes to negative infinity. So, the integrand decays rapidly for large |x|. Therefore, the main contributions are indeed around the maxima at x = ±1.But wait, when I expand around x = 1, I get a Gaussian integral, but I also need to consider the Jacobian or the change of variables. Let me make sure.Let me write x = 1 + t, then dx = dt. So, the integral near x = 1 is approximately ( e^{1} int e^{-4t^2} dt ). Similarly, near x = -1, it's the same.So, the total integral is approximately 2 * e * (sqrt(π)/ (2 sqrt(2))) ) = e * sqrt(π)/sqrt(2).Wait, let me compute that again. The integral of e^{-4t^2} dt from -infty to infty is sqrt(π)/sqrt(4) = sqrt(π)/2. So, the integral near x = 1 is e * sqrt(π)/2. Similarly, near x = -1, it's another e * sqrt(π)/2. So, total is e * sqrt(π).But wait, that seems too large. Because the original integral is over the entire real line, but the main contributions are from around x = ±1, each contributing e * sqrt(π)/2, so total e * sqrt(π).But let me check if I missed any factors. The expansion around x = 1 is f(x) ≈ 1 - 4t^2, so the integral becomes e^{1} ∫ e^{-4t^2} dt. The integral of e^{-4t^2} is sqrt(π)/2, so e * sqrt(π)/2. Similarly for x = -1, so total e * sqrt(π).But wait, the original exponent is -x^4 + 2x^2, which at x = 1 is -1 + 2 = 1, so that's correct. The second derivative is -8, so the Gaussian is e^{-4t^2}, which integrates to sqrt(π)/2. So, two such contributions give e * sqrt(π).But wait, I think I might have missed a factor because the method of steepest descent usually involves a factor of sqrt(2π) / sqrt(|f''(x)|). Let me recall the formula.The Laplace method approximation for an integral ( int e^{M f(x)} dx ) around a maximum at x0 is approximately e^{M f(x0)} * sqrt(2π / (M |f''(x0)|)). But in our case, M is 1, because the exponent is f(x) = -x^4 + 2x^2. So, the approximation would be e^{f(x0)} * sqrt(2π / |f''(x0)|).Wait, so for each maximum at x = ±1, the contribution is e^{1} * sqrt(2π / 8) = e * sqrt(π/4) = e * (sqrt(π)/2). So, two such contributions give e * sqrt(π).Yes, that matches what I had earlier. So, the approximation is e * sqrt(π).But wait, let me check if the original integral is indeed dominated by these contributions. Because the function is symmetric, and the maxima are at ±1, and the function decays rapidly beyond that, this should be a good approximation.So, the answer to the first problem is approximately e * sqrt(π).Now, moving on to the second problem. Given a matrix A in R^{3x3} with eigenvalues λ1 = 1, λ2 = -1, λ3 = 2. We need to determine if A is diagonalizable. If it is, find matrices D and P such that A = PDP^{-1}.Alright, so a matrix is diagonalizable if it has a basis of eigenvectors. For a 3x3 matrix, this requires that there are 3 linearly independent eigenvectors. Since A has 3 distinct eigenvalues, it is diagonalizable. Because if all eigenvalues are distinct, the matrix is diagonalizable.Wait, is that correct? Yes, because for each distinct eigenvalue, there is at least one eigenvector, and since they are distinct, these eigenvectors are linearly independent.So, A is diagonalizable.Now, to find D and P. D is the diagonal matrix with eigenvalues on the diagonal. So, D = diag(1, -1, 2). The order can vary depending on the order of eigenvectors in P.But since we don't have specific eigenvectors, we can't write down P explicitly unless we have more information about A. Wait, but the problem doesn't give us the matrix A, just its eigenvalues. So, we can't find the exact P and D without knowing A.Wait, but perhaps the question is just to state that A is diagonalizable and that D is diag(1, -1, 2), and P is the matrix of eigenvectors. But since we don't have A, we can't find P.Wait, maybe I misread. Let me check. The problem says: "Given a matrix A ∈ R^{3×3} with eigenvalues λ1 = 1, λ2 = -1, and λ3 = 2, determine if A is diagonalizable. If A is diagonalizable, find a diagonal matrix D and an invertible matrix P such that A = PDP^{-1}."So, since A has distinct eigenvalues, it's diagonalizable. So, D is diag(1, -1, 2). But without knowing the eigenvectors, we can't specify P. So, perhaps the answer is that A is diagonalizable, D is diag(1, -1, 2), and P is the matrix whose columns are the eigenvectors of A corresponding to the eigenvalues in D.But since the problem doesn't provide A, we can't compute P explicitly. So, maybe the answer is just that A is diagonalizable with D as diag(1, -1, 2), and P is formed by the eigenvectors.Alternatively, perhaps the problem expects a general form, but without more information, I think that's as far as we can go.Wait, but maybe I'm overcomplicating. Since the eigenvalues are distinct, A is diagonalizable, and D is diag(1, -1, 2). P is the matrix of eigenvectors, but since we don't have A, we can't find P. So, the answer is that A is diagonalizable, and D is diag(1, -1, 2), with P being the matrix of eigenvectors.But perhaps the problem expects more. Let me think again. Maybe it's a trick question. Wait, no, because if a matrix has distinct eigenvalues, it's diagonalizable regardless of the field, as long as the eigenvalues are in the field. Since we're in R, and the eigenvalues are real, it's diagonalizable.So, to sum up, A is diagonalizable, D is diag(1, -1, 2), and P is the matrix of eigenvectors. But since we don't have A, we can't write P explicitly.Wait, but maybe the problem expects us to note that since the eigenvalues are distinct, A is diagonalizable, and D is diag(1, -1, 2). So, perhaps that's sufficient.Alternatively, maybe the problem expects us to construct P and D in terms of the eigenvectors, but without knowing A, we can't do that. So, perhaps the answer is just that A is diagonalizable, and D is diag(1, -1, 2), with P being the matrix whose columns are the corresponding eigenvectors.But to be thorough, let me consider if there's any other condition. For example, if the matrix had repeated eigenvalues, we'd need to check if there are enough eigenvectors. But here, all eigenvalues are distinct, so it's straightforward.So, in conclusion, A is diagonalizable, D is diag(1, -1, 2), and P is the matrix of eigenvectors of A.Wait, but perhaps the problem expects more specific information. Let me think again. Since the eigenvalues are distinct, the matrix is diagonalizable, and D is diag(1, -1, 2). The matrix P is formed by the eigenvectors corresponding to these eigenvalues. So, without knowing A, we can't write P explicitly, but we can state its form.So, I think that's the answer.Now, putting it all together.For problem 1, the integral is approximately e * sqrt(π).For problem 2, A is diagonalizable, D is diag(1, -1, 2), and P is the matrix of eigenvectors.Wait, but let me double-check the integral approximation. I think I might have missed a factor. Let me go back.The method of steepest descent (or Laplace's method) for an integral ( int e^{M f(x)} dx ) is approximated by ( e^{M f(x_0)} sqrt{frac{2pi}{M |f''(x_0)|}} ) when M is large. But in our case, M is 1, so the approximation is ( e^{f(x_0)} sqrt{frac{2pi}{|f''(x_0)|}} ).So, for each maximum at x = ±1, the contribution is ( e^{1} sqrt{frac{2pi}{8}} = e sqrt{frac{pi}{4}} = e frac{sqrt{pi}}{2} ).Since there are two such maxima, the total approximation is ( 2 * e frac{sqrt{pi}}{2} = e sqrt{pi} ).Yes, that's correct. So, the integral is approximately e * sqrt(π).Therefore, the answers are:1. The integral is approximately ( e sqrt{pi} ).2. A is diagonalizable, D is diag(1, -1, 2), and P is the matrix of eigenvectors.