answer:Given ( q = -p^2 + 6p - 5 ), apply the Shoelace Theorem for triangle ( ABC ): [ text{Area} = frac{1}{2} | (0)(3) + (4)(-p^2 + 6p - 5) + (p)(-5) - (-5)(4) - (3)(p) - (-p^2 + 6p - 5)(0) | ] This simplifies to: [ text{Area} = frac{1}{2} | -4p^2 + 24p - 20 - 5p - 12p + 20 | ] [ = frac{1}{2} | -4p^2 + 7p | ] For ( 0 le p le 5 ), find the absolute value as: [ frac{1}{2} | -4p^2 + 7p | = frac{1}{2} | -4p(p - 1.75) | ] To maximize ( | -4p(p - 1.75) | ), we consider critical points where it achieves its maximum within the interval. Maximizing occurs either at ( p = 0 ), ( p = 5 ), or ( p = 1.75 ). Evaluating at these points, we get: [ text{Max Area} = frac{1}{2} max{ |-4(0)(0 - 1.75)|, |-4(5)(5 - 1.75)|, |-4(1.75)(1.75 - 1.75)| } ] [ = frac{1}{2} max{ 0, |-64.5|, 0 } = frac{32.25}{2} = boxed{16.125} ]

answer:Given the sequence {a_{n}} with a_{1}=4 and a_{n}+a_{n+1}=4n+2 for nin mathbb{N}^{*}, we aim to find the maximum value of n such that the sum of the first n terms, S_{n}, is less than 2023. First, we rewrite the given relation as a_{n+1}=-a_{n}+4n+2. To simplify this, we consider a transformation of the form a_{n+1}+k(n+1)+b=-(a_{n}+kn+b), where k and b are real numbers. This leads to the equation a_{n+1}=-a_{n}-2kn-2b-k. Comparing coefficients with the original relation, we get a system of equations: [ left{begin{array}{l} -2k=4 -2b-k=2 end{array}right. ] Solving this system yields k=-2 and b=0. Thus, we have a_{n+1}-2(n+1)=-(a_{n}-2n). Let b_{n}=a_{n}-2n, then b_{n+1}=a_{n+1}-2(n+1), and frac{{b}_{n+1}}{{b}_{n}}=-1. This indicates that {b_{n}} is a geometric sequence with the first term b_{1}=a_{1}-2=2 and a common ratio of -1. Therefore, the general formula for {b_{n}} is b_{n}=2times (-1)^{n-1}. Substituting back, we find a_{n}=2n+2times (-1)^{n-1}. The sum of the first n terms, S_{n}, can be calculated as: [ S_{n}=frac{2+2n}{2}times n+2times frac{1times[1-(-1)^{n}]}{1-(-1)}=n^{2}+n+1-(-1)^{n} ] For odd n, S_{n}=n^{2}+n < 2023, leading to n(n+1) < 2023. The maximum n satisfying this is 43. For even n, S_{n}=n^{2}+n+2 < 2023, leading to n(n+1) < 2021. The maximum n satisfying this is 44. Therefore, the maximum value of n that satisfies S_{n} < 2023 is boxed{44}, corresponding to choice boxed{C}.

answer:1. **Understanding the Path**: The ship travels from point A to point B along a semicircular path centered at Island X. This means that the distance from any point on the semicircle to Island X is the radius of the semicircle, say r. This distance remains constant as the ship moves from A to B. 2. **Graph Representation for AB**: On the graph, the constant distance r as the ship moves from A to B along the semicircle will be represented by a horizontal line segment at height r. 3. **Travel from B to C**: The ship then travels along a straight path from B to C. During this segment of the journey, the distance from the ship to Island X changes. 4. **Analyzing Distance Change from B to C**: - As the ship starts moving from B towards C, it initially moves closer to Island X. - The closest point occurs when the line from X to the ship is perpendicular to the path BC. Let's denote this closest point as Y. - After passing Y, the distance from the ship to Island X begins to increase. 5. **Graph Representation for BC**: The graph will show a decrease in the distance from X as the ship moves from B to Y, reaching a minimum at Y, and then an increase in the distance as the ship moves from Y to C. 6. **Choosing the Correct Graph**: - The graph must show a horizontal line segment at height r for the semicircular path AB. - For the straight path BC, the graph must show a curve that dips down to a minimum and then rises again. - Among the provided choices, the graph that fits this description is option (B). Therefore, the correct graph that represents the ship's distance from Island X as it moves along its course is boxed{text{(B)}}.

answer:To find out how many fish Michael had initially, we need to subtract the number of fish Ben gave him from the total number of fish Michael has now. So, if Michael now has 67 fish and Ben gave him 18.0 fish, we can calculate the initial number of fish Michael had as follows: Initial number of fish = Total number of fish after Ben gave some - Number of fish Ben gave Initial number of fish = 67 - 18.0 Initial number of fish = 49 Therefore, Michael had boxed{49} fish initially.