💡Okay, so I have this problem here about displacement and time. The equation given is s equals negative two t squared plus eight t. They want to know at what moment the velocity is zero. Hmm, velocity is the derivative of displacement with respect to time, right? So I need to find the derivative of s with respect to t and then set it equal to zero to find the time when the velocity is zero.Alright, let's start by writing down the displacement function: s = -2t² + 8t. To find the velocity, I need to take the derivative of s with respect to t. I remember that the derivative of t squared is 2t, so applying that here, the derivative of -2t² should be -4t. Then, the derivative of 8t is just 8. So putting that together, the velocity function v(t) is -4t + 8.Now, I need to find when this velocity is zero. So I'll set up the equation: -4t + 8 = 0. Let me solve for t. If I subtract 8 from both sides, I get -4t = -8. Then, dividing both sides by -4, I find that t equals 2. So, the velocity is zero at t equals 2 seconds.Let me double-check my work. The displacement function is a quadratic, which means the graph is a parabola. Since the coefficient of t squared is negative, the parabola opens downward, meaning the vertex is the maximum point. The vertex occurs at t = -b/(2a). In this case, a is -2 and b is 8, so t = -8/(2*(-2)) = -8/(-4) = 2. That matches what I found earlier. So, the velocity is zero at the peak of the parabola, which makes sense because that's where the object changes direction from moving forward to backward, or vice versa.I think that makes sense. The object starts moving, reaches a maximum displacement at t=2, and then starts moving back. So, at t=2, the instantaneous velocity is zero. Yeah, that seems right. I don't think I made any mistakes in my calculations. The derivative was straightforward, and solving for t gave me a clear answer. Plus, the vertex formula confirmed it. So, I'm pretty confident that t=2 is the correct answer.**Final Answer**The corresponding moment when the velocity is zero is boxed{C}.

💡Okay, so I have this problem about a school that bought some wooden boards to make desks and chairs for labor education. There are two parts to the problem. Let me try to figure out how to solve them step by step.First, part (1) asks how many desks and chairs can be assembled from the wooden boards. The school has 400 type A boards and 500 type B boards. Each desk needs 2 type A and 1 type B, and each chair needs 1 type A and 2 type B. Hmm, so I think I need to set up some equations here.Let me denote the number of desks as x and the number of chairs as y. Then, for type A boards, each desk uses 2 and each chair uses 1, so the total type A used is 2x + y, and that should be less than or equal to 400. Similarly, for type B boards, each desk uses 1 and each chair uses 2, so total type B used is x + 2y, which should be less than or equal to 500.So, I have two inequalities:1. 2x + y ≤ 4002. x + 2y ≤ 500But actually, since the school wants to use all the boards, I think we can set these as equalities:1. 2x + y = 4002. x + 2y = 500Now, I need to solve this system of equations. Maybe I can use substitution or elimination. Let me try elimination. If I multiply the first equation by 2, I get:1. 4x + 2y = 800Now, subtract the second equation from this:(4x + 2y) - (x + 2y) = 800 - 500Which simplifies to:3x = 300So, x = 100Now, plug x = 100 back into the first equation:2(100) + y = 400200 + y = 400y = 200So, the school can assemble 100 desks and 200 chairs. That seems straightforward.Now, moving on to part (2). There are 30 students who need to assemble these desks and chairs. Each student takes 10 minutes to assemble a desk and 7 minutes to assemble a chair. The question is whether they can complete the tasks simultaneously by grouping the students effectively.Hmm, so I need to figure out if there's a way to split the 30 students into two groups: one group making desks and the other making chairs, such that both groups finish at the same time.Let me denote the number of students making desks as a, and the number making chairs as b. Since there are 30 students in total, a + b = 30.Now, the time taken by the desk group to make all the desks is the total number of desks divided by the number of students making desks, multiplied by the time per desk. Similarly for chairs.So, time for desks: (100 desks) / (a students) * 10 minutes per deskTime for chairs: (200 chairs) / (b students) * 7 minutes per chairWe want these times to be equal so that both groups finish simultaneously. So, set them equal:(100 / a) * 10 = (200 / b) * 7Simplify this equation:1000 / a = 1400 / bCross-multiplying:1000b = 1400aDivide both sides by 100:10b = 14aSimplify further by dividing both sides by 2:5b = 7aSo, 5b = 7a. But we also know that a + b = 30. Let me write that down:a + b = 30From 5b = 7a, I can express b in terms of a:b = (7/5)aNow, substitute this into the total number of students:a + (7/5)a = 30Combine like terms:(5/5)a + (7/5)a = 30(12/5)a = 30Multiply both sides by 5:12a = 150Divide both sides by 12:a = 150 / 12a = 12.5Wait, a is 12.5? But the number of students has to be a whole number. You can't have half a student. So, this suggests that it's not possible to split the students into two groups of whole numbers that satisfy both the time equality and the total number of students.Therefore, the tasks cannot be completed simultaneously by grouping the students effectively because we end up with a fractional number of students, which isn't practical.I think that's the conclusion. It makes sense because the ratio of students needed doesn't divide evenly into the total number of students available. So, unless they can somehow split a student, which isn't possible, the tasks can't be finished at the same time.

💡Okay, so I have this problem where I need to find the values of a and b for the function f(x) = x³ + a x² + b x - 4, given that the tangent line at the point P(2, f(2)) is y = x - 4. Then, I also need to find the extreme values of f(x). Hmm, let me try to figure this out step by step.First, I remember that for a tangent line at a point, two conditions must be satisfied: the function must pass through that point, and the derivative of the function at that point must equal the slope of the tangent line. So, in this case, the point is (2, f(2)), and the tangent line is y = x - 4. Let me start by finding f(2). Plugging x = 2 into the function f(x):f(2) = (2)³ + a*(2)² + b*(2) - 4f(2) = 8 + 4a + 2b - 4Simplify that: 8 - 4 is 4, so f(2) = 4 + 4a + 2b.But wait, the tangent line at x = 2 is y = x - 4, so when x = 2, y = 2 - 4 = -2. That means f(2) must equal -2. So:4 + 4a + 2b = -2Let me write that as an equation:4a + 2b = -6I can simplify this by dividing both sides by 2:2a + b = -3 ...(1)Okay, that's one equation involving a and b. Now, I need another equation. That's where the derivative comes in. The slope of the tangent line at x = 2 is equal to the derivative of f at x = 2. The tangent line is y = x - 4, so its slope is 1. Therefore, f'(2) = 1.Let me find the derivative of f(x):f'(x) = d/dx [x³ + a x² + b x - 4] = 3x² + 2a x + bSo, f'(2) = 3*(2)² + 2a*(2) + b = 3*4 + 4a + b = 12 + 4a + bWe know this equals 1:12 + 4a + b = 1Let me write that as another equation:4a + b = 1 - 124a + b = -11 ...(2)Now, I have two equations:(1) 2a + b = -3(2) 4a + b = -11I can solve this system of equations to find a and b. Let me subtract equation (1) from equation (2):(4a + b) - (2a + b) = -11 - (-3)4a + b - 2a - b = -11 + 32a = -8So, a = -8 / 2 = -4Now, plug a = -4 into equation (1):2*(-4) + b = -3-8 + b = -3b = -3 + 8b = 5Alright, so a = -4 and b = 5. That answers part (1).Now, moving on to part (2): finding the extreme values of f(x). Extreme values occur where the derivative is zero or undefined. Since f(x) is a polynomial, its derivative is defined everywhere, so we only need to find where f'(x) = 0.We already found the derivative earlier:f'(x) = 3x² + 2a x + bBut now we know a and b, so plugging them in:f'(x) = 3x² + 2*(-4)x + 5 = 3x² - 8x + 5We need to solve 3x² - 8x + 5 = 0 to find critical points.Let me try factoring this quadratic. Looking for two numbers that multiply to 3*5 = 15 and add up to -8. Hmm, -5 and -3 multiply to 15 and add to -8. So, let's split the middle term:3x² - 5x - 3x + 5 = 0x(3x - 5) -1(3x - 5) = 0(x - 1)(3x - 5) = 0So, the critical points are x = 1 and x = 5/3.Now, to determine if these are maxima or minima, I can use the second derivative test or analyze the sign changes of the first derivative.Let me compute the second derivative:f''(x) = d/dx [3x² - 8x + 5] = 6x - 8Evaluate f''(x) at x = 1:f''(1) = 6*1 - 8 = -2Since f''(1) is negative, the function is concave down at x = 1, so this is a local maximum.Now, evaluate f''(x) at x = 5/3:f''(5/3) = 6*(5/3) - 8 = 10 - 8 = 2Since f''(5/3) is positive, the function is concave up at x = 5/3, so this is a local minimum.Now, let's find the corresponding y-values for these critical points.First, at x = 1:f(1) = (1)³ + (-4)*(1)² + 5*(1) - 4 = 1 - 4 + 5 - 4 = (1 - 4) + (5 - 4) = (-3) + (1) = -2So, f(1) = -2.Next, at x = 5/3:f(5/3) = (5/3)³ + (-4)*(5/3)² + 5*(5/3) - 4Let me compute each term step by step:(5/3)³ = 125 / 27(-4)*(5/3)² = (-4)*(25/9) = -100 / 95*(5/3) = 25 / 3So, putting it all together:f(5/3) = 125/27 - 100/9 + 25/3 - 4To combine these, I need a common denominator, which is 27.Convert each term:125/27 remains as is.-100/9 = -300/2725/3 = 225/27-4 = -108/27Now, add them up:125/27 - 300/27 + 225/27 - 108/27Combine numerators:125 - 300 + 225 - 108 = (125 + 225) - (300 + 108) = 350 - 408 = -58So, f(5/3) = -58/27Therefore, the extreme values are:- Local maximum at x = 1: f(1) = -2- Local minimum at x = 5/3: f(5/3) = -58/27Wait, let me double-check my calculations for f(5/3). I might have made a mistake in the arithmetic.Starting again:f(5/3) = (5/3)³ + (-4)*(5/3)² + 5*(5/3) - 4Compute each term:(5/3)³ = 125 / 27 ≈ 4.6296(-4)*(5/3)² = (-4)*(25/9) = -100 / 9 ≈ -11.11115*(5/3) = 25/3 ≈ 8.3333-4 remains as -4.Now, adding them up:125/27 - 100/9 + 25/3 - 4Convert all to 27 denominator:125/27 - 300/27 + 225/27 - 108/27Now, adding numerators:125 - 300 = -175-175 + 225 = 5050 - 108 = -58So, f(5/3) = -58/27 ≈ -2.1481Yes, that seems correct.So, summarizing:a = -4, b = 5Extreme values:- Local maximum at x = 1: f(1) = -2- Local minimum at x = 5/3: f(5/3) = -58/27I think that's it. Let me just make sure I didn't make any calculation errors.For a and b:From f(2) = -2, we had 2a + b = -3From f'(2) = 1, we had 4a + b = -11Subtracting, 2a = -8, so a = -4, then b = 5. That seems correct.For the derivative, f'(x) = 3x² -8x +5, which factors to (3x -5)(x -1). So critical points at x=1 and x=5/3. Then, using the second derivative test, f''(1) = -2 <0, so maximum; f''(5/3)=2>0, so minimum. Calculated f(1)=-2 and f(5/3)=-58/27. Looks good.Yeah, I think I did it right.

💡First, I need to determine the length of the shoe in size 8 and size 17. I know that each unit increase in size corresponds to a 1/5-inch increase in length. Since size 15 is 10.4 inches long, I can calculate the length of size 8 by subtracting the increase for each size from 8 to 15, and the length of size 17 by adding the increase for each size from 15 to 17.The difference in shoe sizes between size 15 and size 8 is 7 sizes. Therefore, the length of size 8 is:[text{Length of size 8} = 10.4 text{ inches} - (7 times frac{1}{5} text{ inches}) = 10.4 text{ inches} - 1.4 text{ inches} = 9 text{ inches}]The difference in shoe sizes between size 15 and size 17 is 2 sizes. Therefore, the length of size 17 is:[text{Length of size 17} = 10.4 text{ inches} + (2 times frac{1}{5} text{ inches}) = 10.4 text{ inches} + 0.4 text{ inches} = 10.8 text{ inches}]Next, I calculate the percentage difference in length between the largest (size 17) and smallest (size 8) sizes:[text{Percentage difference} = left( frac{10.8 text{ inches} - 9 text{ inches}}{9 text{ inches}} right) times 100% = left( frac{1.8 text{ inches}}{9 text{ inches}} right) times 100% = 0.2 times 100% = 20%]

💡Okay, so I have this problem where I need to construct the midlines of a rectangle using only a ruler that can draw straight lines. Alright, let's break this down step by step. First, I need to understand what midlines are. I think midlines in a rectangle are the lines that connect the midpoints of the opposite sides. So, if I have a rectangle, there should be two midlines: one horizontal and one vertical, intersecting at the center of the rectangle.Alright, so how do I find the midpoints of the sides using just a ruler? Well, if I can find the midpoints, then I can just connect them to form the midlines. But how do I find the midpoints without any measurements? Hmm.Maybe I can use the properties of a rectangle. I remember that in a rectangle, the diagonals bisect each other. That means if I draw both diagonals, they will intersect at the center of the rectangle. So, if I can find the intersection point of the diagonals, that will give me the center. Then, from there, I can somehow find the midpoints of the sides.Let me try to visualize this. Suppose I have a rectangle ABCD with vertices A, B, C, D. If I draw diagonal AC and diagonal BD, they should intersect at point O, which is the center. So, point O is the midpoint of both diagonals. Now, how can I use this point O to find the midpoints of the sides?Maybe I can draw lines from point O to the sides. Wait, but I only have a ruler, so I can only draw straight lines. Perhaps I can use some geometric constructions. Let's think about triangles. If I connect point O to one of the vertices, say A, and then extend that line, maybe I can find some proportional segments.Alternatively, maybe I can use the concept of similar triangles. If I can create triangles that are similar, I can use the properties of similarity to find midpoints. Let me try to recall how similarity works. If two triangles are similar, their corresponding sides are proportional.So, suppose I pick a point S outside the rectangle. Maybe above the rectangle. Then, I can draw lines from S to two adjacent vertices, say A and B. These lines will intersect the opposite side DC at some points P and Q. Now, if I can find the midpoint of PQ, maybe that will help me find the midpoint of AB.Wait, I'm getting a bit confused. Let me try to draw this out mentally. If I have rectangle ABCD, with AB and CD as the top and bottom sides, and AD and BC as the left and right sides. Point O is the center where diagonals AC and BD intersect. If I pick a point S above the rectangle, and draw lines SA and SB, these lines will intersect DC at points P and Q respectively.Now, since S is above the rectangle, the lines SA and SB will intersect DC at some points P and Q. The idea is that the segment PQ is similar to AB, scaled by some factor. If I can find the midpoint of PQ, then I can use that to find the midpoint of AB.But how exactly? Maybe if I can construct a line from S that bisects PQ, that line will also bisect AB. So, if I can find the midpoint of PQ, and then draw a line from S through that midpoint, it should pass through the midpoint of AB.Wait, that sounds promising. So, step by step, I would:1. Draw diagonals AC and BD to find the center O.2. Choose a point S outside the rectangle, say above it.3. Draw lines SA and SB, which intersect DC at P and Q.4. Find the midpoint of PQ by drawing a line perpendicular to PQ through its midpoint.5. Draw a line from S through this midpoint, which should intersect AB at its midpoint.6. Connect this midpoint to O to get one midline.7. Repeat the process for the other sides to get the other midline.But wait, I only have a ruler, so I can't measure or draw perpendicular lines directly. How can I find the midpoint of PQ without a compass or protractor?Maybe I can use the concept of similar triangles again. If I can create a triangle where PQ is one side and AB is another, and they are similar, then the midpoints would correspond.Alternatively, perhaps I can use the center O to help me. Since O is the midpoint of the diagonals, maybe I can use it to find midpoints on the sides.Let me think differently. If I connect O to a vertex, say A, and then extend that line beyond O to the opposite side, it should meet the midpoint of the opposite side. Wait, is that true?No, actually, in a rectangle, the diagonals don't necessarily extend to the midpoints of the opposite sides unless it's a square. Hmm, so that might not work.Maybe I can use the fact that the midlines are parallel to the sides. So, if I can draw a line parallel to AB that passes through O, that should be the horizontal midline. Similarly, a line parallel to AD passing through O would be the vertical midline.But how do I draw a line parallel to AB using only a ruler? I think I can use the concept of corresponding angles. If I can create a transversal that intersects AB and the line I want to draw, ensuring that the corresponding angles are equal.Wait, but without a protractor, it's tricky. Maybe I can use similar triangles to create a parallel line.Let me try this approach:1. Draw diagonal AC and BD to find center O.2. Choose a point S on side AB.3. Draw a line from S through O, extending it to intersect the opposite side CD at point T.4. Since O is the midpoint of the diagonals, ST should be bisected by O.5. Therefore, T is the midpoint of CD.6. Now, connect T to O, which is the midpoint of CD and the center.7. Similarly, repeat for another side to find midpoints.Wait, that might work. So, by choosing a point S on AB, drawing a line through O to intersect CD at T, T becomes the midpoint of CD because O is the midpoint of the diagonal.Therefore, connecting T to O gives me the vertical midline. Similarly, choosing a point on AD and drawing a line through O would give me the horizontal midline.But I'm not sure if this is entirely accurate. Let me verify.Suppose I have rectangle ABCD, with AB at the top, CD at the bottom, and AD and BC as the sides. Diagonals AC and BD intersect at O. If I choose a point S on AB and draw a line SO, it will intersect CD at some point T. Since O is the midpoint of AC and BD, it should also be the midpoint of ST.Therefore, if S is any point on AB, then T must be the corresponding point on CD such that ST is bisected by O. Hence, if S is the midpoint of AB, then T would be the midpoint of CD. But if S is not the midpoint, then T would not be the midpoint either.Wait, so if I choose S as the midpoint of AB, then T would be the midpoint of CD. But how do I find the midpoint of AB without already knowing it?This seems like a circular problem. I need to find the midpoint of AB to find the midpoint of CD, but I can't find the midpoint of AB without already knowing it.Maybe I need a different approach. Let's think about using the properties of the rectangle and the center O.If I can find two points on one side that are equidistant from O, then their midpoint would be the midpoint of that side. But again, without measurements, it's challenging.Alternatively, perhaps I can use the fact that the midlines are the perpendicular bisectors of the sides. But I'm not sure how to construct perpendicular bisectors with just a ruler.Wait, maybe I can use the intersection of the diagonals to help me. Since O is the center, any line through O that is parallel to a side will be a midline.So, if I can draw a line through O that's parallel to AB, that would be the horizontal midline. Similarly, a line through O parallel to AD would be the vertical midline.But how do I draw a line parallel to AB through O with just a ruler?I think I can use the concept of similar triangles or corresponding angles. If I can create a transversal that intersects AB and the line through O, ensuring that the corresponding angles are equal, then the lines will be parallel.Let me try this:1. Draw diagonal AC and BD to find center O.2. Choose a point S on AB.3. Draw a line from S through O, extending it to intersect CD at T.4. Now, draw a line from T parallel to AB. Since AB is a side of the rectangle, this line should be horizontal.5. But wait, I can't draw parallel lines directly. Maybe I can use the fact that AB and CD are already parallel.6. If I can create a triangle where one side is AB and another side is the line through O, ensuring that the angles are equal, then the lines will be parallel.This is getting a bit too abstract. Maybe there's a simpler way.What if I use the fact that the midlines divide the rectangle into smaller rectangles? So, if I can find two points that divide the sides into equal parts, connecting them would give me the midlines.But again, without measurements, it's hard to ensure equal division.Wait, going back to the diagonals. Since O is the midpoint of both diagonals, maybe I can use it to find midpoints on the sides.If I draw a line from O to a vertex, say A, and then extend it beyond O to the opposite side, it should meet the midpoint of that side.Wait, let me test this. If I have rectangle ABCD, and I draw diagonal AC, which intersects BD at O. If I extend AO beyond O to meet CD at T, then T should be the midpoint of CD because O is the midpoint of AC.Yes, that makes sense. Because AO is half of AC, so extending it the same length beyond O would reach the midpoint of CD.Similarly, if I extend BO beyond O to meet AD at S, then S would be the midpoint of AD.Therefore, by extending the diagonals beyond the center O, I can find the midpoints of the opposite sides.So, step by step:1. Draw diagonals AC and BD, intersecting at O.2. Extend AO beyond O to meet CD at T. T is the midpoint of CD.3. Extend BO beyond O to meet AD at S. S is the midpoint of AD.4. Now, connect T and S. This line TS should be the vertical midline.5. Similarly, extend CO beyond O to meet AB at U, and extend DO beyond O to meet BC at V.6. Connect U and V to get the horizontal midline.Wait, but I think I might have made a mistake here. Extending AO beyond O should meet CD at T, which is the midpoint. Similarly, extending BO beyond O meets AD at S, the midpoint. So, connecting T and S gives me the vertical midline.But actually, connecting T and S might not necessarily be vertical. It depends on the rectangle's orientation.Wait, no, in a rectangle, the midlines are perpendicular to each other, one horizontal and one vertical. So, connecting midpoints of opposite sides should give me the midlines.But I'm getting confused again. Maybe I should summarize:- To find the midpoints of the sides, extend the diagonals beyond the center O to meet the opposite sides. These intersection points are the midpoints.- Once I have the midpoints, connect them to form the midlines.Yes, that seems logical. So, the process is:1. Draw both diagonals of the rectangle, which intersect at the center O.2. Extend each diagonal beyond O to meet the opposite side. The points where the extensions meet the opposite sides are the midpoints of those sides.3. Connect these midpoints to form the midlines.This way, using only a ruler to draw straight lines, I can construct the midlines of the rectangle.I think this makes sense. Let me double-check:- Diagonals intersect at O, the center.- Extending AO beyond O meets CD at T, the midpoint of CD.- Similarly, extending BO beyond O meets AD at S, the midpoint of AD.- Connecting T and S gives the vertical midline.- Extending CO beyond O meets AB at U, the midpoint of AB.- Extending DO beyond O meets BC at V, the midpoint of BC.- Connecting U and V gives the horizontal midline.Yes, this should work. So, the key is to use the center O found by the diagonals and then extend those diagonals to find midpoints on the opposite sides, which can then be connected to form the midlines.I think I've got it!

💡First, I need to simplify the complex fraction (frac{1 - 3i}{1 + i}). To do this, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is (1 - i). This step helps eliminate the imaginary unit from the denominator.Next, I'll perform the multiplication in both the numerator and the denominator. In the numerator, multiplying (1 - 3i) by (1 - i) gives (-2 - 4i). In the denominator, multiplying (1 + i) by (1 - i) results in (2).Finally, I'll divide both the real and imaginary parts of the numerator by the denominator to obtain the simplified form of the complex number.