💡Alright, so I have this trapezoid problem here. The bases AB and CD are 155 and 13 units long, respectively. The lateral sides are mutually perpendicular, which means they form right angles where they meet the bases. I need to find the scalar product of the vectors AC and BD. Hmm, okay, let's break this down step by step.First, I should probably draw a diagram to visualize the trapezoid. Let me imagine trapezoid ABCD with AB as the longer base (155 units) and CD as the shorter base (13 units). The legs AD and BC are the lateral sides and are perpendicular to each other. So, points A and B are on the longer base, and points C and D are on the shorter base. Since the legs are perpendicular, this might form some right triangles within the trapezoid.I remember that in a trapezoid, the scalar product of the diagonals can sometimes be found using vector methods. Maybe I can assign coordinates to the vertices and express the vectors AC and BD in terms of these coordinates. That sounds like a plan.Let's assign coordinates to the trapezoid. I'll place point A at the origin (0, 0). Since AB is 155 units long, point B will be at (155, 0). Now, since the legs AD and BC are perpendicular, and assuming the trapezoid is oriented such that AD is vertical, point D would be somewhere along the y-axis, say at (0, h), where h is the height of the trapezoid. Similarly, point C would be somewhere above point B, but shifted left by the length of CD, which is 13 units. Wait, no, actually, since CD is the shorter base, it should be parallel to AB, so point C should be at (155 - 13, h) = (142, h). Hmm, does that make sense?Wait, if AD is vertical, then BC should be horizontal? But the problem says the lateral sides are mutually perpendicular, meaning AD and BC are perpendicular. So, if AD is vertical, then BC must be horizontal. That makes sense because vertical and horizontal lines are perpendicular. So, point D is at (0, h), and point C is at (142, h). Then, the leg BC goes from (155, 0) to (142, h). Let me verify the length of BC.The length of BC can be found using the distance formula: sqrt[(142 - 155)^2 + (h - 0)^2] = sqrt[(-13)^2 + h^2] = sqrt(169 + h^2). Similarly, the length of AD is just h, since it's vertical from (0,0) to (0,h). Since AD and BC are perpendicular, their dot product should be zero. Wait, but AD is a vertical vector and BC is a vector from (155,0) to (142,h). So, the vector AD is (0, h) and the vector BC is (-13, h). Their dot product is (0)(-13) + (h)(h) = h^2. For them to be perpendicular, their dot product should be zero, so h^2 = 0, which implies h = 0. But that can't be right because then the trapezoid would collapse into a line. Hmm, I must have made a mistake here.Wait, maybe I assigned the coordinates incorrectly. If AD and BC are the lateral sides and are perpendicular, perhaps AD is not vertical. Maybe I need to assign the coordinates differently. Let me try another approach.Let me place point A at (0, 0) and point B at (155, 0). Let me assume that the leg AD is from A(0,0) to D(p, q), and the leg BC is from B(155,0) to C(r, s). Since AD and BC are perpendicular, the vectors AD and BC should have a dot product of zero. The vector AD is (p, q) and the vector BC is (r - 155, s). So, their dot product is p(r - 155) + q(s) = 0.Also, since CD is the shorter base of length 13, the distance between points C and D should be 13. So, the distance between D(p, q) and C(r, s) is sqrt[(r - p)^2 + (s - q)^2] = 13. Additionally, since AB and CD are parallel, the slope of AB is zero (since it's along the x-axis), so the slope of CD should also be zero. Therefore, the y-coordinates of points C and D must be the same. So, s = q. That simplifies things a bit.So, points D and C have the same y-coordinate, say h. So, D is at (p, h) and C is at (r, h). The distance between D and C is |r - p| = 13. Since CD is shorter than AB, r - p = 13 or p - r = 13, depending on the orientation. Let's assume r > p, so r = p + 13.Now, the vector AD is (p, h) and the vector BC is (r - 155, h). Their dot product is p(r - 155) + h * h = 0. Substituting r = p + 13, we get p(p + 13 - 155) + h^2 = 0. Simplifying, p(p - 142) + h^2 = 0. So, p^2 - 142p + h^2 = 0.Also, since AB is 155 units and CD is 13 units, the projection of the legs onto the x-axis should account for the difference in the lengths of the bases. The total horizontal difference between the two bases is 155 - 13 = 142 units. This horizontal difference is distributed between the two legs AD and BC. Since AD is from (0,0) to (p, h), its horizontal component is p, and BC is from (155,0) to (r, h), so its horizontal component is r - 155. The sum of these horizontal components should be 142. So, p + (r - 155) = 142. Since r = p + 13, substituting, we get p + (p + 13 - 155) = 142. Simplifying, 2p - 142 = 142, so 2p = 284, which gives p = 142.Wait, if p = 142, then r = p + 13 = 155. So, point D is at (142, h) and point C is at (155, h). But then, the vector BC is from (155,0) to (155, h), which is (0, h). The vector AD is from (0,0) to (142, h), which is (142, h). Their dot product is 142*0 + h*h = h^2. For them to be perpendicular, h^2 must be zero, which again implies h = 0. That can't be right because then the trapezoid would collapse. Hmm, I must be making a mistake in my assumptions.Wait, maybe the horizontal difference isn't 142. Let me think again. The bases are AB = 155 and CD = 13. The difference in their lengths is 155 - 13 = 142. This difference is accounted for by the horizontal projections of the legs AD and BC. So, if AD has a horizontal component of x, then BC must have a horizontal component of 142 - x. But since the legs are perpendicular, the sum of their horizontal components squared plus the sum of their vertical components squared should equal the squares of their lengths. Wait, no, that's not quite right.Actually, since the legs are perpendicular, the vectors AD and BC should satisfy the condition that their dot product is zero. So, if vector AD is (x, y) and vector BC is (u, v), then x*u + y*v = 0. Also, the lengths of AD and BC can be found using the Pythagorean theorem: sqrt(x^2 + y^2) and sqrt(u^2 + v^2), respectively. But I don't know the lengths of the legs, only the lengths of the bases.Wait, maybe I can express the coordinates differently. Let me try placing the trapezoid such that the legs AD and BC are perpendicular. Let me assume that AD is along the y-axis, so point D is at (0, h). Then, since AB is 155 units, point B is at (155, 0). Now, point C must be somewhere such that CD is 13 units and BC is perpendicular to AD. Since AD is vertical, BC must be horizontal. So, point C must be at (155 - 13, h) = (142, h). Therefore, vector BC is from (155,0) to (142,h), which is (-13, h). Vector AD is from (0,0) to (0,h), which is (0,h). The dot product of AD and BC is (0)(-13) + (h)(h) = h^2. For them to be perpendicular, h^2 must be zero, which again gives h = 0. That's not possible.Hmm, this seems like a recurring issue. Maybe my initial assumption of placing AD along the y-axis is causing the problem. Perhaps AD isn't vertical. Let me try a different coordinate system.Let me place point A at (0,0) and point B at (a,0). Then, since the legs AD and BC are perpendicular, I can let AD be a vector (p, q) and BC be a vector (r, s) such that p*r + q*s = 0. The coordinates of D would then be (p, q) and the coordinates of C would be (a + r, s). Since CD is parallel to AB, the y-coordinate of C and D must be the same, so s = q. Therefore, vector BC is (r, q). The dot product of AD and BC is p*r + q*q = 0. Also, the length of CD is 13, so the distance between D(p, q) and C(a + r, q) is |a + r - p| = 13. The length of AB is a = 155.So, we have:1. p*r + q^2 = 02. |155 + r - p| = 13From equation 2, 155 + r - p = ±13. Let's consider both cases.Case 1: 155 + r - p = 13 => r = p - 142Substitute into equation 1: p*(p - 142) + q^2 = 0 => p^2 - 142p + q^2 = 0Case 2: 155 + r - p = -13 => r = p - 168Substitute into equation 1: p*(p - 168) + q^2 = 0 => p^2 - 168p + q^2 = 0Now, we need another equation to solve for p and q. Since the trapezoid has sides AD and BC, which are vectors (p, q) and (r, q), respectively. The lengths of these vectors are sqrt(p^2 + q^2) and sqrt(r^2 + q^2). However, we don't know the lengths of the legs, so we might need another approach.Wait, perhaps we can express the scalar product of AC and BD in terms of p and q. Let's find vectors AC and BD.Vector AC is from A(0,0) to C(a + r, q) = (a + r, q). Since a = 155, it's (155 + r, q).Vector BD is from B(a,0) to D(p, q) = (p - a, q). Since a = 155, it's (p - 155, q).The scalar product of AC and BD is:(155 + r)*(p - 155) + q*qWe need to express this in terms of p and q. From case 1, r = p - 142, so substitute:(155 + p - 142)*(p - 155) + q^2 = (13 + p)*(p - 155) + q^2Expanding:13p - 13*155 + p^2 - 155p + q^2 = p^2 - 142p - 2015 + q^2From equation 1 in case 1: p^2 - 142p + q^2 = 0 => p^2 - 142p + q^2 = 0So, substituting back:0 - 2015 = -2015So, the scalar product is -2015.Wait, that seems too straightforward. Let me check case 2.Case 2: r = p - 168Substitute into equation 1: p*(p - 168) + q^2 = 0 => p^2 - 168p + q^2 = 0Vector AC is (155 + r, q) = (155 + p - 168, q) = (p - 13, q)Vector BD is (p - 155, q)Scalar product:(p - 13)*(p - 155) + q^2 = p^2 - 155p - 13p + 13*155 + q^2 = p^2 - 168p + 2015 + q^2From equation 1 in case 2: p^2 - 168p + q^2 = 0So, substituting back:0 + 2015 = 2015But the scalar product can't be both -2015 and 2015 depending on the case. Hmm, that's confusing. Maybe I made a mistake in the sign when considering the absolute value in equation 2.Wait, in equation 2, |155 + r - p| = 13, so 155 + r - p = ±13. In case 1, I took it as +13, leading to r = p - 142, and in case 2, as -13, leading to r = p - 168. However, when I calculated the scalar product in case 1, I got -2015, and in case 2, I got +2015. But the problem doesn't specify the orientation, so both cases might be possible. However, the scalar product's sign depends on the angle between the vectors. Since the problem doesn't specify the orientation, maybe the magnitude is 2015, but the sign could be negative or positive. However, in the first case, we got -2015, and in the second case, +2015.Wait, but in the first case, when r = p - 142, we had p^2 - 142p + q^2 = 0, and the scalar product became -2015. In the second case, p^2 - 168p + q^2 = 0, and the scalar product became +2015. But the problem states that the lateral sides are mutually perpendicular, so both cases are possible depending on the direction of the vectors. However, in the first case, when r = p - 142, the scalar product is negative, and in the second case, it's positive.But the problem doesn't specify the orientation, so perhaps the answer is -2015 or 2015. However, in the first case, when I considered the horizontal difference as 13, leading to r = p - 142, the scalar product was -2015. In the second case, with the horizontal difference as -13, leading to r = p - 168, the scalar product was +2015. But the problem states that the bases are AB = 155 and CD = 13, so the horizontal difference should be 155 - 13 = 142, which corresponds to case 1. Therefore, the scalar product should be -2015.Wait, but in case 1, when I substituted r = p - 142, the scalar product became -2015, which is consistent with the horizontal difference being 142. So, I think the correct answer is -2015.Let me double-check my calculations. In case 1:Vector AC = (155 + r, q) = (155 + p - 142, q) = (p + 13, q)Vector BD = (p - 155, q)Scalar product = (p + 13)(p - 155) + q^2 = p^2 - 155p + 13p - 2015 + q^2 = p^2 - 142p - 2015 + q^2From equation 1: p^2 - 142p + q^2 = 0 => p^2 - 142p + q^2 = 0So, substituting back:0 - 2015 = -2015Yes, that seems correct. Therefore, the scalar product is -2015.

💡First, convert Bob's hourly wage from dollars to cents. Since 1 equals 100 cents, Bob earns:25 dollars/hour × 100 cents/dollar = 2500 cents/hourNext, calculate 2.5% of 2500 cents to find the local tax deduction:2.5% of 2500 cents = 0.025 × 2500 = 62.5 centsTherefore, Bob pays 62.5 cents per hour in local taxes.

💡Okay, so I have this question about two different sampling methods used in a school survey. Let me try to understand what each method is and then figure out which option matches them.First, the problem says that a school surveyed 200 senior high school students in the humanities about the time they needed to complete their homework. They used two different sampling methods. The first method was the student council randomly selecting 20 students. The second method was the academic affairs office numbering the students from 001 to 200 and selecting those whose last digit of their student number is 2.I need to identify what type of sampling each method is and then choose the correct option from A to D.Let me recall the different sampling methods:1. **Simple Random Sampling (SRS):** Every individual in the population has an equal chance of being selected. It's like drawing names out of a hat or using a random number generator.2. **Stratified Sampling:** The population is divided into subgroups or strata, and then random samples are taken from each stratum. This ensures representation from each subgroup.3. **Systematic Sampling:** The population is ordered in some way, and then every kth individual is selected. For example, selecting every 10th person from a list.4. **Cluster Sampling:** The population is divided into clusters, and then entire clusters are randomly selected.Now, looking back at the problem:- **First Method:** The student council randomly selects 20 students. This sounds like SRS because it's a straightforward random selection without any stratification or systematic approach mentioned.- **Second Method:** The academic affairs office numbers the students from 001 to 200 and selects those whose last digit is 2. This seems like a systematic approach because they're selecting every 10th student (since the last digit being 2 implies selecting every 10th number if we consider 002, 012, 022, etc.). However, I need to be careful here. If they're selecting based on the last digit, it might also be considered as stratified if they're ensuring representation from different parts of the list. But in this case, since it's a fixed interval (every 10th), it's more systematic.Wait, but in systematic sampling, you usually have a starting point and then select every kth individual. Here, they're selecting all individuals where the last digit is 2, which is equivalent to selecting every 10th individual starting from 002. So yes, that's systematic sampling.So, the first method is Simple Random Sampling, and the second is Systematic Sampling. Looking at the options:A: Stratified, Simple Random B: Simple Random, Stratified C: Stratified, Systematic D: Simple Random, SystematicTherefore, the correct answer should be D.

💡First, I'll define the variables:- Let ( B ) represent the number of peaches Benjy has.- Let ( G ) represent the number of peaches Gabrielle has.From the problem, Martine has 6 more than twice as many peaches as Benjy:[text{Martine's peaches} = 2B + 6]Given that Martine has 16 peaches:[16 = 2B + 6]Solving for ( B ):[16 - 6 = 2B 10 = 2B B = frac{10}{2} B = 5]So, Benjy has 5 peaches.Next, Benjy has one-third as many peaches as Gabrielle:[B = frac{1}{3}G]Substituting ( B = 5 ):[5 = frac{1}{3}G G = 5 times 3 G = 15]Therefore, Gabrielle has 15 peaches.

💡Okay, so I need to find the equation of the tangent line to the curve y = f(x) at the point (1, f(1)). The function f(x) is given as an odd function, and when x is less than 0, f(x) is equal to ln(-x) + 3x. Hmm, let me think about how to approach this step by step.First, I remember that an odd function satisfies the property f(-x) = -f(x) for all x in its domain. That means if I know the value of f(x) for negative x, I can find it for positive x by using this property. Since the problem gives me f(x) for x < 0, I can use the odd function property to find f(x) when x > 0.So, for x > 0, let's consider -x, which would be negative. Therefore, f(-x) = ln(-(-x)) + 3(-x) because when x is positive, -x is negative. Simplifying that, f(-x) = ln(x) - 3x. But since f is an odd function, f(-x) = -f(x). So, I can write:f(-x) = ln(x) - 3x = -f(x)Therefore, solving for f(x), I get:f(x) = -ln(x) + 3xOkay, so now I have the expression for f(x) when x > 0. That means at x = 1, which is positive, f(1) should be:f(1) = -ln(1) + 3*1I know that ln(1) is 0 because e^0 = 1. So, f(1) simplifies to:f(1) = 0 + 3 = 3Alright, so the point where I need to find the tangent line is (1, 3). Now, to find the equation of the tangent line, I need two things: the slope of the tangent at that point and a point through which the tangent passes. I already have the point, which is (1, 3). Now, I need to find the derivative f'(x) to get the slope at x = 1.Let's compute the derivative f'(x). Since f(x) for x > 0 is -ln(x) + 3x, I can differentiate term by term.The derivative of -ln(x) with respect to x is -1/x. The derivative of 3x with respect to x is 3. So, putting it together:f'(x) = -1/x + 3Now, I need to evaluate this derivative at x = 1 to get the slope of the tangent line at that point.f'(1) = -1/1 + 3 = -1 + 3 = 2So, the slope of the tangent line at (1, 3) is 2.Now that I have the slope and a point on the line, I can use the point-slope form of the equation of a line, which is:y - y1 = m(x - x1)Where (x1, y1) is the point and m is the slope. Plugging in the values I have:y - 3 = 2(x - 1)Let me simplify this equation to get it into slope-intercept form (y = mx + b).First, distribute the 2 on the right side:y - 3 = 2x - 2Now, add 3 to both sides to solve for y:y = 2x - 2 + 3y = 2x + 1So, the equation of the tangent line is y = 2x + 1.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from f(x) being odd, I correctly used f(-x) = -f(x) to find f(x) for positive x. Then, I calculated f(1) correctly, knowing that ln(1) is 0. The derivative was computed correctly as well, and evaluating it at x = 1 gave me a slope of 2. Then, using point-slope form, I arrived at y = 2x + 1. Everything seems to check out.I think I'm confident with this answer. The tangent line at (1, 3) has a slope of 2, so the equation is y = 2x + 1.**Final Answer**The equation of the tangent line is boxed{y = 2x + 1}.

💡Okay, so I have this problem about skew lines. Let me try to understand what it's asking. We have two skew lines, right? Skew lines are lines that don't lie in the same plane and don't intersect. So, they're kind of like those lines that go off in different directions in 3D space.On one of these skew lines, there are two different points, A and A₁. On the other skew line, there are two different points, B and B₁. The question is asking if the lines AB and A₁B₁ are also skew lines.Hmm, okay. So, AB is a line connecting point A on the first skew line to point B on the second skew line. Similarly, A₁B₁ connects A₁ on the first line to B₁ on the second line. The question is whether these two new lines, AB and A₁B₁, are also skew.First, let me recall what makes two lines skew. They must not be parallel and must not intersect. Also, they shouldn't lie in the same plane. So, if AB and A₁B₁ are not parallel, don't intersect, and aren't in the same plane, then they are skew.But wait, the original lines are skew. So, the lines AA₁ and BB₁ are skew. That means they don't lie in the same plane, they don't intersect, and they aren't parallel.Now, if I connect A to B and A₁ to B₁, what does that do? Let me visualize this. Imagine two skew lines, one with points A and A₁, and the other with points B and B₁. If I connect A to B, that's one line, and A₁ to B₁, that's another line.I wonder if these two new lines could intersect. If they did, that would mean that there's a point where AB and A₁B₁ cross each other. But since the original lines are skew, they don't intersect, so maybe the new lines don't either.Also, are AB and A₁B₁ parallel? If they were, that would mean they never meet and are always the same distance apart. But since the original lines are skew, which aren't parallel, I don't think AB and A₁B₁ would necessarily be parallel.Another thought: could AB and A₁B₁ lie in the same plane? If they did, then they wouldn't be skew. But since the original lines are skew, which means they aren't in the same plane, maybe the new lines aren't either.Wait, but connecting points from two skew lines might create a plane. Let me think. If I have two points on each of the skew lines, could those four points lie in the same plane? If so, then the lines AB and A₁B₁ would lie in that plane, making them not skew.But the original lines are skew, so they don't lie in the same plane. Does that mean the new lines can't lie in the same plane? I'm not sure.Maybe I should consider the properties of skew lines more carefully. Skew lines are non-parallel and non-intersecting, and they don't lie in the same plane. So, if AB and A₁B₁ are to be skew, they must also satisfy these conditions.Let me try to see if AB and A₁B₁ can intersect. Suppose they do intersect at some point. That would mean there's a point where AB and A₁B₁ cross each other. But since A and A₁ are on one skew line, and B and B₁ are on the other, the only way AB and A₁B₁ could intersect is if the original lines intersect, which they don't because they're skew.So, AB and A₁B₁ can't intersect. That's one condition for being skew.Next, are they parallel? If they were, then they would never meet and maintain a constant distance. But since the original lines are skew, which aren't parallel, I don't think AB and A₁B₁ would be parallel either.Finally, do they lie in the same plane? If they did, then they wouldn't be skew. But since the original lines are skew, which aren't in the same plane, it's unlikely that AB and A₁B₁ lie in the same plane.Wait, but connecting points from two skew lines might create a plane. Let me think. If I have two points on each of the skew lines, could those four points lie in the same plane? If so, then the lines AB and A₁B₁ would lie in that plane, making them not skew.But the original lines are skew, so they don't lie in the same plane. Does that mean the new lines can't lie in the same plane? I'm not sure.Maybe I should consider an example. Let's say we have two skew lines in 3D space. Let me assign coordinates to make it concrete. Let’s say the first skew line is along the x-axis, so points on it can be represented as (t, 0, 0) for some parameter t. The second skew line could be along the line (0, s, 1) for some parameter s. These two lines are skew because they don't intersect and aren't parallel.Now, let's take points A and A₁ on the first line. Let’s say A is at (0, 0, 0) and A₁ is at (1, 0, 0). On the second line, let's take B at (0, 0, 1) and B₁ at (0, 1, 1).Now, the line AB connects (0,0,0) to (0,0,1). That's just the line along the z-axis from (0,0,0) to (0,0,1). The line A₁B₁ connects (1,0,0) to (0,1,1). Let me find the parametric equations for these lines.For AB: from (0,0,0) to (0,0,1), so parametric equations are x=0, y=0, z=t where t goes from 0 to 1.For A₁B₁: from (1,0,0) to (0,1,1). The direction vector is (-1,1,1). So parametric equations are x=1 - t, y=0 + t, z=0 + t, where t goes from 0 to 1.Now, are these two lines skew? Let's check if they intersect. Suppose there's a point where AB and A₁B₁ meet. For AB, x=0, y=0, z=t. For A₁B₁, x=1 - t, y=t, z=t.Set them equal: 0 = 1 - t, 0 = t, t = t.From x: 0 = 1 - t => t=1.From y: 0 = t => t=0.But t can't be both 1 and 0 at the same time, so they don't intersect.Are they parallel? The direction vector of AB is (0,0,1), and for A₁B₁ it's (-1,1,1). These are not scalar multiples of each other, so they aren't parallel.Do they lie in the same plane? To check this, we can see if the volume of the tetrahedron formed by the four points is zero. The volume is given by the scalar triple product of vectors AB, AA₁, and AB₁.Wait, maybe it's easier to see if the four points are coplanar. The four points are A(0,0,0), B(0,0,1), A₁(1,0,0), and B₁(0,1,1).To check coplanarity, we can use the determinant of a matrix formed by these points. The formula is:| (B - A) ⋅ [(A₁ - A) × (B₁ - A)] | = 0 if coplanar.Let's compute vectors:B - A = (0,0,1) - (0,0,0) = (0,0,1)A₁ - A = (1,0,0) - (0,0,0) = (1,0,0)B₁ - A = (0,1,1) - (0,0,0) = (0,1,1)Now, compute the cross product (A₁ - A) × (B₁ - A):(1,0,0) × (0,1,1) = determinant of the matrix:i j k1 0 00 1 1= i*(0*1 - 0*1) - j*(1*1 - 0*0) + k*(1*1 - 0*0)= i*(0) - j*(1) + k*(1)= (0, -1, 1)Now, take the dot product of (B - A) with this cross product:(0,0,1) ⋅ (0, -1, 1) = 0*0 + 0*(-1) + 1*1 = 1Since the result is not zero, the four points are not coplanar. Therefore, the lines AB and A₁B₁ do not lie in the same plane.Since AB and A₁B₁ do not intersect, are not parallel, and do not lie in the same plane, they must be skew lines.So, in this example, AB and A₁B₁ are skew. Does this hold in general?I think yes, because if the original lines are skew, then any lines connecting points on them will also be skew. Because if they weren't, it would imply that the original lines lie in the same plane, which contradicts them being skew.Therefore, the answer should be that yes, AB and A₁B₁ are skew lines.