💡First, I note that Catriona has 8 goldfish in her aquarium.Next, she has 4 more angelfish than half the number of goldfish. Half of 8 goldfish is 4, so she has 4 + 4 = 8 angelfish.Then, she has twice as many guppies as the difference between the number of angelfish and goldfish. The difference between 8 angelfish and 8 goldfish is 0, so she has 2 * 0 = 0 guppies.She also has 3 fewer tetras than the square root of the number of goldfish. The square root of 8 is approximately 2.83, which I'll round down to 2. Therefore, she has 2 - 3 = -1 tetras. Since having a negative number of fish isn't possible, I'll assume she has 0 tetras.Lastly, she has 5 more bettas than the number of tetras squared. With 0 tetras, the number of bettas is 0^2 + 5 = 5.Finally, I'll add up all the fish: 8 goldfish + 8 angelfish + 0 guppies + 0 tetras + 5 bettas = 21 fish.

💡Alright, so I've got this problem about a classroom shaped like a regular hexagon with each side being 3 meters long. There's a snore-o-meter in each corner, and each of these devices can detect sleeping students within a 3-meter radius. The sum of all the readings from these devices is 7. I need to figure out how many sleeping students there are in total in the classroom.First, I need to visualize the classroom. A regular hexagon has six equal sides and six equal angles. Each side is 3 meters, so the distance from one corner to the opposite corner (the diameter of the hexagon) should be twice the side length, which is 6 meters. But the snore-o-meters only detect up to 3 meters, which is exactly the side length of the hexagon.Now, each snore-o-meter is at a corner, and it can detect students within a 3-meter radius. Since the side length is 3 meters, this means that each snore-o-meter can detect students in the adjacent corners as well. Wait, no, actually, the radius is 3 meters, which is the same as the side length, so each snore-o-meter can detect students in the adjacent corners and maybe some in the middle.But I need to think about how the detection works. If a student is near one corner, they might be detected by multiple snore-o-meters. For example, a student sitting near an edge might be detected by two snore-o-meters, one at each end of that edge. A student sitting right in the center of the hexagon would be equidistant from all six corners, so they would be detected by all six snore-o-meters.But in this case, the sum of all the snore-o-meter readings is 7. That means if I add up all the counts from each snore-o-meter, it equals 7. However, this doesn't necessarily mean there are 7 students because some students might be counted multiple times by different snore-o-meters.So, I need to figure out how many students there are such that when each snore-o-meter counts the students within its 3-meter radius, the total sum is 7.Let me think about the possible configurations. If there's a student in the center, they would be counted by all six snore-o-meters, contributing 6 to the total sum. If there are students near the edges, each such student would be counted by two snore-o-meters, contributing 2 to the total sum. If there are students near the corners but not in the center, they might be counted by three snore-o-meters, contributing 3 to the total sum.So, I need to find a combination of students in these positions such that the total sum is 7.Let me denote:- C as the number of students in the center (each contributing 6 to the sum)- E as the number of students near the edges (each contributing 2 to the sum)- V as the number of students near the corners (each contributing 3 to the sum)Then, the total sum S is:S = 6C + 2E + 3V = 7I need to find non-negative integers C, E, V that satisfy this equation.Let's consider possible values for C:1. If C = 1: Then, 6(1) + 2E + 3V = 7 So, 2E + 3V = 1 But 2E + 3V = 1 is not possible with non-negative integers E and V because the smallest value 2E + 3V can take is 0 (if E=0 and V=0), and the next is 2 or 3, which are both greater than 1. So, C cannot be 1.2. If C = 0: Then, 6(0) + 2E + 3V = 7 So, 2E + 3V = 7 Now, I need to find non-negative integers E and V such that 2E + 3V = 7.Let's try different values for V:- V = 0: 2E = 7 E = 3.5, which is not an integer. So, V cannot be 0.- V = 1: 2E + 3(1) = 7 2E = 4 E = 2 So, E = 2 and V = 1 is a solution.- V = 2: 2E + 3(2) = 7 2E = 1 E = 0.5, which is not an integer. So, V cannot be 2.- V = 3: 2E + 3(3) = 7 2E = -2, which is not possible because E cannot be negative.So, the only solution when C = 0 is E = 2 and V = 1.Therefore, the total number of students is C + E + V = 0 + 2 + 1 = 3.Wait, but let me double-check. If there are 2 students near the edges and 1 student near a corner, then:- Each edge student is counted by 2 snore-o-meters, so 2 students contribute 2*2 = 4.- The corner student is counted by 3 snore-o-meters, contributing 3.- Total sum = 4 + 3 = 7, which matches the given total.Alternatively, if I consider the possibility of having a student in the center, but as we saw earlier, having C = 1 would require 2E + 3V = 1, which is impossible. So, there must be no students in the center.Another way to think about it is that if there were a student in the center, the total sum would be at least 6, and adding any other students would make the total sum exceed 7. But since the total sum is exactly 7, it's not possible to have a student in the center.Therefore, the only possible configuration is 2 students near the edges and 1 student near a corner, totaling 3 students.I think that makes sense. Each edge student is detected by two devices, and the corner student is detected by three devices, adding up to 7. There's no overlap beyond that because if there were more students, the total sum would be higher.So, the total number of sleeping students in the classroom is 3.

💡First, calculate the total man-hours required to asphalt the 1 km road by the 30 men.Total man-hours for 1 km road = Number of men × Number of days × Number of hours per dayTotal man-hours for 1 km road = 30 men × 12 days × 8 hours/dayTotal man-hours for 1 km road = 30 × 12 × 8Total man-hours for 1 km road = 2880 man-hoursSince the second road is 2 km, it would require twice the man-hours to complete, assuming the work rate is the same.Total man-hours for 2 km road = Total man-hours for 1 km road × 2Total man-hours for 2 km road = 2880 man-hours × 2Total man-hours for 2 km road = 5760 man-hoursNow, we need to find out how many men are required to complete the 2 km road in 20.571428571428573 days, working 14 hours a day.Let's denote the number of men required as "x".Total man-hours for 2 km road = Number of men × Number of days × Number of hours per day5760 man-hours = x men × 20.571428571428573 days × 14 hours/dayNow we can solve for x:x = 5760 man-hours / (20.571428571428573 days × 14 hours/day)x = 5760 / (20.571428571428573 × 14)x = 5760 / 288Now, let's calculate the value of x:x = 5760 / 288x = 20Therefore, 20 men are working on the second road.

💡Alright, so I've got this geometry problem here. Let me try to visualize it first. There's an angle with vertex A, and a circle that's tangent to one side of this angle at point A. The circle also intersects the other side of the angle at some point B. The measure of the angle at A is 40 degrees. Now, there's a point M on the minor arc AB, and I need to find the measure of angle AMB.Hmm, okay. Let me draw a rough sketch in my mind. There's an angle, let's say angle CAB, which is 40 degrees. The circle is tangent to one side, say side AC, at point A. Then, the circle intersects the other side, AB, at point B. So, the circle touches AC at A and cuts AB at B. Point M is somewhere on the minor arc AB, which is the shorter arc connecting A and B on the circle.I remember that when dealing with circles and angles, especially tangents and chords, there are some theorems that might help here. One of them is the Alternate Segment Theorem, which relates the angle between a tangent and a chord to the angle in the alternate segment. Maybe that could be useful here.Let me recall the Alternate Segment Theorem. It states that the angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment. So, in this case, the angle between the tangent at A and the chord AB should be equal to the angle that AB makes with the other side of the circle.Wait, so if I consider the tangent at A, which is side AC, and the chord AB, then the angle between AC and AB (which is given as 40 degrees) should be equal to the angle that AB makes with the other side of the circle. That other angle would be the angle in the alternate segment, which is angle AMB.But hold on, is angle AMB the angle in the alternate segment? Let me think. The Alternate Segment Theorem says that the angle between the tangent and the chord is equal to the angle in the alternate segment. So, in this case, the angle between tangent AC and chord AB is 40 degrees, which should be equal to the angle that AB makes in the alternate segment. That angle would be angle AMB.Wait, no. Actually, angle AMB is subtended by arc AB, but since M is on the minor arc AB, angle AMB is actually the angle subtended by arc AB in the alternate segment. So, according to the theorem, angle between tangent AC and chord AB is equal to angle AMB.Therefore, angle AMB should be equal to 40 degrees. But that seems too straightforward, and I might be missing something here.Wait, let me double-check. The Alternate Segment Theorem says that the angle between the tangent and the chord is equal to the angle in the alternate segment. So, if I have a tangent at A, and a chord AB, then the angle between them (which is 40 degrees) is equal to the angle that AB makes in the alternate segment. That angle in the alternate segment is angle AMB.So, according to that, angle AMB should be 40 degrees. But I'm not entirely sure because sometimes these theorems can be a bit tricky.Alternatively, maybe I should consider the properties of angles subtended by the same arc. Since M is on the minor arc AB, angle AMB is an inscribed angle subtended by arc AB. The measure of an inscribed angle is half the measure of its subtended arc.But wait, what's the measure of arc AB? Since the circle is tangent to side AC at A, and intersects side AB at B, the arc AB would correspond to the angle at the center. But I don't know the measure of that central angle yet.Alternatively, maybe I can find the central angle corresponding to arc AB. If I can find that, then I can find the measure of angle AMB, which is half of that central angle.But how do I find the central angle? Maybe by using the fact that the circle is tangent to side AC at A. The radius at the point of tangency is perpendicular to the tangent. So, if I draw the radius from the center of the circle to point A, it should be perpendicular to side AC.Let me denote the center of the circle as O. So, OA is perpendicular to AC. Since angle CAB is 40 degrees, and OA is perpendicular to AC, then angle OAB would be 90 degrees minus 40 degrees, which is 50 degrees.Wait, is that correct? Let me see. If OA is perpendicular to AC, then angle OAC is 90 degrees. Since angle CAB is 40 degrees, then angle OAB would be 90 degrees minus 40 degrees, which is indeed 50 degrees.So, in triangle OAB, we have OA and OB as radii of the circle, so they are equal in length. Therefore, triangle OAB is isosceles with OA = OB. The base angles of this triangle, angles OAB and OBA, are equal. We already found that angle OAB is 50 degrees, so angle OBA is also 50 degrees.Therefore, the vertex angle at O, which is angle AOB, can be found by subtracting the sum of the base angles from 180 degrees. So, angle AOB = 180 - 50 - 50 = 80 degrees.So, the central angle subtended by arc AB is 80 degrees. Therefore, the inscribed angle subtended by the same arc AB would be half of that, which is 40 degrees. But wait, angle AMB is an inscribed angle subtended by arc AB, so angle AMB should be 40 degrees.But earlier, I thought angle AMB might be 40 degrees based on the Alternate Segment Theorem, and now I'm getting the same result using the inscribed angle theorem. So, that seems consistent.However, I'm a bit confused because sometimes when dealing with circles and tangents, there can be supplementary angles involved. Let me make sure I'm not missing something here.Wait, another way to think about this is to consider the tangent at A and the chord AB. The angle between them is 40 degrees, and according to the Alternate Segment Theorem, this angle is equal to the angle that AB makes in the alternate segment, which is angle AMB. So, that also gives angle AMB as 40 degrees.But hold on, sometimes the angle in the alternate segment can be supplementary to the angle between the tangent and the chord, depending on the position of the point. Is that the case here?No, I think in this scenario, since M is on the minor arc AB, the angle AMB is indeed equal to the angle between the tangent and the chord, which is 40 degrees.Wait, but I also calculated the central angle as 80 degrees, and the inscribed angle as 40 degrees, which matches. So, both methods give me the same result.But I'm still a bit unsure because sometimes these problems can have a twist. Let me try to visualize it again. The circle is tangent to AC at A, so OA is perpendicular to AC. The circle intersects AB at B, and M is on the minor arc AB. So, when I look at angle AMB, it's formed by points A, M, and B on the circumference.Since M is on the minor arc AB, angle AMB is an inscribed angle subtended by arc AB. The measure of this angle is half the measure of arc AB. We found that arc AB corresponds to a central angle of 80 degrees, so the inscribed angle should be 40 degrees.Alternatively, using the Alternate Segment Theorem, the angle between the tangent AC and the chord AB is equal to the angle in the alternate segment, which is angle AMB. Since the angle between AC and AB is 40 degrees, angle AMB is also 40 degrees.Therefore, both methods confirm that angle AMB is 40 degrees.Wait, but I just realized something. The problem says that the circle is tangent to one side of the angle at its vertex A and intersects the other side at point B. So, the angle at A is 40 degrees, which is angle CAB.But in my earlier reasoning, I considered angle OAB as 50 degrees, which is correct because OA is perpendicular to AC, making angle OAC 90 degrees, and angle CAB is 40 degrees, so angle OAB is 50 degrees.Then, in triangle OAB, which is isosceles with OA = OB, the base angles are 50 degrees each, making the central angle AOB equal to 80 degrees. Therefore, the inscribed angle AMB is half of 80 degrees, which is 40 degrees.So, everything seems to check out. Both the Alternate Segment Theorem and the Inscribed Angle Theorem lead me to the conclusion that angle AMB is 40 degrees.But wait, I just thought of another way to verify this. If I consider the power of point A with respect to the circle, but I'm not sure if that's necessary here.Alternatively, maybe I can use the fact that the angle between a tangent and a chord is equal to the angle in the alternate segment, which we've already done.I think I'm confident now that angle AMB is 40 degrees. However, I'm a bit concerned because sometimes these problems can have a different answer due to some property I might have overlooked.Wait, let me think about the position of point M. It's on the minor arc AB, which means it's on the arc that's closer to the vertex A. So, when I consider angle AMB, it's indeed the angle subtended by arc AB in the alternate segment, which is 40 degrees.But hold on, another thought. If I consider the major arc AB, the angle subtended by that arc would be supplementary to the angle subtended by the minor arc AB. So, if the minor arc AB subtends 80 degrees at the center, the major arc AB would subtend 360 - 80 = 280 degrees. But since M is on the minor arc AB, we don't need to consider the major arc.Therefore, I think my earlier conclusion is correct. Angle AMB is 40 degrees.Wait, but I just saw the initial problem again, and it says "the angle is 40 degrees." It doesn't specify whether it's the angle between the two sides or something else. But in the context, it's the angle at vertex A, which is angle CAB, so that's 40 degrees.So, to recap:1. The circle is tangent to side AC at A, so OA is perpendicular to AC.2. Angle CAB is 40 degrees, so angle OAB is 50 degrees.3. Triangle OAB is isosceles with OA = OB, so angles OAB and OBA are 50 degrees each.4. Central angle AOB is 80 degrees.5. Inscribed angle AMB subtended by arc AB is half of 80 degrees, which is 40 degrees.6. Alternatively, by the Alternate Segment Theorem, angle between tangent AC and chord AB is equal to angle AMB, which is 40 degrees.Therefore, angle AMB is 40 degrees.But wait, I just thought of something else. If I consider the tangent at A and the chord AB, the angle between them is 40 degrees. According to the Alternate Segment Theorem, this angle is equal to the angle that AB makes in the alternate segment. So, angle AMB is equal to 40 degrees.Yes, that seems consistent.However, I'm still a bit confused because sometimes the angle in the alternate segment can be supplementary, but in this case, since M is on the minor arc AB, it's not supplementary.Wait, let me think about the Alternate Segment Theorem again. It states that the angle between the tangent and the chord is equal to the angle in the alternate segment. So, in this case, the angle between tangent AC and chord AB is 40 degrees, and the angle in the alternate segment is angle AMB. Therefore, angle AMB is 40 degrees.Yes, that seems correct.But I'm still a bit unsure because sometimes when dealing with circles, especially when the point is inside or outside, the angles can behave differently. But in this case, M is on the circumference, so it's an inscribed angle.Wait, another way to think about it is to consider the properties of cyclic quadrilaterals. If I can form a cyclic quadrilateral, then opposite angles would sum to 180 degrees. But in this case, I don't have a quadrilateral, just triangle AMB.Alternatively, maybe I can consider triangle AMB and see if it's part of a cyclic quadrilateral, but I don't think that's necessary here.I think I've considered multiple approaches, and all of them lead me to the conclusion that angle AMB is 40 degrees. Therefore, I'm confident that the answer is 40 degrees.Wait, but hold on a second. I just realized that the problem says "the angle is 40 degrees," but it doesn't specify whether it's the angle between the two sides of the angle or something else. In the problem statement, it says "the angle is 40 degrees," and it's at vertex A, so angle CAB is 40 degrees.So, everything still checks out. The angle between the tangent and the chord is 40 degrees, which equals angle AMB.Therefore, I think I've thought it through thoroughly, and the answer is 40 degrees.

💡Alright, so I have this problem about an ellipse. Let me try to figure it out step by step. First, part (1) asks for the equation of the ellipse E. The ellipse is centered at the origin, has its foci on the coordinate axes, and passes through points M(2, 1) and N(2√2, 0). Hmm, okay. I remember that the general equation of an ellipse centered at the origin with major and minor axes along the coordinate axes is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. Since the foci are on the coordinate axes, it means the major axis is along one of the axes, either x or y. Looking at the points it passes through, point N is (2√2, 0). That's on the x-axis, so when y=0, x=2√2. Plugging that into the ellipse equation: ( (2√2)² ) / a² + 0 = 1 => (8)/a² = 1 => a² = 8. So, a = 2√2. Now, we also know that the ellipse passes through M(2, 1). Let's plug that into the equation: (2²)/8 + (1²)/b² = 1 => 4/8 + 1/b² = 1 => 1/2 + 1/b² = 1. So, 1/b² = 1 - 1/2 = 1/2. Therefore, b² = 2, so b = √2. So, the equation of the ellipse should be (x²)/8 + (y²)/2 = 1. Let me just double-check that. Plugging in N(2√2, 0): ( (2√2)² ) /8 = 8/8 = 1, which works. Plugging in M(2,1): 4/8 + 1/2 = 0.5 + 0.5 = 1, which also works. Okay, that seems right.Now, moving on to part (2). It says: If a line l parallel to OM intersects the ellipse E at two distinct points A and B, and the slopes of lines MA and MB are k₁ and k₂ respectively, determine whether k₁ + k₂ is a constant. If it is, find this constant; if not, explain why.Hmm, so first, let me understand what's given. OM is the line from the origin O to point M(2,1). So, the slope of OM is (1 - 0)/(2 - 0) = 1/2. Therefore, any line parallel to OM will have the same slope, 1/2. So, the equation of line l can be written as y = (1/2)x + t, where t is some constant.This line intersects the ellipse at points A and B. So, to find the coordinates of A and B, I need to solve the system of equations:1. (x²)/8 + (y²)/2 = 12. y = (1/2)x + tLet me substitute equation 2 into equation 1. So, replacing y with (1/2)x + t in the ellipse equation:(x²)/8 + [ ( (1/2)x + t )² ] / 2 = 1Let me expand this:(x²)/8 + [ ( (1/4)x² + (1/2)t x + t² ) ] / 2 = 1Simplify each term:(x²)/8 + (1/4 x²)/2 + (1/2 t x)/2 + (t²)/2 = 1Which is:(x²)/8 + (x²)/8 + (t x)/4 + (t²)/2 = 1Combine like terms:(2x²)/8 + (t x)/4 + (t²)/2 = 1Simplify:(x²)/4 + (t x)/4 + (t²)/2 = 1Multiply both sides by 4 to eliminate denominators:x² + t x + 2t² = 4So, the quadratic equation in x is:x² + t x + (2t² - 4) = 0Let me denote this as x² + t x + (2t² - 4) = 0. Let's call the roots x₁ and x₂, which correspond to points A and B.From quadratic theory, we know that:x₁ + x₂ = -t (sum of roots)x₁ x₂ = 2t² - 4 (product of roots)Now, the points A and B have coordinates (x₁, y₁) and (x₂, y₂), where y₁ = (1/2)x₁ + t and y₂ = (1/2)x₂ + t.We need to find the slopes k₁ and k₂ of lines MA and MB, respectively.Point M is (2,1). So, the slope k₁ is (y₁ - 1)/(x₁ - 2), and k₂ is (y₂ - 1)/(x₂ - 2).So, let's compute k₁ + k₂:k₁ + k₂ = [ (y₁ - 1)/(x₁ - 2) ] + [ (y₂ - 1)/(x₂ - 2) ]Let me write this as:[ ( (1/2 x₁ + t ) - 1 ) / (x₁ - 2) ] + [ ( (1/2 x₂ + t ) - 1 ) / (x₂ - 2) ]Simplify numerator:(1/2 x₁ + t - 1)/(x₁ - 2) + (1/2 x₂ + t - 1)/(x₂ - 2)Let me denote t - 1 as a constant term, say c. So, c = t - 1.So, expression becomes:(1/2 x₁ + c)/(x₁ - 2) + (1/2 x₂ + c)/(x₂ - 2)Let me combine these two fractions:[ (1/2 x₁ + c)(x₂ - 2) + (1/2 x₂ + c)(x₁ - 2) ] / [ (x₁ - 2)(x₂ - 2) ]Let me compute the numerator:(1/2 x₁ x₂ - x₁ + c x₂ - 2c) + (1/2 x₂ x₁ - x₂ + c x₁ - 2c)Combine like terms:1/2 x₁ x₂ + 1/2 x₁ x₂ = x₁ x₂- x₁ - x₂c x₂ + c x₁ = c(x₁ + x₂)-2c -2c = -4cSo, numerator is:x₁ x₂ - (x₁ + x₂) + c(x₁ + x₂) - 4cFactor terms:x₁ x₂ + ( -1 + c )(x₁ + x₂) - 4cNow, we know from earlier that x₁ + x₂ = -t and x₁ x₂ = 2t² - 4.Also, c = t - 1.So, substitute these into numerator:(2t² - 4) + ( -1 + (t - 1) )*(-t) - 4(t - 1)Simplify term by term:First term: 2t² - 4Second term: ( -1 + t - 1 )*(-t) = (t - 2)*(-t) = -t(t - 2) = -t² + 2tThird term: -4(t - 1) = -4t + 4So, putting it all together:2t² - 4 - t² + 2t -4t + 4Combine like terms:2t² - t² = t²-4 + 4 = 02t -4t = -2tSo, numerator simplifies to t² - 2tNow, let's compute the denominator: (x₁ - 2)(x₂ - 2)Expand this:x₁ x₂ - 2x₁ - 2x₂ + 4Again, using x₁ + x₂ = -t and x₁ x₂ = 2t² - 4:(2t² - 4) -2(x₁ + x₂) + 4 = 2t² -4 -2*(-t) +4 = 2t² -4 +2t +4 = 2t² + 2tSo, denominator is 2t² + 2tTherefore, k₁ + k₂ = (t² - 2t)/(2t² + 2t)Factor numerator and denominator:Numerator: t(t - 2)Denominator: 2t(t + 1)So, k₁ + k₂ = [ t(t - 2) ] / [ 2t(t + 1) ] We can cancel t from numerator and denominator (assuming t ≠ 0, which is fine because if t=0, the line l would pass through the origin, but since it's parallel to OM, which has slope 1/2, and passes through origin, but in that case, it would intersect the ellipse at two points, but let's not worry about t=0 for now).So, after canceling t:(t - 2)/(2(t + 1))Hmm, so k₁ + k₂ = (t - 2)/(2(t + 1)). Wait, that seems like it depends on t, which is the y-intercept of line l. So, unless this expression simplifies to a constant, k₁ + k₂ is not a constant. But wait, maybe I made a mistake in calculation.Let me double-check the numerator and denominator.Numerator after substitution was t² - 2t.Denominator was 2t² + 2t.So, (t² - 2t)/(2t² + 2t) = [t(t - 2)]/[2t(t + 1)] = (t - 2)/(2(t + 1)).Hmm, so unless (t - 2)/(2(t + 1)) is a constant, which it's not, because it varies with t. So, does that mean k₁ + k₂ is not a constant?Wait, but the problem says "determine whether k₁ + k₂ is a constant." So, according to my calculation, it's not a constant because it depends on t. But the initial thought was that it might be zero. Maybe I made a mistake.Wait, let me go back to the numerator calculation. Let me re-examine:Numerator after expansion:x₁ x₂ - (x₁ + x₂) + c(x₁ + x₂) - 4cSubstituting:(2t² - 4) + ( -1 + c )*(-t) -4cWait, c = t -1, so:(2t² -4) + ( -1 + t -1 )*(-t) -4(t -1)Simplify:2t² -4 + (t -2)*(-t) -4t +4Which is:2t² -4 -t² +2t -4t +4Combine like terms:2t² - t² = t²-4 +4 = 02t -4t = -2tSo, numerator is t² -2t.Denominator was 2t² + 2t.So, yeah, same as before.Wait, but in the initial problem statement, the user's solution said that k₁ + k₂ = 0. So, maybe I did something wrong.Wait, let me think differently. Maybe instead of parametrizing the line as y = (1/2)x + t, I should use a different parameter. Or maybe there's a property of ellipses that makes this sum zero.Alternatively, perhaps the numerator simplifies to zero. Wait, in my calculation, numerator was t² -2t, which is not zero unless t=0 or t=2. So, unless t is fixed, it's not zero.But in the problem, line l is any line parallel to OM, so t can vary. So, unless for all t, (t² -2t)/(2t² + 2t) is zero, which is not the case.Wait, but maybe I made a mistake in the numerator. Let me re-examine the numerator:After expanding, we had:x₁ x₂ - (x₁ + x₂) + c(x₁ + x₂) - 4cWait, c = t -1, so:x₁ x₂ - (x₁ + x₂) + (t -1)(x₁ + x₂) -4(t -1)Which is:x₁ x₂ + [ -1 + t -1 ](x₁ + x₂) -4t +4Which is:x₁ x₂ + (t -2)(x₁ + x₂) -4t +4Substituting x₁ + x₂ = -t and x₁ x₂ = 2t² -4:(2t² -4) + (t -2)(-t) -4t +4Compute each term:2t² -4(t -2)(-t) = -t² +2t-4t +4So, adding all together:2t² -4 -t² +2t -4t +4Simplify:(2t² - t²) = t²(-4 +4) = 0(2t -4t) = -2tSo, numerator is t² -2tDenominator is 2t² +2tSo, same result.Wait, but in the initial solution, the user said that the numerator equals zero, which would make k₁ + k₂ =0. So, maybe I missed something.Wait, let me check the initial user's solution:They had:Numerator: x₁x₂ + (t -2)(x₁ + x₂) -4(t -1) = 2t² -4 + (t -2)(-2t) -4(t -1) = 2t² -4 -2t² +4t -4t +4 =0Wait, how did they get (t -2)(-2t)? Because in my calculation, x₁ + x₂ = -t, but in their calculation, they have (t -2)(-2t). Wait, that seems different.Wait, let me see:In their solution, they had:Numerator: (1/2 x₁ + t -1)(x₂ -2) + (1/2 x₂ + t -1)(x₁ -2)Then, they expanded it to:x₁x₂ + (t -2)(x₁ + x₂) -4(t -1)But in my calculation, I had:x₁x₂ - (x₁ + x₂) + c(x₁ + x₂) -4c, where c = t -1Which is x₁x₂ + (c -1)(x₁ + x₂) -4cWhich is x₁x₂ + (t -2)(x₁ + x₂) -4(t -1)So, same as user's solution.Then, substituting x₁ + x₂ = -2t and x₁x₂ = 2t² -4.Wait, wait, in my calculation, x₁ + x₂ = -t, but in the user's solution, they have x₁ + x₂ = -2t. That's the key difference.Wait, why is that? Let me go back to the quadratic equation.We had:x² + t x + (2t² -4) =0So, sum of roots x₁ + x₂ = -tProduct x₁x₂ = 2t² -4But in the user's solution, they have x₁ + x₂ = -2t, which is incorrect. So, that must be the mistake.Wait, in the user's solution, they wrote:From the system:y = (1/2)x + tx² +4y² -8 =0Substituting y gives x² + 2tx + 2t² -4=0Wait, let me check that substitution.Original ellipse equation: x²/8 + y²/2 =1Multiply both sides by 8: x² +4y² =8So, substituting y = (1/2)x + t:x² +4*( (1/2 x + t )² ) =8Compute:x² +4*( (1/4 x² + t x + t² ) ) =8Which is:x² + x² +4t x +4t² =8Combine like terms:2x² +4t x +4t² -8=0Divide both sides by 2:x² +2t x +2t² -4=0So, quadratic equation is x² +2t x + (2t² -4)=0Therefore, sum of roots x₁ + x₂ = -2tProduct x₁x₂ =2t² -4Ah, so in my initial calculation, I made a mistake in the substitution. I incorrectly substituted into the ellipse equation without multiplying through by 8 first. So, the correct quadratic equation is x² +2t x +2t² -4=0, leading to x₁ + x₂ = -2t, not -t.Therefore, in the numerator calculation, using x₁ + x₂ = -2t and x₁x₂ =2t² -4.So, let's redo the numerator:Numerator: x₁x₂ + (t -2)(x₁ + x₂) -4(t -1)Substitute:(2t² -4) + (t -2)(-2t) -4(t -1)Compute each term:2t² -4(t -2)(-2t) = -2t² +4t-4(t -1) = -4t +4Now, add them together:2t² -4 -2t² +4t -4t +4Simplify:2t² -2t² =0-4 +4=04t -4t=0So, numerator is 0.Therefore, k₁ +k₂ =0/(denominator)=0.So, k₁ +k₂ is zero, a constant.Ah, so my initial mistake was not correctly substituting into the ellipse equation, leading to an incorrect quadratic. Once corrected, the numerator becomes zero, making k₁ +k₂=0.Therefore, the answer is that k₁ +k₂ is a constant, zero.**Final Answer**(1) The equation of the ellipse is boxed{dfrac{x^2}{8} + dfrac{y^2}{2} = 1}.(2) The sum (k_1 + k_2) is a constant, specifically boxed{0}.

💡Okay, so I have this inequality to solve: (x + 3)/2 - (5x - 1)/5 ≥ 0. Hmm, let's see. I remember that when dealing with inequalities involving fractions, it's usually a good idea to eliminate the denominators first to make things simpler. Alright, so the denominators here are 2 and 5. The least common multiple of 2 and 5 is 10. Maybe I can multiply both sides of the inequality by 10 to get rid of the fractions. Let me try that.Multiplying each term by 10, I get: 10*(x + 3)/2 - 10*(5x - 1)/5 ≥ 10*0. Simplifying each term, 10 divided by 2 is 5, so the first term becomes 5*(x + 3). Similarly, 10 divided by 5 is 2, so the second term becomes 2*(5x - 1). The right side is just 0 because 10*0 is 0.So now the inequality looks like: 5*(x + 3) - 2*(5x - 1) ≥ 0. Let me expand these terms. For the first part, 5*(x + 3) is 5x + 15. For the second part, 2*(5x - 1) is 10x - 2. So putting it all together, the inequality is now 5x + 15 - 10x + 2 ≥ 0.Wait, I think I made a mistake with the signs. The second term is subtracted, so it should be minus 10x plus 2. Let me double-check that. Yes, because it's -2*(5x - 1), which is -10x + 2. So combining like terms, 5x - 10x is -5x, and 15 + 2 is 17. So the inequality simplifies to -5x + 17 ≥ 0.Now, I need to solve for x. Let's move the 17 to the other side by subtracting 17 from both sides: -5x ≥ -17. To isolate x, I'll divide both sides by -5. But wait, I remember that when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips. So dividing both sides by -5, the inequality becomes x ≤ 17/5.Let me just make sure I did everything correctly. Starting from the original inequality, I found a common denominator, multiplied through, expanded, combined like terms, and solved for x, remembering to flip the inequality sign when dividing by a negative. Yeah, that seems right. So the solution should be x is less than or equal to 17/5.