answer：Okay, so I have this problem here where a line l passes through the point M(-2, 0) and intersects an ellipse given by the equation x² + 2y² = 2 at two points, P₁ and P₂. The midpoint of the segment P₁P₂ is called P. I need to find the product of the slopes k₁ and k₂, where k₁ is the slope of line l and k₂ is the slope of the line OP, with O being the origin.Hmm, let me start by understanding the problem step by step. First, the ellipse equation is x² + 2y² = 2. I can rewrite this in standard form by dividing both sides by 2, which gives (x²)/2 + y² = 1. So, this is an ellipse centered at the origin, with semi-major axis √2 along the x-axis and semi-minor axis 1 along the y-axis.Next, the line l passes through M(-2, 0). Since it's a line, I can express it in the slope-intercept form. But since it passes through a specific point, maybe it's better to use the point-slope form. Let me denote the slope of line l as k₁. So, the equation of line l can be written as y = k₁(x + 2), because it passes through (-2, 0).Now, this line intersects the ellipse at points P₁ and P₂. To find these points, I can solve the system of equations consisting of the ellipse and the line. Substituting y from the line equation into the ellipse equation:x² + 2[k₁(x + 2)]² = 2.Let me expand this:x² + 2[k₁²(x + 2)²] = 2x² + 2k₁²(x² + 4x + 4) = 2x² + 2k₁²x² + 8k₁²x + 8k₁² = 2.Now, combine like terms:(1 + 2k₁²)x² + 8k₁²x + (8k₁² - 2) = 0.This is a quadratic equation in terms of x. Let me denote this as:A x² + B x + C = 0,where A = 1 + 2k₁²,B = 8k₁²,C = 8k₁² - 2.Since the line intersects the ellipse at two points P₁ and P₂, the quadratic equation will have two solutions, x₁ and x₂, corresponding to the x-coordinates of P₁ and P₂.The midpoint P of segment P₁P₂ will have coordinates ( (x₁ + x₂)/2, (y₁ + y₂)/2 ). Let me denote P as (x, y). So,x = (x₁ + x₂)/2,y = (y₁ + y₂)/2.From the quadratic equation, I know that the sum of the roots x₁ + x₂ = -B/A. So,x = (-B/A)/2 = (-8k₁²)/(2(1 + 2k₁²)) = (-4k₁²)/(1 + 2k₁²).Wait, hold on. Let me verify that. The sum of roots x₁ + x₂ = -B/A, so x = (x₁ + x₂)/2 = (-B/A)/2 = (-8k₁²)/(2(1 + 2k₁²)) = (-4k₁²)/(1 + 2k₁²). Yeah, that seems correct.Now, for the y-coordinate of P, since P lies on the line l, we can substitute x into the line equation:y = k₁(x + 2) = k₁[ (-4k₁²)/(1 + 2k₁²) + 2 ].Let me compute that:First, simplify inside the brackets:(-4k₁²)/(1 + 2k₁²) + 2 = [ -4k₁² + 2(1 + 2k₁²) ] / (1 + 2k₁²)= [ -4k₁² + 2 + 4k₁² ] / (1 + 2k₁²)= 2 / (1 + 2k₁²).Therefore, y = k₁ * [ 2 / (1 + 2k₁²) ] = 2k₁ / (1 + 2k₁²).So, the coordinates of P are:x = (-4k₁²)/(1 + 2k₁²),y = 2k₁ / (1 + 2k₁²).Now, the slope k₂ of line OP is the slope from the origin (0,0) to point P(x, y). So,k₂ = y / x = [2k₁ / (1 + 2k₁²)] / [ (-4k₁²)/(1 + 2k₁²) ].Simplify this:The denominators are the same, so they cancel out:k₂ = (2k₁) / (-4k₁²) = (2k₁) / (-4k₁²) = - (2k₁) / (4k₁²) = - (1) / (2k₁).So, k₂ = -1/(2k₁).Therefore, the product k₁k₂ is:k₁ * (-1/(2k₁)) = -1/2.Wait, so k₁k₂ = -1/2. That seems straightforward. Let me just check my steps to make sure I didn't make a mistake.First, I found the equation of the line l correctly: y = k₁(x + 2). Then, substituting into the ellipse equation, I expanded and combined terms correctly to get the quadratic in x. Then, I found the sum of roots x₁ + x₂ = -B/A, which is correct. Then, I found x as half of that, which is correct.For the y-coordinate, I used the line equation, substituted x, and correctly simplified it. Then, I found k₂ as y/x, which is the slope from the origin to P. The calculation there seems correct, leading to k₂ = -1/(2k₁).Multiplying k₁ and k₂ gives -1/2, which is a constant, independent of k₁. That makes sense because the problem states that k₁ ≠ 0, so we don't have any undefined expressions here.I think that's solid. So, the value of k₁k₂ is -1/2.

answer：Okay, so I need to find the arithmetic mean of the numbers 17, 25, and 38. Hmm, arithmetic mean is like the average, right? I remember that to find the average, you add up all the numbers and then divide by how many numbers there are. Let me see if I can do this step by step.First, I'll add the numbers together. So, 17 plus 25. Let me do that in my head. 17 plus 20 is 37, and then plus 5 more is 42. Okay, so 17 plus 25 equals 42. Now, I need to add 38 to that. So, 42 plus 38. Hmm, 40 plus 30 is 70, and 2 plus 8 is 10, so 70 plus 10 is 80. So, altogether, 17 plus 25 plus 38 equals 80.Now, there are three numbers here: 17, 25, and 38. So, to find the arithmetic mean, I need to divide the total sum, which is 80, by the number of numbers, which is 3. So, 80 divided by 3. Let me think about how to do that. 3 goes into 8 two times, which is 6, and then there's a remainder of 2. So, 2 and 2/3, which is approximately 2.666...Wait, but I think it's better to write it as a fraction. So, 80 divided by 3 is the same as 80 over 3, which is 80/3. That's an improper fraction, but it's also equal to 26 and 2/3 when converted to a mixed number. So, the arithmetic mean is 80/3 or approximately 26.666...Let me double-check my addition to make sure I didn't make a mistake. 17 plus 25 is 42, and 42 plus 38 is 80. Yeah, that seems right. And dividing by 3 gives me 80/3. Okay, I think that's correct.So, the arithmetic mean of 17, 25, and 38 is 80/3, which is approximately 26.666...

answer：First, I need to convert each of the base 4 numbers to base 10 to perform the arithmetic operations more easily.For (203_4):[2 times 4^2 + 0 times 4^1 + 3 times 4^0 = 32 + 0 + 3 = 35_{10}]For (321_4):[3 times 4^2 + 2 times 4^1 + 1 times 4^0 = 48 + 8 + 1 = 57_{10}]For (112_4):[1 times 4^2 + 1 times 4^1 + 2 times 4^0 = 16 + 4 + 2 = 22_{10}]Next, I perform the arithmetic operations in base 10:[35_{10} + 57_{10} - 22_{10} = 92_{10} - 22_{10} = 70_{10}]Finally, I convert (70_{10}) back to base 4:[70 div 4 = 17 text{ remainder } 2 quad text{(least significant digit)}][17 div 4 = 4 text{ remainder } 1][4 div 4 = 1 text{ remainder } 0][1 div 4 = 0 text{ remainder } 1 quad text{(most significant digit)}]Thus, (70_{10} = 1012_4).

answer：First, I need to determine the total rainfall recorded by Bonnie's science class over the three days: Monday, Tuesday, and Wednesday.I'll start by listing the rainfall amounts for each day:- **Monday:** 0.17 centimeters- **Tuesday:** 0.42 centimeters- **Wednesday:** 0.08 centimetersNext, I'll add these amounts together to find the total rainfall.Adding Monday and Tuesday:0.17 cm + 0.42 cm = 0.59 cmThen, adding Wednesday's rainfall:0.59 cm + 0.08 cm = 0.67 cmSo, the total rainfall recorded by the class over the three days is 0.67 centimeters.

answer：Okay, so I have this problem where I need to find the value of ( y ) such that the mean, median, and unique mode of the numbers 2, 3, 4, 5, 5, 6, and ( y ) are all equal. Hmm, that sounds a bit tricky, but let me break it down step by step.First, let me recall what mean, median, and mode mean. The mean is the average of all the numbers, the median is the middle number when the numbers are arranged in order, and the mode is the number that appears most frequently. The problem says that all three of these should be equal, which means I need to find a value of ( y ) that makes the mean, median, and mode all the same.Let me start by looking at the mode because it seems the simplest. The mode is the number that appears most often. In the given list, I see that the number 5 appears twice, and all the other numbers appear only once. So, currently, the mode is 5. For the mode to remain 5, ( y ) should either be 5 or another number that doesn't appear more than once. If ( y ) is 5, then 5 will appear three times, which would make it the unique mode. If ( y ) is any other number, it will only appear once, so 5 will still be the mode. So, ( y ) could be 5 or any other number, but if it's any other number, it shouldn't create a new mode. That means ( y ) can't be 2, 3, 4, or 6 because those already appear once, and if ( y ) is one of them, that number would appear twice, tying with 5 for the mode. But the problem says the mode is unique, so ( y ) can't be 2, 3, 4, or 6. Therefore, ( y ) must be 5 or a number that isn't already in the list. But wait, if ( y ) is a number not in the list, like 7 or 10, then 5 remains the unique mode. So, ( y ) can be 5 or any number not in the list, but we need to check if that affects the median and mean.Next, let's look at the median. The median is the middle number when the numbers are arranged in order. Since there are seven numbers in total, the median will be the fourth number. Let me arrange the numbers in order: 2, 3, 4, 5, 5, 6, and ( y ). Depending on the value of ( y ), the order might change.If ( y ) is less than or equal to 5, the order would be: ( y ), 2, 3, 4, 5, 5, 6. Wait, no, that's not right. If ( y ) is less than or equal to 5, it would be placed somewhere in the first four positions. Let me think again. The original list without ( y ) is 2, 3, 4, 5, 5, 6. Adding ( y ), if ( y ) is less than or equal to 5, it would be placed before or at the position of 5. So, the sorted list would be: 2, 3, 4, ( y ), 5, 5, 6 if ( y ) is 4 or less. Wait, no, if ( y ) is 5, it would be 2, 3, 4, 5, 5, 5, 6. So, the fourth number is 5 in this case. If ( y ) is greater than 5, say 6 or more, the sorted list would be 2, 3, 4, 5, 5, 6, ( y ), making the fourth number still 5. So, regardless of whether ( y ) is less than or greater than 5, the median remains 5. Therefore, the median is 5 regardless of ( y )'s value, as long as ( y ) doesn't disrupt the order in a way that changes the middle number. But since there are seven numbers, the fourth one is always the median, and as we saw, it's 5 whether ( y ) is 5 or not.Now, onto the mean. The mean is the average of all the numbers. So, I need to calculate the sum of all the numbers and divide by 7 (since there are seven numbers). The sum of the given numbers is 2 + 3 + 4 + 5 + 5 + 6. Let me add those up: 2 + 3 is 5, plus 4 is 9, plus 5 is 14, plus another 5 is 19, plus 6 is 25. So, the sum without ( y ) is 25. Adding ( y ), the total sum becomes 25 + ( y ). The mean is then ( frac{25 + y}{7} ). We want the mean to be equal to the median and the mode, which we've established is 5. So, we set up the equation:[frac{25 + y}{7} = 5]To solve for ( y ), I'll multiply both sides by 7:[25 + y = 35]Then, subtract 25 from both sides:[y = 10]So, ( y ) must be 10 to make the mean equal to 5. But wait, earlier I thought that if ( y ) is 10, which is not in the list, then 5 remains the unique mode. Let me verify that. If ( y ) is 10, the list becomes 2, 3, 4, 5, 5, 6, 10. The mode is still 5 because it appears twice, and all others appear once. The median is still the fourth number, which is 5, and the mean is 5 as calculated. So, all three measures are equal to 5 when ( y ) is 10.But hold on, earlier I considered that if ( y ) is 5, the mode would still be 5, and the median would still be 5. Let me check the mean in that case. If ( y ) is 5, the sum becomes 25 + 5 = 30, and the mean is ( frac{30}{7} ), which is approximately 4.2857, not 5. So, the mean wouldn't be 5 if ( y ) is 5. Therefore, ( y ) can't be 5 because that would make the mean not equal to 5. So, even though 5 is the mode and median, the mean wouldn't match. Therefore, ( y ) must be 10 to satisfy all three conditions.Wait, but the options given are A) 5, B) 6, C) 10, D) 11, E) None of these. So, 10 is an option, which is C. But let me double-check everything to make sure I didn't make a mistake.If ( y = 10 ), the list is 2, 3, 4, 5, 5, 6, 10. Sorted, it's 2, 3, 4, 5, 5, 6, 10. The mode is 5, the median is 5, and the mean is ( frac{2 + 3 + 4 + 5 + 5 + 6 + 10}{7} = frac{35}{7} = 5 ). So, all three are indeed 5. Therefore, ( y = 10 ) is the correct answer.But wait, the initial thought was that ( y ) could be 5, but that didn't satisfy the mean. So, 10 is the correct value. Therefore, the answer should be C) 10.However, looking back at the problem, it says "unique mode." If ( y = 10 ), the mode is still 5, which is unique. If ( y ) were 5, the mode would still be 5, but the mean wouldn't be 5. So, 10 is the only value that satisfies all three conditions. Therefore, the correct answer is C) 10.

answer：Okay, so I need to find the area of the function y = sin(3x) on the interval [0, 2π/3]. Hmm, I remember that the area under a curve can be found using integrals. But wait, the problem mentions something about the area of y = sin(nx) on [0, π/n] being 2/n. Maybe I can use that information to help me solve this problem without having to compute the integral from scratch.Let me think. If n is a positive integer, then for y = sin(nx), the area on [0, π/n] is 2/n. So, in this case, we have y = sin(3x), which means n = 3. Therefore, the area on [0, π/3] should be 2/3. But the interval we're interested in is [0, 2π/3], which is twice as long as [0, π/3]. Does that mean the area is just double?Wait, let me make sure. The function sin(3x) has a period of 2π/3 because the period of sin(kx) is 2π/k. So, over [0, 2π/3], it completes one full period. The area over one period of a sine function is twice the area over half a period, right? Because the sine function is positive in the first half and negative in the second half, but since we're talking about the area (which is always positive), we take the absolute value.So, if the area from 0 to π/3 is 2/3, then from π/3 to 2π/3, the function sin(3x) is negative, but the area would still be the same as the positive part. Therefore, the total area from 0 to 2π/3 would be 2*(2/3) = 4/3. That makes sense.But just to be thorough, maybe I should compute the integral directly to verify. The area is the integral of |sin(3x)| from 0 to 2π/3. Let's compute that.First, let's find the integral of sin(3x) from 0 to 2π/3. The integral of sin(3x) dx is (-1/3)cos(3x) + C. Evaluating from 0 to 2π/3:At 2π/3: (-1/3)cos(3*(2π/3)) = (-1/3)cos(2π) = (-1/3)(1) = -1/3.At 0: (-1/3)cos(0) = (-1/3)(1) = -1/3.So, the integral from 0 to 2π/3 is (-1/3) - (-1/3) = 0. But that's the net area, not the actual area. Since the function is positive and negative over the interval, we need to take the absolute value.So, let's find where sin(3x) crosses the x-axis between 0 and 2π/3. Setting sin(3x) = 0, we get 3x = nπ, so x = nπ/3. For x in [0, 2π/3], n can be 0, 1, or 2. So, the zeros are at x = 0, π/3, and 2π/3.Therefore, the function is positive from 0 to π/3 and negative from π/3 to 2π/3. So, the area is the integral from 0 to π/3 of sin(3x) dx plus the integral from π/3 to 2π/3 of -sin(3x) dx.Let's compute the first integral:Integral from 0 to π/3 of sin(3x) dx = [(-1/3)cos(3x)] from 0 to π/3.At π/3: (-1/3)cos(π) = (-1/3)(-1) = 1/3.At 0: (-1/3)cos(0) = (-1/3)(1) = -1/3.So, the first integral is 1/3 - (-1/3) = 2/3.Now, the second integral:Integral from π/3 to 2π/3 of -sin(3x) dx = -[(-1/3)cos(3x)] from π/3 to 2π/3.Compute at 2π/3: (-1/3)cos(2π) = (-1/3)(1) = -1/3.Compute at π/3: (-1/3)cos(π) = (-1/3)(-1) = 1/3.So, the integral is -[(-1/3) - (1/3)] = -[(-2/3)] = 2/3.Adding both areas: 2/3 + 2/3 = 4/3.Okay, so that confirms it. The area is indeed 4/3. So, the answer is 4/3.I think I understand now. The key was recognizing that the area over one period is twice the area over half a period because the sine function is symmetric and positive in the first half, negative in the second half. So, by knowing the area over half a period, I can just double it to get the total area over the full period. That's a useful shortcut.But it's also good to verify by computing the integral directly, especially since I initially thought about the net area being zero, which confused me for a moment. Taking the absolute value into account by splitting the integral at the zero crossing was the right approach.So, in summary, the area of y = sin(3x) on [0, 2π/3] is 4/3.