answer：Okay, so I have this cubic equation here: (x^3 - 15x^2 + 75x - 50 = 0). I need to find the product of its roots. Hmm, I remember something about Vieta's formulas from algebra class. Let me try to recall how that works.First, Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a general cubic equation like (ax^3 + bx^2 + cx + d = 0), there are three roots, say (p), (q), and (r). Vieta's formulas tell us that:1. The sum of the roots (p + q + r = -frac{b}{a}).2. The sum of the products of the roots taken two at a time (pq + pr + qr = frac{c}{a}).3. The product of the roots (pqr = -frac{d}{a}).So, in my equation, the coefficients are:- (a = 1) (since the coefficient of (x^3) is 1),- (b = -15),- (c = 75),- (d = -50).I need the product of the roots, which is the third formula: (pqr = -frac{d}{a}).Plugging in the values from my equation:(pqr = -frac{-50}{1}).Wait, that simplifies to (pqr = 50). So, the product of the roots should be 50.Let me double-check to make sure I didn't make a mistake. The formula says the product is (-d/a). Since (d) is (-50), then (-d) would be (50), and dividing by (a = 1) doesn't change anything. Yep, that seems right.I guess another way to think about it is if I factor the cubic equation. If the equation can be factored as ((x - p)(x - q)(x - r) = 0), then expanding this gives (x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr = 0). Comparing this to my original equation:- The coefficient of (x^2) is (-15), which should be equal to (-(p + q + r)). So, (p + q + r = 15).- The coefficient of (x) is (75), which should be equal to (pq + pr + qr).- The constant term is (-50), which should be equal to (-pqr). So, (pqr = 50).That matches what I got earlier using Vieta's formula. So, I'm pretty confident that the product of the roots is indeed 50.Just to be thorough, maybe I can try to find the roots explicitly and multiply them. But solving a cubic equation can be a bit tricky. Let me see if I can factor this equation.Looking at the equation (x^3 - 15x^2 + 75x - 50 = 0), I can try to find rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. Since the constant term is (-50) and the leading coefficient is 1, the possible rational roots are (pm1, pm2, pm5, pm10, pm25, pm50).Let me test (x = 5):(5^3 - 15(5)^2 + 75(5) - 50 = 125 - 375 + 375 - 50 = 75). Not zero.How about (x = 2):(2^3 - 15(2)^2 + 75(2) - 50 = 8 - 60 + 150 - 50 = 48). Not zero.(x = 1):(1 - 15 + 75 - 50 = -9). Not zero.(x = 10):(1000 - 1500 + 750 - 50 = 200). Not zero.Hmm, maybe (x = 5) again? Wait, I did that. Maybe I made a mistake. Let me recalculate (x = 5):(125 - 375 + 375 - 50 = 125 - 375 is -250, plus 375 is 125, minus 50 is 75. Yeah, still 75.What about (x = 10):(1000 - 1500 is -500, plus 750 is 250, minus 50 is 200. Still not zero.Maybe (x = 25):That's too big, probably not a root.Wait, maybe I should try (x = 5) again, but perhaps I made a mistake in calculation. Let me do it step by step:(5^3 = 125),(-15(5)^2 = -15*25 = -375),(75*5 = 375),(-50).So, 125 - 375 = -250,-250 + 375 = 125,125 - 50 = 75. Yeah, still 75.Hmm, maybe there are no rational roots. That complicates things. Maybe I need to use the cubic formula or something, but that's pretty complicated. Since I already know from Vieta's formula that the product is 50, maybe I don't need to find the actual roots.Alternatively, maybe I can factor by grouping. Let me try that.Looking at (x^3 - 15x^2 + 75x - 50), I can group the first two terms and the last two terms:((x^3 - 15x^2) + (75x - 50)).Factor out (x^2) from the first group: (x^2(x - 15)).Factor out 25 from the second group: (25(3x - 2)).So, we have (x^2(x - 15) + 25(3x - 2)). Hmm, that doesn't seem to help much because the terms inside the parentheses aren't the same.Maybe another way of grouping? Let me try grouping the first term with the third term and the second term with the fourth term:((x^3 + 75x) + (-15x^2 - 50)).Factor out (x) from the first group: (x(x^2 + 75)).Factor out (-5) from the second group: (-5(3x^2 + 10)).So, we have (x(x^2 + 75) - 5(3x^2 + 10)). Again, the terms inside the parentheses aren't the same, so factoring doesn't help here.Maybe I need to try a different approach. Since factoring isn't working easily, perhaps I can use synthetic division or look for irrational roots. But that might be time-consuming.Alternatively, since I already know the product of the roots is 50 from Vieta's formula, maybe I can just stick with that answer. It seems consistent with both the formula and the expanded form of the factored cubic.Wait, just to make sure, let me think about the relationship between the coefficients and the roots again. If I have (x^3 - 15x^2 + 75x - 50 = 0), and I write it as ((x - p)(x - q)(x - r) = 0), then expanding gives:(x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr = 0).Comparing coefficients:- Coefficient of (x^3): 1 = 1, which matches.- Coefficient of (x^2): -15 = -(p + q + r), so p + q + r = 15.- Coefficient of (x): 75 = pq + pr + qr.- Constant term: -50 = -pqr, so pqr = 50.Yep, that all checks out. So, even though I couldn't factor the equation easily, Vieta's formula gives me the product of the roots directly as 50.I think I'm confident enough with this answer. It's always good to verify using multiple methods, but in this case, since both Vieta's formula and the expanded factored form lead me to the same conclusion, I don't see any reason to doubt it.Just to recap:1. Identified the type of equation (cubic).2. Remembered Vieta's formulas for cubic equations.3. Applied the formula for the product of the roots.4. Verified by expanding the factored form and comparing coefficients.5. Attempted factoring but found it complicated, so relied on Vieta's result.6. Double-checked the reasoning to ensure no mistakes were made.So, after all that, I'm pretty sure the product of the roots is 50.

answer：First, identify the total number of balls in the box, which is 4.Next, determine the number of favorable outcomes, which is drawing the ball numbered 2. There is only 1 such ball.Finally, calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

answer：Alright, so I've got this problem about Olya and Masha coloring a 2022x2022 board with 2x2 squares using red and blue. The rules are that each cell can be painted no more than once in blue and no more than once in red. If a cell is painted both colors, it becomes purple. After all the cells are painted, they count how many are purple. The question is asking what possible counts of purple cells they could have, and the options are 2022×2022, 2022×2021, 2022×2020, and 2020×2020.Okay, first, let me understand the problem. They're taking turns coloring 2x2 squares, so each move affects four cells. Each cell can be painted once red and once blue, but no more than that. So, a cell can be painted red, then blue, making it purple, or blue then red, also purple. If a cell is painted only once, it remains red or blue. The goal is to figure out how many cells can end up being purple after the entire board is painted.Let me think about the total number of cells. The board is 2022x2022, so that's 2022 multiplied by 2022, which is 4,088,484 cells. Each 2x2 square covers four cells, so each move paints four cells. Now, since each cell can be painted twice at most, once red and once blue, the maximum number of times a cell can be painted is two.But wait, the problem says they take turns coloring 2x2 squares. So, each turn, a player paints a 2x2 square, either red or blue. They alternate turns, right? So, Olya might go first, painting a 2x2 square red, then Masha paints a 2x2 square blue, and so on.Now, the key here is that each cell can be painted at most once red and once blue. So, a cell can be painted red in one move and blue in another move, making it purple. If a cell is painted red twice, that's not allowed, and similarly for blue. So, the maximum number of times a cell can be painted is two, once red and once blue.So, the total number of paints across the entire board would be 2022x2022 times 2, which is 8,176,968 paints. But since each move paints four cells, the total number of moves would be 8,176,968 divided by 4, which is 2,044,242 moves.Wait, but that's the maximum number of moves if every cell is painted twice. But in reality, some cells might be painted only once, so the total number of moves could be less. But the problem says "once all cells are painted," so I think that means every cell has been painted at least once. But they can be painted twice, making them purple.So, the total number of paints is variable, depending on how many cells are painted twice. Let me denote the number of purple cells as x. Each purple cell is painted twice, so that's 2x paints. The remaining cells, which are painted only once, are 2022x2022 - x, and that's 1 paint each, so total paints would be 2022x2022 - x + 2x = 2022x2022 + x.But each move paints four cells, so the total number of paints must be divisible by 4. Therefore, 2022x2022 + x must be divisible by 4.Let me compute 2022 modulo 4. 2022 divided by 4 is 505 with a remainder of 2, so 2022 ≡ 2 mod 4. Therefore, 2022x2022 ≡ 2x2 ≡ 4 ≡ 0 mod 4. So, 2022x2022 is divisible by 4. Therefore, x must also be divisible by 4 for 2022x2022 + x to be divisible by 4.So, x must be a multiple of 4. Now, looking at the options, 2022x2022 is 4,088,484, which is divisible by 4, but is that possible? If all cells are purple, that would mean every cell was painted twice, but that would require that every cell was covered by two different 2x2 squares, which might not be possible because of the way 2x2 squares overlap.Wait, can every cell be covered by two different 2x2 squares? Let me think. In a 2022x2022 board, each cell can be part of multiple 2x2 squares. For example, a cell in the middle can be part of four different 2x2 squares: one starting at its top-left, one at top-right, one at bottom-left, and one at bottom-right. So, in theory, each cell can be painted multiple times, but the problem restricts each cell to being painted at most once red and once blue.But if every cell is painted twice, once red and once blue, then every cell would be purple. Is that possible? Let me see. If Olya and Masha take turns painting 2x2 squares, alternating red and blue, is it possible for every cell to be covered exactly once by red and once by blue?I think that might not be possible because of the parity of the board. Let me consider a smaller board, say 2x2. If they paint the entire board, each cell is painted once, so no purple cells. If they paint it twice, each cell is painted twice, so all cells are purple. But in a 2x2 board, painting it twice would mean painting the same 2x2 square twice, but the problem says they take turns, so Olya paints first, then Masha, and so on. So, in a 2x2 board, if Olya paints it red, then Masha paints it blue, then all cells are purple. So, in that case, all cells are purple.But in a larger board, like 2022x2022, which is even in both dimensions, can they paint the entire board such that every cell is painted exactly once red and once blue? That would require that the entire board is covered by red and blue 2x2 squares in such a way that every cell is covered exactly once by red and once by blue.This sounds similar to a domino tiling, but with 2x2 squares. I'm not sure if that's possible. Maybe it is, but I'm not certain. If it is possible, then x could be 2022x2022, but I'm not sure.Wait, but the problem says "once all cells are painted," so they might not necessarily paint every cell twice. They just paint until all cells are painted at least once, but some could be painted twice, making them purple.But the problem is asking for possible counts of purple cells, not necessarily the maximum or minimum. So, x could range from 0 to 2022x2022, but with the constraint that x must be a multiple of 4.Looking at the options, 2022x2022 is 4,088,484, which is divisible by 4. 2022x2021 is 4,088,484 - 2022 = 4,086,462, which is not divisible by 4 because 2022 is 2 mod 4, so 4,088,484 - 2022 = 4,086,462, which is 4,086,462 mod 4. 4,086,462 divided by 2 is 2,043,231, which is odd, so 4,086,462 is 2 mod 4, not divisible by 4. So, 2022x2021 is not possible.2022x2020 is 2022*2020. Let's compute that. 2022*2020 = (2000 + 22)*(2000 + 20) = 2000*2000 + 2000*20 + 22*2000 + 22*20 = 4,000,000 + 40,000 + 44,000 + 440 = 4,084,440. Is that divisible by 4? 4,084,440 divided by 4 is 1,021,110, which is an integer, so yes, it's divisible by 4.Similarly, 2020x2020 is 2020 squared, which is 4,080,400. Divided by 4 is 1,020,100, which is an integer, so that's also divisible by 4.So, both 2022x2020 and 2020x2020 are possible counts of purple cells because they are multiples of 4. The other options are not.But wait, is 2022x2022 possible? Earlier, I thought maybe not because of the way 2x2 squares overlap, but mathematically, since 2022x2022 is divisible by 4, it's possible in terms of the total paints. But practically, can every cell be painted twice? I'm not sure. Maybe it's possible, but I'm not certain. However, the problem doesn't specify that they have to paint until all cells are painted, but rather that they continue until all cells are painted, which could mean that some cells are painted twice, but not necessarily all.So, given that, the possible counts are the ones that are multiples of 4, which are 2022x2020 and 2020x2020.Wait, but 2022x2022 is also a multiple of 4, but I'm not sure if it's possible to have all cells painted twice. Maybe it's not possible because of the way the 2x2 squares overlap. For example, in a 4x4 board, can you paint every cell twice with 2x2 squares? Let's see. If you paint the entire board with red 2x2 squares, then paint it again with blue 2x2 squares, then every cell would be purple. So, in that case, yes, it's possible. So, maybe in a 2022x2022 board, it's also possible.But the problem is that Olya and Masha are taking turns, so they might not necessarily paint the entire board twice. They could stop once all cells are painted at least once. So, the number of purple cells could vary depending on how they paint.But the problem is asking for possible counts, not necessarily the maximum or minimum. So, as long as the count is a multiple of 4, it's possible. Therefore, both 2022x2020 and 2020x2020 are possible, as well as 2022x2022 if it's possible to paint every cell twice.But I'm not sure if 2022x2022 is possible because of the way the 2x2 squares overlap. Maybe it's not possible to cover every cell exactly twice with 2x2 squares, but I'm not certain. Given that, I'll go with the options provided, which are 2022x2020 and 2020x2020, both of which are multiples of 4 and possible counts.

answer：First, I need to determine how many weeks Eva should eat an apple every day if she needs to buy 14 apples.I know that there are 7 days in a week.Since Eva eats one apple each day, the number of apples she needs per week is equal to the number of days in a week.To find out how many weeks she needs 14 apples, I can divide the total number of apples by the number of apples she consumes per week.So, 14 apples divided by 7 apples per week equals 2 weeks.Therefore, Eva should eat an apple every day for 2 weeks.

answer：First, identify the weight of James's bag, which is 18 kg.Next, determine the weight of one of Oliver's bags by calculating one-sixth of James's bag:18 kg × (1/6) = 3 kg.Finally, calculate the combined weight of both of Oliver's bags by multiplying the weight of one bag by 2:3 kg × 2 = 6 kg.

answer：Okay, so I have this problem where I need to find the equation of an ellipse that shares a common focus with the ellipse given by the equation ( x^2 + 4y^2 = 4 ) and passes through the point ( A(2,1) ). Hmm, let me try to figure this out step by step.First, I remember that the standard form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a ) and ( b ) are the semi-major and semi-minor axes, respectively. The foci of an ellipse are located at a distance ( c ) from the center, where ( c = sqrt{a^2 - b^2} ) for an ellipse that is wider along the x-axis, or ( c = sqrt{b^2 - a^2} ) if it's taller along the y-axis.Looking at the given ellipse equation ( x^2 + 4y^2 = 4 ), I can rewrite it in standard form by dividing both sides by 4:[frac{x^2}{4} + frac{y^2}{1} = 1]So here, ( a^2 = 4 ) and ( b^2 = 1 ), which means ( a = 2 ) and ( b = 1 ). Since ( a > b ), this ellipse is wider along the x-axis. The distance to the foci ( c ) is calculated as:[c = sqrt{a^2 - b^2} = sqrt{4 - 1} = sqrt{3}]So, the foci of the given ellipse are at ( (pm sqrt{3}, 0) ).Now, the problem says that the desired ellipse shares a common focus with this ellipse. That means the foci of the desired ellipse are also at ( (pm sqrt{3}, 0) ). Therefore, for the desired ellipse, ( c = sqrt{3} ) as well.Let me denote the equation of the desired ellipse as ( frac{x^2}{A^2} + frac{y^2}{B^2} = 1 ). Since it shares the same foci, its ( c ) is also ( sqrt{3} ). So, for this ellipse, ( c = sqrt{A^2 - B^2} ) if it's wider along the x-axis, or ( c = sqrt{B^2 - A^2} ) if it's taller along the y-axis.But I don't know yet whether the desired ellipse is wider along the x-axis or the y-axis. Hmm, maybe I can figure that out based on the point it passes through, which is ( A(2,1) ).Let me assume first that the desired ellipse is wider along the x-axis, so ( A > B ). Then, ( c = sqrt{A^2 - B^2} = sqrt{3} ), which gives:[A^2 - B^2 = 3]So, ( B^2 = A^2 - 3 ). Therefore, the equation of the ellipse can be written as:[frac{x^2}{A^2} + frac{y^2}{A^2 - 3} = 1]Now, since the ellipse passes through the point ( A(2,1) ), substituting ( x = 2 ) and ( y = 1 ) into the equation should satisfy it:[frac{2^2}{A^2} + frac{1^2}{A^2 - 3} = 1][frac{4}{A^2} + frac{1}{A^2 - 3} = 1]Now, I have an equation in terms of ( A^2 ). Let me denote ( A^2 = k ) for simplicity, so the equation becomes:[frac{4}{k} + frac{1}{k - 3} = 1]To solve for ( k ), I can find a common denominator:[frac{4(k - 3) + k}{k(k - 3)} = 1][frac{4k - 12 + k}{k(k - 3)} = 1][frac{5k - 12}{k(k - 3)} = 1]Now, cross-multiplying:[5k - 12 = k(k - 3)][5k - 12 = k^2 - 3k]Bring all terms to one side:[k^2 - 3k - 5k + 12 = 0][k^2 - 8k + 12 = 0]Now, solving this quadratic equation:[k = frac{8 pm sqrt{64 - 48}}{2} = frac{8 pm sqrt{16}}{2} = frac{8 pm 4}{2}]So, ( k = frac{8 + 4}{2} = 6 ) or ( k = frac{8 - 4}{2} = 2 ).But ( k = A^2 ), and since ( B^2 = A^2 - 3 ), ( A^2 ) must be greater than 3. So, ( k = 6 ) is acceptable, but ( k = 2 ) would give ( B^2 = -1 ), which is not possible. Therefore, ( k = 6 ).So, ( A^2 = 6 ) and ( B^2 = 6 - 3 = 3 ).Therefore, the equation of the desired ellipse is:[frac{x^2}{6} + frac{y^2}{3} = 1]Wait, let me double-check if this ellipse actually passes through the point ( A(2,1) ):Substituting ( x = 2 ) and ( y = 1 ):[frac{4}{6} + frac{1}{3} = frac{2}{3} + frac{1}{3} = 1]Yes, that works.Just to be thorough, I should also consider the case where the desired ellipse might be taller along the y-axis, meaning ( B > A ). In that case, ( c = sqrt{B^2 - A^2} = sqrt{3} ), so:[B^2 - A^2 = 3]But then, the equation would be:[frac{x^2}{A^2} + frac{y^2}{A^2 + 3} = 1]Substituting ( A(2,1) ):[frac{4}{A^2} + frac{1}{A^2 + 3} = 1]Let me denote ( A^2 = m ), so:[frac{4}{m} + frac{1}{m + 3} = 1]Common denominator:[frac{4(m + 3) + m}{m(m + 3)} = 1][frac{4m + 12 + m}{m(m + 3)} = 1][frac{5m + 12}{m(m + 3)} = 1]Cross-multiplying:[5m + 12 = m^2 + 3m][m^2 + 3m - 5m - 12 = 0][m^2 - 2m - 12 = 0]Solving:[m = frac{2 pm sqrt{4 + 48}}{2} = frac{2 pm sqrt{52}}{2} = frac{2 pm 2sqrt{13}}{2} = 1 pm sqrt{13}]Since ( m = A^2 ) must be positive, ( m = 1 + sqrt{13} ) is acceptable, but ( m = 1 - sqrt{13} ) is negative, which is not possible.So, ( A^2 = 1 + sqrt{13} ), and ( B^2 = A^2 + 3 = 4 + sqrt{13} ).Therefore, the equation would be:[frac{x^2}{1 + sqrt{13}} + frac{y^2}{4 + sqrt{13}} = 1]But wait, this seems more complicated, and the first solution I found was much simpler. Also, let me check if this second ellipse actually passes through ( A(2,1) ):Substituting ( x = 2 ) and ( y = 1 ):[frac{4}{1 + sqrt{13}} + frac{1}{4 + sqrt{13}} ]This is approximately:[frac{4}{4.6055} + frac{1}{6.6055} approx 0.868 + 0.151 approx 1.019]Which is approximately 1, but not exactly. Hmm, maybe due to the approximation. Let me calculate it more accurately.But actually, since we derived it algebraically, it should satisfy the equation exactly. Let me compute:First term: ( frac{4}{1 + sqrt{13}} )Multiply numerator and denominator by ( sqrt{13} - 1 ):[frac{4(sqrt{13} - 1)}{(1 + sqrt{13})(sqrt{13} - 1)} = frac{4(sqrt{13} - 1)}{13 - 1} = frac{4(sqrt{13} - 1)}{12} = frac{sqrt{13} - 1}{3}]Second term: ( frac{1}{4 + sqrt{13}} )Multiply numerator and denominator by ( sqrt{13} - 4 ):[frac{sqrt{13} - 4}{(4 + sqrt{13})(sqrt{13} - 4)} = frac{sqrt{13} - 4}{13 - 16} = frac{sqrt{13} - 4}{-3} = frac{4 - sqrt{13}}{3}]Adding both terms:[frac{sqrt{13} - 1}{3} + frac{4 - sqrt{13}}{3} = frac{(sqrt{13} - 1) + (4 - sqrt{13})}{3} = frac{3}{3} = 1]Ah, so it does satisfy the equation exactly. So, there are actually two ellipses that satisfy the given conditions: one wider along the x-axis and one taller along the y-axis.But the problem doesn't specify whether the major axis is along the x-axis or y-axis, so both are possible. However, the first solution I found was simpler, and often in such problems, the major axis is assumed to be along the x-axis unless specified otherwise.But just to be safe, I should probably present both solutions. However, looking back at the problem statement, it just says "the ellipse," implying perhaps a unique solution. Maybe I missed something.Wait, the given ellipse is ( x^2 + 4y^2 = 4 ), which is wider along the x-axis. So, if the desired ellipse shares the same foci, which are along the x-axis, it's more natural that the desired ellipse is also wider along the x-axis. So, the first solution is likely the intended one.Therefore, the equation of the desired ellipse is:[frac{x^2}{6} + frac{y^2}{3} = 1]I think that's the answer.**Final Answer**The equation of the ellipse is boxed{dfrac{x^2}{6} + dfrac{y^2}{3} = 1}.