answer：Okay, so I have this problem here: there's a vat of orange juice that contains 502.75 pints, and I need to pour this into 21 glasses, making sure each glass has the same amount. The question is asking how many pints would be in each glass.Alright, let's break this down. First, I need to figure out how to distribute 502.75 pints equally among 21 glasses. That sounds like a division problem because I'm splitting a total amount into equal parts. So, I think I need to divide 502.75 by 21 to find out how much each glass will have.But wait, let me make sure I'm approaching this correctly. Division is definitely the right operation here because we're dealing with equal distribution. If I have a total quantity and I want to split it into a certain number of equal parts, division is the way to go. So, 502.75 divided by 21 should give me the amount per glass.Now, let's think about how to perform this division. 502.75 is a decimal number, and 21 is a whole number. I can set this up as a long division problem. Alternatively, I could use a calculator for precision, but since I'm trying to understand the process, doing it manually might be better for learning.So, let's set up the division: 502.75 ÷ 21.First, I'll consider how many times 21 fits into 502.75. To make this easier, I can think of 502.75 as 50275 hundredths, which might help with the decimal placement later.But perhaps it's simpler to just proceed with the division step by step.21 goes into 50 twice because 21 times 2 is 42, which is less than 50. 21 times 3 is 63, which is too big. So, I'll write down 2 as the first digit of the quotient.Then, I'll subtract 42 from 50, which gives me 8. Bring down the next digit, which is 2, making it 82.Now, how many times does 21 go into 82? Let's see: 21 times 3 is 63, 21 times 4 is 84. 84 is too big, so it goes in 3 times. So, I'll write down 3 next to the 2, making the quotient so far 23.Subtract 63 from 82, which leaves me with 19. Bring down the next digit, which is 7, making it 197.Now, how many times does 21 go into 197? Let's see: 21 times 9 is 189, and 21 times 10 is 210, which is too big. So, it goes in 9 times. I'll write down 9 next to the 23, making the quotient 239.Subtract 189 from 197, which leaves me with 8. Bring down the next digit, which is 5, making it 85.Now, how many times does 21 go into 85? 21 times 4 is 84, which is less than 85, and 21 times 5 is 105, which is too big. So, it goes in 4 times. I'll write down 4 next to the 239, making the quotient 2394.Subtract 84 from 85, which leaves me with 1. Since there are no more digits to bring down, I can add a decimal point and a zero to continue the division.So, I'll write down a decimal point after 2394 and bring down a zero, making it 10.How many times does 21 go into 10? It doesn't, so I'll write down 0 and bring down another zero, making it 100.Now, how many times does 21 go into 100? 21 times 4 is 84, and 21 times 5 is 105, which is too big. So, it goes in 4 times. I'll write down 4 next to the 0, making the quotient 2394.4.Subtract 84 from 100, which leaves me with 16. Bring down another zero, making it 160.How many times does 21 go into 160? 21 times 7 is 147, and 21 times 8 is 168, which is too big. So, it goes in 7 times. I'll write down 7 next to the 4, making the quotient 2394.47.Subtract 147 from 160, which leaves me with 13. Bring down another zero, making it 130.How many times does 21 go into 130? 21 times 6 is 126, and 21 times 7 is 147, which is too big. So, it goes in 6 times. I'll write down 6 next to the 7, making the quotient 2394.476.Subtract 126 from 130, which leaves me with 4. Bring down another zero, making it 40.How many times does 21 go into 40? 21 times 1 is 21, and 21 times 2 is 42, which is too big. So, it goes in 1 time. I'll write down 1 next to the 6, making the quotient 2394.4761.Subtract 21 from 40, which leaves me with 19. Bring down another zero, making it 190.How many times does 21 go into 190? 21 times 9 is 189, and 21 times 10 is 210, which is too big. So, it goes in 9 times. I'll write down 9 next to the 1, making the quotient 2394.47619.Subtract 189 from 190, which leaves me with 1. At this point, I notice that I'm starting to repeat the earlier steps, as I had a remainder of 1 before. This suggests that the decimal is repeating.So, putting it all together, the quotient is approximately 23.94047619... and so on. Since the decimal repeats, I can indicate this by placing a bar over the repeating digits. However, for practical purposes, I might round this to a reasonable number of decimal places, say two or three, depending on the context.In this case, since we're dealing with pints of orange juice, and typically measurements are taken to two decimal places, I'll round 23.94047619... to 23.94 pints per glass.But just to double-check, let's multiply 21 glasses by 23.94 pints per glass to see if it gets us back to approximately 502.75 pints.21 * 23.94 = ?First, 20 * 23.94 = 478.8Then, 1 * 23.94 = 23.94Adding them together: 478.8 + 23.94 = 502.74Hmm, that's very close to 502.75, but not exact. The slight difference is due to rounding. If I use more decimal places, say 23.94047619, and multiply by 21, I should get exactly 502.75.Let's check:23.94047619 * 21 = ?23.94047619 * 20 = 478.809523823.94047619 * 1 = 23.94047619Adding them together: 478.8095238 + 23.94047619 = 502.75Perfect, so 23.94047619 pints per glass multiplied by 21 glasses gives us exactly 502.75 pints.Therefore, the exact amount per glass is 23.94047619 pints, which can be rounded to 23.94 pints for practical purposes.I think that's a thorough breakdown of how to approach and solve this problem. It involved setting up the division, performing long division step by step, handling the decimal points, recognizing repeating decimals, and verifying the result by multiplication. This ensures accuracy and understanding of the process.

answer：Okay, so I need to find the standard equation of a circle. Hmm, I remember that the standard form of a circle's equation is something like (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Let me make sure I got that right. Yeah, I think that's correct. Alright, the problem says the center is C(1, -2). So, h is 1 and k is -2. That means in the equation, it should be (x - 1)^2 + (y - (-2))^2. Wait, subtracting a negative is like adding, so that would be (x - 1)^2 + (y + 2)^2. Got it so far.Now, the radius is given as 3. So, r is 3, and r squared is 9. Putting it all together, the equation should be (x - 1)^2 + (y + 2)^2 = 9. Let me double-check that. If I plug in the center coordinates into the equation, it should satisfy the equation. So, plugging x=1 and y=-2, we get (1 - 1)^2 + (-2 + 2)^2 = 0 + 0 = 0, which is less than 9. Wait, that doesn't seem right. Shouldn't the center satisfy the equation?Oh, no, I think I made a mistake. The center is the point (1, -2), and in the standard equation, when you plug in the center coordinates, you should get the radius squared on the right side. So, actually, (1 - 1)^2 + (-2 + 2)^2 = 0 + 0 = 0, which is not equal to 9. That means the center isn't on the circle itself, which makes sense because the center is the center, not a point on the circumference. So, I think my equation is correct after all.Let me visualize it. The circle is centered at (1, -2) and has a radius of 3. So, it should extend 3 units up, down, left, and right from the center. That would mean the circle touches points like (1 + 3, -2) = (4, -2), (1 - 3, -2) = (-2, -2), (1, -2 + 3) = (1, 1), and (1, -2 - 3) = (1, -5). If I plug in one of these points into the equation, it should satisfy it.Let's take (4, -2). Plugging into the equation: (4 - 1)^2 + (-2 + 2)^2 = (3)^2 + (0)^2 = 9 + 0 = 9, which matches the right side. Good. How about (1, 1)? (1 - 1)^2 + (1 + 2)^2 = 0 + (3)^2 = 9. That works too. And (1, -5): (1 - 1)^2 + (-5 + 2)^2 = 0 + (-3)^2 = 9. Perfect.Wait, what if I pick a point that's not on the circle? Like (1, -2), the center. Plugging in: (1 - 1)^2 + (-2 + 2)^2 = 0 + 0 = 0, which is less than 9. That means the center is inside the circle, which is correct because the radius is 3. So, everything checks out.I think I've got it right. The standard equation is (x - 1)^2 + (y + 2)^2 = 9. I don't see any mistakes in my reasoning. It makes sense because the standard form directly uses the center coordinates and the radius squared. So, I'm confident that's the correct equation.

answer：Alright, so I have these two combinatorial sums to work on. Let me start with part (a). The problem is to prove that the alternating sum of even binomial coefficients equals (2^{n / 2} cos frac{n pi}{4}). Hmm, okay.I remember that binomial coefficients can be related to complex numbers, especially when dealing with alternating signs. Maybe I can use the binomial theorem with a complex number. Let me recall that ((1 + i)^n) can be expanded using the binomial theorem, and since (i) is involved, it might relate to trigonometric functions because of Euler's formula.So, let's write out ((1 + i)^n). Using the binomial theorem, this would be the sum from (k = 0) to (n) of (binom{n}{k} i^k). Now, since (i^k) cycles every four powers: (i^0 = 1), (i^1 = i), (i^2 = -1), (i^3 = -i), (i^4 = 1), and so on. So, the even powers of (i) will be real numbers, and the odd powers will be imaginary.Therefore, the real part of ((1 + i)^n) is the sum of the even binomial coefficients multiplied by ((-1)^{k/2}), which is exactly the sum we're supposed to find: (C_n^0 - C_n^2 + C_n^4 - ldots). The imaginary part would be the sum of the odd binomial coefficients multiplied by (i(-1)^{(k-1)/2}), which is related to part (b).Now, let's compute ((1 + i)^n) using polar form. The complex number (1 + i) has a magnitude of (sqrt{2}) and an angle of (pi/4). So, in polar form, (1 + i = sqrt{2} left( cos frac{pi}{4} + i sin frac{pi}{4} right)). Raising this to the power of (n) gives:[(1 + i)^n = (sqrt{2})^n left( cos frac{npi}{4} + i sin frac{npi}{4} right) = 2^{n/2} left( cos frac{npi}{4} + i sin frac{npi}{4} right)]So, the real part of this expression is (2^{n/2} cos frac{npi}{4}), which matches the sum we're supposed to prove. Therefore, part (a) is proven.Moving on to part (b), we need to calculate the sum (C_n^1 - C_n^3 + C_n^5 - ldots). From the previous part, we saw that the imaginary part of ((1 + i)^n) is related to this sum. Specifically, the imaginary part is:[2^{n/2} sin frac{npi}{4}]But in the expansion of ((1 + i)^n), the imaginary part is also equal to:[C_n^1 - C_n^3 + C_n^5 - ldots]However, each term in the imaginary part is multiplied by (i), so to get the actual sum, we need to divide by (i). But since (i) is a unit imaginary number, dividing by (i) is equivalent to multiplying by (-i). However, since we're only interested in the magnitude, the sum itself is equal to:[2^{n/2} sin frac{npi}{4}]Wait, but let me double-check that. If the imaginary part is (2^{n/2} sin frac{npi}{4}), and the imaginary part of the expansion is (i(C_n^1 - C_n^3 + C_n^5 - ldots)), then equating these gives:[i(C_n^1 - C_n^3 + C_n^5 - ldots) = 2^{n/2} sin frac{npi}{4}]Therefore, to solve for the sum, we can divide both sides by (i):[C_n^1 - C_n^3 + C_n^5 - ldots = 2^{n/2} sin frac{npi}{4} / i]But dividing by (i) is the same as multiplying by (-i), so:[C_n^1 - C_n^3 + C_n^5 - ldots = -i cdot 2^{n/2} sin frac{npi}{4}]Hmm, that introduces an imaginary unit, which doesn't make sense because the sum of real numbers should be real. I must have made a mistake here.Let me think again. The imaginary part of ((1 + i)^n) is indeed (2^{n/2} sin frac{npi}{4}), and this is equal to the sum (C_n^1 - C_n^3 + C_n^5 - ldots) multiplied by (i). Therefore, to get the sum, we should equate the imaginary parts:[C_n^1 - C_n^3 + C_n^5 - ldots = 2^{n/2} sin frac{npi}{4}]Wait, no, because the imaginary part is (i) times the sum, so actually:[i cdot (C_n^1 - C_n^3 + C_n^5 - ldots) = 2^{n/2} sin frac{npi}{4}]Therefore, solving for the sum:[C_n^1 - C_n^3 + C_n^5 - ldots = frac{2^{n/2} sin frac{npi}{4}}{i}]But (frac{1}{i} = -i), so:[C_n^1 - C_n^3 + C_n^5 - ldots = -i cdot 2^{n/2} sin frac{npi}{4}]This still leaves an imaginary unit, which contradicts the fact that the sum is real. I must be missing something here.Wait, perhaps I should consider that the imaginary part is equal to the sum multiplied by (i), so to get the sum, I should take the imaginary part and divide by (i), but since the sum is real, the imaginary unit should cancel out. Maybe I need to take the real part of the expression after dividing by (i).Alternatively, perhaps I should consider the complex conjugate. Let me think differently.If I take ((1 - i)^n), its expansion would be similar but with alternating signs for the imaginary parts. Maybe I can use both ((1 + i)^n) and ((1 - i)^n) to separate the real and imaginary parts.Let me compute ((1 + i)^n + (1 - i)^n). This would give me twice the real part, which is (2 cdot 2^{n/2} cos frac{npi}{4}), matching part (a). Similarly, ((1 + i)^n - (1 - i)^n) would give me twice the imaginary part, which is (2i cdot 2^{n/2} sin frac{npi}{4}).Therefore, the imaginary part is (2^{n/2} sin frac{npi}{4}), and since the imaginary part of ((1 + i)^n) is (i) times the sum (C_n^1 - C_n^3 + C_n^5 - ldots), we have:[i cdot (C_n^1 - C_n^3 + C_n^5 - ldots) = 2^{n/2} sin frac{npi}{4}]Thus, solving for the sum:[C_n^1 - C_n^3 + C_n^5 - ldots = frac{2^{n/2} sin frac{npi}{4}}{i} = -i cdot 2^{n/2} sin frac{npi}{4}]But this still leaves an imaginary unit, which doesn't make sense. I must be making a mistake in handling the imaginary unit.Wait, perhaps I should consider that the sum (C_n^1 - C_n^3 + C_n^5 - ldots) is actually equal to the imaginary part divided by (i), but since the imaginary part is a real number, dividing by (i) introduces a factor of (-i), which would make the sum imaginary. This contradicts the fact that the sum is real.I think I need to reconsider my approach. Maybe instead of directly equating the imaginary parts, I should use another method, such as generating functions or considering the expansion of ((1 + i)^n) and ((1 - i)^n) together.Let me try adding and subtracting ((1 + i)^n) and ((1 - i)^n):[(1 + i)^n + (1 - i)^n = 2 cdot text{Re}((1 + i)^n) = 2 cdot 2^{n/2} cos frac{npi}{4}][(1 + i)^n - (1 - i)^n = 2i cdot text{Im}((1 + i)^n) = 2i cdot 2^{n/2} sin frac{npi}{4}]From the first equation, we get the sum of even binomial coefficients with alternating signs, which is part (a). From the second equation, we get the sum of odd binomial coefficients with alternating signs multiplied by (2i). Therefore, dividing both sides by (2i), we get:[frac{(1 + i)^n - (1 - i)^n}{2i} = 2^{n/2} sin frac{npi}{4}]But the left-hand side is also equal to the sum (C_n^1 - C_n^3 + C_n^5 - ldots), because when you expand ((1 + i)^n - (1 - i)^n), the even terms cancel out, and the odd terms double up, with the imaginary unit (i) factored out.Therefore, the sum (C_n^1 - C_n^3 + C_n^5 - ldots) is equal to (2^{n/2} sin frac{npi}{4}).Wait, but earlier I thought there was an issue with the imaginary unit. Maybe I was overcomplicating it. Since the left-hand side is the sum multiplied by (i), and the right-hand side is (2^{n/2} sin frac{npi}{4}), then dividing both sides by (i) gives the sum as (-i cdot 2^{n/2} sin frac{npi}{4}), which is imaginary. But the sum itself is real, so perhaps I need to take the real part or consider the absolute value.Wait, no, because when I take ((1 + i)^n - (1 - i)^n), the result is purely imaginary, as the real parts cancel out and the imaginary parts double. Therefore, dividing by (2i) gives a real number, which is the sum (C_n^1 - C_n^3 + C_n^5 - ldots). So, in fact, the sum is equal to (2^{n/2} sin frac{npi}{4}), which is real.Yes, that makes sense. So, the sum is indeed (2^{n/2} sin frac{npi}{4}).Okay, so to summarize:For part (a), by expanding ((1 + i)^n) and taking the real part, we find that the alternating sum of even binomial coefficients equals (2^{n/2} cos frac{npi}{4}).For part (b), by considering the imaginary part of ((1 + i)^n) and using the relationship between the expansion and the sine function, we find that the alternating sum of odd binomial coefficients equals (2^{n/2} sin frac{npi}{4}).I think that clears up my confusion. The key was to recognize that the imaginary part, when properly handled, gives the desired real sum without any imaginary components.

answer：First, I need to determine the total volume of the mixture after adding 20% water to 1 liter of milk.20% of 1 liter is 0.2 liters.So, the total volume of the mixture is 1 liter (milk) + 0.2 liters (water) = 1.2 liters.Next, I'll calculate the selling price per liter of the mixture.The selling price for 1.2 liters is ₹15.Therefore, the selling price per liter is ₹15 / 1.2 liters = ₹12.50 per liter.Now, I'll determine the cost price per liter of milk.The cost price for 1 liter of milk is ₹12.Finally, I'll calculate the percentage gain.Profit per liter = Selling Price per liter - Cost Price per literProfit per liter = ₹12.50 - ₹12.00 = ₹0.50Percentage Gain = (Profit per liter / Cost Price per liter) * 100Percentage Gain = (₹0.50 / ₹12.00) * 100 ≈ 4.17%So, the percentage gain is approximately 4.17%.

answer：Okay, so I have this problem here: the function is given as f(x) = a sin x + b x³ + 4, where a and b are real numbers. I need to find the value of f(2016) + f(-2016) + f'(2017) - f'(-2017). The options are A: 0, B: 2016, C: 2017, D: 8. Hmm, let me try to figure this out step by step.First, I remember that to find f'(x), I need to take the derivative of f(x). So, let me compute that. The derivative of a sin x is a cos x, right? Because the derivative of sin x is cos x. Then, the derivative of b x³ is 3b x², using the power rule. And the derivative of the constant term, 4, is 0. So, putting it all together, f'(x) should be a cos x + 3b x².Okay, so f'(x) = a cos x + 3b x². Now, I need to evaluate f'(2017) and f'(-2017). Let me write those down:f'(2017) = a cos(2017) + 3b (2017)² f'(-2017) = a cos(-2017) + 3b (-2017)²Wait, I remember that cosine is an even function, which means cos(-x) = cos x. So, cos(-2017) is equal to cos(2017). Also, (-2017)² is the same as (2017)² because squaring a negative number makes it positive. So, f'(-2017) simplifies to a cos(2017) + 3b (2017)², which is exactly the same as f'(2017). Therefore, f'(2017) - f'(-2017) would be [a cos(2017) + 3b (2017)²] - [a cos(2017) + 3b (2017)²] = 0. So, that part of the expression cancels out. That's helpful; now I only need to compute f(2016) + f(-2016).Let me write down f(2016) and f(-2016):f(2016) = a sin(2016) + b (2016)³ + 4 f(-2016) = a sin(-2016) + b (-2016)³ + 4Hmm, I remember that sine is an odd function, so sin(-x) = -sin x. So, sin(-2016) is equal to -sin(2016). Also, (-2016)³ is equal to - (2016)³ because when you cube a negative number, it remains negative.So, substituting these into f(-2016):f(-2016) = a (-sin(2016)) + b (- (2016)³) + 4 = -a sin(2016) - b (2016)³ + 4Now, let's add f(2016) and f(-2016) together:f(2016) + f(-2016) = [a sin(2016) + b (2016)³ + 4] + [-a sin(2016) - b (2016)³ + 4]Let me simplify this term by term:- The a sin(2016) and -a sin(2016) cancel each other out. - Similarly, b (2016)³ and -b (2016)³ also cancel each other out. - The constants 4 and 4 add up to 8.So, f(2016) + f(-2016) = 8.Earlier, we found that f'(2017) - f'(-2017) = 0. Therefore, the entire expression f(2016) + f(-2016) + f'(2017) - f'(-2017) simplifies to 8 + 0 = 8.Looking back at the options, D is 8. So, the answer should be D.Wait, let me just double-check my steps to make sure I didn't make any mistakes. I took the derivative correctly, right? The derivative of sin x is cos x, so a sin x becomes a cos x. The derivative of b x³ is 3b x², and the derivative of 4 is 0. So, f'(x) is correct.Then, for f'(2017) and f'(-2017), I used the fact that cosine is even, so cos(-x) = cos x, and (-x)² is x². So, f'(2017) and f'(-2017) are equal, which makes their difference zero. That seems right.For f(2016) and f(-2016), I used the fact that sine is odd, so sin(-x) = -sin x, and (-x)³ is -x³. Then, when I added f(2016) and f(-2016), the terms with a and b canceled out, leaving me with 4 + 4 = 8. That makes sense.So, yeah, I think my steps are correct. The answer is 8, which is option D.**Final Answer**boxed{D}

answer：Okay, so I need to evaluate this infinite series: 2/4^1 + 4/4^2 + 6/4^3 + ... + 2k/4^k + ... Hmm, it looks like each term has a numerator that's an even number increasing by 2 each time, and the denominator is 4 raised to the power of the term's position. First, maybe I can write this series in a more general form. Let me denote the sum as S. So,S = 2/4^1 + 4/4^2 + 6/4^3 + 8/4^4 + ... I notice that each term is of the form (2k)/4^k where k starts from 1 and goes to infinity. So, S can be written as the sum from k=1 to infinity of (2k)/4^k. Hmm, this seems similar to a geometric series, but it's not exactly the same because of the 2k in the numerator. I remember that there are techniques to handle series with terms that involve k, like arithmetic-geometric series. Maybe I can manipulate this series to find a pattern or use a known formula.Let me try multiplying both sides by 1/4 to see if that helps. If I do that, I get:(1/4)S = 2/4^2 + 4/4^3 + 6/4^4 + 8/4^5 + ... Now, if I subtract this new series from the original series S, perhaps some terms will cancel out. Let's write that out:S - (1/4)S = (2/4^1 + 4/4^2 + 6/4^3 + 8/4^4 + ...) - (2/4^2 + 4/4^3 + 6/4^4 + 8/4^5 + ...)When I subtract these, most of the terms should cancel except for the first term of S and the terms that don't have a corresponding term in (1/4)S. Let's see:S - (1/4)S = 2/4^1 + (4/4^2 - 2/4^2) + (6/4^3 - 4/4^3) + (8/4^4 - 6/4^4) + ... Simplifying each of these terms:= 2/4 + (2)/4^2 + (2)/4^3 + (2)/4^4 + ... So, the left side is (3/4)S, and the right side is a geometric series where each term is 2/4^k starting from k=1. Let me write that:(3/4)S = 2/4 + 2/4^2 + 2/4^3 + 2/4^4 + ... This is a geometric series with the first term a = 2/4 = 1/2 and common ratio r = 1/4. The sum of an infinite geometric series is a / (1 - r), provided that |r| < 1, which it is here since r = 1/4.So, the sum on the right side is (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) * (4/3) = 2/3.Therefore, (3/4)S = 2/3. To find S, I can multiply both sides by 4/3:S = (2/3) * (4/3) = 8/9.Wait, let me double-check that. If (3/4)S = 2/3, then S = (2/3) * (4/3) = 8/9. Yeah, that seems right.So, the sum of the series is 8/9. I think that makes sense because each term is getting smaller and smaller, so the series should converge to a finite value, and 8/9 is a reasonable number for that.I don't see any mistakes in my steps. I transformed the series by multiplying by 1/4, subtracted to get a geometric series, summed it up, and solved for S. Everything seems to check out.**Final Answer**The sum of the series is boxed{dfrac{8}{9}}.