💡First, I need to substitute the given complex numbers into the expression ( A - B + C - D ).Next, I'll handle the subtraction of complex numbers by distributing the negative sign appropriately.Then, I'll combine like terms, separating the real and imaginary parts.Finally, I'll simplify the expression to find the final result.

💡First, I need to convert all the time durations into hours to ensure consistency in the final calculation.The bus ride takes 75 minutes. To convert this into hours:75 minutes ÷ 60 = 1.25 hours.Next, Luke walks for 15 minutes to the train center. Converting this into hours:15 minutes ÷ 60 = 0.25 hours.He waits twice the walking time for the train to arrive, which is 15 minutes × 2 = 30 minutes. Converting this into hours:30 minutes ÷ 60 = 0.5 hours.Finally, the train ride to London takes 6 hours.Adding up all the durations:1.25 hours (bus) + 0.25 hours (walk) + 0.5 hours (waiting) + 6 hours (train) = 8 hours.Therefore, Luke's entire trip takes 8 hours.

💡Okay, so I have this vector problem here. It says that the sum of vectors a, b, and c is zero. So, a + b + c = 0. The magnitudes are given: |a| is 1, and |b| and |c| are both 2. I need to find the value of the expression a·b + b·c + c·a.Hmm, let me think. I remember that when dealing with vectors, sometimes squaring both sides of an equation can help because it relates to the dot product. Since a + b + c = 0, maybe I can square both sides of this equation to get some useful information.So, if I square both sides, I get (a + b + c)·(a + b + c) = 0·0, which is 0. Expanding the left side, it should be a·a + b·b + c·c + 2(a·b + a·c + b·c) = 0.Let me write that out:|a|² + |b|² + |c|² + 2(a·b + a·c + b·c) = 0.I know the magnitudes of the vectors, so I can substitute those values in. |a| is 1, so |a|² is 1. |b| and |c| are both 2, so |b|² and |c|² are both 4. Plugging those in:1 + 4 + 4 + 2(a·b + a·c + b·c) = 0.Adding those numbers up: 1 + 4 is 5, plus another 4 is 9. So:9 + 2(a·b + a·c + b·c) = 0.Now, I can solve for the expression I need. Let me subtract 9 from both sides:2(a·b + a·c + b·c) = -9.Then, divide both sides by 2:a·b + a·c + b·c = -9/2.Wait, that seems straightforward. But let me double-check to make sure I didn't make a mistake. So, starting from the beginning: a + b + c = 0, so squaring both sides gives the sum of the squares plus twice the sum of the dot products equals zero. Plugging in the magnitudes, I get 1 + 4 + 4, which is 9, plus twice the dot products equals zero. So, 9 + 2*(dot products) = 0, so 2*(dot products) = -9, hence dot products = -9/2.Hmm, that seems right. I don't think I made any calculation errors. So, the value of a·b + b·c + c·a is -9/2.**Final Answer**The value of overrightarrow{a}cdot overrightarrow{b}+overrightarrow{b}cdot overrightarrow{c}+overrightarrow{c}cdot overrightarrow{a} is boxed{-dfrac{9}{2}}.

💡First, I need to choose a president from the 10 people. There are 10 possible choices for this role.After selecting the president, I have 9 people remaining to form the 3-person committee. Since the order of the committee members doesn't matter, I'll use combinations to determine the number of ways to choose 3 people from these 9.The combination formula for selecting 3 people out of 9 is (binom{9}{3}), which equals 84.Finally, to find the total number of ways to choose both the president and the committee, I multiply the number of choices for the president by the number of ways to form the committee: (10 times 84 = 840).

💡Okay, so I need to figure out which digit is in the 573rd place after the decimal point in the decimal representation of 11 divided by 37. Hmm, I remember that when you divide two numbers, sometimes the decimal repeats. So, first, I should probably do the division of 11 by 37 to see what the decimal looks like.Let me try dividing 11 by 37. I know that 37 goes into 11 zero times, so I write down 0. and then proceed with the division. 37 goes into 110 two times because 2 times 37 is 74. Subtracting 74 from 110 gives me 36. Then, I bring down another 0, making it 360. 37 goes into 360 nine times because 9 times 37 is 333. Subtracting 333 from 360 leaves me with 27. Bringing down another 0 makes it 270. 37 goes into 270 seven times because 7 times 37 is 259. Subtracting 259 from 270 gives me 11. Now, I'm back to where I started with 11, so the decimal repeats.So, the decimal representation of 11/37 is 0.297297297..., where "297" repeats indefinitely. That means the decimal is periodic with a period of 3 digits. Now, I need to find the 573rd digit after the decimal point.Since the decimal repeats every 3 digits, I can think of the digits as being in cycles of 3. To find out which digit corresponds to the 573rd place, I can use modular arithmetic. Specifically, I can divide 573 by 3 and look at the remainder. If there's no remainder, it means the 573rd digit is the last digit of the repeating cycle.Let me calculate 573 divided by 3. 3 times 191 is 573, so 573 divided by 3 is exactly 191 with no remainder. That means the 573rd digit is the last digit of the repeating sequence "297". The last digit is 7.Wait, let me double-check that. If I list out the positions:1: 22: 93: 74: 25: 96: 7...So, every multiple of 3 (3rd, 6th, 9th, etc.) is a 7. Since 573 is a multiple of 3, the 573rd digit should indeed be 7.I think that makes sense. So, the digit in the 573rd place after the decimal point in the decimal representation of 11/37 is 7.

💡Alright, so I have this integral to calculate: the integral of 1 over (z minus a) to the power of n, around a circle L that's centered at a with radius R, and it's being traversed counterclockwise. Okay, let's break this down step by step.First, I remember that when dealing with integrals around circles in complex analysis, it's often helpful to parameterize the circle. So, if the circle is centered at a with radius R, any point z on this circle can be written as a plus R times e to the i times phi, where phi ranges from 0 to 2pi. That makes sense because e to the i times phi gives us points on the unit circle, and multiplying by R scales it to radius R, and then adding a shifts it to be centered at a.So, z equals a plus R e^{i phi}. Now, to find dz, I need to take the derivative of z with respect to phi. The derivative of a is zero, the derivative of R e^{i phi} is i R e^{i phi}, so dz equals i R e^{i phi} d phi. Got that.Now, substituting z and dz into the integral, I get the integral from 0 to 2pi of (i R e^{i phi} d phi) divided by (z minus a)^n. But z minus a is just R e^{i phi}, so (z minus a)^n is (R e^{i phi})^n. Therefore, the integrand becomes (i R e^{i phi}) divided by (R^n e^{i n phi}) times d phi.Simplifying that, I can factor out the constants: i R^{1 - n} times e^{i phi (1 - n)} d phi. So the integral becomes i R^{1 - n} times the integral from 0 to 2pi of e^{i phi (1 - n)} d phi.Now, this integral depends on the value of n. If n is 1, then the exponent becomes e^{i phi (1 - 1)} which is e^0, which is 1. So the integral becomes i R^{0} times the integral from 0 to 2pi of 1 d phi, which is just 2pi. So in that case, the integral is 2pi i.But what if n is not 1? Then the exponent is e^{i phi (1 - n)}, which is e^{i k phi} where k is an integer not equal to zero. I remember that the integral of e^{i k phi} over 0 to 2pi is zero because the positive and negative parts cancel out over the full circle. So, in that case, the integral is zero.Wait, let me make sure I'm not missing anything here. Is there any condition on R or anything else? Well, the circle is centered at a, and we're integrating around it, so as long as the function is analytic inside the circle, which it is except when n is 1, because then we have a pole at z equals a. But since the circle is centered at a, that point is on the contour, so actually, for n equals 1, we have a simple pole at z equals a, and by the residue theorem, the integral is 2pi i times the residue, which is 1 in this case. So that matches up with what I got earlier.For other values of n, if n is greater than 1, then (z - a)^n is analytic everywhere inside and on the contour, so by Cauchy's theorem, the integral is zero. If n is less than 1, say n is zero, then we're integrating 1 over (z - a)^0, which is just 1, so the integral is the integral of 1 around the circle, which is zero because it's a closed contour and the integral of a constant is zero. If n is negative, say n equals -1, then we have (z - a)^{-1}, which is similar to the n equals 1 case but with a negative exponent. Wait, no, actually, for n equals -1, it's (z - a)^{-1}, which is the same as 1/(z - a), which is analytic everywhere except at z equals a, so again, by the residue theorem, the integral would be 2pi i times the residue at z equals a, which is 1. But wait, in our original problem, n is an integer, so n could be positive or negative. Hmm, but in the problem statement, n is an integer, but it doesn't specify whether it's positive or negative. So, for n equals 1, it's 2pi i, for n equals -1, it's also 2pi i, but for other values, it's zero.Wait, no, actually, for n equals -1, (z - a)^{-1} is 1/(z - a), which is the same as the n equals 1 case, so the integral is 2pi i. But for n equals 2, it's (z - a)^2, which is analytic everywhere, so the integral is zero. For n equals -2, it's 1/(z - a)^2, which is analytic everywhere except at z equals a, but the residue at z equals a for 1/(z - a)^2 is zero because the residue is the coefficient of (z - a)^{-1}, which is zero in this case. So, the integral would be zero.Wait, hold on, I think I made a mistake there. For n equals -1, the function is 1/(z - a), which has a simple pole at z equals a, so the residue is 1, hence the integral is 2pi i. For n equals -2, the function is 1/(z - a)^2, which has a pole of order 2 at z equals a. The residue for a pole of order m is given by the limit as z approaches a of the (m-1)th derivative of (z - a)^m f(z) divided by (m - 1)!. So, for m equals 2, it's the limit as z approaches a of the first derivative of (z - a)^2 * (1/(z - a)^2) divided by 1!, which simplifies to the derivative of 1, which is zero. So, the residue is zero, hence the integral is zero.Similarly, for any n not equal to 1, the integral is zero. So, to summarize, the integral is 2pi i when n equals 1, and zero otherwise.But wait, let me double-check for n equals 0. If n equals 0, then the integrand is 1/(z - a)^0, which is 1. So, integrating 1 around the circle, which is a closed contour, should give zero because the integral of a constant around a closed contour is zero. That makes sense.What about n equals 2? Then the integrand is 1/(z - a)^2, which is analytic everywhere except at z equals a. But since it's a pole of order 2, and as we saw earlier, the residue is zero, so the integral is zero.Similarly, for n equals -1, it's 1/(z - a), which has a simple pole with residue 1, so the integral is 2pi i. For n equals -2, it's 1/(z - a)^2, which has a pole of order 2 with residue zero, so the integral is zero.Okay, so it seems consistent that only when n equals 1 do we get a non-zero integral, and in that case, it's 2pi i.I think I've covered all the cases. So, the final answer is that the integral is 2pi i when n equals 1, and zero otherwise.