answer：Okay, so I have this problem here about vectors. It says that given two unit vectors, a and b, which satisfy the condition that the magnitude of a + 2b is equal to the square root of 3. I need to find the magnitude of a - 2b, and the options are given as A: square root of 2, B: square root of 5, C: square root of 3, and D: square root of 7.Alright, let me start by recalling some vector properties. Since a and b are unit vectors, their magnitudes are both 1. That means |a| = 1 and |b| = 1. The problem gives me |a + 2b| = sqrt(3). I need to find |a - 2b|.I remember that the magnitude squared of a vector is equal to the dot product of the vector with itself. So, |v|^2 = v · v. Maybe I can use this property to find the dot product of a and b, which might help me find |a - 2b|.Let me write down the given condition:|a + 2b| = sqrt(3)If I square both sides, I get:|a + 2b|^2 = (sqrt(3))^2|a + 2b|^2 = 3Now, expanding the left side using the dot product:(a + 2b) · (a + 2b) = 3Let me compute this dot product step by step. The dot product distributes over addition, so:a · a + a · 2b + 2b · a + 2b · 2b = 3Simplify each term:a · a is |a|^2, which is 1 because a is a unit vector.Similarly, 2b · 2b is 4|b|^2, which is 4*1 = 4.Then, the cross terms: a · 2b and 2b · a. Since the dot product is commutative, a · 2b = 2(a · b) and similarly 2b · a = 2(a · b). So together, they add up to 4(a · b).Putting it all together:1 + 4(a · b) + 4 = 3Combine like terms:1 + 4 + 4(a · b) = 35 + 4(a · b) = 3Subtract 5 from both sides:4(a · b) = 3 - 54(a · b) = -2Divide both sides by 4:a · b = -2/4a · b = -1/2Okay, so the dot product of a and b is -1/2. That tells me something about the angle between them. The dot product formula is a · b = |a||b|cos(theta), where theta is the angle between them. Since both |a| and |b| are 1, this simplifies to cos(theta) = -1/2. So the angle between a and b is 120 degrees because cos(120°) = -1/2. But maybe I don't need that information directly.Now, I need to find |a - 2b|. Let me use the same approach as before. Compute the magnitude squared and then take the square root.|a - 2b|^2 = (a - 2b) · (a - 2b)Again, expanding this dot product:a · a - a · 2b - 2b · a + (2b) · (2b)Simplify each term:a · a is |a|^2 = 1.- a · 2b is -2(a · b).- 2b · a is another -2(a · b).(2b) · (2b) is 4|b|^2 = 4*1 = 4.Putting it all together:1 - 2(a · b) - 2(a · b) + 4Combine like terms:1 + 4 - 4(a · b)Which is:5 - 4(a · b)We already found that a · b = -1/2, so substitute that in:5 - 4*(-1/2) = 5 + 2 = 7Therefore, |a - 2b|^2 = 7, so |a - 2b| = sqrt(7).Looking back at the options, D is sqrt(7). So the answer should be D.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from |a + 2b| = sqrt(3), squared to get 3. Expanded to 1 + 4(a · b) + 4 = 3, which simplifies to 5 + 4(a · b) = 3, leading to a · b = -1/2. Then, for |a - 2b|, squared is 1 - 4(a · b) + 4, which is 5 - 4(a · b). Plugging in a · b = -1/2 gives 5 + 2 = 7, so sqrt(7). Yeah, that seems correct.I think I did everything right. So the answer is D: sqrt(7).**Final Answer**boxed{D}

answer：First, I need to determine the total score for Korean and Mathematics based on their average.Given that the average score for Korean and Mathematics is 86 points, I can calculate the total score by multiplying the average by the number of subjects:Total score for Korean and Mathematics = 86 points × 2 = 172 pointsNext, I need to find the average score for the three subjects, including English. The English score is provided as 98 points.To find the average score of the three subjects, I will add the total score for Korean and Mathematics to the English score and then divide by the number of subjects:Average score = (Total score for Korean and Mathematics + English score) / 3Average score = (172 points + 98 points) / 3Average score = 270 points / 3Average score = 90 pointsTherefore, the average score of the three subjects is 90 points.

answer：First, I need to determine the areas of the four concentric circles with radii of 2, 4, 6, and 8. Using the formula for the area of a circle, ( A = pi r^2 ), I calculate each area:- The innermost circle (radius = 2): ( A_1 = pi times 2^2 = 4pi ).- The second circle (radius = 4): ( A_2 = pi times 4^2 = 16pi ).- The third circle (radius = 6): ( A_3 = pi times 6^2 = 36pi ).- The fourth circle (radius = 8): ( A_4 = pi times 8^2 = 64pi ).Next, I identify the areas of the black and white regions:- **Black Areas**: - The innermost circle: ( 4pi ). - The third ring (between radii 4 and 6): ( 36pi - 16pi = 20pi ). - Total black area: ( 4pi + 20pi = 24pi ).- **White Areas**: - The second ring (between radii 2 and 4): ( 16pi - 4pi = 12pi ). - The fourth ring (between radii 6 and 8): ( 64pi - 36pi = 28pi ). - Total white area: ( 12pi + 28pi = 40pi ).Finally, I calculate the ratio of the black area to the white area:[text{Ratio} = frac{24pi}{40pi} = frac{24}{40} = frac{3}{5}]

answer：Okay, so I need to multiply these two matrices together. Let me just write them down first to make sure I have them right. The first matrix is:[begin{pmatrix} 2 & 3 & -1 1 & -2 & 5 0 & 6 & 1 end{pmatrix}]And the second matrix is:[begin{pmatrix} 1 & 0 & 4 3 & 2 & -1 0 & 4 & -2 end{pmatrix}]Alright, so matrix multiplication involves taking the dot product of the rows of the first matrix with the columns of the second matrix. I remember that the resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix. Since both matrices are 3x3, the result should also be a 3x3 matrix.Let me start by recalling how the dot product works. For two vectors, you multiply corresponding components and then add them up. So, for example, if I have a row from the first matrix and a column from the second matrix, I multiply each element of the row by the corresponding element of the column and sum them all up. That gives me one element of the resulting matrix.Let me try to visualize this. The first element of the resulting matrix is obtained by multiplying the first row of the first matrix with the first column of the second matrix. So, let's compute that.First row of the first matrix: 2, 3, -1First column of the second matrix: 1, 3, 0So, the dot product is 2*1 + 3*3 + (-1)*0. Let me calculate that:2*1 = 23*3 = 9(-1)*0 = 0Adding them up: 2 + 9 + 0 = 11Okay, so the first element of the resulting matrix is 11.Now, moving on to the first row, second column of the resulting matrix. That means I need to take the first row of the first matrix and the second column of the second matrix.First row: 2, 3, -1Second column: 0, 2, 4Dot product: 2*0 + 3*2 + (-1)*4Calculating each term:2*0 = 03*2 = 6(-1)*4 = -4Adding them up: 0 + 6 - 4 = 2So, the first row, second column of the result is 2.Next, first row, third column. First row of the first matrix and third column of the second matrix.First row: 2, 3, -1Third column: 4, -1, -2Dot product: 2*4 + 3*(-1) + (-1)*(-2)Calculating:2*4 = 83*(-1) = -3(-1)*(-2) = 2Adding them up: 8 - 3 + 2 = 7So, the first row, third column is 7.Alright, first row done. Now, moving on to the second row of the resulting matrix. Let's compute each element.Second row, first column: second row of the first matrix and first column of the second matrix.Second row: 1, -2, 5First column: 1, 3, 0Dot product: 1*1 + (-2)*3 + 5*0Calculating:1*1 = 1(-2)*3 = -65*0 = 0Adding up: 1 - 6 + 0 = -5So, second row, first column is -5.Second row, second column: second row of the first matrix and second column of the second matrix.Second row: 1, -2, 5Second column: 0, 2, 4Dot product: 1*0 + (-2)*2 + 5*4Calculating:1*0 = 0(-2)*2 = -45*4 = 20Adding up: 0 - 4 + 20 = 16So, second row, second column is 16.Second row, third column: second row of the first matrix and third column of the second matrix.Second row: 1, -2, 5Third column: 4, -1, -2Dot product: 1*4 + (-2)*(-1) + 5*(-2)Calculating:1*4 = 4(-2)*(-1) = 25*(-2) = -10Adding up: 4 + 2 - 10 = -4So, second row, third column is -4.Alright, second row done. Now, onto the third row of the resulting matrix.Third row, first column: third row of the first matrix and first column of the second matrix.Third row: 0, 6, 1First column: 1, 3, 0Dot product: 0*1 + 6*3 + 1*0Calculating:0*1 = 06*3 = 181*0 = 0Adding up: 0 + 18 + 0 = 18So, third row, first column is 18.Third row, second column: third row of the first matrix and second column of the second matrix.Third row: 0, 6, 1Second column: 0, 2, 4Dot product: 0*0 + 6*2 + 1*4Calculating:0*0 = 06*2 = 121*4 = 4Adding up: 0 + 12 + 4 = 16So, third row, second column is 16.Third row, third column: third row of the first matrix and third column of the second matrix.Third row: 0, 6, 1Third column: 4, -1, -2Dot product: 0*4 + 6*(-1) + 1*(-2)Calculating:0*4 = 06*(-1) = -61*(-2) = -2Adding up: 0 - 6 - 2 = -8So, third row, third column is -8.Putting it all together, the resulting matrix is:First row: 11, 2, 7Second row: -5, 16, -4Third row: 18, 16, -8So, the product of the two matrices is:[begin{pmatrix} 11 & 2 & 7 -5 & 16 & -4 18 & 16 & -8 end{pmatrix}]I think that's it. Let me just double-check a couple of the calculations to make sure I didn't make any mistakes.For example, let's check the element in the second row, third column. It was calculated as 1*4 + (-2)*(-1) + 5*(-2) = 4 + 2 - 10 = -4. That seems correct.Another one: third row, third column. 0*4 + 6*(-1) + 1*(-2) = 0 - 6 - 2 = -8. That also looks right.I think I did it correctly. So, the final result is the matrix I wrote above.

answer：Alright, so I need to find the first year after 2010 where the sum of its digits equals 4. Hmm, okay, let's break this down step by step. First, I know that years are four-digit numbers, and since we're looking for years after 2010, we'll be dealing with years from 2011 onwards. The goal is to find the smallest year in this range where the sum of its digits is exactly 4.Let me think about how to approach this. Maybe I can start by considering the structure of a year. A year can be represented as ABCD, where A, B, C, and D are digits. For years after 2010, A is 2, and B is 0. So, the year looks like 20CD. Now, I need the sum of the digits A + B + C + D to equal 4. Since A is 2 and B is 0, the sum simplifies to 2 + 0 + C + D = 4. That means C + D should equal 2 because 2 + 0 + C + D = 4 implies C + D = 2.Okay, so I need to find the smallest year 20CD where C and D are digits such that their sum is 2. Let's list out the possible combinations for C and D:1. C = 0, D = 2 → Year: 20022. C = 1, D = 1 → Year: 20113. C = 2, D = 0 → Year: 2020Wait a minute, 2002 is actually before 2010, so it doesn't count. That leaves us with 2011 and 2020. Between these two, 2011 is the smaller year. But hold on, is 2011 the first year after 2010 where the sum of the digits equals 4? Let me double-check. In 2011, the digits are 2, 0, 1, and 1. Adding them up: 2 + 0 + 1 + 1 = 4. Yep, that works. Just to be thorough, let's check the years immediately after 2010 to ensure there isn't a smaller year that fits the criteria. - 2011: Sum = 4 (as calculated)- 2012: Sum = 2 + 0 + 1 + 2 = 5- 2013: Sum = 2 + 0 + 1 + 3 = 6- 2014: Sum = 2 + 0 + 1 + 4 = 7- 2015: Sum = 2 + 0 + 1 + 5 = 8- 2016: Sum = 2 + 0 + 1 + 6 = 9- 2017: Sum = 2 + 0 + 1 + 7 = 10- 2018: Sum = 2 + 0 + 1 + 8 = 11- 2019: Sum = 2 + 0 + 1 + 9 = 12- 2020: Sum = 2 + 0 + 2 + 0 = 4Okay, so after checking, 2011 is indeed the first year after 2010 where the sum of the digits equals 4. 2020 also fits the criteria, but it comes later. I think I've covered all the necessary steps and checked my work to ensure accuracy. It seems like 2011 is the correct answer.

answer：Okay, so I need to find the arithmetic mean of two fractions: 3/7 and 6/11. Hmm, arithmetic mean is just the average of the two numbers, right? So that means I have to add them together and then divide by 2. Let me write that down.Arithmetic mean = (3/7 + 6/11) / 2Alright, first step is to add 3/7 and 6/11. But wait, to add fractions, they need a common denominator. What's the denominator here? 7 and 11. I remember that to find a common denominator, I can multiply the two denominators together if they don't have any common factors. Since 7 and 11 are both prime numbers, their least common multiple (LCM) should be 7 times 11, which is 77. So, 77 is the common denominator.Now, I need to convert both fractions to have 77 as the denominator. Let me start with 3/7. To get from 7 to 77, I multiply by 11. So, I have to do the same to the numerator. That means 3 times 11 is 33. So, 3/7 is equivalent to 33/77.Next, 6/11. To get from 11 to 77, I multiply by 7. So, I multiply the numerator by 7 as well. 6 times 7 is 42. Therefore, 6/11 is equivalent to 42/77.Now, I can add them together since they have the same denominator. So, 33/77 + 42/77. Adding the numerators: 33 + 42 is 75. So, the sum is 75/77.Wait, that seems right. 33 plus 42 is definitely 75, and the denominator remains 77. Okay, so now I have the sum of the two fractions, which is 75/77. But I need the arithmetic mean, which means I have to divide this sum by 2.Dividing by 2 is the same as multiplying by 1/2. So, 75/77 multiplied by 1/2. Let me do that. 75 times 1 is 75, and 77 times 2 is 154. So, the result is 75/154.Is that the simplest form? Let me check if 75 and 154 have any common factors. 75 factors into 3 times 5 squared, which is 3, 5, 5. 154 factors into 2 times 7 times 11. I don't see any common factors between 75 and 154. So, 75/154 is already in its simplest form.Therefore, the arithmetic mean of 3/7 and 6/11 is 75/154.Wait, let me just go through that again to make sure I didn't make any mistakes. Starting with adding 3/7 and 6/11. Converted both to 77 denominators: 33/77 and 42/77. Added them to get 75/77. Then divided by 2, which is 75/154. Yep, that seems correct.I think I'm confident with that answer. So, the arithmetic mean is 75/154.