answer：First, I recognize that we're dealing with an arithmetic sequence where the first term ( a_1 = 2 ) and the fifth term ( a_5 ) is three times the third term ( a_3 ).I recall that in an arithmetic sequence, each term can be expressed as:[a_n = a_1 + (n-1)d]where ( d ) is the common difference.Using this formula, I can express ( a_5 ) and ( a_3 ):[a_5 = a_1 + 4d][a_3 = a_1 + 2d]Given that ( a_5 = 3a_3 ), I substitute the expressions:[a_1 + 4d = 3(a_1 + 2d)]Plugging in ( a_1 = 2 ):[2 + 4d = 3(2 + 2d)]Simplifying the equation:[2 + 4d = 6 + 6d][-2d = 4][d = -2]Now that I have the common difference ( d = -2 ), I can find the sum of the first 9 terms using the sum formula for an arithmetic sequence:[S_n = frac{n}{2} times (2a_1 + (n-1)d)]Substituting ( n = 9 ), ( a_1 = 2 ), and ( d = -2 ):[S_9 = frac{9}{2} times (4 + 8 times (-2))][S_9 = frac{9}{2} times (4 - 16)][S_9 = frac{9}{2} times (-12)][S_9 = 9 times (-6)][S_9 = -54]Therefore, the sum of the first 9 terms is (-54).

answer：Okay, so I have this problem here about vectors a and b. Both vectors have a magnitude of 2, which is given as |a| = |b| = 2. There's also this dot product condition: a · (b - a) = -2. I need to find the magnitude of the vector 2a - b, which is written as |2a - b|. The options are A: 2, B: 2√3, C: 4, D: 8.Alright, let me break this down step by step. First, I remember that the dot product of two vectors can be expressed in terms of their magnitudes and the cosine of the angle between them. So, a · b = |a||b|cosθ, where θ is the angle between vectors a and b. But I'm not sure if I need to find the angle here or if there's a more straightforward way.The given condition is a · (b - a) = -2. Maybe I can expand this expression first. Let's see: a · (b - a) is equal to a · b - a · a. So, that simplifies to a · b - |a|² because a · a is just the magnitude of a squared. We know |a| is 2, so |a|² is 4. Therefore, the equation becomes a · b - 4 = -2.Solving for a · b, I can add 4 to both sides: a · b = 2. Okay, so the dot product of a and b is 2. That's useful information.Now, I need to find |2a - b|. I remember that the magnitude of a vector can be found using the dot product as well. Specifically, |v| = √(v · v). So, if I let v = 2a - b, then |2a - b| = √((2a - b) · (2a - b)).Let me compute that dot product. Expanding (2a - b) · (2a - b), I get:(2a) · (2a) - (2a) · b - b · (2a) + b · b.Simplifying each term:- (2a) · (2a) is 4(a · a) which is 4|a|².- (2a) · b is 2(a · b).- Similarly, b · (2a) is also 2(a · b).- And b · b is |b|².Putting it all together:4|a|² - 2(a · b) - 2(a · b) + |b|².Simplify the middle terms: -2(a · b) - 2(a · b) is -4(a · b).So, the expression becomes:4|a|² - 4(a · b) + |b|².Now, substitute the known values. We know |a| = |b| = 2, so |a|² = 4 and |b|² = 4. Also, we found earlier that a · b = 2.Plugging these in:4*4 - 4*2 + 4.Calculating each term:4*4 = 16,4*2 = 8,and the last term is 4.So, 16 - 8 + 4 = 12.Therefore, |2a - b|² = 12. To find |2a - b|, take the square root of 12.√12 simplifies to 2√3 because 12 = 4*3 and √4 is 2.So, |2a - b| = 2√3.Looking back at the options, B is 2√3, so that must be the correct answer.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded a · (b - a) correctly to a · b - |a|².2. Substituted |a|² = 4, so a · b = 2.3. Expanded |2a - b|² correctly to 4|a|² - 4(a · b) + |b|².4. Substituted the values correctly: 4*4 = 16, 4*2 = 8, and |b|² = 4.5. Calculated 16 - 8 + 4 = 12.6. Took the square root of 12 to get 2√3.Everything seems to check out. I don't see any errors in my calculations or reasoning. So, I'm confident that the answer is B: 2√3.

answer：Okay, so I have this cubic equation: ( ax^3 + bx^2 + cx + d = 0 ). They told me that two of the roots are 4 and -3. I need to find ( frac{b + c}{a} ). Hmm, let's see.First, I remember that for a cubic equation, if I know some of the roots, I can factor the equation. Since it's a cubic, there must be three roots in total. Let me denote the third root as ( r ). So, the equation can be written as ( a(x - 4)(x + 3)(x - r) = 0 ). That makes sense because if I plug in x = 4 or x = -3, the equation will equal zero.Now, I need to expand this factored form to get it into the standard cubic form. Let me do that step by step. First, multiply ( (x - 4) ) and ( (x + 3) ). Using the distributive property:( (x - 4)(x + 3) = x^2 + 3x - 4x - 12 = x^2 - x - 12 ).Okay, so now I have ( (x^2 - x - 12)(x - r) ). Let me multiply these two:( x^2(x - r) - x(x - r) - 12(x - r) ).Calculating each term:1. ( x^2 cdot x = x^3 )2. ( x^2 cdot (-r) = -r x^2 )3. ( -x cdot x = -x^2 )4. ( -x cdot (-r) = r x )5. ( -12 cdot x = -12x )6. ( -12 cdot (-r) = 12r )Now, combine all these terms:( x^3 - r x^2 - x^2 + r x - 12x + 12r ).Combine like terms:- The ( x^3 ) term remains: ( x^3 ).- The ( x^2 ) terms: ( -r x^2 - x^2 = -(r + 1) x^2 ).- The ( x ) terms: ( r x - 12x = (r - 12) x ).- The constant term: ( 12r ).So, putting it all together, the expanded form is:( x^3 - (r + 1) x^2 + (r - 12) x + 12r ).But remember, the original equation is ( ax^3 + bx^2 + cx + d = 0 ). So, if I factor out the 'a' from my expanded form, it should match:( a(x^3 - (r + 1) x^2 + (r - 12) x + 12r) = 0 ).Therefore, comparing coefficients:- ( a ) is just the coefficient of ( x^3 ), which is already factored out.- ( b = -a(r + 1) )- ( c = a(r - 12) )- ( d = 12a r )But I don't know the value of ( r ), the third root. Hmm, how can I find ( r )? Maybe I can use Vieta's formula. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a cubic equation ( ax^3 + bx^2 + cx + d = 0 ), the sum of the roots is ( -frac{b}{a} ), the sum of the products of the roots two at a time is ( frac{c}{a} ), and the product of the roots is ( -frac{d}{a} ).Given that two roots are 4 and -3, let me denote them as ( alpha = 4 ), ( beta = -3 ), and the third root is ( gamma = r ).So, according to Vieta's formulas:1. Sum of roots: ( alpha + beta + gamma = -frac{b}{a} )2. Sum of products two at a time: ( alphabeta + alphagamma + betagamma = frac{c}{a} )3. Product of roots: ( alphabetagamma = -frac{d}{a} )But I don't know ( d ) or ( r ), so maybe I can express ( b ) and ( c ) in terms of ( r ) and then find ( frac{b + c}{a} ).From the first Vieta's formula:( 4 + (-3) + r = -frac{b}{a} )Simplify:( 1 + r = -frac{b}{a} )So, ( b = -a(1 + r) ). That's consistent with what I had earlier.From the second Vieta's formula:( (4)(-3) + (4)(r) + (-3)(r) = frac{c}{a} )Calculate each term:( 4 times -3 = -12 )( 4 times r = 4r )( -3 times r = -3r )So, combining:( -12 + 4r - 3r = frac{c}{a} )Simplify:( -12 + r = frac{c}{a} )Therefore, ( c = a(r - 12) ). Again, consistent with my earlier expansion.Now, I need ( frac{b + c}{a} ). Let's compute ( b + c ):( b + c = -a(1 + r) + a(r - 12) )Factor out 'a':( = a[ - (1 + r) + (r - 12) ] )Simplify inside the brackets:( -1 - r + r - 12 )The ( -r ) and ( +r ) cancel out:( -1 - 12 = -13 )So, ( b + c = a(-13) )Therefore, ( frac{b + c}{a} = -13 ).Wait, so I didn't even need to know the third root? That's interesting. It seems that regardless of the value of ( r ), the combination ( frac{b + c}{a} ) is always -13. Let me double-check my steps to make sure I didn't make a mistake.1. Factored the cubic as ( a(x - 4)(x + 3)(x - r) ).2. Expanded it step by step, got the coefficients in terms of ( r ).3. Used Vieta's formulas to express ( b ) and ( c ) in terms of ( r ).4. Added ( b ) and ( c ), simplified, and found that ( r ) cancels out, leaving -13.Yes, that seems correct. So, the answer is -13.**Final Answer**boxed{-13}

answer：Okay, so I need to find the remainder when the product of the numbers 3, 13, 23, 33, all the way up to 183 and 193 is divided by 5. Hmm, that sounds a bit complicated at first glance, but maybe I can break it down step by step.First, let me understand the sequence of numbers we're dealing with. It starts at 3 and each subsequent number increases by 10. So, 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 103, 113, 123, 133, 143, 153, 163, 173, 183, 193. Let me count how many numbers that is. Starting from 3 and going up by 10 each time until 193. So, the first term is 3, and the last term is 193. To find the number of terms, I can use the formula for the nth term of an arithmetic sequence: a_n = a_1 + (n-1)d, where d is the common difference. Here, a_1 is 3, d is 10, and a_n is 193. So, 193 = 3 + (n-1)*10. Subtracting 3 from both sides gives 190 = (n-1)*10. Dividing both sides by 10 gives 19 = n-1, so n = 20. So, there are 20 numbers in this sequence.Now, I need to find the product of all these 20 numbers and then find the remainder when this product is divided by 5. That seems like a huge number, so calculating it directly isn't feasible. Instead, I should look for a pattern or a property that can simplify this calculation.I remember that when dealing with remainders and multiplication, modular arithmetic can be really helpful. Specifically, if I can find the remainder of each number when divided by 5, then multiply those remainders together, and then take the result modulo 5, that should give me the same remainder as the original product divided by 5. This is because of the property that (a * b) mod m = [(a mod m) * (b mod m)] mod m.So, let's apply this. First, let's find the remainder of each number in the sequence when divided by 5.Starting with 3: 3 divided by 5 is 0 with a remainder of 3. So, 3 mod 5 is 3.Next, 13: 13 divided by 5 is 2 with a remainder of 3. So, 13 mod 5 is 3.Similarly, 23: 23 divided by 5 is 4 with a remainder of 3. So, 23 mod 5 is 3.Continuing this pattern, each number in the sequence is of the form 10k + 3, where k is an integer starting from 0. So, 10k + 3 divided by 5 is 2k + 0 with a remainder of 3. Therefore, every number in this sequence leaves a remainder of 3 when divided by 5.So, each of the 20 numbers in the sequence has a remainder of 3 modulo 5. Therefore, the product of all these numbers modulo 5 is equivalent to 3^20 mod 5.Now, I need to compute 3^20 mod 5. Calculating 3^20 directly would be tedious, but perhaps there's a pattern or a theorem that can help simplify this.I recall that Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since 3 and 5 are coprime, φ(5) is 4 because 5 is prime. So, Euler's theorem tells us that 3^4 ≡ 1 mod 5.Let me verify that: 3^1 = 3 mod 5 is 3, 3^2 = 9 mod 5 is 4, 3^3 = 27 mod 5 is 2, 3^4 = 81 mod 5 is 1. Yes, that's correct. So, 3^4 ≡ 1 mod 5.Therefore, 3^20 can be written as (3^4)^5. Since 3^4 ≡ 1 mod 5, then (3^4)^5 ≡ 1^5 mod 5, which is 1 mod 5.So, 3^20 ≡ 1 mod 5.Therefore, the product of all these numbers modulo 5 is 1.Wait, let me double-check my reasoning. Each number is congruent to 3 mod 5, so the product is 3^20 mod 5. Since 3^4 ≡ 1 mod 5, 3^20 = (3^4)^5 ≡ 1^5 ≡ 1 mod 5. That seems correct.Alternatively, I can think about the cyclicity of the powers of 3 modulo 5. The powers of 3 modulo 5 cycle every 4 exponents: 3, 4, 2, 1, and then repeat. So, 3^1 ≡ 3, 3^2 ≡ 4, 3^3 ≡ 2, 3^4 ≡ 1, 3^5 ≡ 3, and so on. Since 20 is a multiple of 4, 3^20 would be at the end of a cycle, which is 1 mod 5.Yes, that confirms my earlier conclusion.So, putting it all together, the remainder when the product 3 × 13 × 23 × ... × 193 is divided by 5 is 1.

answer：First, I need to determine the sampling ratio used for the freshmen. There are 210 freshmen, and 7 were selected. By dividing the total number of freshmen by the number selected, I can find the sampling ratio.Next, I will apply this sampling ratio to the number of sophomores. There are 270 sophomores, so multiplying this number by the sampling ratio will give me the number of sophomores that should be selected.Finally, I will calculate the number of sophomores selected based on the determined sampling ratio.

answer：Okay, so I'm trying to figure out the score range for level B in this math exam. The scores follow a normal distribution with a mean of 76 and a standard deviation of 16. The levels are divided into A, B, C, and D, with proportions of 15%, 35%, 35%, and 15% respectively. Students scoring 83 or above are in level A.First, I need to understand what the normal distribution parameters mean. The mean is 76, which is the average score, and the standard deviation is 16, which tells me how spread out the scores are. A normal distribution is symmetric around the mean, so half of the students scored above 76, and half scored below.Level A is the top 15%, so that's the highest scoring 15% of students. Since the distribution is normal, I can use z-scores to find the cutoff points for each level. A z-score tells me how many standard deviations away from the mean a particular score is.I know that for a normal distribution, about 68% of the data lies within one standard deviation of the mean, 95% within two, and 99.7% within three. But since we're dealing with specific percentages like 15% and 35%, I might need to use a z-table or a calculator to find the exact z-scores corresponding to these percentages.Given that level A is 15%, that means the cutoff score for level A is the score that separates the top 15% from the rest. I can find the z-score that corresponds to the 85th percentile (since 100% - 15% = 85%) because that's where the top 15% begins.Looking up the 85th percentile in a z-table, I find that the z-score is approximately 1.04. This means that the cutoff score for level A is 1.04 standard deviations above the mean.Calculating that: 1.04 * 16 = 16.64. Adding that to the mean: 76 + 16.64 = 92.64. Wait, but the problem says that level A starts at 83, not 92.64. That doesn't match up. Maybe I made a mistake.Oh, I see. The standard deviation given is 16, but in the initial problem, the normal distribution is written as N(76,16). Wait, actually, in the standard notation, N(μ, σ²), so the second parameter is the variance, not the standard deviation. So, if it's N(76,16), that means the variance is 16, so the standard deviation is the square root of 16, which is 4. Okay, that makes more sense.So, the standard deviation is 4, not 16. That changes things. Let me recalculate.The z-score for the 85th percentile is still approximately 1.04. So, the cutoff score for level A is 1.04 * 4 = 4.16. Adding that to the mean: 76 + 4.16 = 80.16. But the problem states that level A starts at 83, not 80.16. Hmm, there's still a discrepancy.Wait, maybe I need to find the z-score that corresponds to the cumulative probability of 0.85, which is indeed around 1.04. But if the cutoff is at 83, let's find the z-score for 83.Calculating z = (83 - 76) / 4 = 7 / 4 = 1.75. So, the z-score is 1.75. Looking at the z-table, a z-score of 1.75 corresponds to a cumulative probability of about 0.9599, which is 95.99%. That means only about 4.01% of students scored above 83, but the problem says 15% are in level A. That doesn't add up.Wait, maybe I misinterpreted the problem. It says the scores follow N(76,16), which is mean 76 and variance 16, so standard deviation 4. But if the top 15% is above 83, that would mean that 83 is the 85th percentile. But according to the z-score calculation, 83 is the 95.99th percentile, which contradicts the given information.This suggests that either the standard deviation is different, or the cutoff for level A is incorrect. But the problem states that level A is 15%, so maybe I need to adjust my approach.Perhaps instead of assuming the standard deviation is 4, I need to find the correct standard deviation that makes 83 the 85th percentile. Let's denote the standard deviation as σ.Given that P(X >= 83) = 0.15, which means P(X < 83) = 0.85. The z-score corresponding to 0.85 is about 1.04. So, z = (83 - 76) / σ = 1.04. Solving for σ: σ = (83 - 76) / 1.04 ≈ 7 / 1.04 ≈ 6.73.But the problem states that the distribution is N(76,16), which implies σ = 4. There's a conflict here. Maybe the problem has a typo, or I'm misunderstanding the parameters.Alternatively, perhaps the standard deviation is indeed 16, not 4. Let's try that. If σ = 16, then z = (83 - 76) / 16 = 7 / 16 ≈ 0.4375. Looking up z = 0.4375 in the z-table, the cumulative probability is about 0.6700, meaning 67% scored below 83, and 33% scored above. But the problem says 15% are in level A, so this doesn't match either.This is confusing. Maybe the problem uses a different method to assign levels, not strictly based on the normal distribution percentiles. Perhaps they are using quartiles or some other method.Given that levels A, B, C, D are in proportions 15%, 35%, 35%, 15%, it's symmetric around the mean. So, level A is top 15%, level D is bottom 15%, and levels B and C are each 35%. Since the distribution is normal, which is symmetric, this makes sense.Given that, the cutoff for level A is at 83, which is the 85th percentile. To find the cutoff for level B, which is the next 35%, I need to find the score that separates the top 15% (level A) from the next 35% (level B).So, the cumulative probability for level A is 0.15, and for level B, it's 0.15 + 0.35 = 0.50. Wait, that would mean the cutoff for level B is at the median, which is the mean, 76. But that can't be right because level B should be above level C, which is also 35%.Wait, no. Actually, if level A is top 15%, then level B is the next 35%, which would be from the 15th percentile to the 50th percentile. But since the distribution is symmetric, the 15th percentile is below the mean, and the 50th percentile is the mean. So, level B would be from the 15th percentile to the mean.But the problem states that level A starts at 83, which is above the mean. So, perhaps the levels are divided as follows:- Level A: top 15% (above 83)- Level B: next 35% (from some lower score to 83)- Level C: next 35% (from some lower score to the lower cutoff)- Level D: bottom 15% (below some score)But this contradicts the initial thought that levels are symmetric around the mean. Maybe the levels are not symmetric, but rather, the top 15% is level A, then the next 35% is level B, then the next 35% is level C, and the bottom 15% is level D.Given that, the cutoff for level A is 83 (top 15%), then the cutoff for level B would be the score that separates the top 15% + 35% = 50%, which is the mean, 76. So, level B would be from 76 to 83.But that seems too simplistic because in a normal distribution, the mean is the median, so 50% score below and 50% above. If level A is 15% above 83, then level B would be from 76 to 83, which is 35% of the students. That makes sense.Wait, but if the mean is 76, and level B is from 76 to 83, that's 35% of the students. Then level C would be from some lower score to 76, and level D below that.But the problem doesn't specify the lower cutoffs, only asking for level B. So, based on this reasoning, level B is from 76 to 83.But let me double-check. If level A is 15% above 83, then the cumulative probability up to 83 is 85%. The z-score for 85% is about 1.04, so z = 1.04. Given that, the score at 1.04 standard deviations above the mean is 76 + 1.04*4 = 76 + 4.16 = 80.16. But the problem says level A starts at 83, which is higher than 80.16. This inconsistency suggests that either the standard deviation is different or the cutoff is not based on the 85th percentile.Alternatively, maybe the problem uses a different method to assign levels, not strictly based on percentiles. Perhaps they are using quartiles or some other method.Given the confusion, I think the safest assumption is that level B is from the mean (76) to the cutoff for level A (83), which would be 76 to 83. Therefore, the score range for level B is [76, 83).