answer：First, I need to determine the total number of seats on the airplane. The problem provides the number of First Class seats and the proportions of Business Class and Economy Class seats relative to the total.I'll start by defining the total number of seats as ( s ).According to the problem:- First Class has 36 seats.- Business Class has ( frac{1}{3} ) of the total seats.- Economy Class has ( frac{1}{2} ) of the total seats.The sum of all these seats should equal the total number of seats:[36 + frac{1}{3}s + frac{1}{2}s = s]Next, I'll combine the terms involving ( s ):[frac{1}{3}s + frac{1}{2}s = s - 36]To simplify, I'll find a common denominator for the fractions:[frac{2}{6}s + frac{3}{6}s = frac{6}{6}s - 36][frac{5}{6}s = s - 36]Now, I'll isolate ( s ) by moving all ( s ) terms to one side:[frac{5}{6}s - s = -36][-frac{1}{6}s = -36]Finally, I'll solve for ( s ) by multiplying both sides by -6:[s = 216]Therefore, the total number of seats on the airplane is 216.

answer：First, identify the number of legs one bee has, which is 6.Next, determine the number of bees, which is 8.Multiply the number of legs per bee by the number of bees: 6 legs/bee × 8 bees = 48 legs.Therefore, 8 bees have a total of 48 legs.

answer：Okay, so I have this problem here where the sum of the first n terms of a sequence {a_n} is given by S_n = (3/2)a_n - 3. I need to find the general formula for the sequence, and the options are A, B, C, D. Hmm, let's see.First, I remember that the sum of the first n terms, S_n, is related to the nth term a_n. Specifically, S_n = a_1 + a_2 + ... + a_n. So, if I can express a_n in terms of S_n, maybe I can find a recursive relationship or something.Given S_n = (3/2)a_n - 3. Hmm, that seems a bit tricky because S_n is expressed in terms of a_n, which is part of the sum. Maybe I can find a relationship between S_n and S_{n-1} to get a_n.I know that a_n = S_n - S_{n-1}. Let me write that down:a_n = S_n - S_{n-1}Given that S_n = (3/2)a_n - 3, then S_{n-1} would be (3/2)a_{n-1} - 3, right?So, substituting these into the equation for a_n:a_n = [(3/2)a_n - 3] - [(3/2)a_{n-1} - 3]Let me simplify this:a_n = (3/2)a_n - 3 - (3/2)a_{n-1} + 3The -3 and +3 cancel out, so:a_n = (3/2)a_n - (3/2)a_{n-1}Hmm, let's get all the a_n terms on one side. Subtract (3/2)a_n from both sides:a_n - (3/2)a_n = - (3/2)a_{n-1}Simplify the left side:(1 - 3/2)a_n = - (3/2)a_{n-1}Which is:(-1/2)a_n = - (3/2)a_{n-1}Multiply both sides by -2 to eliminate the fractions:a_n = 3a_{n-1}Oh, so this is a geometric sequence where each term is 3 times the previous term. That means the common ratio r is 3. So, if I can find the first term, I can write the general formula.Let me find a_1. When n=1, S_1 = a_1. So, plug n=1 into the given formula:S_1 = (3/2)a_1 - 3But S_1 is also equal to a_1, so:a_1 = (3/2)a_1 - 3Let me solve for a_1:a_1 - (3/2)a_1 = -3(-1/2)a_1 = -3Multiply both sides by -2:a_1 = 6Okay, so the first term is 6, and each subsequent term is 3 times the previous one. So, the general formula for a geometric sequence is a_n = a_1 * r^{n-1}Plugging in the values:a_n = 6 * 3^{n-1}Hmm, let me see if that can be simplified. 6 is 2*3, so:a_n = 2*3*3^{n-1} = 2*3^nSo, a_n = 2*3^nLooking at the options, D is a_n = 2*3^n, so that must be the answer.Wait, let me double-check to make sure I didn't make any mistakes. Let's test n=1:a_1 = 2*3^1 = 6, which matches what I found earlier. Good.Now, let's check n=2. From the sum formula:S_2 = (3/2)a_2 - 3But S_2 is also a_1 + a_2 = 6 + a_2So,6 + a_2 = (3/2)a_2 - 3Let me solve for a_2:6 + a_2 = (3/2)a_2 - 3Multiply both sides by 2 to eliminate fractions:12 + 2a_2 = 3a_2 - 6Bring variables to one side:12 + 6 = 3a_2 - 2a_218 = a_2From my formula, a_2 = 2*3^2 = 2*9 = 18. Perfect, that matches.Let me try n=3. From the formula, a_3 = 2*3^3 = 54From the sum formula:S_3 = (3/2)a_3 - 3 = (3/2)*54 - 3 = 81 - 3 = 78But S_3 should also be a_1 + a_2 + a_3 = 6 + 18 + 54 = 78. That's correct.Okay, so it seems my formula is working for the first few terms. I think I did it right.So, the answer is D: a_n = 2*3^n.**Final Answer**boxed{D}

answer：Okay, so I have this problem where positive numbers ( x, y, z ) satisfy the condition:[ xy + yz + zx + 2xyz = 1. ]And I need to prove that:[ 4x + y + z ge 2. ]Hmm, let me try to figure this out step by step. First, the condition given seems a bit complicated with the combination of products and the term ( 2xyz ). Maybe I can find a substitution or some inequality that can help simplify this.I remember that sometimes in problems with symmetric conditions, substituting variables using fractions can help. Maybe I can let ( x = frac{a}{b+c} ), ( y = frac{b}{c+a} ), and ( z = frac{c}{a+b} ). I think this substitution is often used in symmetric inequalities because it can turn complicated expressions into something more manageable.Let me try substituting these into the condition:[ xy + yz + zx + 2xyz = 1. ]So, substituting each term:- ( xy = frac{a}{b+c} cdot frac{b}{c+a} = frac{ab}{(b+c)(c+a)} )- ( yz = frac{b}{c+a} cdot frac{c}{a+b} = frac{bc}{(c+a)(a+b)} )- ( zx = frac{c}{a+b} cdot frac{a}{b+c} = frac{ca}{(a+b)(b+c)} )- ( 2xyz = 2 cdot frac{a}{b+c} cdot frac{b}{c+a} cdot frac{c}{a+b} = 2 cdot frac{abc}{(a+b)(b+c)(c+a)} )Adding all these up:[ frac{ab}{(b+c)(c+a)} + frac{bc}{(c+a)(a+b)} + frac{ca}{(a+b)(b+c)} + 2 cdot frac{abc}{(a+b)(b+c)(c+a)} = 1. ]Hmm, let me see if I can factor or simplify this expression. If I factor out ( frac{1}{(a+b)(b+c)(c+a)} ), the equation becomes:[ frac{ab(a+b) + bc(b+c) + ca(c+a) + 2abc}{(a+b)(b+c)(c+a)} = 1. ]Let me expand the numerator:- ( ab(a+b) = a^2b + ab^2 )- ( bc(b+c) = b^2c + bc^2 )- ( ca(c+a) = c^2a + ca^2 )- ( 2abc ) remains as is.So the numerator is:[ a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 2abc. ]Wait a minute, I think this is equal to ( (a+b)(b+c)(c+a) ). Let me check that:[ (a+b)(b+c)(c+a) = (a+b)(b c + a c + b^2 + a b) ]Wait, no, let me multiply it step by step:First, multiply ( (a+b)(b+c) ):[ (a+b)(b+c) = ab + a c + b^2 + b c. ]Then multiply by ( (c+a) ):[ (ab + a c + b^2 + b c)(c + a) ]Let me expand this:- ( ab cdot c = abc )- ( ab cdot a = a^2b )- ( a c cdot c = a c^2 )- ( a c cdot a = a^2 c )- ( b^2 cdot c = b^2 c )- ( b^2 cdot a = a b^2 )- ( b c cdot c = b c^2 )- ( b c cdot a = a b c )So adding all these terms:[ a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 2abc. ]Which is exactly the numerator we had earlier. So, the numerator is indeed ( (a+b)(b+c)(c+a) ). Therefore, the equation simplifies to:[ frac{(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)} = 1, ]which is true. So, the substitution works.Now, I need to prove that:[ 4x + y + z ge 2. ]Substituting ( x, y, z ) with the expressions in terms of ( a, b, c ):[ 4 cdot frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ge 2. ]Hmm, okay. So, I need to show that:[ frac{4a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ge 2. ]I wonder if I can use some inequality here, like the AM-GM inequality or Cauchy-Schwarz. Let me think.The AM-GM inequality states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply it to the terms ( frac{4a}{b+c} ), ( frac{b}{c+a} ), and ( frac{c}{a+b} ).But wait, there are three terms, and I need to relate them to 2. Maybe I can pair them or find a way to bound them.Alternatively, I can consider using the Cauchy-Schwarz inequality, which is often useful in such situations. The Cauchy-Schwarz inequality in the form:[ left( sum frac{u_i^2}{v_i} right) ge frac{(sum u_i)^2}{sum v_i}. ]But I'm not sure if that directly applies here.Wait, another thought: Maybe I can use the substitution ( a = frac{1}{2} ), ( b = c = 1 ) or something like that to test the inequality. But since the problem is general, I need a more systematic approach.Let me try to manipulate the expression ( frac{4a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ).I notice that ( frac{4a}{b+c} ) is similar to ( frac{2a}{b+c} + frac{2a}{b+c} ). Maybe I can split it into two terms and apply some inequality.Alternatively, perhaps I can use the substitution ( p = b + c ), ( q = c + a ), ( r = a + b ). But I'm not sure if that helps.Wait, another idea: Maybe I can use the fact that ( frac{a}{b+c} ) is related to the tangent of an angle in a triangle, but that might be overcomplicating.Alternatively, perhaps I can use the substitution ( x = frac{a}{b+c} ), ( y = frac{b}{c+a} ), ( z = frac{c}{a+b} ) and then express the inequality in terms of ( x, y, z ). But I'm not sure.Wait, going back, the original condition is ( xy + yz + zx + 2xyz = 1 ). Maybe I can express ( 4x + y + z ) in terms of this condition.Let me denote ( S = 4x + y + z ). I need to show ( S ge 2 ).Alternatively, maybe I can express ( y + z ) in terms of ( x ) using the condition.From the condition:[ xy + yz + zx + 2xyz = 1. ]Let me factor out ( y + z ):Wait, let's see:[ xy + yz + zx = y(x + z) + x z. ]But I'm not sure if that helps.Alternatively, let me solve for ( y + z ). Let me denote ( y + z = t ) and ( y z = k ). Then, the condition becomes:[ x t + k + 2x k = 1. ]But I'm not sure if that helps either.Wait, perhaps I can use the substitution ( x = frac{a}{b+c} ), ( y = frac{b}{c+a} ), ( z = frac{c}{a+b} ) and then express ( 4x + y + z ) in terms of ( a, b, c ).So, ( 4x + y + z = frac{4a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ).I need to show this is at least 2.Let me consider the function ( f(a, b, c) = frac{4a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ).I wonder if I can find its minimum value under the condition that ( a, b, c ) are positive real numbers.Alternatively, maybe I can use the method of Lagrange multipliers, but that might be too advanced for this problem.Wait, another idea: Maybe I can assume without loss of generality that ( a + b + c = 1 ) or some other normalization to simplify the problem.But I'm not sure if that's necessary.Alternatively, perhaps I can use the Cauchy-Schwarz inequality on the terms ( frac{4a}{b+c} ), ( frac{b}{c+a} ), ( frac{c}{a+b} ).Let me recall that for positive real numbers, ( frac{a}{b} + frac{b}{c} + frac{c}{a} ge 3 ) by AM-GM, but this is a different expression.Wait, let's try applying Cauchy-Schwarz on the terms ( frac{4a}{b+c} ), ( frac{b}{c+a} ), ( frac{c}{a+b} ).The Cauchy-Schwarz inequality states that:[ left( sum frac{u_i^2}{v_i} right) ge frac{(sum u_i)^2}{sum v_i}. ]But I'm not sure how to apply it here directly.Alternatively, maybe I can use the Titu's lemma, which is a specific case of Cauchy-Schwarz:[ sum frac{a_i^2}{b_i} ge frac{(sum a_i)^2}{sum b_i}. ]But again, I'm not sure.Wait, perhaps I can consider the terms as follows:Let me write ( frac{4a}{b+c} = frac{2a}{b+c} + frac{2a}{b+c} ). So, I have two terms of ( frac{2a}{b+c} ), plus ( frac{b}{c+a} ) and ( frac{c}{a+b} ).Now, I can think of four terms:[ frac{2a}{b+c}, frac{2a}{b+c}, frac{b}{c+a}, frac{c}{a+b}. ]Maybe I can apply the AM-GM inequality on these four terms.The AM-GM inequality for four terms states that:[ frac{frac{2a}{b+c} + frac{2a}{b+c} + frac{b}{c+a} + frac{c}{a+b}}{4} ge sqrt[4]{frac{2a}{b+c} cdot frac{2a}{b+c} cdot frac{b}{c+a} cdot frac{c}{a+b}}. ]Simplifying the right-hand side:[ sqrt[4]{frac{4a^2 cdot b cdot c}{(b+c)^2 (c+a)(a+b)}}. ]So, the inequality becomes:[ frac{4x + y + z}{4} ge sqrt[4]{frac{4a^2 b c}{(b+c)^2 (c+a)(a+b)}}. ]But I'm not sure if this helps me directly, because I need to relate this to 2.Alternatively, maybe I can find a lower bound for the product inside the fourth root.Wait, let me think differently. Maybe I can use the substitution ( a = b = c ). Let me test this case.If ( a = b = c ), then ( x = y = z = frac{a}{2a} = frac{1}{2} ).Substituting into the condition:[ xy + yz + zx + 2xyz = 3 cdot frac{1}{2} cdot frac{1}{2} + 2 cdot frac{1}{2} cdot frac{1}{2} cdot frac{1}{2} = frac{3}{4} + frac{1}{4} = 1. ]So, the condition is satisfied.Now, ( 4x + y + z = 4 cdot frac{1}{2} + frac{1}{2} + frac{1}{2} = 2 + 1 = 3 ), which is greater than 2. So, in this case, the inequality holds.But I need to show it's always at least 2, not just in this specific case.Wait, maybe I can consider the case where ( a ) is very small compared to ( b ) and ( c ). Let me see what happens.Suppose ( a ) approaches 0, then ( x = frac{a}{b+c} ) approaches 0, ( y = frac{b}{c+a} ) approaches ( frac{b}{c} ), and ( z = frac{c}{a+b} ) approaches ( frac{c}{b} ).Substituting into the condition:[ xy + yz + zx + 2xyz approx 0 + frac{b}{c} cdot frac{c}{b} + 0 + 0 = 1. ]So, the condition is satisfied.Now, ( 4x + y + z approx 0 + frac{b}{c} + frac{c}{b} ).By AM-GM, ( frac{b}{c} + frac{c}{b} ge 2 ), so in this case, ( 4x + y + z ge 2 ).So, in this limit, the inequality holds with equality when ( b = c ).Hmm, interesting. So, when ( a ) approaches 0 and ( b = c ), we get ( 4x + y + z = 2 ).This suggests that the minimum value of ( 4x + y + z ) is 2, achieved when ( a ) approaches 0 and ( b = c ).So, maybe the inequality is tight in this case.Now, how can I generalize this?Perhaps I can use the method of Lagrange multipliers to find the minimum of ( 4x + y + z ) subject to the constraint ( xy + yz + zx + 2xyz = 1 ).But that might be a bit involved.Alternatively, maybe I can use substitution to express ( y ) and ( z ) in terms of ( x ), and then find the minimum.Let me try that.From the condition:[ xy + yz + zx + 2xyz = 1. ]Let me factor out ( y + z ):Wait, let me rearrange the equation:[ xy + yz + zx = 1 - 2xyz. ]Hmm, not sure.Alternatively, let me solve for ( y + z ).Let me denote ( S = y + z ) and ( P = y z ).Then, the condition becomes:[ x S + P + 2x P = 1. ]So,[ x S + P (1 + 2x) = 1. ]But I also know that for two variables ( y ) and ( z ), ( S^2 ge 4P ) by AM-GM.So, ( P le frac{S^2}{4} ).Substituting into the equation:[ x S + frac{S^2}{4} (1 + 2x) le 1. ]Wait, but I'm not sure if this helps me find the minimum of ( 4x + S ).Alternatively, maybe I can express ( S ) in terms of ( x ) and ( P ).From the condition:[ x S + P (1 + 2x) = 1. ]So,[ P = frac{1 - x S}{1 + 2x}. ]Since ( P le frac{S^2}{4} ), we have:[ frac{1 - x S}{1 + 2x} le frac{S^2}{4}. ]Multiplying both sides by ( 1 + 2x ) (which is positive since ( x > 0 )):[ 1 - x S le frac{S^2}{4} (1 + 2x). ]Rearranging:[ 1 le x S + frac{S^2}{4} (1 + 2x). ]Hmm, this seems complicated. Maybe I can consider this as a quadratic in ( S ).Let me rewrite it:[ frac{S^2}{4} (1 + 2x) + x S - 1 ge 0. ]Multiplying both sides by 4 to eliminate the fraction:[ S^2 (1 + 2x) + 4x S - 4 ge 0. ]This is a quadratic in ( S ):[ (1 + 2x) S^2 + 4x S - 4 ge 0. ]Let me denote ( A = 1 + 2x ), ( B = 4x ), ( C = -4 ).The quadratic equation ( A S^2 + B S + C = 0 ) has solutions:[ S = frac{-B pm sqrt{B^2 - 4AC}}{2A}. ]Substituting the values:[ S = frac{-4x pm sqrt{(4x)^2 - 4(1 + 2x)(-4)}}{2(1 + 2x)}. ]Simplifying inside the square root:[ 16x^2 - 4(1 + 2x)(-4) = 16x^2 + 16(1 + 2x) = 16x^2 + 16 + 32x. ]So,[ S = frac{-4x pm sqrt{16x^2 + 16 + 32x}}{2(1 + 2x)}. ]Factor out 16 inside the square root:[ sqrt{16(x^2 + 2x + 1)} = 4(x + 1). ]So,[ S = frac{-4x pm 4(x + 1)}{2(1 + 2x)}. ]Let me consider both signs:1. With the plus sign:[ S = frac{-4x + 4x + 4}{2(1 + 2x)} = frac{4}{2(1 + 2x)} = frac{2}{1 + 2x}. ]2. With the minus sign:[ S = frac{-4x - 4x - 4}{2(1 + 2x)} = frac{-8x - 4}{2(1 + 2x)} = frac{-4(2x + 1)}{2(1 + 2x)} = -2. ]But since ( S = y + z > 0 ), we discard the negative solution.So, the quadratic inequality ( (1 + 2x) S^2 + 4x S - 4 ge 0 ) holds when ( S ge frac{2}{1 + 2x} ).Therefore, ( y + z ge frac{2}{1 + 2x} ).Now, our target expression is ( 4x + y + z ge 4x + frac{2}{1 + 2x} ).So, we need to show that:[ 4x + frac{2}{1 + 2x} ge 2. ]Let me define ( f(x) = 4x + frac{2}{1 + 2x} ). I need to show that ( f(x) ge 2 ) for ( x > 0 ).Let me compute the derivative of ( f(x) ) to find its minimum.First, ( f(x) = 4x + frac{2}{1 + 2x} ).Compute ( f'(x) ):[ f'(x) = 4 - frac{4}{(1 + 2x)^2}. ]Set ( f'(x) = 0 ):[ 4 - frac{4}{(1 + 2x)^2} = 0 ][ frac{4}{(1 + 2x)^2} = 4 ][ frac{1}{(1 + 2x)^2} = 1 ][ (1 + 2x)^2 = 1 ][ 1 + 2x = pm 1 ]But since ( x > 0 ), ( 1 + 2x > 1 ), so ( 1 + 2x = 1 ) is not possible. Therefore, the only critical point is when ( 1 + 2x = 1 ), which gives ( x = 0 ), but ( x > 0 ).Wait, this suggests that ( f(x) ) is increasing for all ( x > 0 ), because ( f'(x) = 4 - frac{4}{(1 + 2x)^2} ). Since ( (1 + 2x)^2 > 1 ) for ( x > 0 ), ( frac{4}{(1 + 2x)^2} < 4 ), so ( f'(x) > 0 ). Therefore, ( f(x) ) is increasing on ( x > 0 ).Thus, the minimum of ( f(x) ) occurs as ( x ) approaches 0.Compute ( lim_{x to 0^+} f(x) = 0 + frac{2}{1 + 0} = 2 ).Therefore, ( f(x) ge 2 ) for all ( x > 0 ), with equality when ( x to 0 ).So, combining this with our earlier result ( y + z ge frac{2}{1 + 2x} ), we have:[ 4x + y + z ge 4x + frac{2}{1 + 2x} ge 2. ]Therefore, the inequality ( 4x + y + z ge 2 ) holds, with equality when ( x to 0 ) and ( y = z ).So, the proof is complete.

answer：First, I need to understand the relationship between the percentage increase in diameter and the percentage increase in area for a circular logo.I know that the area of a circle is proportional to the square of its diameter. This means that if the diameter increases by a certain percentage, the area will increase by the square of that percentage.Given that the area has increased by 143.36%, the total new area is 243.36% of the original area. To find the factor by which the diameter has increased, I take the square root of 2.4336, which is approximately 1.56.This indicates that the diameter has increased by 56%.

answer：First, I need to evaluate the expression (7^3 - 4 times 7^2 + 6 times 7 - 2).I'll start by calculating each term individually.1. Calculate (7^3): [ 7^3 = 7 times 7 times 7 = 343 ]2. Calculate (4 times 7^2): [ 4 times 7^2 = 4 times (7 times 7) = 4 times 49 = 196 ]3. Calculate (6 times 7): [ 6 times 7 = 42 ]Now, substitute these values back into the original expression:[343 - 196 + 42 - 2]Next, perform the addition and subtraction step by step:[343 - 196 = 147][147 + 42 = 189][189 - 2 = 187]Therefore, the final result of the expression is 187.