answer：First, I need to understand what it means for two lines to be symmetric with respect to the y-axis. Symmetry about the y-axis implies that for every point (x, y) on the original line ( l_1 ), there is a corresponding point (-x, y) on the line ( l_2 ).Given the equation of ( l_1 ): ( y = 2x + 3 ), I can identify the slope and the y-intercept. The slope of ( l_1 ) is 2, and it crosses the y-axis at (0, 3).To find the equation of ( l_2 ), which is symmetric to ( l_1 ) with respect to the y-axis, I need to determine how the slope changes. Reflecting a line over the y-axis changes the sign of the slope. Therefore, the slope of ( l_2 ) will be -2.Next, I consider the y-intercept. Since the reflection over the y-axis does not affect the y-intercept, ( l_2 ) will also cross the y-axis at (0, 3).Using the slope-intercept form of a line, ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept, I can write the equation of ( l_2 ) as:[y = -2x + 3]Finally, I can conclude that the equation of ( l_2 ) is ( y = -2x + 3 ).

answer：First, I need to determine how much punch was in the bowl before Mark added the final 12 gallons to fill it up. Since the bowl can hold 16 gallons, subtracting the 12 gallons Mark added gives me 4 gallons remaining before the final addition.Next, I consider that Sally drank 2 gallons before Mark added those 12 gallons. So, before Sally drank, there must have been 4 gallons plus the 2 gallons she drank, totaling 6 gallons.Mark added 4 gallons after his cousin drank half the punch. Therefore, before Mark added those 4 gallons, there were 6 gallons minus 4 gallons, which equals 2 gallons in the bowl.Since the cousin drank half of the punch, the initial amount Mark added must have been twice the amount that remained after the cousin drank. So, if there were 2 gallons left after the cousin drank half, Mark initially added 2 gallons multiplied by 2, which equals 4 gallons.

answer：Okay, so I have this problem about an acute triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively. It's given that 2a sin B equals sqrt(3) times b. I need to find the size of angle A first, and then, given that a is 6 and b plus c is 8, find the area of triangle ABC.Starting with part (1). I remember from the Law of Sines that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, that means a/sin A equals b/sin B. Maybe I can use that here.Given 2a sin B = sqrt(3) b. Let me write that down:2a sin B = sqrt(3) b.From the Law of Sines, I know that a/sin A = b/sin B. So, maybe I can express sin B in terms of a, b, and sin A.From a/sin A = b/sin B, cross-multiplying gives a sin B = b sin A. So, sin B = (b sin A)/a.Wait, but in the given equation, I have 2a sin B. Let's substitute sin B from the Law of Sines into that equation.So, 2a sin B = 2a * (b sin A)/a = 2b sin A.But the given equation is 2a sin B = sqrt(3) b. So, substituting, I get:2b sin A = sqrt(3) b.Hmm, I can divide both sides by b, assuming b is not zero, which it can't be in a triangle.So, 2 sin A = sqrt(3).Therefore, sin A = sqrt(3)/2.Okay, so sin A is sqrt(3)/2. I know that sin 60 degrees is sqrt(3)/2, and sin 120 degrees is also sqrt(3)/2. But since the triangle is acute, all angles are less than 90 degrees. So, angle A must be 60 degrees, which is pi/3 radians.So, angle A is 60 degrees or pi/3 radians.Alright, that seems straightforward. I think that's part (1) done.Now, moving on to part (2). Given that a = 6 and b + c = 8, find the area of triangle ABC.I remember that the area of a triangle can be found using several formulas. One common formula is (1/2)ab sin C, but in this case, I don't know angle C. Alternatively, I can use Heron's formula, but for that, I need all three sides. I only know a and that b + c = 8, so I don't have b and c individually. Maybe I can find bc somehow?Alternatively, since I know angle A, maybe I can use the formula involving sides and the sine of an angle. Let me think.I know that area = (1/2) * b * c * sin A. Since I know angle A is 60 degrees, sin A is sqrt(3)/2. So, if I can find the product bc, I can compute the area.So, how can I find bc? I know that a = 6, and from the Law of Cosines, I can relate a, b, c, and angle A.Law of Cosines says that a^2 = b^2 + c^2 - 2bc cos A.Given that angle A is 60 degrees, cos A is 0.5. So, plugging in:6^2 = b^2 + c^2 - 2bc * 0.5Simplify:36 = b^2 + c^2 - bcOkay, so I have the equation b^2 + c^2 - bc = 36.I also know that b + c = 8. Maybe I can square both sides of this equation to get another relation.So, (b + c)^2 = 8^2 = 64.Expanding the left side:b^2 + 2bc + c^2 = 64.So, now I have two equations:1) b^2 + c^2 - bc = 362) b^2 + 2bc + c^2 = 64If I subtract equation 1 from equation 2, I can eliminate b^2 + c^2.So, subtracting:(b^2 + 2bc + c^2) - (b^2 + c^2 - bc) = 64 - 36Simplify:(2bc + bc) = 28So, 3bc = 28Therefore, bc = 28/3.Alright, so bc is 28/3. Now, going back to the area formula:Area = (1/2) * bc * sin A = (1/2) * (28/3) * (sqrt(3)/2)Let me compute that step by step.First, multiply 1/2 and 28/3:(1/2) * (28/3) = 14/3.Then, multiply that by sqrt(3)/2:14/3 * sqrt(3)/2 = (14 sqrt(3))/6.Simplify 14/6 to 7/3:(7 sqrt(3))/3.So, the area is 7 sqrt(3)/3.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from the Law of Cosines:a^2 = b^2 + c^2 - 2bc cos A36 = b^2 + c^2 - bcThen, from b + c = 8, squaring gives b^2 + 2bc + c^2 = 64.Subtracting the first equation from the second:(b^2 + 2bc + c^2) - (b^2 + c^2 - bc) = 64 - 36Which simplifies to 3bc = 28, so bc = 28/3.Then, area = (1/2) * bc * sin A = (1/2) * (28/3) * (sqrt(3)/2) = (14/3) * (sqrt(3)/2) = (14 sqrt(3))/6 = (7 sqrt(3))/3.Yes, that seems correct.So, the area is 7 sqrt(3)/3.I think that's the answer.**Final Answer**(1) The size of angle ( A ) is (boxed{dfrac{pi}{3}}).(2) The area of triangle ( ABC ) is (boxed{dfrac{7sqrt{3}}{3}}).

answer：First, I need to calculate the quotient for each division expression provided.For option A: 254 divided by 5 equals 50.8.For option B: 400 divided by 6 is approximately 66.67.For option C: 492 divided by 7 is approximately 70.29.Now, I need to determine which of these quotients is closest to 70.Comparing the results:- 50.8 is 19.2 less than 70.- 66.67 is 3.33 less than 70.- 70.29 is 0.29 more than 70.Among these, 70.29 is the closest to 70.Therefore, the quotient closest to 70 is from option C.

answer：Alright, so I've got this problem here about a positive sequence {a_n} where the sum of the first n terms is S_n, and it's given that 2S_n equals a_n squared plus a_n for all positive integers n. There are two parts: first, finding the general formula for {a_n}, and second, dealing with another sequence {b_n} defined in terms of {a_n} and finding the sum of its first n terms, denoted as T_n.Starting with part (1). I need to find a general formula for {a_n}. The given relation is 2S_n = a_n² + a_n. Hmm, okay. Since S_n is the sum of the first n terms, S_n = a_1 + a_2 + ... + a_n. So, 2 times that sum equals a_n squared plus a_n. That feels like a recursive relation, but maybe I can find a pattern or figure out what kind of sequence this is.Let me try plugging in small values of n to see if I can spot a pattern. For n=1, S_1 is just a_1, so 2a_1 = a_1² + a_1. Simplifying that, 2a_1 - a_1 = a_1², so a_1 = a_1². That implies a_1(a_1 - 1) = 0. Since it's a positive sequence, a_1 can't be zero, so a_1 must be 1.Okay, so a_1 is 1. Now, let's try n=2. S_2 = a_1 + a_2, so 2(a_1 + a_2) = a_2² + a_2. We already know a_1 is 1, so 2(1 + a_2) = a_2² + a_2. That simplifies to 2 + 2a_2 = a_2² + a_2. Bringing everything to one side: a_2² - a_2 - 2 = 0. Solving this quadratic equation, the discriminant is 1 + 8 = 9, so a_2 = [1 ± 3]/2. Since it's positive, a_2 = (1 + 3)/2 = 2.Alright, so a_2 is 2. Let's do n=3. S_3 = a_1 + a_2 + a_3 = 1 + 2 + a_3 = 3 + a_3. Then, 2S_3 = 2(3 + a_3) = 6 + 2a_3. This should equal a_3² + a_3. So, 6 + 2a_3 = a_3² + a_3. Rearranging: a_3² - a_3 - 6 = 0. The discriminant is 1 + 24 = 25, so a_3 = [1 ± 5]/2. Positive solution is (1 + 5)/2 = 3. So, a_3 is 3.Hmm, I see a pattern here. a_1=1, a_2=2, a_3=3. It seems like a_n = n. Let me test this hypothesis for n=4. S_4 = 1 + 2 + 3 + a_4 = 6 + a_4. Then, 2S_4 = 12 + 2a_4. This should equal a_4² + a_4. So, 12 + 2a_4 = a_4² + a_4. Rearranging: a_4² - a_4 - 12 = 0. Discriminant is 1 + 48 = 49, so a_4 = [1 ± 7]/2. Positive solution is (1 + 7)/2 = 4. Yep, a_4 is 4. So, it seems like a_n = n.But I should verify this for a general n. Let's assume that a_k = k for all k up to n. Then, S_n = 1 + 2 + 3 + ... + n = n(n + 1)/2. Then, 2S_n = n(n + 1). On the other hand, a_n² + a_n = n² + n. So, 2S_n = n² + n, which is the same as a_n² + a_n. So, the formula holds if a_n = n.Therefore, the general formula for {a_n} is a_n = n.Moving on to part (2). We have b_n defined as a_1/3^n + a_2/3^{n-1} + a_3/3^{n-2} + ... + a_{n-1}/3² + a_n/3. So, it's a sum where each term is a_k divided by 3^{n - (k - 1)} or something like that. Wait, let me parse it correctly.Looking at the expression: b_n = a_1/3^n + a_2/3^{n-1} + a_3/3^{n-2} + ... + a_{n-1}/3² + a_n/3. So, each term is a_k divided by 3^{n - (k - 1)}. For k=1, it's a_1/3^n, for k=2, it's a_2/3^{n-1}, and so on until k=n, which is a_n/3.Since we've established that a_k = k, we can substitute that in. So, b_n = 1/3^n + 2/3^{n-1} + 3/3^{n-2} + ... + (n-1)/3² + n/3.Now, we need to find T_n, the sum of the first n terms of {b_n}. So, T_n = b_1 + b_2 + ... + b_n.First, let's write out what each b_k looks like:b_1 = 1/3^1 = 1/3b_2 = 1/3^2 + 2/3^1 = 1/9 + 2/3b_3 = 1/3^3 + 2/3^2 + 3/3^1 = 1/27 + 2/9 + 3/3And so on, up to b_n.So, T_n is the sum of these b_k's. This seems a bit complicated, but maybe we can find a pattern or a way to sum them up.Alternatively, perhaps we can find a general expression for b_n first and then sum over n.Looking back at b_n: b_n = sum_{k=1}^n (k)/3^{n - (k - 1)} = sum_{k=1}^n k / 3^{n - k + 1}.Wait, that exponent can be rewritten as 3^{(n - k + 1)}. Let me change variables to make it easier. Let m = n - k + 1. When k=1, m = n; when k=n, m=1. So, b_n = sum_{m=1}^n (n - m + 1)/3^m.So, b_n = sum_{m=1}^n (n - m + 1)/3^m.That might be easier to handle. So, b_n = sum_{m=1}^n (n + 1 - m)/3^m.We can split this into two sums: b_n = (n + 1) sum_{m=1}^n 1/3^m - sum_{m=1}^n m/3^m.So, b_n = (n + 1) * S1 - S2, where S1 is the sum of 1/3^m from m=1 to n, and S2 is the sum of m/3^m from m=1 to n.We can compute these sums separately.First, S1 = sum_{m=1}^n 1/3^m. This is a finite geometric series with first term 1/3 and ratio 1/3. The sum is S1 = (1/3)(1 - (1/3)^n)/(1 - 1/3) = (1/3)(1 - 1/3^n)/(2/3) = (1/3)*(3/2)(1 - 1/3^n) = (1/2)(1 - 1/3^n).Next, S2 = sum_{m=1}^n m/3^m. This is a finite arithmetic-geometric series. There's a formula for this. The sum from m=1 to infinity of m x^m is x/(1 - x)^2, but since we have a finite sum, it's a bit more involved.The formula for sum_{m=1}^n m x^m is x(1 - (n + 1)x^n + n x^{n + 1}) / (1 - x)^2.In our case, x = 1/3. So, S2 = (1/3)(1 - (n + 1)(1/3)^n + n(1/3)^{n + 1}) / (1 - 1/3)^2.Simplify the denominator: (1 - 1/3)^2 = (2/3)^2 = 4/9.So, S2 = (1/3)(1 - (n + 1)/3^n + n/3^{n + 1}) / (4/9) = (1/3)*(9/4)(1 - (n + 1)/3^n + n/3^{n + 1}) = (3/4)(1 - (n + 1)/3^n + n/3^{n + 1}).Simplify further: (3/4) - (3/4)(n + 1)/3^n + (3/4)(n)/3^{n + 1}.Simplify the last term: (3/4)(n)/3^{n + 1} = (3/4)(n)/(3*3^n) = (1/4)(n)/3^n.So, S2 = (3/4) - (3/4)(n + 1)/3^n + (1/4)n/3^n.Combine the terms with 1/3^n: - (3/4)(n + 1)/3^n + (1/4)n/3^n = [ -3(n + 1)/4 + n/4 ] / 3^n = [ -3n - 3 + n ] / 4 / 3^n = [ -2n - 3 ] / 4 / 3^n.So, S2 = 3/4 - (2n + 3)/(4*3^n).Therefore, going back to b_n:b_n = (n + 1)S1 - S2 = (n + 1)(1/2)(1 - 1/3^n) - [3/4 - (2n + 3)/(4*3^n)].Let's expand this:= (n + 1)/2 - (n + 1)/(2*3^n) - 3/4 + (2n + 3)/(4*3^n).Combine like terms:= (n + 1)/2 - 3/4 + [ - (n + 1)/(2*3^n) + (2n + 3)/(4*3^n) ].Simplify the constants: (n + 1)/2 - 3/4 = (2(n + 1) - 3)/4 = (2n + 2 - 3)/4 = (2n - 1)/4.Now, the terms with 1/3^n:= [ - (n + 1)/(2*3^n) + (2n + 3)/(4*3^n) ] = [ -2(n + 1) + (2n + 3) ] / (4*3^n) = [ -2n - 2 + 2n + 3 ] / (4*3^n) = (1)/(4*3^n).So, putting it all together:b_n = (2n - 1)/4 + 1/(4*3^n).Therefore, b_n = (2n - 1)/4 + 1/(4*3^n).Now, we need to find T_n = sum_{k=1}^n b_k.So, T_n = sum_{k=1}^n [ (2k - 1)/4 + 1/(4*3^k) ] = (1/4) sum_{k=1}^n (2k - 1) + (1/4) sum_{k=1}^n 1/3^k.Let's compute each sum separately.First sum: sum_{k=1}^n (2k - 1) = 2 sum_{k=1}^n k - sum_{k=1}^n 1 = 2*(n(n + 1)/2) - n = n(n + 1) - n = n^2 + n - n = n^2.Second sum: sum_{k=1}^n 1/3^k. This is a finite geometric series with first term 1/3 and ratio 1/3. The sum is S = (1/3)(1 - (1/3)^n)/(1 - 1/3) = (1/3)(1 - 1/3^n)/(2/3) = (1/3)*(3/2)(1 - 1/3^n) = (1/2)(1 - 1/3^n).Therefore, T_n = (1/4)*n^2 + (1/4)*(1/2)(1 - 1/3^n) = (n^2)/4 + (1/8)(1 - 1/3^n).So, T_n = n²/4 + 1/8 - 1/(8*3^n).We can write this as T_n = (n²)/4 + (1 - 1/3^n)/8.Alternatively, combining the constants:T_n = (n²)/4 + 1/8 - 1/(8*3^n).So, that's the expression for T_n.**Final Answer**(1) The general formula for the sequence is boxed{a_n = n}.(2) The sum of the first ( n ) terms of the sequence ({b_n}) is boxed{T_n = frac{n^2}{4} + frac{1}{8} - frac{1}{8 cdot 3^n}}.

answer：First, I'll assign variables to represent the ages of A, B, and C as ( a ), ( b ), and ( c ) respectively.The problem states that the total age of A and B is 20 years more than the total age of B and C. This can be written as:[a + b = (b + c) + 20]To find out how much younger C is compared to A, I'll rearrange the equation. Subtract ( b ) from both sides:[a = c + 20]This equation shows that C is 20 years younger than A.