answer:1. **Objective**: We aim to maximize the fraction frac{A+B}{C+D} under the constraint that C+D > 1 and with A, B, C, D as distinct digits from 0 to 9. 2. **Maximizing A+B**: - Choose the largest available distinct digits for A and B. Opting for A = 9 and B = 7 (since using 8 again as in the original solution but trying a different combination). - This gives A+B = 9 + 7 = 16. 3. **Minimizing C+D with a constraint of C+D > 1**: - Choose the smallest available distinct digits greater than 1 for C and D that are unused by A and B. Let's use C = 2 and D = 3. - This yields C+D = 2 + 3 = 5. 4. **Calculating the fraction**: - We now calculate frac{A+B}{C+D} = frac{16}{5}. However, this is not an integer. - Adjust A or B to find an integer ratio: Instead using A = 9 and B = 6. - Now A+B = 9 + 6 = 15, and frac{A+B}{C+D} = frac{15}{5} = 3, which is an integer and likely maximized given our constraints. 5. **Conclusion**: - The highest possible integer ratio under the given constraints with C+D > 1 is achieved with A+B = 15. - Therefore, the value of A+B is 15. The final answer is boxed{15} (Choice D).

answer:**Solution**: For option A, the concentration of lemon juice in the mixture is frac{2}{15} , while in Mix A, the concentration of lemon juice is frac{1}{6} , and in Mix B, the concentration of lemon juice is frac{3}{12} . It is not possible to achieve the corresponding integer part solution, so it cannot be adjusted to this ratio, which does not meet the requirements of the problem; For option B, the concentration of lemon juice in the mixture is frac{4}{15} , while in Mix A, the concentration of lemon juice is frac{1}{6} , and in Mix B, the concentration of lemon juice is frac{3}{12} . It is not possible to achieve the corresponding integer part solution, so it cannot be adjusted to this ratio, which does not meet the requirements of the problem; For option C, the concentration of lemon juice in the mixture is frac{3}{15} , while in Mix A, the concentration of lemon juice is frac{1}{6} , and in Mix B, the concentration of lemon juice is frac{3}{12} . Let's take x parts of Mix A and y parts of Mix B. frac{1}{6}x + frac{3}{12}y = frac{3}{15}(x+y) Solving this equation gives x = 1.5y, so we can take 3 parts of Mix A and 2 parts of Mix B to make the mixture, which meets the requirements of the problem; For option D, the concentration of lemon juice in the mixture is frac{5}{18} , while in Mix A, the concentration of lemon juice is frac{1}{6} , and in Mix B, the concentration of lemon juice is frac{3}{12} . It is not possible to achieve the corresponding integer part solution, so it cannot be adjusted to this ratio, which does not meet the requirements of the problem; Therefore, the correct answer is boxed{text{C}}.

answer:1. **Understanding the Problem:** We need to find the maximum number of elements in a set ( B ) such that for any two distinct elements ( a ) and ( b ) in ( B ), the difference ( a - b ) does not divide the sum ( a + b ). 2. **Initial Observations:** - If ( a ) and ( b ) are two elements in ( B ), then ( a neq b ). - We need to ensure that ( (a - b) not| (a + b) ). 3. **Considering Differences:** - If ( a ) and ( b ) are consecutive integers, say ( a = k ) and ( b = k+1 ), then ( a - b = -1 ) and ( a + b = 2k + 1 ). Clearly, (-1) divides ( 2k + 1 ), so we cannot have consecutive integers in ( B ). - If ( a ) and ( b ) differ by 2, say ( a = k ) and ( b = k+2 ), then ( a - b = -2 ) and ( a + b = 2k + 2 ). Clearly, (-2) divides ( 2k + 2 ), so we cannot have integers differing by 2 in ( B ). 4. **Finding an Upper Bound:** - To avoid differences of 1 or 2, we can consider numbers that are congruent to the same residue modulo 3. - Let’s choose numbers that are ( 1 mod 3 ). These numbers are of the form ( 3k + 1 ). 5. **Constructing the Set:** - The numbers ( 1, 4, 7, 10, ldots ) are all ( 1 mod 3 ). - The maximum number of such numbers in the range ( 1, 2, ldots, n ) is given by ( leftlceil frac{n}{3} rightrceil ). 6. **Verification:** - For any two numbers ( a ) and ( b ) in this set, ( a equiv 1 mod 3 ) and ( b equiv 1 mod 3 ). - Therefore, ( a - b equiv 0 mod 3 ) and ( a + b equiv 2 mod 3 ). - Since ( a - b ) is a multiple of 3 and ( a + b ) is not, ( (a - b) not| (a + b) ). Thus, the maximum number of elements in set ( B ) is ( leftlceil frac{n}{3} rightrceil ). The final answer is ( boxed{ leftlceil frac{n}{3} rightrceil } ).

answer:Given that acute angles alpha and beta satisfy sinalpha=frac{4}{5}, we start by finding cosalpha using the Pythagorean identity for sine and cosine: cosalpha = sqrt{1-sin^2alpha} = sqrt{1-left(frac{4}{5}right)^2} = sqrt{1-frac{16}{25}} = sqrt{frac{9}{25}} = frac{3}{5}. Given that cos(alpha+beta)=-frac{12}{13} and knowing that alpha + beta is in the interval (0,pi), which implies that alpha + beta is an obtuse angle, we find sin(alpha+beta) using the Pythagorean identity: sin(alpha+beta) = sqrt{1-cos^2(alpha+beta)} = sqrt{1-left(-frac{12}{13}right)^2} = sqrt{1-frac{144}{169}} = sqrt{frac{25}{169}} = frac{5}{13}. Now, we can find cosbeta using the angle subtraction formula for cosine, which is cos(alpha-beta) = cosalphacosbeta + sinalphasinbeta: cosbeta = cos[(alpha+beta)-alpha] = cos(alpha+beta)cosalpha + sin(alpha+beta)sinalpha = left(-frac{12}{13}right)left(frac{3}{5}right) + left(frac{5}{13}right)left(frac{4}{5}right) = -frac{36}{65} + frac{20}{65} = -frac{16}{65}. Therefore, the correct answer is boxed{B}.