answer:Let's denote the number of sit-ups Carrie can do per minute as ( C ) and the number of sit-ups Jerrie can do per minute as ( J ). According to the problem, Jerrie can do 5 more sit-ups per minute than Carrie, so we have: ( J = C + 5 ) We know that Barney can do 45 sit-ups in one minute. The total number of sit-ups performed by all three is 510. We can write this as: ( 45 times 1 + C times 2 + J times 3 = 510 ) Substituting ( J ) with ( C + 5 ), we get: ( 45 + 2C + 3(C + 5) = 510 ) Expanding the equation, we get: ( 45 + 2C + 3C + 15 = 510 ) Combining like terms, we get: ( 5C + 60 = 510 ) Subtracting 60 from both sides, we get: ( 5C = 450 ) Dividing both sides by 5, we get: ( C = 90 ) So, Carrie can do 90 sit-ups per minute. Now, we need to find the ratio of the number of sit-ups Carrie can do per minute to the number of sit-ups Barney can do per minute. Since Barney can do 45 sit-ups per minute, the ratio is: ( frac{C}{45} = frac{90}{45} ) Simplifying the ratio, we get: ( frac{90}{45} = 2 ) Therefore, the ratio of the number of sit-ups Carrie can do per minute to the number of sit-ups Barney can do per minute is boxed{2:1} .

answer:To solve the given problem, we start with the given condition that a+b=3 and the inequality a^{5}+b^{5}geqslant lambda ab. We aim to find the range of lambda for which this inequality always holds. First, we rewrite the given inequality in a more manageable form. We have: [ frac{a^5}{b} + frac{b^5}{a} geq lambda ] Given that a+b=3, we can manipulate the left-hand side of the inequality as follows: [ frac{a^5}{b} + frac{b^5}{a} = frac{(frac{a^5}{b} + frac{b^5}{a})(a+b)}{3} = frac{frac{a^5}{b} + frac{b^5}{a} + a^4 + b^4}{3} ] Applying the AM-GM inequality, we get: [ geq frac{2sqrt{frac{a^5}{b} cdot frac{b^5}{a}} + a^4 + b^4}{3} = frac{a^4 + b^4 + 2a^2b^2}{3} = frac{(a^2 + b^2)^2}{3} ] Further simplifying, we use the fact that the arithmetic mean is greater than or equal to the geometric mean: [ = frac{(frac{a^2 + b^2}{2} + frac{a^2 + b^2}{2})^2}{3} geq frac{(frac{a^2 + b^2}{2} + frac{2ab}{2})^2}{3} = frac{(a+b)^4}{12} = frac{27}{4} ] Thus, we have shown that frac{a^4}{b} + frac{b^4}{a} geq frac{27}{4}, and equality holds when a = b = frac{3}{2}. This means that the range of lambda for which the original inequality always holds is (-infty, frac{27}{4}]. Therefore, the correct answer is boxed{D}.

answer:First, find the divisors of 60. 60 = 2^2 cdot 3 cdot 5, so the divisors are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. These can be paired off into six pairs, where the product of each pair is 60: begin{align*} {1&, 60}, {2&, 30}, {3&, 20}, {4&, 15}, {5&, 12}, text{and} {6&, 10}. end{align*} Therefore, B = 60^6. Simplifying, [ 60^6 = (2^2 cdot 3 cdot 5)^6 = 2^{12} cdot 3^6 cdot 5^6. ] B has three distinct prime factors, which are 2, 3, and 5. Conclusion: B has boxed{3} distinct prime factors.

answer:According to the steps and methods of proving mathematical propositions by contradiction, we should first assume that the negation of the proposition is true. The negation of the proposition "neither a nor b can be divided by 5" is "at least one of a and b can be divided by 5", Therefore, the correct choice is boxed{text{C}}.