answer:1. The contrapositive of "If xy=0, then x=0 and y=0" is "If x neq 0 or y neq 0, then xy neq 0". Since the original statement is incorrect, its contrapositive is also incorrect. 2. The negation of "A square is a rhombus" is "A quadrilateral that is not a square is not a rhombus". According to the definition of a rhombus, this negation is incorrect. 3. The converse of "If ac^2 > bc^2, then a > b" is "If a > b, then ac^2 > bc^2". This does not hold when c=0, so the converse is false. 4. If "m > 2, then the solution set of the inequality x^2 - 2x + m > 0 is mathbb{R}". When m > 2, Delta = 4 - 4m < 0, so the solution set of the inequality x^2 - 2x + m > 0 is indeed mathbb{R}. This proposition is true. In summary, only proposition 4 is true. Therefore, the correct answer is boxed{text{B}}.

answer:Let the roots of (x - sqrt[3]{25})(x - sqrt[3]{75})(x - sqrt[3]{125}) = 0 be alpha = sqrt[3]{25}, beta = sqrt[3]{75}, and gamma = sqrt[3]{125}. Using Vieta's formulas, the sums and products of the roots u, v, and w can be expressed as: begin{align*} u + v + w &= alpha + beta + gamma, uv + vw + uw &= alpha beta + beta gamma + gamma alpha, uvw &= alpha beta gamma + frac{1}{5}. end{align*} From the factorization of the cubic sum, we have: [u^3 + v^3 + w^3 - 3uvw = (u + v + w)((u + v + w)^2 - 3(uv + vw + uw))] Thus, [u^3 + v^3 + w^3 = (u + v + w)((u + v + w)^2 - 3(uv + vw + uw)) + 3uvw.] Substituting the values, [u^3 + v^3 + w^3 = (alpha + beta + gamma)((alpha + beta + gamma)^2 - 3(alpha beta + beta gamma + gamma alpha)) + 3(alpha beta gamma + frac{1}{5}).] Since the cubes of alpha, beta, gamma are alpha^3=25, beta^3=75, gamma^3=125, and the product alpha beta gamma =125 cdot 5 = 625, begin{align*} u^3 + v^3 + w^3 &= (5 + 15 + 25)((5 + 15 + 25)^2 - 3(75 + 375 + 1875)) + 3 times (625 + frac{1}{5}) &= 45(45^2 - 2325) + 3 times 625 + frac{3}{5} &= 45(2025 - 2325) + 1875 + 0.6 &= 45(-300) + 1875.6 &= -13500 + 1875.6 &= boxed{524.4}. end{align*}

answer:Given that this is an arithmetic sequence, we can express the terms as follows: a_1, a_1 + d, a_1 + 2d, ..., where d is the common difference. From the given, we have: 1. a_1 + a_4 + a_7 = a_1 + (a_1 + 3d) + (a_1 + 6d) = 39 2. a_2 + a_5 + a_8 = (a_1 + d) + (a_1 + 4d) + (a_1 + 7d) = 33 For a_3 + a_6 + a_9, it can be expressed as (a_1 + 2d) + (a_1 + 5d) + (a_1 + 8d). Notice that the sum of the coefficients of d in each equation forms an arithmetic sequence: 9, 12, 15. Similarly, the sum of the constants (which is just a_1 three times) forms an arithmetic sequence: 3a_1, 3a_1, 3a_1. Given the pattern, since the difference between 39 and 33 is 6, and the sequence of sums is decreasing, the next sum in the sequence would decrease by 6 again. Therefore, a_3 + a_6 + a_9 = 33 - 6 = 27. The correct answer is boxed{text{B: }27}.

answer:Given that u and v have a ratio of 3:5 and their sum is 16, we proceed as follows: 1. **Expressing u and v in terms of a common variable k**: [ frac{u}{v} = frac{3}{5} quad text{implies} quad u = 3k quad text{and} quad v = 5k ] 2. **Using the sum of u and v**: [ u + v = 16 quad Rightarrow quad 3k + 5k = 16 quad Rightarrow quad 8k = 16 quad Rightarrow quad k = 2 ] 3. **Substituting back to find u and v**: [ u = 3k = 3 times 2 = 6 quad text{and} quad v = 5k = 5 times 2 = 10 ] 4. **Determining the smaller value**: Since u = 6 and v = 10, we conclude that u is smaller than v. 5. **Conclusion**: The smaller of u and v is u, which is 6. Thus, the correct answer is 6. The final answer is boxed{6}.