answer:1. **Given expressions for u and v:** [ u = 3 + 3^q quad text{and} quad v = 3 + 3^{-q} ] 2. **Express 3^q in terms of u:** [ u = 3 + 3^q implies 3^q = u - 3 ] 3. **Substitute 3^q in the expression for v:** [ v = 3 + 3^{-q} = 3 + frac{1}{3^q} ] Using the expression for 3^q from step 2: [ v = 3 + frac{1}{u-3} ] 4. **Combine terms under a common denominator:** [ v = frac{3(u-3) + 1}{u-3} = frac{3u - 9 + 1}{u-3} = frac{3u - 8}{u-3} ] 5. **Conclusion:** The expression for v in terms of u is frac{3u-8}{u-3}. [ frac{3u-8{u-3}} ] The final answer is boxed{text{B}}

answer:To address the false proposition and provide a counterexample, let's analyze the statement and its implications step by step: 1. The proposition states: "A triangle with two acute angles is an acute triangle." 2. We know that triangles can be classified based on their angles into three types: acute triangles (all angles are acute), right triangles (one angle is exactly 90^circ), and obtuse triangles (one angle is obtuse, i.e., more than 90^circ). 3. For a triangle to be an acute triangle, all its angles must be acute. However, the proposition only mentions two acute angles, leaving the third angle's nature unspecified. 4. Considering the nature of right triangles, they have exactly one right angle and two acute angles. This configuration contradicts the proposition since it presents a scenario where a triangle has two acute angles but is not an acute triangle. 5. Therefore, a right triangle serves as a counterexample to the proposition. It has two acute angles but is not classified as an acute triangle due to its right angle. Hence, the counterexample to the false proposition "A triangle with two acute angles is an acute triangle" is: boxed{text{A right triangle has two acute angles.}}

answer:1. **Finding ( s_1 ):** The sequence ( A ) has the first term ( a_1 = 5 ) and common difference ( d_1 = 10 - 5 = 5 ). The sum of the first ( n ) terms is given by: [ s_1 = frac{n}{2}(2a_1 + (n - 1)d_1) ] Substituting the given values: [ s_1 = frac{n}{2}(2 times 5 + (n - 1) times 5) = frac{n}{2}(10 + 5n - 5) = frac{n}{2}(5n + 5) = frac{5n(n + 1)}{2} ] 2. **Finding ( s_2 ):** The sequence ( B ) has the first term ( a_2 = 20 ) and common difference ( d_2 = 25 - 20 = 5 ). The sum formula becomes: [ s_2 = frac{n}{2}(2a_2 + (n - 1)d_2) ] Substituting the given values: [ s_2 = frac{n}{2}(2 times 20 + (n - 1) times 5) = frac{n}{2}(40 + 5n - 5) = frac{n}{2}(5n + 35) = frac{5n(n + 7)}{2} ] 3. **Equating ( s_1 ) and ( s_2 )**: [ frac{5n(n + 1)}{2} = frac{5n(n + 7)}{2} ] Simplifying, we get: [ 5n(n + 1) = 5n(n + 7) ] [ n + 1 = n + 7 ] [ 1 = 7 ] This is absurd. Therefore, there are no values of ( n ) that satisfy ( s_1 = s_2 ). [ text{No value of n} ] The final answer is boxed{(A) no value of n}

answer:Using the given values, calculate (left(frac{4}{7}right)^2 left(frac{5}{6}right)^{-2}). This can be rewritten using the properties of exponents as: [ left(frac{4}{7}right)^2 left(left(frac{5}{6}right)^{-1}right)^2 = left(frac{4}{7}right)^2 left(frac{6}{5}right)^2. ] Expanding this using the rule (left(frac{a}{b}right)^n = frac{a^n}{b^n}), we get: [ frac{4^2}{7^2} cdot frac{6^2}{5^2} = frac{16}{49} cdot frac{36}{25}. ] Combining the fractions, we have: [ frac{16 cdot 36}{49 cdot 25} = frac{576}{1225}. ] Thus, the final answer is (boxed{frac{576}{1225}}).