answer:Let's start solving the problem step-by-step by simplifying each logarithmic expression. Given: [ frac{log_{2} 80}{log_{40} 2} - frac{log_{2} 160}{log_{20} 2} ] 1. **Convert the logarithms using the change of base formula.** The change of base formula states: [ log_b a = frac{log_k a}{log_k b} ] Therefore, we can rewrite the fractions involving logarithms as follows: [ frac{log_{2} 80}{log_{40} 2} = frac{frac{log 80}{log 2}}{frac{log 2}{log 40}} ] [ frac{log_{2} 160}{log_{20} 2} = frac{frac{log 160}{log 2}}{frac{log 2}{log 20}} ] 2. **Simplify the fractions:** Using the property that (frac{a}{frac{b}{c}} = a cdot frac{c}{b}), [ frac{frac{log 80}{log 2}}{frac{log 2}{log 40}} = frac{log 80}{log 2} cdot frac{log 40}{log 2} = frac{log 80 cdot log 40}{log^2 2} ] [ frac{frac{log 160}{log 2}}{frac{log 2}{log 20}} = frac{log 160}{log 2} cdot frac{log 20}{log 2} = frac{log 160 cdot log 20}{log^2 2} ] 3. **Combine the simplified fractions:** Thus, [ frac{log_{2} 80}{log_{40} 2} - frac{log_{2} 160}{log_{20} 2} = frac{log 80 cdot log 40 - log 160 cdot log 20}{log^2 2} ] 4. **Use the properties of logarithms to simplify further:** Logarithms can be expressed as: [ log 80 = log (2^4 cdot 5) = 4 log 2 + log 5 ] [ log 40 = log (2^3 cdot 5) = 3 log 2 + log 5 ] [ log 160 = log (2^5 cdot 5) = 5 log 2 + log 5 ] [ log 20 = log (2^2 cdot 5) = 2 log 2 + log 5 ] 5. **Substitute these expressions:** Now substitute these back into our combined expression: [ frac{(4 log 2 + log 5) cdot (3 log 2 + log 5) - (5 log 2 + log 5) cdot (2 log 2 + log 5)}{log^2 2} ] 6. **Expand the terms:** Expand both products in the numerator: [ (4 log 2 + log 5)(3 log 2 + log 5) = 12 log^2 2 + 4 log 2 log 5 + 3 log 2 log 5 + log^2 5 = 12 log^2 2 + 7 log 2 log 5 + log^2 5 ] Similarly, [ (5 log 2 + log 5)(2 log 2 + log 5) = 10 log^2 2 + 5 log 2 log 5 + 2 log 2 log 5 + log^2 5 = 10 log^2 2 + 7 log 2 log 5 + log^2 5 ] 7. **Subtract the expanded terms:** [ frac{(12 log^2 2 + 7 log 2 log 5 + log^2 5) - (10 log^2 2 + 7 log 2 log 5 + log^2 5)}{log^2 2} ] Simplify the numerator: [ = frac{12 log^2 2 + 7 log 2 log 5 + log^2 5 - 10 log^2 2 - 7 log 2 log 5 - log^2 5}{log^2 2} ] [ = frac{2 log^2 2}{log^2 2} ] 8. **Simplify the fraction:** [ = 2 ] # Conclusion: Thus, the final answer to the problem is: [ boxed{2} ]

answer:Let's denote the number of men on the team as M and the number of women as W. According to the information given: W = M + 6 (since there are 6 more women than men) We also know that the total number of players on the team is 24: M + W = 24 Now we can substitute the first equation into the second equation: M + (M + 6) = 24 2M + 6 = 24 2M = 24 - 6 2M = 18 M = 18 / 2 M = 9 Now that we know there are 9 men on the team, we can find the number of women: W = M + 6 W = 9 + 6 W = 15 So, there are 9 men and 15 women on the team. The ratio of men to women is: Men : Women 9 : 15 To simplify the ratio, we can divide both numbers by the greatest common divisor, which is 3: 9/3 : 15/3 3 : 5 Therefore, the ratio of men to women on the team is boxed{3:5} .

answer:Since sin x - cos x = frac{1}{2}, We square both sides to get 1 - sin 2x = frac{1}{4}, which simplifies to sin 2x = frac{3}{4}. Here's the step-by-step breakdown: 1. Square both sides of the equation: (sin x - cos x)^2 = (frac{1}{2})^2 2. Apply the double angle identity for sine: sin^2 x + cos^2 x - 2sin x cos x = frac{1}{4} 3. Recall that sin^2 x + cos^2 x = 1, so the equation becomes: 1 - 2sin x cos x = frac{1}{4} 4. The double angle identity for sine can also be written as sin 2x = 2sin x cos x, so we can substitute this into the equation: 1 - sin 2x = frac{1}{4} 5. Solving for sin 2x gives us: sin 2x = frac{3}{4} Therefore, the correct answer is A: boxed{frac{3}{4}}. This problem primarily tests the application of the double angle identity for sine and can be classified as a basic problem.

answer:Since Q(x) has real coefficients, the roots must include both cos alpha + i sin alpha, cos alpha - i sin alpha and cos(alpha + frac{pi}{6}) + i sin(alpha + frac{pi}{6}), cos(alpha + frac{pi}{6}) - i sin(alpha + frac{pi}{6}). These roots form two pairs of complex conjugates. [ Q(x) = (x - (cos alpha + i sin alpha))(x - (cos alpha - i sin alpha))(x - (cos(alpha + frac{pi}{6}) + i sin(alpha + frac{pi}{6})))(x - (cos(alpha + frac{pi}{6}) - i sin(alpha + frac{pi}{6}))) ] Expanding the products of conjugate pairs: [ = (x^2 - 2xcosalpha + 1)(x^2 - 2xcos(alpha + frac{pi}{6}) + 1) ] Sum of roots = 2(cosalpha + cos(alpha + frac{pi}{6})), To find Q(0): [ Q(0) = 1 cdot 1 = 1 ] So the area must be frac{1}{4}. We check if the sum of roots cos alpha + cos(alpha + frac{pi}{6}) gives a condition consistent with an area of frac{1}{4}. Conclusion: Given the final expression, we simplify the sum using the cosine sum formula. Assuming alpha was chosen adequately: [ boxed{2(cosalpha + cos(alpha + frac{pi}{6})) = text{sum of roots}}, boxed{Q(0) = 1} ]