answer:First, we convert the base of the last term to base 2 using the change of base formula: 2^{log_{4} 3} = 2^{frac{log_{2} 3}{log_{2} 4}} = 2^{frac{log_{2} 3}{2}} = sqrt{2^{log_{2} 3}} = sqrt{3}. Now we rewrite the expression using the properties of logarithms: lg frac{5}{2} + 2 lg 2 + 2^{log_{4} 3} = lg frac{5}{2} + lg 2^2 + sqrt{3} = lg frac{5}{2} + lg 4 + sqrt{3}. Next, we combine the first two terms using the product rule for logarithms: lg frac{5}{2} + lg 4 = lg left(frac{5}{2} times 4 right) = lg 10. Finally, we substitute this result back into the expression and simplify: lg 10 + sqrt{3} = 1 + sqrt{3}. Thus, the final answer is boxed{1 + sqrt{3}}.

answer:Let's start by calculating the total amount raised from the SUVs, trucks, and motorcycles that we know were washed. For the SUVs: 3 SUVs * 12 per SUV = 36 For the trucks: 8 trucks * 10 per truck = 80 For the motorcycles: 5 motorcycles * 15 per motorcycle = 75 Now, let's add up the amounts from the SUVs, trucks, and motorcycles: 36 (SUVs) + 80 (trucks) + 75 (motorcycles) = 191 We know that the total amount raised was 500, so let's subtract the amount raised from the known vehicles from the total: 500 (total) - 191 (known vehicles) = 309 This 309 is the amount raised from the cars and buses. Let's denote the number of cars as C and the number of buses as B. We have the following equations based on the prices for cars and buses and the total amount raised from them: 7C (for cars) + 18B (for buses) = 309 We also know that the total number of vehicles cannot exceed 15, and we already have 3 SUVs, 8 trucks, and 5 motorcycles, which adds up to 16 vehicles. This is not possible, so there must be a mistake in the given information. However, if we proceed with the assumption that the total number of vehicles is 15, we have: 3 (SUVs) + 8 (trucks) + 5 (motorcycles) + C (cars) + B (buses) = 15 Since we already have boxed{16} vehicles, we cannot add any more cars or buses without exceeding the total vehicle limit. Therefore, there seems to be an inconsistency in the provided information. Please check the details and provide the correct information so we can proceed with the calculation.

answer:To find the roots of the quadratic 5x^2 + 7x + k, we apply the quadratic formula: x = frac{-b pm sqrt{b^2 - 4ac}}{2a} For this specific equation: x = frac{-7 pm sqrt{7^2 - 4 cdot 5 cdot k}}{10} We simplify this to: x = frac{-7 pm sqrt{49 - 20k}}{10} Given that the roots are frac{-7 pm isqrt{191}}{10}, equate the expressions inside the square root: sqrt{49 - 20k} = isqrt{191} Square both sides to remove the square root: 49 - 20k = -191 Solve for k: 49 + 191 = 20k 240 = 20k k = frac{240}{20} = boxed{12}

answer:1. **Equation Setup Based on Given Mean:** - Given the mean of ( x ) and ( y ) is 55, we have: [ frac{x + y}{2} = 55 ] - Simplifying, we find: [ x + y = 110 ] 2. **Objective to Maximize ( frac{x}{y} ):** - To maximize ( frac{x}{y} ), ( x ) should be the maximum odd integer within the limits and ( y ) should be the minimum odd integer to remain within constraints. - Since ( x ) and ( y ) are confined to be odd, the maximum odd two-digit integer is ( 99 ). Setting ( x = 99 ), we then solve for ( y ): [ 99 + y = 110 implies y = 11 ] - Here, ( y = 11 ) satisfies the conditions of being an odd two-digit integer, and is the smallest possible value maintaining ( y ) as a two-digit number. 3. **Calculate the Ratio:** - With ( x = 99 ) and ( y = 11 ), compute ( frac{x}{y} ): [ frac{x}{y} = frac{99}{11} = 9 ] 4. **Conclusion:** - The maximum value of the ratio ( frac{x}{y} ), given the constraints of ( x ) and ( y ) being odd two-digit integers whose mean is 55, is ( 9 ). - The final answer is ( 9 ). The final answer is boxed{(textbf{(D)} 9)}