answer:Start by converting each repeating decimal to its respective fraction: 1. 0.overline{6} becomes frac{6}{9} = frac{2}{3}, since 6 = 2 times 3 and 9 = 3 times 3. 2. 0.overline{2} is written as frac{2}{9}. 3. 0.overline{4} becomes frac{4}{9}. 4. 0.overline{7} is expressed as frac{7}{9}. Next, sum these fractions: frac{2}{3} + frac{2}{9} - frac{4}{9} - frac{7}{9}. Now, finding a common denominator (9) for all terms: 1. Convert frac{2}{3} to frac{6}{9} to match the denominator of the other terms. 2. Combine all terms: frac{6}{9} + frac{2}{9} - frac{4}{9} - frac{7}{9}. Perform the arithmetic on the numerators: (6+2-4-7) = -3. Therefore, the combined fraction is frac{-3}{9}, which simplifies to frac{-1}{3}. Thus, the final result is boxed{frac{-1}{3}}.

answer:Let's call the number we're looking for "N". According to the problem, one fourth of one third of two fifths of N is 16. We can write this as an equation: (1/4) * (1/3) * (2/5) * N = 16 Now, let's solve for N: (1/4) * (1/3) * (2/5) * N = 16 (2/60) * N = 16 (1/30) * N = 16 Now, multiply both sides by 30 to solve for N: N = 16 * 30 N = 480 Now that we have the value of N, we can find 40% of it: 40% of N = 0.40 * N 40% of N = 0.40 * 480 40% of N = 192 So, 40% of the number is boxed{192} .

answer:Let t = sin x. Since x in (0, pi), we have t in (0, 1]. The function f(x) = sin^3 x + acos^2 x can be rewritten using t as follows: begin{equation} f(x) = t^3 + a(1 - t^2) = t^3 - at^2 + a. end{equation} Taking the derivative with respect to t gives us: begin{equation} y' = frac{df}{dt} = 3t^2 - 2at = t(3t - 2a). end{equation} Setting y' = 0 yields the critical points t = 0 or t = frac{2a}{3}. Since f(x) = sin^3 x + acos^2 x has a minimum in the interval (0, pi), the corresponding function of t, y = t^3 - at^2 + a, must be decreasing or first decrease then increase over the interval (0, 1]. This requires frac{2a}{3} > 0 and thus a > 0. Therefore, the range of the real number a is (0, +infty), So, the correct choice is: boxed{D}. To solve this problem, we used the property of the sine function to determine the range of t given x in (0, pi), substituted t into f(x), took the derivative to find the critical points, and then determined the monotonicity of the function on the given interval. The inequality came from the relationship between the derivative and monotonicity of the function, which helped us find the range of the real number a. The problem tests the understanding of the properties of the sine function, the relationship between derivatives and monotonicity of functions, and the application of the constructive method and substitution method, assessing simplification and transformation skills.

answer:To determine the number of zeros at the end of 345!, we need to count the number of times 5 is a factor in the numbers from 1 to 345. This is because each pair of factors of 2 and 5 contributes a terminal zero, and there are always more factors of 2 than 5 in factorials. 1. Count factors of 5: leftlfloor frac{345}{5} rightrfloor = 69 2. Count factors of 5^2 = 25: leftlfloor frac{345}{25} rightrfloor = 13 3. Count factors of 5^3 = 125: leftlfloor frac{345}{125} rightrfloor = 2 4. No factors from 5^4 = 625 as 625 > 345. Adding these together gives the total number of factors of 5: 69 + 13 + 2 = 84. Thus, 345! ends in boxed{84} zeros.