answer:Let's start with the original formula: a = (2r)^n Now, if r is halved, we replace r with r/2: a' = [2(r/2)]^n a' = (r)^n We are told that when r is halved, a is multiplied by 0.125. This means that: a' = 0.125 * a Substituting the expressions for a and a' from above, we get: (r)^n = 0.125 * (2r)^n Now, we can divide both sides by (r)^n to isolate the constants: 1 = 0.125 * (2)^n Now, we need to solve for n. We can rewrite 0.125 as a power of 2 because 0.125 is 1/8, and 1/8 is 2^-3: 1 = (2^-3) * (2)^n Using the property of exponents that a^m * a^n = a^(m+n), we can combine the exponents on the right side: 1 = 2^(n-3) Since the left side is 1, which is 2^0, we can equate the exponents: 0 = n - 3 Solving for n: n = 3 So the value of n is boxed{3} .

answer:Since f(x) is an even function, we have f(|x+2|)=f(x+2), thus, f(x+2) < 5 can be transformed into f(|x+2|) < 5, which is |x+2|^{2}-4|x+2| < 5, (|x+2|+1)(|x+2|-5) < 0, therefore, |x+2| < 5, solving this gives -7 < x < 3, hence, the solution set of the inequality f(x+2) < 5 is (-7,3). Therefore, the answer is: boxed{(-7,3)}. By the property of even functions, we have: f(|x+2|)=f(x+2), then f(x+2) < 5 can be transformed into f(|x+2|) < 5, substituting the given expression to represent the inequality, first solve the range of |x+2|, then find the range of x. This problem examines the properties of even and odd functions and the solution method for quadratic inequalities. Utilizing the property of even functions to specify the inequality is the key to solving this problem.

answer:Given that the boxes are distinguishable but the balls are not, we need to consider various cases for the distribution of the balls across 4 boxes: 1. **Case (7,0,0,0)**: Here we have 4 ways (choosing which box gets all the balls). 2. **Case (6,1,0,0)**: For this distribution, we select one box to get 6 balls (4 choices), and one box to get 1 ball from the remaining 3 boxes (3 choices), yielding 4 times 3 = 12 combinations. 3. **Case (5,2,0,0)**: We choose one box for 5 balls (4 ways), and one box for 2 balls from the remaining 3 boxes (3 ways), making 4 times 3 = 12 combinations. 4. **Case (5,1,1,0)**: One box receives 5 balls (4 choices), and two out of the remaining three boxes each get 1 ball (binom{3}{2} = 3 choices), for a total of 4 times 3 = 12 combinations. 5. **Case (4,3,0,0)**: Choose one box for 4 balls (4 ways) and one box for 3 balls from the remaining 3 boxes (3 ways), amounting to 4 times 3 = 12 combinations. 6. **Case (4,2,1,0) and (4,1,1,1)**: Choose one for 4 balls (4 ways). For (4,2,1,0), select one box for 2 balls from the remaining (3 ways) and one for 1 ball from the remaining (2 ways), giving 4 times 3 times 2 = 24 combinations. For (4,1,1,1), after choosing the box for 4 balls, the remaining distribution is fixed, so 4 combinations. 7. **Case (3,3,1,0) and (3,2,2,0) and (3,2,1,1) and (2,2,2,1)**: Each of these cases is symmetrical: - (3,3,1,0): Choose a box for 3 balls (binom{4}{2} = 6 ways for the pairs), and one for 1 ball from 2 remaining (2 ways), total 6 times 2 = 12 combos. - (3,2,2,0): Choose a box for 3 balls (4 ways), two boxes get 2 balls binom{3}{2} = 3 ways, yielding 4 times 3 = 12 combinations. - (3,2,1,1): Choose a box for 3 balls (4 ways), box for 2 balls from remaining three (3 ways), fixed other boxes, 4 times 3 = 12 combinations. - (2,2,2,1): Choose a box for 1 ball (4 ways), distribute 2-2-2 to rest, binom{3}{3} = 1 way, so 4 combos. Summing these, we have 4 + 12 + 12 + 12 + 12 + 24 + 12 + 12 + 12 + 12 + 4 = boxed{128} arrangements.

answer:1. **Analyze the multiplication equation:** [ begin{array}{cc} & A B times & C A hline & D A end{array} ] From the units column: Bcdot A = A (mod 10). For this to be possible, B cdot A equiv A pmod{10} implies either B equiv 1 or A = 0. 2. **Examine the subtraction equation:** [ begin{array}{cc} & A B - & C A hline & B end{array} ] From the units column, B - A = B pmod{10}, which is always true. Thus, no constraints here on A, B. No borrowing implies C leq A. 3. **Determine the range of C:** Since CA's tens digit affects D, CA must be a multiple of 10. Substituting B = 1 (since other cases lead to contradictions like B = 0 which is less suitable due to multiplication), we find: If A = 8 and C = 1, then 18 times 18 = 324, then D = 2. 4. **Substitute back to find B:** Since the preferences highly suggest (A, B) = (8, 1), and testing C = 1 gives D = 2 + 3 = 5. Conclusively D = 5. Final answer: 5. The final answer is boxed{textbf{(D)} 5}