answer:To solve this problem, we need to find all the geometric sequences that satisfy the conditions. These sequences are: 1, 2, 4; 4, 2, 1; 2, 4, 8; 8, 4, 2; 1, 3, 9; 9, 3, 1; 4, 6, 9; and 9, 6, 4, making a total of boxed{8} sequences. To reach the conclusion, we simply enumerate all the geometric sequences that meet the criteria. This question tests knowledge of permutations and combinations and the two fundamental principles, as well as the definition of a geometric sequence. It is considered a basic question.

answer:1. Given that (a, b, c) are non-negative real numbers such that (a + b + c = 1), we need to prove: [ frac{3}{16} leq left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 leq frac{1}{4}. ] 2. **Proof of the left inequality:** Let's construct the function: [ f(x) = left( frac{x}{1 + x} right)^2, quad x in [0, 1]. ] 3. Calculate the derivative of ( f(x) ): [ f'(x) = frac{d}{dx} left( frac{x^2}{(1 + x)^2} right). ] Using the quotient rule: [ f'(x) = frac{2x(1 + x)^2 - 2x^2(1 + x)}{(1 + x)^4} = frac{2x + 2x^2 - 2x^2}{(1 + x)^3} = frac{2x}{(1 + x)^3}. ] 4. Consider the point (left(frac{1}{3}, frac{1}{16} right)). Calculate ( f'left(frac{1}{3}right) ): [ f'left(frac{1}{3}right) = frac{2 cdot frac{1}{3}}{left(1 + frac{1}{3}right)^3} = frac{frac{2}{3}}{left(frac{4}{3}right)^3} = frac{frac{2}{3}}{frac{64}{27}} = frac{2 cdot 27}{3 cdot 64} = frac{54}{192} = frac{9}{32}. ] 5. The tangent line to ( f(x) ) at (left(frac{1}{3}, frac{1}{16} right) ) is: [ y = f'left(frac{1}{3}right) left(x - frac{1}{3} right) + fleft(frac{1}{3}right) = frac{9}{32} left(x - frac{1}{3} right) + frac{1}{16}. ] Simplifying this: [ y = frac{9}{32}x - frac{9}{32} cdot frac{1}{3} + frac{1}{16} = frac{9}{32}x - frac{3}{32} + frac{1}{16} = frac{9}{32}x - frac{3}{32} + frac{2}{32} = frac{9}{32}x - frac{1}{32}. ] 6. Establish the inequality: [ left( frac{x}{1 + x} right)^2 geq frac{9}{32} x - frac{1}{32}, quad x in [0, 1]. ] This can be simplified to: [ (x - 1)(3x - 1)^2 leq 0, quad x in [0, 1]. ] Since this inequality is satisfied for ( x in [0, 1] ). 7. Replace ( x ) with ( a, b, c ) respectively: [ left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 geq frac{9}{32}(a + b + c) - frac{3}{32}. ] 8. Using ( a + b + c = 1 ): [ frac{9}{32}(a + b + c) - frac{3}{32} = frac{9 cdot 1}{32} - frac{3}{32} = frac{9 - 3}{32} = frac{6}{32} = frac{3}{16}. ] Hence: [ left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 geq frac{3}{16}. ] 9. **Proof of the Right Inequality:** Construct the function again: [ f(x) = left( frac{x}{1 + x} right)^2, quad x in [0, 1]. ] 10. Consider the points ( P(0, 0) ) and ( Q left(1, frac{1}{4} right) ). 11. The line passing through ( P ) and ( Q ): [ y = frac{1}{4} x. ] 12. Establish the inequality: [ left( frac{x}{1 + x} right)^2 leq frac{1}{4} x, quad x in [0, 1]. ] Simplify this to: [ (x - 1)^2 x geq 0, quad x in [0, 1]. ] Since this inequality is satisfied for ( x in [0, 1] ). 13. Replace ( x ) with ( a, b, c ) respectively: [ left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 leq frac{1}{4} (a + b + c). ] 14. Using ( a + b + c = 1 ): [ frac{1}{4} (a + b + c) = frac{1 cdot 1}{4} = frac{1}{4}. ] Hence: [ left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 leq frac{1}{4}. ] # Conclusion: [ boxed{frac{3}{16} leq left( frac{a}{1 + a} right)^2 + left( frac{b}{1 + b} right)^2 + left( frac{c}{1 + c} right)^2 leq frac{1}{4}} ]

answer:To calculate 2 # 3 using the given operation, we substitute a with 2 and b with 3 into the formula: 2 # 3 = (2 * 3) - 3 + (3 ^ 2) First, we perform the multiplication and exponentiation: = (6) - 3 + (9) Then we perform the subtraction and addition: = 6 - 3 + 9 = 3 + 9 = 12 Therefore, 2 # 3 = boxed{12} .

answer:1. **Pattern Identification**: The width of the (n)-th rectangle is (2n) tiles, and the height is (n) tiles. 2. **Calculate the number of tiles in the eighth and ninth rectangles**: - The number of tiles in one rectangle is equal to its width times its height. - The eighth rectangle is (16 times 8 = 128) tiles because width = (2 times 8) and height = (8). - The ninth rectangle is (18 times 9 = 162) tiles because width = (2 times 9) and height = (9). 3. **Determine the difference in the number of tiles between the ninth and eighth rectangles**: - The difference is (162 - 128 = 34). 4. **Conclusion**: The ninth rectangle requires 34 more tiles than the eighth rectangle. 34 The final answer is boxed{C. 34}