answer:Since the function f(x)= begin{cases} log_{2}x & x > 0 3^{x} & xleqslant 0 end{cases}, we have f(0)=3^{0}=1, Therefore, f[f(0)]=f(1)=log_{2}1=0, Hence, the answer is boxed{0}. The value of f(0) is obtained from the function's definition, and then the value of f[f(0)] is determined. This question mainly tests the ability to use piecewise functions to find the value of a function, and it is a basic question.

answer:1. **Identify and Define Given Areas**: - The area of the large square is given as (16 , text{cm}^2). - The area of each small corner square is (1 , text{cm}^2). 2. **Calculate the Side Lengths of Squares**: - Let's denote the side length of the large square as (s_{text{large}}). By definition of area for a square: [ s_{text{large}}^2 = 16 , text{cm}^2 implies s_{text{large}} = sqrt{16} = 4 , text{cm} ] - Similarly, the side length of each small square, (s_{text{small}}), can be calculated as: [ s_{text{small}}^2 = 1 , text{cm}^2 implies s_{text{small}} = sqrt{1} = 1 , text{cm} ] 3. **Determine the Geometry and Dimensions of Triangles**: - Each of the four triangles is formed between the centers of adjacent small squares along the sides of the large square. - The base of each triangle is thus equivalent to the side length of adjacent small squares: ( s_{text{small}} + s_{text{small}} = 1 , text{cm} + 1 , text{cm} = 2 , text{cm} ). - The height of each triangle, as the diagonal distance from center to center of the small squares, lying halfway along the side of the large square, is also (2 , text{cm} ). 4. **Calculate the Area of One Triangle**: - The area (A) of each of these right triangles can be calculated using: [ A_{triangle} = frac{1}{2} times text{base} times text{height} ] [ A_{triangle} = frac{1}{2} times 2 , text{cm} times 2 , text{cm} = 2 , text{cm}^2 ] 5. **Sum the Areas of All Triangles**: - There are four such triangles. Thus, the total area of these triangles is: [ 4 times 2 , text{cm}^2 = 8 , text{cm}^2 ] 6. **Calculate the Shaded Area**: - The total area occupied by shapes other than the shaded area is combination of the triangles and the small squares: [ text{Total non-shaded area} = text{Area of 4 triangles} + text{Area of 4 small squares} ] [ text{Total non-shaded area} = 8 , text{cm}^2 + (4 times 1 , text{cm}^2) = 8 , text{cm}^2 + 4 , text{cm}^2 = 12 , text{cm}^2 ] - Consequently, the shaded area (A_{text{shaded}}) is the difference between the area of the large square and the non-shaded areas: [ A_{text{shaded}} = text{Area of large square} - text{Total non-shaded area} ] [ A_{text{shaded}} = 16 , text{cm}^2 - 12 , text{cm}^2 = 4 , text{cm}^2 ] Conclusion: [ boxed{4 , text{cm}^2} ]

answer:First, let's find the volume of one cube with an edge of 3 units. The volume of a cube is found by cubing the length of one of its edges: Volume of one cube = edge^3 = 3^3 = 3 * 3 * 3 = 27 cubic units Since 48 such cubes can be inserted into the material, the total volume of the material is: Total volume = Volume of one cube * Number of cubes = 27 * 48 Now, let's calculate that: Total volume = 27 * 48 = 1296 cubic units The volume of the material is also given as the product of three numbers, two of which are 18 and 6. So, the volume can also be expressed as: Total volume = first number * 18 * 6 We know the total volume is 1296 cubic units, so we can set up the equation: 1296 = first number * 18 * 6 Now, let's solve for the first number: first number = 1296 / (18 * 6) first number = 1296 / 108 first number = 12 So, the first number in the product is boxed{12} .

answer:1. Since a_1 + a_4 = 14, we have 2a_1 + 3d = 14, (eq. 1). And, since a_2^2 = a_1a_7, we have (a_1 + d)^2 = a_1(a_1 + 6d), (eq. 2). Solving equations (eq. 1) and (eq. 2), we get d^2 = 4a_1d. As d neq 0, we have d = 4a_1. Substituting this back into (eq. 1), we find d = 4 and a_1 = 1. Therefore, the general term formula is a_n = a_1 + (n - 1)d = 4n - 3, and the sum formula for the first n terms is S_n = frac{n(1 + 4n - 3)}{2} = 2n^2 - n. 2. From part 1, we know that b_n = frac{2n^2 - n}{n + k}. Since {b_n} is an arithmetic sequence, we have 2b_2 = b_1 + b_3, which gives us 2 cdot frac{6}{2 + k} = frac{1}{1 + k} + frac{15}{3 + k}. Solving this equation, we find two possible values for k: k = -frac{1}{2} or k = 0. Case 1: If k = -frac{1}{2}, then b_n = 2n, and frac{1}{b_n b_{n + 1}} = frac{1}{4}(frac{1}{n} - frac{1}{n + 1}). Therefore, T_n = frac{1}{4}(1 - frac{1}{n + 1}) = boxed{frac{n}{4(n + 1)}}. Case 2: If k = 0, then b_n = 2n - 1, and frac{1}{b_n b_{n + 1}} = frac{1}{2}(frac{1}{2n - 1} - frac{1}{2n + 1}). Therefore, T_n = frac{1}{2}(1 - frac{1}{2n + 1}) = boxed{frac{n}{2n + 1}}. In both cases, it is easy to verify that T_n leqslant frac{1}{2}.