answer:# Problem: A cube (ABCD-A_1B_1C_1D_1) has a side length of 1. Find the distance between the line segments (A_1C_1) and (BD_1). 1. First, let's visualize the cube and identify the relevant lines. The cube is represented with vertices: - Bottom face: (A, B, C, D) - Top face: (A_1, B_1, C_1, D_1) The lines in question are (A_1C_1) and (BD_1). 2. Consider the plane of the square (BB_1D_1D). Since the side length of the cube is 1, each edge has length 1. 3. To find the distance between the line segment (A_1C_1) and (BD_1), we need to determine the perpendicular distance between these lines. 4. Identify the midpoint (O) of the line segment (B_1D_1). Since (B_1D_1) is a diagonal of a square face with vertices (B_1, C_1, D_1), and this square face has a side length of 1: [ text{Length of } B_1D_1 = sqrt{1^2 + 1^2} = sqrt{2} ] 5. The midpoint (O) of (B_1D_1) can be found. Let (B_1 = (1, 0, 1)) and (D_1 = (0, 1, 1)). The coordinates of the midpoint (O) are calculated as: [ O = left(frac{1+0}{2}, frac{0+1}{2}, frac{1+1}{2} right) = left(frac{1}{2}, frac{1}{2}, 1right) ] 6. Now, construct the perpendicular line from (O) to (BD_1). The line (BD_1) has endpoints at: [ B = (1, 0, 0) quad text{and} quad D_1 = (0, 1, 1) ] 7. We need the perpendicular distance from (O) to (BD_1). 8. To find this distance, use the distance formula for a point to a line in 3D. Let (O) be the point (left(frac{1}{2}, frac{1}{2}, 1right)) and the line defined by points (B = (1, 0, 0)) and (D_1 = (0, 1, 1)). The vector (B to D_1) is: [ overrightarrow{BD_1} = (-1, 1, 1) ] 9. The coordinates of (O) can be seen as a point whose distance to the line is calculated by projecting onto the directional vector of (BD_1): 10. The distance between the point (O) and the line (BD_1) can be found using the formula for the distance of a point from a line in 3D. The distance (d) is given by: [ d = frac{| overrightarrow{OA} times overrightarrow{BD_1} |}{| overrightarrow{BD_1} |} ] where (overrightarrow{OA}) is the vector from point (O) to point (A) and (overrightarrow{BD_1}) is the directional vector of the line. 11. Calculating (overrightarrow{OA}): [ overrightarrow{OA} = O - B = left(frac{1}{2}, frac{1}{2}, 1right) - (1, 0, 0) = left(-frac{1}{2}, frac{1}{2}, 1right) ] 12. The cross product (overrightarrow{OA} times overrightarrow{BD_1}) is: [ overrightarrow{OA} times overrightarrow{BD_1} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} -frac{1}{2} & frac{1}{2} & 1 -1 & 1 & 1 end{vmatrix} = mathbf{i}(0 - 1) - mathbf{j}(-frac{1}{2} + 1) + mathbf{k}(-frac{1}{2} - frac{1}{2}) = -mathbf{i} - mathbf{j}left(frac{1}{2}right) - mathbf{k}(-1) = -mathbf{i} - mathbf{j}left(frac{1}{2}right) - mathbf{k}(1) ] [ overrightarrow{OA} times overrightarrow{BD_1} = (-1, -frac{1}{2}, -1) ] 13. The magnitude of this vector (| overrightarrow{OA} times overrightarrow{BD_1} |) is: [ | overrightarrow{OA} times overrightarrow{BD_1} | = sqrt{(-1)^2 + left(-frac{1}{2}right)^2 + (-1)^2} = sqrt{1 + frac{1}{4} + 1} = sqrt{frac{9}{4}} = frac{3}{2} ] 14. Finally, the magnitude of (| overrightarrow{BD_1} |): [ | overrightarrow{BD_1} | = sqrt{(-1)^2 + 1^2 + 1^2} = sqrt{3} ] 15. Plugging into the distance formula: [ d = frac{| overrightarrow{OA} times overrightarrow{BD_1} |}{| overrightarrow{BD_1} |} = frac{frac{3}{2}}{sqrt{3}} = frac{3}{2sqrt{3}} = frac{sqrt{3}}{2} = frac{sqrt{6}}{6} ] # Conclusion: (boxed{frac{sqrt{6}}{6}}) The distance between the line segments (A_1C_1) and (BD_1) is (frac{sqrt{6}}{6}).

answer:1. Set x = y = 0 in the given equation: [f(0 + 0) = frac{f(0)f(0)}{f(0cdot 0)} = frac{f(0)^2}{f(0)}.] Assuming f(0) neq 0 to avoid division by zero, simplify to get: [f(0) = f(0).] 2. The equation is valid, but it does not give constraints on f(0). Assume a non-zero constant solution, (f(x) = c neq 0) for all x. Plugging this into the function equation: [c = frac{ccdot c}{c} = c^2 / c = c.] 3. Check if c = 0 works. If f(x) = 0 for all x, then: [f(x + y) = frac{0 times 0}{0}.] This case leads to an undefined situation unless specified as zero. Given the above, the possible solutions are either f(x) is always zero or a nonzero constant. Examination of different constant cases didn't simplify further due to the undefined division when f(0) = 0. Thus, assuming the typical non-zero scenario, [boxed{f(x) = c} text{ where } c text{ is any non-zero constant.}]

answer:Let's denote the number of tricycles as T. We know that each bicycle has 2 wheels and each tricycle has 3 wheels. There are 16 bicycles, so the total number of wheels from bicycles is 16 * 2. The total number of wheels from tricycles would be 3 * T, where T is the number of tricycles. We are given that there are 53 wheels in total. So, the equation to represent this situation is: 16 * 2 + 3 * T = 53 Now, let's solve for T: 32 + 3T = 53 Subtract 32 from both sides: 3T = 53 - 32 3T = 21 Now, divide both sides by 3 to solve for T: T = 21 / 3 T = 7 So, there are boxed{7} tricycles in the storage area.

answer:We start by analyzing the problem. Let the larger side of the rectangle be ( k ) cm, where ( k > 7 ), and let the side of the resulting square be ( n ) cm. According to the problem, the other side of the rectangle is ( k - 7 ) cm. 1. **Area Equality:** Since the areas of the rectangle and the square must be equal, we have: [ k(k - 7) = n^2 ] 2. **Express ( n ) in terms of ( k ):** Given that ( n < k ), let ( n = k - m ) where ( m geq 1 ). Substitute ( n ) into the area equality: [ k(k - 7) = (k - m)^2 ] 3. **Expand and Simplify:** [ k^2 - 7k = k^2 - 2mk + m^2 ] [ 7k = 2mk - m^2 ] [ m(2k - m) = 7k ] 4. **Divisibility Argument:** For the above equation to hold, at least one of ( m ) or ( 2k - m ) must be divisible by 7. 5. **Case 1: ( m = 7s ):** Let's first try ( m = 7s ): [ 7s(2k - 7s) = 7k ] [ 2ks - 7s^2 = k ] [ k = frac{14s^2}{2s - 1} ] Thus, ( k ) must be an integer, implying that ( frac{14s^2}{2s - 1} ) needs to be an integer. [ k = 14s^2 / (2s - 1) = 7 left( frac{2s^2}{2s - 1} right) ] To determine ( s ): [ 4k = 2s(14s) / (2s-1) Rightarrow 4k = 14s (s - 0.5) Rightarrow k = frac{14s / s - 1.5} = frac{14. downarrow applying factors, ] [ 4(If s = 1 :) ] [ m = 7s = 7 which results to the 1 m = 7, k meaning (). resulting to computation which is ( invalid and thus no real anwser), 6. **Case ( s = 4 ):** Replace: [ implies k ] [ m = 7s = 28, k Hence ( boxed{24} ``` The way attribute all we find is only univalxbe end verifying: sideSquare 12 be 12 as supplement (boxed ) . Summary which consistently it all (chartExactly: boxed.