answer: Step-by-Step Solution# Part 1: Relationship between the water fee y and the water usage xGiven the water usage fee standards:1. For the first 10 tons, the price is 2.2 yuan per ton.2. For any usage above 10 tons, the price is 3 yuan per ton.For a household that uses x tons of water, where x > 10, the total fee y can be calculated as follows:- The fee for the first 10 tons is 2.2 times 10 yuan.- The fee for the excess water usage (above 10 tons) is (x - 10) times 3 yuan.Therefore, the total fee y is:y = 2.2 times 10 + (x - 10) times 3Simplifying this expression:y = 22 + 3x - 30y = 3x - 8Thus, the relationship between the water fee y and the water usage x is:boxed{y = 3x - 8}# Part 2: Calculating the water usage for Xiao Qiang's familyGiven that Xiao Qiang's family paid 67 yuan for water, we substitute y = 67 into the relationship found in part (1):67 = 3x - 8Solving for x:3x = 67 + 83x = 75x = frac{75}{3}x = 25Therefore, Xiao Qiang's family used boxed{25} tons of water.

answer:Each method is uniquely determined by the number of summands. Indeed, let 2004 be split into n summands, r of which are equal to q+1, and the rest are equal to q(0 leq r<n). Then 2004=q n+r. Thus, the numbers q and r are the quotient and remainder of the division of 2004 by n, they are uniquely determined by the choice of n. Therefore, by choosing any n from 1 to 2004, we get a unique partition of 2004 into n approximately equal summands.## Answer2004 ways.

answer:1. Let ( s ) be the number of square faces and ( t ) be the number of triangular faces.2. According to the problem, no two square faces share an edge, and no two triangular faces share an edge. This implies that each edge of the polyhedron is shared by one triangular face and one square face.3. Each square face has 4 edges, and each triangular face has 3 edges. Since each edge is shared by one triangular face and one square face, the total number of edges contributed by the square faces must equal the total number of edges contributed by the triangular faces.4. Therefore, the total number of edges contributed by the square faces is ( 4s ), and the total number of edges contributed by the triangular faces is ( 3t ).5. Since these edges are shared, we have the equation: [ 4s = 3t ]6. To find the ratio of the number of triangular faces to the number of square faces, we solve for ( frac{t}{s} ): [ frac{t}{s} = frac{4}{3} ]7. Therefore, the ratio of the number of triangular faces to the number of square faces is ( 4:3 ).The final answer is (boxed{4:3}).

answer:Example 2's proof: From (70,111)=1 and Example 8's left(12371^{56}+right. 34)^{28} equiv 70(bmod 111), we get left(12371^{56}+34,111right)=1. Since 111=3 times 37 and Lemma 14, we have varphi(111)=2 times 36=72. Since c is a positive integer and by Theorem 1, we have left(12371^{56}+right. 34)^{72 c} equiv 1(bmod 111). Therefore, we obtainleft(12371^{56}+34right)^{72 c+28} equiv 70(bmod 111)

answer:The coordinates corresponding to z_1=2+i are (2,1).Since the complex numbers z_1 and z_2 correspond to points in the complex plane that are symmetric about the imaginary axis,the coordinates of the point symmetric to (2,1) about the imaginary axis are (-2,1).Therefore, the corresponding complex number, z_2=-2+i.Then, z_1z_2=(2+i)(-2+i)=i^2-4=-1-4=-5.Hence, the correct choice is: boxed{A}Analysis: By understanding the geometric meaning of complex numbers, we can find z_2 and thus reach the conclusion.

answer:A number is divisible by six if and only if it is divisible by two and simultaneously by three, i.e., if and only if it is even and the sum of its digits is divisible by three. Therefore, on all the cards, there must be even digits, and their sum must be divisible by three. The triplets of digits that meet these two requirements are (disregarding the order):2,2,2, quad 2,2,8, quad 2,4,6,2,8,8, quad 4,4,4, quad 4,6,8,6,6,6,8,8,8.Now, it is necessary to test each triplet to see if it can form a three-digit number divisible by 11 when the digits are arranged in some way. This does not happen except for the third triplet, and for this triplet, two of the six possible arrangements give a number divisible by 11:462: 11=42, quad 264: 11=24The digits on the cards can only be 2, 4, and 6.Note. When verifying divisibility by 11, it is advantageous to use the following criterion: a number is divisible by 11 if and only if the difference between the sum of the digits in even positions and the sum of the digits in odd positions is divisible by 11. (Here 4-6+2=0 and 2-6+4=0 are divisible by 11, but for example, 2-4+6=4 is not.)