answer:Let A(1,1). According to the problem, the locus of point M is a circle with AP as the diameter. The coordinates of the circle center C are (0,2), and the radius is sqrt {2}. Since |CQ|= sqrt {2^{2}+2^{2}}=2 sqrt {2}, we have |CQ|- sqrt {2} leq |MQ| leq |CQ|+ sqrt {2}, which means sqrt {2} leq |MQ| leq 3 sqrt {2}. Therefore, the answer is boxed{[sqrt {2}, 3 sqrt {2}]}. First, we obtain the locus of M as a circle, then subtract the radius of the circle from |CQ| to get the minimum value of |MQ|, and add the radius to get the maximum value of |MQ|. This problem tests the formula for the distance between two points and is of medium difficulty.

answer:To solve for the nth monomial in the sequence, let's analyze the given monomials step by step: 1. For the first monomial, -2a^{2}, we can rewrite it as: [ -2a^{2} = (-1)^{1} times 2a^{2} = (-1)^{1} times (1+1)a^{2 times 1} ] 2. For the second monomial, 3a^{4}, it can be rewritten as: [ 3a^{4} = (-1)^{2} times 3a^{4} = (-1)^{2} times (2+1)a^{2 times 2} ] 3. For the third monomial, -4a^{6}, it follows the pattern: [ -4a^{6} = (-1)^{3} times 4a^{6} = (-1)^{3} times (3+1)a^{2 times 3} ] 4. Similarly, for the fourth monomial, 5a^{8}, we have: [ 5a^{8} = (-1)^{4} times 5a^{8} = (-1)^{4} times (4+1)a^{2 times 4} ] From these observations, we can deduce the pattern for the nth monomial in the sequence. The coefficient alternates in sign, which is represented by (-1)^{n}, and it increases by 1 each time, which is represented by (n+1). The exponent of a doubles the position of the term in the sequence, represented by 2n. Therefore, the nth monomial is given by: [ boxed{(-1)^{n}(n+1)a^{2n}} ] Thus, the correct answer is boxed{D}.

answer:1. **Introduction and Context**: - Consider a graph ( G ). - Let ( A, B subseteq V(G) ). - We need to demonstrate that ( G ) contains a collection (mathcal{P}) of vertex-disjoint ( A-B ) paths and an ( A-B ) separation ( S ) such that the separation ( S ) is located on (mathcal{P}). 2. **Key Definitions**: - **Wave**: An ( A rightarrow B ) wave (mathcal{W}) is a set of disjoint paths starting from ( A ) and ending in a set ( Z ) such that ( Z ) can separate ( A ) from ( B ) in ( G ), even if some paths are infinite or end before reaching ( B ). - **Proper Wave**: A wave is "proper" if at least one path is nontrivial or if all paths are trivial, but its boundary ( X ) is a proper subset of ( A ). 3. **Constructing Waves**: - Define a wave, starting from ( A ) and constructing as many vertex-disjoint paths as possible towards ( B ). 4. **Mathematical Details and Generalization**: - Given a set ( X subseteq V(G) ), denote ( G_{X rightarrow B} ) as a subgraph of ( G ) formed by ( X ) and all components of ( G-X ) intersecting with ( B ). - Define a family of disjoint paths ( mathcal{W} = { W_a | a in A} ), where each path ( W_a ) starts from an element ( a in A ). 5. **Iterative Construction**: - If no ( A rightarrow B ) waves exist that partition ( G ), consider the iterative construction of these paths. - Explore different graph substructures and use recursive arguments to generate infinite ( A rightarrow B ) path families, ensuring each subset size remains finite. 6. **Lemma and Zorn’s Lemma Application**: - Use Theorem 3.3.2 and 3.3.3 to argue for finitely generated subgraphs despite potential infinite vertex and edge sets. - Apply Zorn’s Lemma to assert the existence of a maximum ( A rightarrow B ) wave (mathcal{W}) in ( G ). - Denote ( X ) as the boundary set of (mathcal{W}). 7. **Final Separation**: - By examining the subgraph ( G_{X rightarrow B} ): - If we can find disjoint paths connecting ( B ) and ( X ), then ( X ) acts as a separating set, proving the existence of such a wave. - If not, ( G_{X rightarrow B} ) contains no proper ( X rightarrow B ) waves, implying ( X ) is an ( A-B ) separation set. 8. **Conclusion**: - Using the limit and boundary sets defined: - Identify the disjoint paths and separation set to meet the criteria of the theorem. - Hence, the conclusion asserts there exists a collection of vertex-disjoint ( A-B ) paths and an ( A-B ) separation set on these paths. boxed{text{The theorem is thus proved.}}

answer:To calculate the profit percentage, you first need to determine the profit in dollars, and then divide that by the cost price, and finally multiply by 100 to get the percentage. Profit in dollars = Selling price - Cost price Profit in dollars = 63 - 60 Profit in dollars = 3 Now, calculate the profit percentage: Profit percentage = (Profit in dollars / Cost price) * 100 Profit percentage = (3 / 60) * 100 Profit percentage = 0.05 * 100 Profit percentage = 5% The profit percentage is boxed{5%} .