answer:Let's calculate the total number of dog toys Daisy would have if her owner found all the lost ones, including the new ones. On Monday, Daisy started with 5 dog toys. On Tuesday, she had 3 left, which means she lost 5 - 3 = 2 dog toys. Her owner then bought her 3 more, so she had 3 + 3 = 6 dog toys. On Wednesday, all of Daisy's dog toys were missing, so her owner bought her 5 more. If her owner found all the lost dog toys, we would add the original number of toys, the ones bought on Tuesday, and the ones bought on Wednesday. Original toys from Monday: 5 Lost and then found from Tuesday: 2 Bought on Tuesday: 3 Bought on Wednesday: 5 Total dog toys Daisy would have: 5 (original) + 2 (found from Tuesday) + 3 (bought on Tuesday) + 5 (bought on Wednesday) = boxed{15} dog toys.

answer:To determine if the natural number solutions (x, y,) and (z) of the equation (x^3 + y^4 = z^5) form a finite or infinite set, we proceed as follows: 1. We begin by verifying that there exists at least one set of solutions that satisfies (x^3 + y^4 = z^5): [ 2^{24} + 2^{24} = 2^{25} ] This can be verified by directly calculating each term: [ 2^{24} + 2^{24} = 2 times 2^{24} = 2^{25} ] Thus, ( 2^{24} + 2^{24} = 2^{25} ). 2. Recognizing (2^{8}, 2^{6},) and (2^{5}) as the cube, fourth power, and fifth root respectively, we express the above identity in terms of powers. [ left(2^{8}right)^3 + left(2^{6}right)^4 = left(2^{5}right)^5 ] When equated: [ left(2^{8}right)^3 + left(2^{6}right)^4 = left(2^{5}right)^5 ] We get: [ left(256right)^3 + left(64right)^4 = left(32right)^5 ] Hence: [ left(256right)^3 = 2^{24}, quad left(64right)^4 = 2^{24}, quad left(32right)^5 = 2^{25} ] Thus, we have: [ (256)^{3} + (64)^{4} = (32)^{5} ] Therefore: [ left{ begin{array}{l} x = 2^{8} = 256, y = 2^{6} = 64, z = 2^{5} = 32 end{array} right. ] This confirms the existence of one solution set ( (x, y, z) = (256, 64, 32) ). 3. Consider another set of solutions in general form. Assume (left(x_{0}, y_{0}, z_{0}right)) is one solution: [ x_0^3 + y_0^4 = z_0^5 ] For any integer (k), the following manipulations remain in the form required by the original equation: [ left{ begin{array}{l} x = k^{20} x_0, y = k^{15} y_0, z = k^{12} z_0 end{array} right. ] Explanation: [ (k^{20} x_0)^3 + (k^{15} y_0)^4 = k^{60} x_0^3 + k^{60} y_0^4 = k^{60} (x_0^3 + y_0^4) = k^{60} z_0^5 = (k^{12} z_0)^5 ] 4. Thus, for any integer (k), [ (k^{20} x_0)^3 + (k^{15} y_0)^4 = (k^{12} z_0)^5 ] is a valid solution. 5. Since (k) can be any integer, the equation (x^3 + y^4 = z^5) possesses infinitely many solutions. Conclusively, the set of natural number solutions to the equation (x^3 + y^4 = z^5) is infinite. boxed{text{Infinite}}

answer:# Problem: Given (n) points on a plane ((n geq 3)), we can perform the following operation: for any two points (A) and (B), rotate them about their midpoint by a certain angle to obtain two new points. Prove that if it is possible to transform these (n) points into the vertices of a regular (n)-gon through a finite number of such operations, then the center and size of this regular (n)-gon are unique. 1. **Notation**: - Let ( A_{1}, A_{2}, ldots, A_{n} ) denote the ( n ) points. - Let ( P ) be an arbitrary point in the plane. 2. **Transformation**: - Suppose an operation transforms ( A_{i} ) and ( A_{j} ) into ( A_{i}^{prime} ) and ( A_{j}^{prime} ). - Let ( M ) be the midpoint of ( A_{i} ) and ( A_{j} ). 3. **Vector Equations**: - The vector equation for the transformation through midpoint rotation around ( P ) can be described as: [ overrightarrow{P A}_{i} + overrightarrow{P A}_{j} = 2 overrightarrow{P M} = overrightarrow{P A}_{i}^{prime} + overrightarrow{P A}_{j}^{prime} ] - Therefore, the sum of the vectors from ( P ) to the points remains invariant: [ sum_{i=1}^{n} overrightarrow{P A}_{i} quad text{is preserved by such operations}. ] 4. **Sum of Vectors**: - If the ( n ) points are transformed into vertices of a regular ( n )-gon centered at ( O ), then: [ overrightarrow{P O} = frac{1}{n} sum_{i=1}^{n} overrightarrow{P A}_{i} ] - Hence, ( O ) is uniquely determined by the set of points and their configuration. 5. **Invariance of Summed Distances** ( T ): - Consider the total squared distance from any point ( O ) to all points ( A_{i} ): [ T = sum_{i=1}^{n} left| O A_{i} right|^{2} ] - By the midpoint theorem (considering transformations involving midpoint ( M )): [ left| O A_{i} right|^{2} + left| O A_{j} right|^{2} = 2 |O M|^{2} - frac{1}{2} left| A_{i} A_{j} right|^{2} = 2 |O M|^{2} - frac{1}{2} left| A_{i}^{prime} A_{j}^{prime} right|^{2} ] [ = left| O A_{i}^{prime} right|^{2} + left| O A_{j}^{prime} right|^{2} ] 6. **Distance Sum Invariance**: - Therefore, [ T = sum_{i=1}^{n} left| O A_{i} right|^{2} ] - remains unchanged by the operations. # Conclusion: Since the vector sum and the total squared distances remain invariant regardless of the operations, the center ( O ) and the size of the regular ( n )-gon are unique. (blacksquare)

answer:When 3 fair coins are tossed, each coin has 2 possible outcomes: heads (H) or tails (T). Therefore, there are a total of 2^3 = 8 possible outcomes when tossing 3 coins. These outcomes are: 1. HHH 2. HHT 3. HTH 4. HTT 5. THH 6. THT 7. TTH 8. TTT We are interested in the outcomes with exactly 2 tails. Looking at the list above, we can see that there are 3 such outcomes: 1. H T T 2. T H T 3. T T H So there are 3 favorable outcomes out of 8 possible outcomes. The probability of getting exactly 2 tails when tossing 3 fair coins is therefore: P(2 tails) = Number of favorable outcomes / Total number of possible outcomes P(2 tails) = 3 / 8 So the probability of getting 2 tails when 3 fair coins are tossed is boxed{3/8} .