answer:Given that ABCD is a square with side length 2 and E is the midpoint of AB, we can analyze the problem step by step. 1. **Identify the relationship between vectors overrightarrow{EB} and overrightarrow{EA}**: Since E is the midpoint of AB, overrightarrow{EB} and overrightarrow{EA} are equal in magnitude but opposite in direction. This implies that overrightarrow{EB}cdotoverrightarrow{EA}=-|overrightarrow{EB}|^2. Given that EB=EA=1 (half the side length of the square), we have overrightarrow{EB}cdotoverrightarrow{EA}=-1. 2. **Analyze perpendicular vectors**: We know that overrightarrow{EB}perpoverrightarrow{AD} and overrightarrow{EA}perpoverrightarrow{BC}. The dot product of two perpendicular vectors is 0, so overrightarrow{EB}cdotoverrightarrow{AD}=0 and overrightarrow{EA}cdotoverrightarrow{BC}=0. 3. **Calculate the dot product of overrightarrow{BC} and overrightarrow{AD}**: Since both vectors have the same magnitude (2) and are parallel, their dot product is 2times 2=4. 4. **Find overrightarrow{EC}cdotoverrightarrow{ED}**: We can express overrightarrow{EC} as overrightarrow{EB}+overrightarrow{BC} and overrightarrow{ED} as overrightarrow{EA}+overrightarrow{AD}. Therefore, overrightarrow{EC}cdotoverrightarrow{ED} can be expanded as follows: [ overrightarrow{EC}cdotoverrightarrow{ED} = (overrightarrow{EB}+overrightarrow{BC})cdot(overrightarrow{EA}+overrightarrow{AD}) = overrightarrow{EB}cdotoverrightarrow{EA} + overrightarrow{EB}cdotoverrightarrow{AD} + overrightarrow{EA}cdotoverrightarrow{BC} + overrightarrow{BC}cdotoverrightarrow{AD} ] Given the values calculated in steps 1, 2, and 3, we substitute them into the equation: [ overrightarrow{EC}cdotoverrightarrow{ED} = -1 + 0 + 0 + 4 = 3 ] Therefore, the correct answer is boxed{3}, which corresponds to choice B.

answer:1. **Understanding the Problem:** We need to find all possible values of ( f(2024) ) given that ( f ) is a function from the set of positive integers to itself, and for any cardinal set ( mathcal{S} ), the set ( f(mathcal{S}) ) is also cardinal. A set ( mathcal{S} ) is cardinal if it contains the integer ( |mathcal{S}| ), where ( |mathcal{S}| ) denotes the number of distinct elements in ( mathcal{S} ). 2. **Initial Observations:** - For ( mathcal{S} = {1} ), ( f({1}) = {f(1)} ) must be cardinal, implying ( f(1) = 1 ). - For ( mathcal{S} = {1, 2} ), ( f({1, 2}) = {1, f(2)} ) must be cardinal, implying ( f(2) = 1 ) or ( f(2) = 2 ). 3. **Defining a Set ( T ):** Define ( T = {k : f(k) = 2} ). We need to show that ( |T| leq min(T) - 1 ). 4. **Proof of Lemma 1:** Suppose ( |T| geq min(T) ). Choose ( V subset T ) such that ( min(T) in V ) and ( |V| = min(T) ). Since ( V ) is cardinal, ( f(V) = {2} ) must also be cardinal, which is a contradiction because ( {2} ) is not cardinal. Hence, ( |T| leq min(T) - 1 ). 5. **Case 1: ( f(2) = 1 ):** - The codomain of ( f ) is restricted to ( {1, 2} ). - Suppose there exists ( x ) such that ( f(x) geq 3 ). Then ( f({2, x}) = {1, f(x)} ) must be cardinal, which is a contradiction because ( {1, f(x)} ) is not cardinal. Hence, ( f(x) ) cannot be greater than 2. 6. **Case 2: ( f(2) = 2 ):** - By Lemma 1, ( f(3) neq 2 ). - For ( mathcal{S} = {1, 2, 3} ), ( f({1, 2, 3}) = {1, 2, f(3)} ) must be cardinal, implying ( f(3) = 1 ) or ( f(3) = 3 ). 7. **Subcase 1: ( f(3) = 1 ):** - The codomain of ( f ) is restricted to ( {1, 2} ). - Suppose there exists ( x ) such that ( f(x) geq 3 ). Then ( f({1, 3, x}) = {1, f(x)} ) must be cardinal, which is a contradiction because ( {1, f(x)} ) is not cardinal. Hence, ( f(x) ) cannot be greater than 2. 8. **Subcase 2: ( f(3) = 3 ):** - We claim that ( f(n) = n ) for all ( n in mathbb{N} ). - Base cases: ( f(1) = 1 ), ( f(2) = 2 ), and ( f(3) = 3 ). - Inductive step: Assume ( f(i) = i ) for all ( i in {1, 2, dots, k} ). - **Lemma 2:** ( f(k+1) leq k+1 ). - Suppose ( f(k+1) > k+1 ). Then ( f({1, 2, dots, k, k+1}) = {1, 2, dots, k, f(k+1)} ) must be cardinal, which is a contradiction because ( {1, 2, dots, k, f(k+1)} ) is not cardinal. - If ( f(k+1) = l ) where ( l in {1, 2, dots, k} ): - If ( l = 1 ), ( f({1, 2, dots, k-1, k, k+1}) = {1, 2, dots, k-1, k} ) is not cardinal. - If ( l = 2 ), ( f({2, k+1}) = {2} ) is not cardinal. - If ( 3 leq l leq k ), ( f({1, 2, dots, l-2, l, k+1}) = {1, 2, dots, l-2, l} ) is not cardinal. - Therefore, ( f(k+1) = k+1 ) is forced, and by induction, ( f(n) = n ) for all ( n in mathbb{N} ). 9. **Conclusion:** - If ( T = emptyset ), then ( f(n) = 1 ) for all ( n in mathbb{N} ), and ( f(2024) = 1 ). - If ( T = {2024} ), then ( f(n) = 1 ) for all ( n in mathbb{N} - {2024} ), and ( f(2024) = 2 ). - If ( f(n) = n ) for all ( n in mathbb{N} ), then ( f(2024) = 2024 ). The final answer is ( boxed{ f(2024) in {1, 2, 2024} } ).

answer:After the hike: - Northward movement: (15 text{ miles north} - 9 text{ miles south} = 6 text{ miles north}) - Eastward movement: (8 text{ miles east} + 2 text{ miles east} = 10 text{ miles east}) To find the total distance from the starting point, we note that this scenario forms a right triangle where the legs are: - North-South displacement: 6 miles - East-West displacement: 10 miles We apply the Pythagorean theorem: [ text{Displacement} = sqrt{(6^2) + (10^2)} = sqrt{36 + 100} = sqrt{136} ] Since (136 = 4(34)), and factoring under the square root: [ sqrt{136} = 2sqrt{34} ] Thus, the hiker is (boxed{2sqrt{34}}) miles away from the starting point.

answer:Since ab - a - b + 1 = 0, we have (a - 1)(b - 1) = 0. Given that a is an irrational number, it follows that a - 1 neq 0, thus b - 1 = 0, which means b = 1. Therefore, b is a positive rational number. Hence, the correct option is boxed{text{B}}.