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Let \( U=\{d, f, g, h, k, m, n\} \), and \( A=\{d, k\} \). Determine if the statement below is true or false. \( A \subset U \) Choose the correct answer below. A. The statement is true because every element of set \( A \) is contained in \( U \) and \( A \neq U \). B. The statement is true because every element of set \( A \) is contained in \( U \) and \( A=U \). C. The statement is false because not every element of set \( A \) is contained in \( U \). The statement is false because every element of set \( A \) is contained in \( U \) but \( A \neq U \).

Ask by Mejia Wyatt. in the United States
Mar 09,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The statement \( A \subset U \) is true because every element of set \( A \) is contained in \( U \) and \( A \) is not equal to \( U \).

Solution

1. We are given the universal set \[ U=\{d, f, g, h, k, m, n\}, \] and the set \[ A=\{d, k\}. \] 2. To determine if \[ A \subset U, \] we check whether every element of \( A \) is an element of \( U \). 3. The elements of \( A \) are \( d \) and \( k \). Both \( d \) and \( k \) are in \( U \). 4. Therefore, the statement \( A \subset U \) is true. 5. Note that \( A \neq U \) because \( U \) contains additional elements (namely \( f, g, h, m, n \)) that are not in \( A \). 6. Comparing with the provided choices, the correct answer is: **A. The statement is true because every element of set \( A \) is contained in \( U \) and \( A \neq U \).**

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A. The statement is true because every element of set \( A \) is contained in \( U \) and \( A \neq U \). Every element in set \( A \) (which contains \( d \) and \( k \)) is indeed found in \( U \), which includes all elements of both sets. Thus, \( A \) is a subset of \( U \), confirming that the statement is true since \( A \) does not equal \( U \). As for a common mistake, it's easy to assume that two sets are the same just because they share some elements. Remember, a subset simply needs to hold some (or all) of its elements in the larger set without requiring equality!

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