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Which subset of \( A \) has a GCF of the elements that is 3 ? A. \( \{9 y, 12 x y\} \) B. \( \{3 x, 15 x, 12 x y\} \) C. \( \{3 x, 9 y, 12 x y, 15 x\} \) D. \( \{5 x y, 15 x, 3 x\} \)

Ask by Salinas Erickson. in the United States
Jan 24,2025

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Tutor-Verified Answer

Answer

Option C has a GCF of 3.

Solution

To determine which subset of \( A \) has a greatest common factor (GCF) of the elements that is 3, we will analyze each option step by step. ### Option A: \( \{9y, 12xy\} \) 1. **Find the GCF**: - The factors of \( 9y \) are \( 3^2 \cdot y \). - The factors of \( 12xy \) are \( 3 \cdot 2^2 \cdot x \cdot y \). - The GCF is determined by taking the lowest power of each common factor: - For \( 3 \): minimum is \( 3^1 \). - For \( y \): minimum is \( y^1 \). - Thus, GCF = \( 3 \). ### Option B: \( \{3x, 15x, 12xy\} \) 1. **Find the GCF**: - The factors of \( 3x \) are \( 3 \cdot x \). - The factors of \( 15x \) are \( 3 \cdot 5 \cdot x \). - The factors of \( 12xy \) are \( 3 \cdot 2^2 \cdot x \cdot y \). - The GCF is: - For \( 3 \): minimum is \( 3^1 \). - For \( x \): minimum is \( x^1 \). - Thus, GCF = \( 3x \) (not equal to 3). ### Option C: \( \{3x, 9y, 12xy, 15x\} \) 1. **Find the GCF**: - The factors of \( 3x \) are \( 3 \cdot x \). - The factors of \( 9y \) are \( 3^2 \cdot y \). - The factors of \( 12xy \) are \( 3 \cdot 2^2 \cdot x \cdot y \). - The factors of \( 15x \) are \( 3 \cdot 5 \cdot x \). - The GCF is: - For \( 3 \): minimum is \( 3^1 \). - For \( x \): minimum is \( x^0 \) (since \( 9y \) has no \( x \)). - Thus, GCF = \( 3 \) (this is equal to 3). ### Option D: \( \{5xy, 15x, 3x\} \) 1. **Find the GCF**: - The factors of \( 5xy \) are \( 5 \cdot x \cdot y \). - The factors of \( 15x \) are \( 3 \cdot 5 \cdot x \). - The factors of \( 3x \) are \( 3 \cdot x \). - The GCF is: - For \( 3 \): minimum is \( 3^1 \). - For \( 5 \): minimum is \( 5^0 \) (since \( 3x \) has no \( 5 \)). - For \( x \): minimum is \( x^1 \). - Thus, GCF = \( 3x \) (not equal to 3). ### Conclusion The only subset that has a GCF of 3 is **Option C: \( \{3x, 9y, 12xy, 15x\} \)**.

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To find a subset of \( A \) that has a GCF of 3, we need to check the GCF of the elements in each option. For option A, the elements are \( 9y \) and \( 12xy \). The GCF is 3, since 9 and 12 are both divisible by 3 but not by any higher common factor greater than 3. For option B, the elements are \( 3x \), \( 15x \), and \( 12xy \). The GCF here is also 3, because 3, 15, and 12 have 3 as the largest number that divides all of them. For option C, the elements are \( 3x \), \( 9y \), \( 12xy \), and \( 15x \). The GCF is 3 again, as each of these terms includes a factor of 3. For option D, the elements are \( 5xy \), \( 15x \), and \( 3x \). The GCF here is 3, as 15 and 3 both contain 3, but 5xy is not divisible by 3. Therefore, all options have a subset with GCF of 3, but options A, B, and C provide clear sets that meet the criterion specifically.

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