Let's go step-by-step

We are given: \[ 6. 24 \] --- **Step 1: Write it as a fraction** \[ 6. 24 = 6 + 0. 24 \] \[ 0. 24 = \frac{24}{100} \] \[ 6. 24 = \frac{624}{100} \] --- **Step 2: Simplify the fraction** Divide numerator and denominator by their greatest common divisor (GCD). GCD of 624 and 100: - \( 624 = 2^4 \times 3 \times 13 \) - \( 100 = 2^2 \times 5^2 \) Common factor: \( 2^2 = 4 \). \[ \frac{624 \div 4}{100 \div 4} = \frac{156}{25} \] --- **Step 3: Check if it’s a terminating decimal in base 10** \( \frac{156}{25} = \frac{156}{5^2} \), denominator only has prime factors 5 and no 2’s? Wait, check: Actually \( 156/25 \) — denominator \( 25 = 5^2 \), but for a fraction to be a terminating decimal in base 10, denominator after simplification must be of the form \( 2^a 5^b \). Here denominator is \( 25 = 5^2 \) which is not of the form \( 2^a 5^b \) unless we factor in the numerator’s factors of 2. Let’s check: \( \frac{156}{25} \) — 156 = \( 4 \times 39 = 2^2 \times 39 \). So \( \frac{156}{25} = \frac{2^2 \times 39}{5^2} \). Multiply numerator and denominator by \( 2^0 \) (already has enough 2’s? No, to make it \( 2^a 5^b \), we need \( a = b \) for terminating decimal? Actually, for terminating decimal, denominator after simplifying must be \( 2^m 5^n \) with m, n ≥ 0. But our denominator is \( 5^2 \), which is already of the form \( 2^0 5^2 \), so it is of the form \( 2^a 5^b \). Yes, \( a=0, b=2 \). So it’s terminating. --- **Step 4: Already in simplest form** \[ 6. 24 = \frac{156}{25} \] --- **Step 5: Convert to percent** Multiply by 100: \[ 6. 24 \times 100\% = 624\% \] --- **Step 6: Mixed number** \[ \frac{156}{25} = 6 \frac{6}{25} \] because \( 156 \div 25 = 6\) remainder \(6\), so \( 6 + \frac{6}{25} \). --- **Final answers in different forms:** 1. As a fraction: \( \frac{156}{25} \) 2. As a mixed number: \( 6\frac{6}{25} \) 3. As a percent: \( 624\% \) \[ \boxed{\frac{156}{25}} \] (or \( 6\frac{6}{25} \), or \( 624\% \), depending on what’s asked)