| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the determinant of the given matrix: \[ \left( \begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array} \right) \] we use the formula for the determinant of a 2x2 matrix: \[ \text{det} = ad - bc \] where the matrix is: \[ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \] For the given matrix: \[ a = \frac{34}{15}, \quad b = \frac{2}{5}, \quad c = \frac{47}{15}, \quad d = \frac{53}{15} \] First, we calculate \( ad \): \[ ad = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \] Next, we calculate \( bc \): \[ bc = \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} \] To subtract these fractions, we need a common denominator. The common denominator for 225 and 75 is 225: \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225} \] Now, we can subtract: \[ ad - bc = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \approx 6.8 \] Thus, the determinant of the matrix is: \[ \boxed{6.8} \] |
\frac{304}{45} |
| deepseek-r1 | To find the determinant of the matrix \[ \begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{pmatrix}, \] we use the formula for the determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), which is \(ad - bc\). First, compute the product of the elements on the main diagonal: \[ \frac{34}{15} \times \frac{53}{15} = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225}. \] Next, compute the product of the off-diagonal elements: \[ \frac{2}{5} \times \frac{47}{15} = \frac{2 \times 47}{5 \times 15} = \frac{94}{75}. \] Convert \(\frac{94}{75}\) to a fraction with the same denominator as \(\frac{1802}{225}\): \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225}. \] Subtract the two fractions: \[ \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225}. \] Simplify the resulting fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5: \[ \frac{1520}{225} = \frac{1520 \div 5}{225 \div 5} = \frac{304}{45}. \] Thus, the determinant of the matrix is \[ \boxed{\dfrac{304}{45}}. \] |
\frac{304}{45} |
| gpto1 | To find the determinant of the 2x2 matrix: \[ \begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{pmatrix} \] We use the formula for the determinant of a 2x2 matrix: \[ \text{det} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc \] First, compute the products in the determinant formula: \[ \begin{align*} ad &= \left( \frac{34}{15} \right) \left( \frac{53}{15} \right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \\ bc &= \left( \frac{2}{5} \right) \left( \frac{47}{15} \right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} = \frac{282}{225} \quad (\text{after adjusting denominator to 225}) \end{align*} \] Next, subtract \( bc \) from \( ad \): \[ \text{det} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD), which is 5: \[ \frac{1520}{225} = \frac{1520 \div 5}{225 \div 5} = \frac{304}{45} \] Thus, the determinant of the matrix is: \[ \boxed{\dfrac{304}{45}} \] \(\boxed{\dfrac{304}{45}}\) |
\frac{304}{45} |
| gpt-4o-2024-08-06 | To find the determinant of a \(2 \times 2\) matrix \[ \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \] we use the formula: \[ \text{det} = ad - bc. \] For the given matrix \[ \begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{pmatrix}, \] we identify \(a = \frac{34}{15}\), \(b = \frac{2}{5}\), \(c = \frac{47}{15}\), and \(d = \frac{53}{15}\). First, calculate \(ad\): \[ ad = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15}. \] Calculate \(34 \times 53\): \[ 34 \times 53 = 34 \times (50 + 3) = 34 \times 50 + 34 \times 3 = 1700 + 102 = 1802. \] Thus, \[ ad = \frac{1802}{225}. \] Next, calculate \(bc\): \[ bc = \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15}. \] Calculate \(2 \times 47\): \[ 2 \times 47 = 94. \] Thus, \[ bc = \frac{94}{75}. \] Now, find the determinant: \[ \text{det} = ad - bc = \frac{1802}{225} - \frac{94}{75}. \] To subtract these fractions, find a common denominator. The least common multiple of 225 and 75 is 225. Convert \(\frac{94}{75}\) to a fraction with denominator 225: \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225}. \] Now subtract: \[ \text{det} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225}. \] Simplify \(\frac{1520}{225}\). The greatest common divisor of 1520 and 225 is 5. Divide both the numerator and the denominator by 5: \[ \frac{1520}{225} = \frac{1520 \div 5}{225 \div 5} = \frac{304}{45}. \] Thus, the determinant of the matrix is \[ \boxed{\frac{304}{45}}. \] |
\frac{304}{45} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the determinant of the given 2x2 matrix, we use the formula for the determinant of a 2x2 matrix \(\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\), which is \(ad - bc\). Given matrix: \[ \left(\begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array}\right) \] Here, \(a = \frac{34}{15}\), \(b = \frac{2}{5}\), \(c = \frac{47}{15}\), and \(d = \frac{53}{15}\). Step-by-step calculation: 1. Calculate \(ad\): \[ ad = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \] 2. Calculate \(bc\): \[ bc = \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} \] 3. Subtract \(bc\) from \(ad\): \[ \text{Determinant} = ad - bc = \frac{1802}{225} - \frac{94}{75} \] 4. To subtract these fractions, we need a common denominator. The least common multiple of 225 and 75 is 225. Convert \(\frac{94}{75}\) to a fraction with a denominator of 225: \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225} \] 5. Now subtract the fractions: \[ \text{Determinant} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] 6. Simplify the fraction \(\frac{1520}{225}\): \[ \frac{1520}{225} = \frac{304}{45} \] Thus, the determinant of the given matrix is: \[ \boxed{\frac{304}{45}} \] |
\frac{304}{45} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Recall the formula for the determinant of a 2x2 matrix The determinant of a 2x2 matrix $\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right)$ is given by the formula $ad - bc$. ## Step 2: Apply the formula to the given matrix Using the formula, we can calculate the determinant of the given matrix as $\left(\frac{34}{15}\right)\left(\frac{53}{15}\right) - \left(\frac{2}{5}\right)\left(\frac{47}{15}\right)$. ## Step 3: Perform the multiplication First, multiply the fractions in the first term: $\left(\frac{34}{15}\right)\left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225}$. ## Step 4: Perform the multiplication in the second term Next, multiply the fractions in the second term: $\left(\frac{2}{5}\right)\left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75}$. ## Step 5: Subtract the second term from the first term Now, subtract the result of the second term from the result of the first term: $\frac{1802}{225} - \frac{94}{75}$. ## Step 6: Find a common denominator for the fractions To subtract the fractions, we need a common denominator, which is 225. Convert the second fraction to have a denominator of 225: $\frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225}$. ## Step 7: Subtract the fractions Now subtract the fractions: $\frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225}$. ## Step 8: Simplify the fraction if necessary The fraction $\frac{1520}{225}$ can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 5. $\frac{1520}{225} = \frac{304}{45}$. The final answer is: $\boxed{\frac{304}{45}}$ |
\frac{304}{45} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Recall the formula for the determinant of a 2x2 matrix The determinant of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by the formula $ad - bc$. ## Step 2: Identify the elements of the given matrix The given matrix is $\begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{pmatrix}$. So, $a = \frac{34}{15}$, $b = \frac{2}{5}$, $c = \frac{47}{15}$, and $d = \frac{53}{15}$. ## Step 3: Apply the determinant formula Substitute the values of $a$, $b$, $c$, and $d$ into the determinant formula: $\left(\frac{34}{15}\right)\left(\frac{53}{15}\right) - \left(\frac{2}{5}\right)\left(\frac{47}{15}\right)$. ## Step 4: Perform the multiplication Calculate the products: $\frac{34 \times 53}{15 \times 15} - \frac{2 \times 47}{5 \times 15}$. ## Step 5: Simplify the fractions Simplify the fractions: $\frac{1802}{225} - \frac{94}{75}$. ## Step 6: Find a common denominator The common denominator for the fractions is 225. Convert $\frac{94}{75}$ to have a denominator of 225: $\frac{94 \times 3}{75 \times 3} = \frac{282}{225}$. ## Step 7: Subtract the fractions Subtract the fractions: $\frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225}$. ## Step 8: Perform the subtraction Calculate the difference: $\frac{1520}{225}$. ## Step 9: Simplify the result The fraction $\frac{1520}{225}$ can be simplified by dividing both the numerator and the denominator by 5: $\frac{304}{45}$. The final answer is: $\boxed{\frac{304}{45}}$ |
\frac{304}{45} |
| Qwen2-72B-Instruct | To find the determinant of a 2x2 matrix, you can use the formula: $$\text{det}(A) = ad - bc$$ where $A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$. Given the matrix $A = \left( \begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{array} \right)$, we have $a = \frac{34}{15}$, $b = \frac{2}{5}$, $c = \frac{47}{15}$, and $d = \frac{53}{15}$. Substituting these values into the formula gives: $$\text{det}(A) = \left(\frac{34}{15}\right)\left(\frac{53}{15}\right) - \left(\frac{2}{5}\right)\left(\frac{47}{15}\right)$$ Simplifying this expression: $$\text{det}(A) = \frac{1802}{225} - \frac{94}{75} = \frac{1802}{225} - \frac{188}{225} = \frac{1614}{225} = \frac{538}{75}$$ Therefore, the determinant of the given matrix is $\boxed{\frac{538}{75}}$. |
\frac{304}{45} |
| baichuan4 | To find the determinant of a $2 \times 2$ matrix, we use the formula: $$\text{det}\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = ad - bc$$ Given the matrix: $$\left(\begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array}\right)$$ We can identify the elements as: $$a = \frac{34}{15}, \quad b = \frac{2}{5}, \quad c = \frac{47}{15}, \quad d = \frac{53}{15}$$ Now, we calculate the determinant: $$\text{det} = \left(\frac{34}{15}\right)\left(\frac{53}{15}\right) - \left(\frac{2}{5}\right)\left(\frac{47}{15}\right)$$ First, we multiply the diagonal elements: $$\left(\frac{34}{15}\right)\left(\frac{53}{15}\right) = \frac{34 \cdot 53}{15 \cdot 15} = \frac{1802}{225}$$ Then, we multiply the off-diagonal elements: $$\left(\frac{2}{5}\right)\left(\frac{47}{15}\right) = \frac{2 \cdot 47}{5 \cdot 15} = \frac{94}{75}$$ Now, we subtract these two products: $$\text{det} = \frac{1802}{225} - \frac{94}{75}$$ In order to subtract these fractions, we need to find a common denominator. We can see that $225$ is $3$ times $75$, so we multiply the second fraction by $\frac{3}{3}$ to get the same denominator: $$\text{det} = \frac{1802}{225} - \frac{94 \cdot 3}{75 \cdot 3} = \frac{1802}{225} - \frac{282}{225}$$ Now, we can subtract the numerators: $$\text{det} = \frac{1802 - 282}{225} = \frac{1520}{225}$$ This fraction can be simplified by dividing both the numerator and denominator by $15$: $$\text{det} = \frac{1520 \div 15}{225 \div 15} = \frac{101 \cdot 15}{15 \cdot 15} = \frac{101}{15}$$ Therefore, the determinant of the given matrix is: $$\boxed{\frac{101}{15}}$$ |
\frac{304}{45} |
| doubao | The determinant of a 2x2 matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is $ad-bc$. For the given matrix $\begin{pmatrix}\frac{34}{15}&\frac{2}{5}\\\frac{47}{15}&\frac{53}{15}\end{pmatrix}$, the determinant is: \[ \begin{align*} &\frac{34}{15}\times\frac{53}{15}-\frac{2}{5}\times\frac{47}{15}\\ =&\frac{1782}{225}-\frac{94}{75}\\ =&\frac{1782}{225}-\frac{94\times3}{75\times3}\\ =&\frac{1782}{225}-\frac{282}{225}\\ =&\frac{1500}{225}\\ =&\frac{20}{3} \end{align*} \] So the answer is $\boxed{\frac{20}{3}}$. |
\frac{304}{45} |
| gpto1 | To find the determinant of the 2x2 matrix: \[ \begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{pmatrix} \] We use the formula for the determinant of a 2x2 matrix: \[ \text{det} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc \] First, compute the products in the determinant formula: \[ \begin{align*} ad &= \left( \frac{34}{15} \right) \left( \frac{53}{15} \right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \\ bc &= \left( \frac{2}{5} \right) \left( \frac{47}{15} \right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} = \frac{282}{225} \quad (\text{after adjusting denominator to 225}) \end{align*} \] Next, subtract \( bc \) from \( ad \): \[ \text{det} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD), which is 5: \[ \frac{1520}{225} = \frac{1520 \div 5}{225 \div 5} = \frac{304}{45} \] Thus, the determinant of the matrix is: \[ \boxed{\dfrac{304}{45}} \] \(\boxed{\dfrac{304}{45}}\) |
\frac{304}{45} |
| kimi | To find the determinant of the given 2x2 matrix, we will use the formula for the determinant of a 2x2 matrix: \[ \text{det}\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = ad - bc \] Given the matrix: \[ \left(\begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array}\right) \] We identify the elements as follows: - \( a = \frac{34}{15} \) - \( b = \frac{2}{5} \) - \( c = \frac{47}{15} \) - \( d = \frac{53}{15} \) Now, we substitute these values into the determinant formula: \[ \text{det} = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) - \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) \] First, calculate \( ad \): \[ ad = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \] Next, calculate \( bc \): \[ bc = \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} \] To subtract these fractions, we need a common denominator. The least common multiple of 225 and 75 is 225. Convert \( \frac{94}{75} \) to a fraction with a denominator of 225: \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225} \] Now, subtract the fractions: \[ \text{det} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] Simplify the fraction by finding the greatest common divisor (GCD) of 1520 and 225. The GCD is 5. \[ \frac{1520 \div 5}{225 \div 5} = \frac{304}{45} \] Thus, the determinant of the matrix is: \[ \boxed{\frac{304}{45}} \] |
\frac{304}{45} |
| minimax | To find the determinant of a 2x2 matrix, we use the formula: $$ \text{det}\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = ad - bc $$ For the given matrix: $$ \left(\begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array}\right) $$ The determinant is: $$ \text{det} = \left(\frac{34}{15}\right)\left(\frac{53}{15}\right) - \left(\frac{2}{5}\right)\left(\frac{47}{15}\right) $$ Now, we calculate the products and subtract: $$ \text{det} = \frac{34 \cdot 53}{15 \cdot 15} - \frac{2 \cdot 47}{5 \cdot 15} $$ $$ \text{det} = \frac{1802}{225} |
\frac{304}{45} |
| wenxinyiyan | To find the determinant of a 2x2 matrix, we use the formula: $$\text{Determinant} = (a \times d) - (b \times c)$$ For the given matrix: $$\left( \begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\\ \frac{47}{15} & \frac{53}{15} \\\ \end{array} \right)$$ We have: $a = \frac{34}{15}$ $b = \frac{2}{5}$ $c = \frac{47}{15}$ $d = \frac{53}{15}$ Plugging these values into the determinant formula, we get: $$\text{Determinant} = \left(\frac{34}{15} \times \frac{53}{15}\right) - \left(\frac{2}{5} \times \frac{47}{15}\right)$$ $$\text{Determinant} = \frac{1802}{225} - \frac{94}{75}$$ To subtract these fractions, we find a common denominator: $$\text{Determinant} = \frac{1802}{225} - \frac{282}{225}$$ $$\text{Determinant} = \frac{1520}{225}$$ Simplifying this fraction, we get: $$\text{Determinant} = \frac{304}{45}$$ So, the determinant of the given matrix is $\boxed{\frac{304}{45}}$. |
\frac{304}{45} |
| zhipu | To find the determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), we use the formula: \[ \text{det} = ad - bc \] Given the matrix: \[ \begin{pmatrix} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \end{pmatrix} \] we identify \(a = \frac{34}{15}\), \(b = \frac{2}{5}\), \(c = \frac{47}{15}\), and \(d = \frac{53}{15}\). First, compute \(ad\): \[ ad = \left(\frac{34}{15}\right) \left(\frac{53}{15}\right) = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \] Next, compute \(bc\): \[ bc = \left(\frac{2}{5}\right) \left(\frac{47}{15}\right) = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} \] To subtract these fractions, we need a common denominator. The common denominator of 225 and 75 is 225. Convert \(\frac{94}{75}\) to a fraction with a denominator of 225: \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225} \] Now, subtract \(bc\) from \(ad\): \[ ad - bc = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] Simplify \(\frac{1520}{225}\): \[ \frac{1520}{225} = \frac{1520 \div 5}{225 \div 5} = \frac{304}{45} \] Further simplify \(\frac{304}{45}\): \[ \frac{304}{45} = \frac{304 \div 1}{45 \div 1} = \frac{304}{45} \] Since 304 and 45 have no common factors other than 1, \(\frac{304}{45}\) is in its simplest form. Thus, the determinant of the matrix is: \[ \boxed{\frac{304}{45}} \] |
\frac{304}{45} |
| deepseek-v3 | To find the determinant of the given \( 2 \times 2 \) matrix: \[ \left(\begin{array}{cc} \frac{34}{15} & \frac{2}{5} \\ \frac{47}{15} & \frac{53}{15} \\ \end{array}\right) \] we use the formula for the determinant of a \( 2 \times 2 \) matrix: \[ \text{det} = ad - bc \] where the matrix is: \[ \left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) \] **Step 1: Identify the elements of the matrix.** \[ a = \frac{34}{15}, \quad b = \frac{2}{5}, \quad c = \frac{47}{15}, \quad d = \frac{53}{15} \] **Step 2: Apply the determinant formula.** \[ \text{det} = \left(\frac{34}{15} \times \frac{53}{15}\right) - \left(\frac{2}{5} \times \frac{47}{15}\right) \] **Step 3: Calculate each product separately.** \[ \frac{34}{15} \times \frac{53}{15} = \frac{34 \times 53}{15 \times 15} = \frac{1802}{225} \] \[ \frac{2}{5} \times \frac{47}{15} = \frac{2 \times 47}{5 \times 15} = \frac{94}{75} \] **Step 4: Express both fractions with a common denominator.** The least common denominator (LCD) of 225 and 75 is 225. \[ \frac{94}{75} = \frac{94 \times 3}{75 \times 3} = \frac{282}{225} \] **Step 5: Subtract the two fractions.** \[ \text{det} = \frac{1802}{225} - \frac{282}{225} = \frac{1802 - 282}{225} = \frac{1520}{225} \] **Step 6: Simplify the fraction.** Divide the numerator and the denominator by their greatest common divisor (GCD). The GCD of 1520 and 225 is 5. \[ \frac{1520 \div 5}{225 \div 5} = \frac{304}{45} \] **Final Answer:** \[ \boxed{\dfrac{304}{45}} \] |
\frac{304}{45} |