Answer:
To evaluate the expression {(3!)²×(3Σk=2 + cos 0)} - (7√5)² - √x² for f(x) when x = -3, we can follow these steps: 1. Let's simplify the expression step by step. 2. Starting with the term (3!), which represents the factorial of 3. The factorial of 3 is calculated as 3! = 3 × 2 × 1 = 6. 3. Next, we have (3!)², which means we need to square the result of the factorial: (3!)² = 6² = 36. 4. Moving on to the term (3Σk=2 + cos 0). Here, Σ represents the summation operator, and k=2 indicates that we start the summation from k = 2. - Since we are given that cos 0 = 1, the term simplifies to (3Σk=2 + 1). - The summation (Σ) involves adding the values of k for the given range, which is from k = 2 to k = 3. - Adding 2 and 3 gives us 5, so the term becomes (3 × 5 + 1) = 16. 5. Continuing with the expression, we have (7√5)², which is the square of the square root of 5 multiplied by 7. Evaluating this term, we get (7√5)² = (7 × √5)² = (7 × √5) × (7 × √5) = 49 × 5 = 245. 6. Finally, we have -√x², where x = -3. Plugging in the value of x, we get -√(-3)² = -√9 = -3. 7. Now, we can substitute all the simplified values back into the original expression: 36 × 16 - 245 - 3. 8. Multiplying 36 by 16 gives us 576. 9. Subtracting 245 from 576 gives us 331. 10. Finally, subtracting 3 from 331 gives us the final result: 328.
Therefore, when x = -3, the value of the expression {(3!)²×(3Σk=2 + cos 0)} - (7√5)² - √x² for f(x) is 328. I hope this step-by-step explanation helps you understand how to evaluate the given expression. If you have any further questions, feel free to ask.
Hope this is right