answer:To solve the given equation by inserting the appropriate operators between the numbers to achieve the result of 3, let's check the proposed solutions one by one. 1. 1 + 2 - 3 + 4 + 5 - 6 = 3 First, we evaluate the expression step by step: [ 1 + 2 - 3 + 4 + 5 - 6 ] [ 1 + 2 = 3 ] [ 3 - 3 = 0 ] [ 0 + 4 = 4 ] [ 4 + 5 = 9 ] [ 9 - 6 = 3 ] Therefore, the first expression is correct. 2. 1 + 2 + 3 - 4 - 5 + 6 = 3 Now, we evaluate the second expression step by step: [ 1 + 2 + 3 - 4 - 5 + 6 ] [ 1 + 2 = 3 ] [ 3 + 3 = 6 ] [ 6 - 4 = 2 ] [ 2 - 5 = -3 ] [ -3 + 6 = 3 ] Therefore, the second expression is also correct. # Conclusion: Both proposed expressions satisfy the condition: [ 1 + 2 - 3 + 4 + 5 - 6 = 3; quad 1 + 2 + 3 - 4 - 5 + 6 = 3 ] Thus, the conclusion is correct. boxed{1 + 2 - 3 + 4 + 5 - 6 = 3; quad 1 + 2 + 3 - 4 - 5 + 6 = 3}

answer:Given that the function fleft(xright)=ax^{2}+bx+1 is an even function with the domain left[2a,1-aright], we can derive the values of a and b as follows: 1. **Deducing the value of a from the domain left[2a,1-aright]:** - The domain suggests that the function is defined from 2a to 1-a. For an even function, the center of the domain should be 0, because an even function is symmetric about the y-axis. Thus, the midpoint of the domain left[2a,1-aright] must be 0. We find the midpoint by averaging the endpoints: frac{2a + (1-a)}{2} = 0. Simplifying this: [ frac{2a + 1 - a}{2} = 0 frac{1 + a}{2} = 0 1 + a = 0 a = -1 ] 2. **Finding the value of b through the property of even functions:** - For f(x) to be an even function, fleft(-xright)=fleft(xright) must hold. By applying this property to the given function: [ ax^{2}+bx+1 = ax^{2}-bx+1 ] - By comparing the coefficients of x on both sides, we conclude that b = -b, which implies b = 0. Combining the values of a and b, we find: [ a + b = -1 + 0 ] [ a + b = -1 ] Therefore, the correct answer is boxed{A}.

answer:We again consider different seating arrangements based on Alice's position, as her position affects the seating of the others. 1. **Alice sits in the center chair (3rd position):** - The 2nd and 4th chairs must be occupied by Bob and Carla in either order because Alice cannot sit next to Derek or Eric. - The 1st and 5th chairs, the only ones left, must be occupied by Derek and Eric in either order. - There are 2! = 2 ways to arrange Bob and Carla, and 2! = 2 ways to arrange Derek and Eric. - Total ways for this case: 2 times 2 = 4. 2. **Alice sits in one of the end chairs (1st or 5th position):** - The chair next to Alice (either 2nd or 4th) must be occupied by either Bob or Carla, as Derek and Eric cannot sit next to Alice. - The center chair (3rd position) can be occupied by either Derek or Eric (as Carla refuses to sit next to Derek). - The remaining two people (one not chosen for the 2nd or 4th position and the remaining from Derek/Eric) fill the remaining two chairs, ensuring Carla and Derek are not adjacent. - There are 2 choices for Alice's position (1st or 5th), 2 choices for the person next to Alice (Bob or Carla), 1 choice for the person in the center chair (considering the avoidance between Carla and Derek), and 2 ways to arrange the last two people. - Total ways for this case: 2 times 2 times 1 times 2 = 8. 3. **Alice sits in one of the next-to-end chairs (2nd or 4th position):** - The chairs next to Alice (either 1st and 3rd or 3rd and 5th) must be occupied by Bob and Carla in either order. - The remaining two chairs (either 1st and 5th or 2nd and 5th if Alice is in 2nd or 4th respectively) must be occupied by Derek and Eric in either order. - There are 2 choices for Alice's position (2nd or 4th), 2! = 2 ways to arrange Bob and Carla next to her, and 2! = 2 ways to arrange Derek and Eric in the remaining chairs. - Total ways for this case: 2 times 2 times 2 = 8. Adding all cases together, we get: [ 4 + 8 + 8 = 20. ] Thus, the total number of ways they can be seated under these conditions is 20. The final answer is boxed{textbf{(C)} 20}

answer:The key step for solving this problem is identifying the angle theta such that cos theta = frac{1}{2}. From trigonometric values, we know that cos frac{pi}{3} = frac{1}{2}. Thus, using the definition of arccos, which gives the principal value (the value of theta in the range 0 to pi), [ arccos frac{1}{2} = frac{pi}{3} ] So, the solution to the problem is boxed{frac{pi}{3}}.