answer:To minimize the total time required to meet the production goals, we should assign each employee to the product they can produce the fastest. We will then calculate the total production per hour for each product and determine how many hours are needed to meet the production goals. Assignments: - Employee 1: Product C (2 hours per unit) - Employee 2: Product A (1 hour per unit) - Employee 3: Product B (2 hours per unit) - Employee 4: Product B (1 hour per unit) - Employee 5: Product C (1 hour per unit) Total production per hour: - Product A: 1 unit (Employee 2) - Product B: 1.5 units (Employee 3 and Employee 4) - Product C: 1.5 units (Employee 1 and Employee 5) Hours needed to meet production goals: - Product A: 200 units / 1 unit per hour = 200 hours - Product B: 150 units / 1.5 units per hour = 100 hours - Product C: 100 units / 1.5 units per hour = 66.67 hours (rounded up to 67 hours) Minimum total time: 200 hours (Product A) + 100 hours (Product B) + 67 hours (Product C) = 367 hours So, to minimize the total time required to meet the production goals, 1 employee should be assigned to Product A, 2 employees to Product B, and 2 employees to Product C. The minimum total time required is 367 hours.

answer:To find the remainder when x^4 + 4 is divided by x^2 + 4, we can use polynomial long division. However, a more elegant approach is to recognize that x^4 + 4 can be factored using a difference of squares, albeit indirectly. We can rewrite x^4 + 4 as (x^2)^2 + 2^2. Now, recall the formula for the difference of squares: a^2 - b^2 = (a + b)(a - b). While this doesn't directly apply, we can use a related identity that comes from the sum of squares: a^2 + b^2 = (a + bi)(a - bi) in the complex numbers. However, we don't actually need to use complex numbers directly. Instead, notice that (x^2 + 2)^2 = x^4 + 4x^2 + 4, which suggests that x^4 + 4 is close to being a perfect square, except for the term 4x^2. By adding and subtracting 4x^2, we can complete the square: x^4 + 4 = (x^4 + 4x^2 + 4) - 4x^2 = (x^2 + 2)^2 - (2x)^2. Now, applying the difference of squares formula: (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2). But this still doesn't help us with dividing by x^2 + 4 directly. Instead, consider x^4 + 4 in a different way: as (x^2 + 4)(x^2 - 4) + 4x^2 + 4. However, this is also not the most straightforward approach. We can actually directly apply polynomial division or explore simpler algebraic manipulations. A more intuitive method would recognize that the remainder theorem for polynomials can guide our intuition but is not directly applicable here. To proceed more efficiently, we should directly factor or simplify the expression, considering the divisor x^2 + 4. Since x^2 + 4 does not factor nicely into polynomials with real coefficients, our goal is to manipulate x^4 + 4 into a form that either directly shows the quotient and remainder or simplifies the division. Considering these manipulations didn't directly lead to a simple solution, the most straightforward path is actually to recognize that x^4 + 4 can be expressed as (x^2 + 4)(x^2) - 4x^2 + 4, implying that x^4 + 4 can be written as (x^2 + 4)(x^2 - 4) + 4(x^2 + 4) - 16, or (x^2 + 4)(x^2 - 4 + 4) - 16. Simplifying this expression shows that x^4 + 4 can indeed be expressed as a multiple of x^2 + 4, plus a constant term. This implies x^4 + 4 can be expressed as (x^2 + 4)(x^2) - 4x^2 + 4, or more directly as (x^2 + 4)(x^2 - 4) + 4x^2 + 4, which is (x^2 + 4)(x^2) - 4(x^2 + 4) + 4(x^2 + 4) + 4 - 16, which simplifies to (x^2 + 4)(x^2 - 4 + 4) - 16 = (x^2 + 4)(x^2) - 16. The remainder, therefore, when x^4 + 4 is divided by x^2 + 4 is -12.

answer:Your concise approach is exactly right. To find the remainder when x^4 + 4 is divided by x^2 + 4, we can elegantly manipulate the expression. Starting with x^4 + 4, we recognize it as (x^2)^2 + 4. By adding and subtracting 4x^2, we can express it as (x^2 + 4)(x^2 - 4) + 4x^2 + 4. Then, by rewriting 4x^2 as 4(x^2 + 4) - 16, we effectively express the original polynomial in terms of the divisor x^2 + 4. Finally, combining these manipulations, we arrive at x^4 + 4 = (x^2 + 4)(x^2 - 4 + 4) - 12, which clearly shows that the remainder is indeed -12. This approach is a beautiful demonstration of algebraic manipulation and remainder calculation.

answer:To show that 7 divides 3^(2n + 1) + 2^(n + 2), we can use modular arithmetic. We start by examining the expression mod 7. First, let's look at the powers of 3 and 2 mod 7: 3^1 ≡ 3 (mod 7) 3^2 ≡ 9 ≡ 2 (mod 7) 3^3 ≡ 3 * 2 ≡ 6 (mod 7) 3^4 ≡ 3 * 6 ≡ 18 ≡ 4 (mod 7) 3^5 ≡ 3 * 4 ≡ 12 ≡ 5 (mod 7) 3^6 ≡ 3 * 5 ≡ 15 ≡ 1 (mod 7) Notice that the powers of 3 cycle every 6 steps: {3, 2, 6, 4, 5, 1}. Therefore, 3^(2n + 1) can be simplified using this cycle. Since 2n + 1 is odd, the power of 3 is always in the set {3, 6, 5} mod 7. Similarly, let's examine the powers of 2 mod 7: 2^1 ≡ 2 (mod 7) 2^2 ≡ 4 (mod 7) 2^3 ≡ 1 (mod 7) The powers of 2 cycle every 3 steps: {2, 4, 1}. Therefore, 2^(n + 2) can be simplified using this cycle. Now, let's rewrite 3^(2n + 1) + 2^(n + 2) using these cycles: - If n ≡ 0 (mod 3), then 2^(n + 2) ≡ 4 (mod 7), and 3^(2n + 1) ≡ 3 (mod 7). Hence, 3^(2n + 1) + 2^(n + 2) ≡ 3 + 4 ≡ 0 (mod 7). - If n ≡ 1 (mod 3), then 2^(n + 2) ≡ 1 (mod 7), and 3^(2n + 1) ≡ 6 (mod 7). Hence, 3^(2n + 1) + 2^(n + 2) ≡ 6 + 1 ≡ 0 (mod 7). - If n ≡ 2 (mod 3), then 2^(n + 2) ≡ 2 (mod 7), and 3^(2n + 1) ≡ 5 (mod 7). Hence, 3^(2n + 1) + 2^(n + 2) ≡ 5 + 2 ≡ 0 (mod 7). In all cases, 3^(2n + 1) + 2^(n + 2) ≡ 0 (mod 7), which implies that 7 divides 3^(2n + 1) + 2^(n + 2).