answer:**Answer**: A, Generally, the prevalence of infectious diseases among prey decreases with the increase in predator population density, so A is incorrect; B, The variety of grassland producers does not decrease with the increase in the density of herbivorous animal populations, so B is incorrect; C, After eradicating predators and large beasts, the environmental carrying capacity for ptarmigans in the area increased, so C is correct; D, In the coevolution of predators and prey, the evolution of prey cannot completely avoid predation, so D is incorrect. Therefore, the correct choice is boxed{C}. **Analysis**: 1. Based on the information provided in the problem: "Norway, in an effort to protect the ptarmigan, eradicated predators and large beasts in the 19th century, but the result was multiple mass deaths of ptarmigans," it indicates that in the absence of natural enemies, factors such as outbreaks of diseases led to the mass deaths of ptarmigans, demonstrating the important role of natural enemies in maintaining ecological balance. 2. Factors affecting population changes include climate, food, natural enemies, infectious diseases, space, human impact, and other ecological factors. Therefore, the numbers of most populations are always fluctuating.

answer:Given ( x, y, z ) are real numbers greater than 1 such that (frac{1}{x} + frac{1}{y} + frac{1}{z} = 2), we need to prove that: [ sqrt{x-1} + sqrt{y-1} + sqrt{z-1} leq sqrt{x+y+z}. ] 1. **Express the given condition in a useful form:** [ frac{1}{x} + frac{1}{y} + frac{1}{z} = 2. ] This implies: [ frac{x-1}{x} + frac{y-1}{y} + frac{z-1}{z} = 3 - 2 = 1. ] 2. **Use the Cauchy-Schwarz inequality:** The Cauchy-Schwarz inequality states that for any real numbers (a_i) and (b_i): [ left( sum_{i=1}^n a_i b_i right)^2 leq left( sum_{i=1}^n a_i^2 right) left( sum_{i=1}^n b_i^2 right). ] Let (a_i = sqrt{x_i - 1}) and (b_i = frac{1}{sqrt{x_i}}) for (i = 1, 2, 3). Then: [ left( sum_{i=1}^3 sqrt{x_i - 1} cdot frac{1}{sqrt{x_i}} right)^2 leq left( sum_{i=1}^3 (x_i - 1) right) left( sum_{i=1}^3 frac{1}{x_i} right). ] 3. **Simplify the left-hand side:** [ left( sum_{i=1}^3 frac{sqrt{x_i - 1}}{sqrt{x_i}} right)^2 = left( sum_{i=1}^3 sqrt{frac{x_i - 1}{x_i}} right)^2. ] 4. **Simplify the right-hand side:** [ left( sum_{i=1}^3 (x_i - 1) right) left( sum_{i=1}^3 frac{1}{x_i} right) = (x + y + z - 3) cdot 2. ] 5. **Combine the results:** [ left( sqrt{frac{x-1}{x}} + sqrt{frac{y-1}{y}} + sqrt{frac{z-1}{z}} right)^2 leq 2(x + y + z - 3). ] 6. **Use the fact that (frac{x-1}{x} + frac{y-1}{y} + frac{z-1}{z} = 1):** [ left( sqrt{frac{x-1}{x}} + sqrt{frac{y-1}{y}} + sqrt{frac{z-1}{z}} right)^2 leq 2(x + y + z - 3). ] 7. **Apply the AM-GM inequality:** [ sqrt{frac{x-1}{x}} leq sqrt{x-1}, quad sqrt{frac{y-1}{y}} leq sqrt{y-1}, quad sqrt{frac{z-1}{z}} leq sqrt{z-1}. ] 8. **Combine the inequalities:** [ sqrt{x-1} + sqrt{y-1} + sqrt{z-1} leq sqrt{2(x + y + z - 3)}. ] 9. **Simplify the right-hand side:** [ sqrt{2(x + y + z - 3)} leq sqrt{x + y + z}. ] 10. **Conclude the proof:** Since (x, y, z > 1), we have (x + y + z > 3). Therefore: [ sqrt{x-1} + sqrt{y-1} + sqrt{z-1} leq sqrt{x + y + z}. ] (blacksquare)

answer:Spiders have 8 legs each, and ants have 6 legs each. For the spiders: 8 spiders * 8 legs each = 64 legs For the ants: 12 ants * 6 legs each = 72 legs To find the total number of legs in Monroe's collection, we add the number of spider legs to the number of ant legs: 64 legs (spiders) + 72 legs (ants) = 136 legs So, the total number of legs in Monroe's collection of ants and spiders is boxed{136} legs.

answer:First, we solve z^4 = 1, which factors as (z^2 + 1)(z^2 - 1) = 0. The nonreal roots are obtained from z^2 + 1 = 0, hence z = pm i. Let omega = i. To find |aomega + b| = 1, we set: [ |ai + b|^2 = 1. ] Expanding this, we get: [ (a i + b)(-ai + b) = 1 implies a^2 + b^2 = 1. ] We need integer solutions (a, b) such that a^2 + b^2 = 1. The possible integer values for a and b are -1, 0, 1. Testing these, we find: - If a = 0, then b^2 = 1, giving pairs (0, 1) and (0, -1). - If a = 1, then b^2 = 0, giving the pair (1, 0). - If a = -1, then b^2 = 0, giving the pair (-1, 0). Therefore, the possible pairs (a, b) are (0, 1), (0, -1), (1, 0), and (-1, 0). Thus, there are boxed{4} possible pairs (a, b).