answer:Given the inequality x^{2}+(a-4)x+4 > 0 holds for all x in [0,3], let's analyze it step by step: 1. **For x=0:** Substituting x=0 into the inequality, we get 4 > 0. This condition is always true, indicating that a can be any real number based on this condition alone. 2. **For x in (0,3]:** The inequality x^{2}+(a-4)x+4 > 0 must hold for all x in this interval. This implies that the expression on the left must be positive for all x in this range. 3. **Analyzing the inequality a > -x-frac{4}{x}+4:** This inequality is derived from rearranging the original inequality to solve for a. We need to find the minimum value of the expression on the right to ensure the inequality holds for all x in (0,3]. 4. **Using AM-GM inequality:** The expression -x-frac{4}{x}+4 can be rewritten and analyzed using the Arithmetic Mean-Geometric Mean inequality. Specifically, -x-frac{4}{x}+4 leq -2sqrt{xcdotfrac{4}{x}}+4 = 0, where the equality holds when x=2. This means the minimum value of the expression on the right is 0 when x=2. 5. **Conclusion for a:** Since the minimum value of -x-frac{4}{x}+4 is 0, for the original inequality to always hold, we must have a > 0. Therefore, combining the conditions, we conclude that the range of values for a is boxed{(0,+infty)}.

answer:Solution: For example, 6:3=2 If we enlarge the antecedent of the ratio to 3 times its original, it becomes 18, and reduce the consequent to frac{1}{3} of its original, it becomes 1, then the ratio is 18; Therefore, if we enlarge the antecedent of a ratio to 3 times its original and reduce the consequent to frac{1}{3} of its original, the value of this ratio will be enlarged by 9 times. Hence, the answer is: enlarged by 9 times. According to the properties of ratios, it is known that "enlarging the antecedent of a ratio to 3 times its original and reducing the consequent to frac{1}{3} of its original", the value of this ratio will be enlarged by 3 times 3 = 9 times, which can be verified by a specific example. This question examines the application of the properties of ratios: the value of a ratio remains unchanged only if both the antecedent and the consequent are multiplied or divided by the same number (excluding zero). Therefore, the final answer is boxed{text{enlarged by 9 times}}.

answer:1. **Define Midpoints and Triangle Properties**: - M is the midpoint of BC, and N is the midpoint of AC. This implies AM = MC and AN = NC. - Assuming BC = a, M is at frac{a}{2}. With AC = 20, N is at 10 (midpoint). 2. **Triangle Geometry and Midpoint Analysis**: - BM = MC = frac{a}{2}. - As both M and N are midpoints, MN is parallel to AB. 3. **Using Coordinate System (assuming B as origin)**: - Coordinates: B = (0, 0), C = (a, 0), M = left(frac{a}{2}, 0right), A = (0, 20), N = (0, 10). - Since N and M are midpoints and line CL bisects MNB, let's locate L. As AB = 15, use the triangle's properties to estimate a. 4. **Determine BC and MN**: - Following similar triangle properties, if triangle ABC sim triangle MNC, then frac{AB}{AC} = frac{BM}{MC} = frac{15}{20} = frac{3}{4}. - Solving for a, a = frac{20}{3/4} = frac{20 cdot 4}{3}. Calculate frac{20 times 4}{3} = frac{80}{3}. 5. **Calculation of MN**: - Given MN parallel AB and the similarity ratio, MN = frac{1}{2} times BC as N and M are midpoints. - We get MN = frac{1}{2} times frac{80}{3} = frac{40}{3}. 6. **Concluding with Final Answer**: - frac{40{3}}. The final answer is A) boxed{frac{40}{3}}

answer:To solve this problem, let's break it down step by step: 1. First, we calculate the total number of seashells Henry and Paul collected together. Henry collected 11 seashells, and Paul collected 24 seashells. So, the total number of seashells collected by Henry and Paul is: [11 + 24 = 35] 2. The total number of seashells initially collected by Henry, Paul, and Leo is 59. To find out how many seashells Leo collected, we subtract the total seashells collected by Henry and Paul from the initial total. Thus, the number of seashells Leo collected is: [59 - 35 = 24] 3. Leo gave away a quarter of his seashells to a younger kid. To find out how many seashells this is, we calculate a quarter of Leo's collection: [24 times frac{1}{4} = 6] 4. To find out how many seashells they have now, we subtract the number of seashells Leo gave away from the initial total: [59 - 6 = 53] Therefore, after giving away a quarter of his seashells, the total number of seashells they have now is boxed{53}.