answer:To solve this, first simplify 8frac{5}{3} - 3. Convert the mixed number into an improper fraction: [ 8frac{5}{3} = 8 + frac{5}{3} = frac{24}{3} + frac{5}{3} = frac{29}{3} ] Subtract 3 (which is frac{9}{3} as a fraction): [ frac{29}{3} - frac{9}{3} = frac{20}{3} ] Now, ask "how many three-eighths are there in frac{20}{3}?" This is equivalent to calculating: [ frac{20}{3} div frac{3}{8} ] Using the reciprocal of frac{3}{8}, which is frac{8}{3}, we have: [ frac{20}{3} cdot frac{8}{3} = frac{160}{9} ] So, there are boxed{frac{160}{9}} or boxed{17frac{7}{9}} three-eighths in 8frac{5}{3} - 3.

answer:1. **Prime factorizations of the new LCMs**: - 180 = 2^2 cdot 3^2 cdot 5 - 450 = 2 cdot 3^2 cdot 5^2 - 675 = 3^3 cdot 5^2 2. **Constraints on a**: - From text{lcm}(a, b) = 180, a must include at least 2^2 or 3^2, and possibly 5. - From text{lcm}(a, c) = 450, a must include at least 3^2 and 5. Therefore, any 2's in a would depend on b. - Therefore, a must be 45 or 90. 3. **Constraints on b**: - From text{lcm}(a, b) = 180 and a = 45 or 90, b must compensate for 2^2 not in a. - When a = 45, b must be 180 to satisfy the lcm condition, and when a = 90, b could be 90 or 180. 4. **Constraints on c**: - From text{lcm}(a, c) = 450 and text{lcm}(b, c) = 675. - Possible c values include 450 (if b = 180) since it has to satisfy 5^2 and 3^2 from 450's factorization. 5. **Combining the constraints**: - Check combinations: - (a, b, c) = (45, 180, 450) - (a, b, c) = (90, 90, 450) - (a, b, c) = (90, 180, 450) 6. **Counting the valid triples**: - There are a total of 3 valid triples. The number of ordered triples (a, b, c) that satisfy the given conditions is 3. The final answer is boxed{textbf{(C)} 3}

answer:First, factor out the common term of 5^2 from both the numerator and the denominator: [ frac{5^5 + 5^3}{5^4 - 5^2} = frac{5^3(5^2 + 1)}{5^2(5^2 - 1)} ] Simplify by canceling out 5^2 from the numerator and the denominator: [ frac{5^3(5^2 + 1)}{5^2(5^2 - 1)} = frac{5^3(25 + 1)}{5^2(25 - 1)} = frac{5^3 cdot 26}{5^2 cdot 24} ] Cancel out 5^2: [ frac{5^3 cdot 26}{5^2 cdot 24} = frac{5 cdot 26}{24} = frac{130}{24} ] Simplify frac{130}{24} by dividing both numerator and denominator by their greatest common divisor, which is 2: [ frac{130}{24} = frac{65}{12} ] Hence, the simplified fraction is boxed{frac{65}{12}}.

answer:To find the number of ordered triples ((A, B, C)) of sets such that (A cup B cup C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}), where each of the elements from (1) to (10) belongs to at least one of the sets (A), (B), or (C), we can proceed as follows: 1. **Understand the Requirement:** Each element must be in at least one of the sets (A), (B), or (C). 2. **Consider the Placement of One Element (e.g., 1):** For any particular element (say 1), it has three sets to potentially belong to: (A), (B), or (C). - It can belong to (A). - It can belong to (B). - It can belong to (C). However, it cannot be excluded from all three sets concurrently. 3. **Calculate the Total Ways for One Element Including Exclusion:** Each element has: [ 2^3 = 8 quad text{(since each element can either be in or not be in each subset A, B, or C)} ] ways to be placed in sets (A, B), or (C). But one of these ways includes being in none of the sets, which is not allowed. 4. **Adjust for the Exclusion-inclusively-Exclusive Constraint:** Specifically, since each element needs to be in at least one of the sets, [ 8 - 1 = 7 quad text{ways for correctly placing the element} ] 5. **Apply the Calculation to All Elements in the Universal Set ({1, 2, ldots, 10}):** Since there are 10 elements, each having 7 valid configurations, the total number of ways to distribute all 10 elements into the sets (A, B), and (C) while following the rule is: [ 7^{10} ] # Conclusion: The number of ordered triples ((A, B, C)) is (7^{10}). Thus, the answer is: [ boxed{7^{10}} ]