answer:In triangle ABC, we have a=3, A=60^{circ}, and b=sqrt{6}. By the Sine Rule (Law of Sines), we have: frac{a}{sin A} = frac{b}{sin B} Substituting the given values: frac{3}{sin 60^{circ}} = frac{sqrt{6}}{sin B} Solving for sin B: sin B = frac{sqrt{6} cdot sin 60^{circ}}{3} = frac{sqrt{6} cdot frac{sqrt{3}}{2}}{3} = frac{sqrt{2}}{2} Since b < a, we know that angle B is smaller than angle A. Therefore, B in (0^{circ}, 60^{circ}). Given that sin B = frac{sqrt{2}}{2}, we find that B = 45^{circ}. So the answer is: boxed{B = 45^{circ}}. This problem primarily tests the understanding and application of the Sine Rule and the relationship between the lengths of sides and their corresponding angles in a triangle.

answer:To factor the expression 75x + 45, we start by identifying the greatest common factor (GCF) of the coefficients 75 and 45, and the variable term x. - The GCF of 75 and 45 is 15. - Factor 15 out of both terms: [ 75x + 45 = 15 cdot 5x + 15 cdot 3 ] [ = 15(5x + 3) ] Thus, the factored expression is boxed{15(5x+3)}.

answer:1. **Identify the elements of (S) and the multiples of 2:** The set (S) is given by: [ S = {1, 2, 3, ldots, 49, 50} ] The multiples of 2 within this set are: [ 2, 4, 6, ldots, 50 ] Since 50 is the largest number in (S), and every second number is a multiple of 2, we can count the multiples of 2 by noting there are half as many as there are total numbers (since every second number is a multiple of 2): [ text{Number of multiples of 2} = frac{50}{2} = 25 ] 2. **Remove the multiples of 2 from (S):** Removing these 25 multiples of 2 from (S) leaves the set of odd numbers from 1 to 49: [ {1, 3, 5, ldots, 49} ] The count of odd integers is: [ 50 - 25 = 25 ] 3. **Identify and remove the odd multiples of 3:** Next, we remove the multiples of 3 from this set of 25 odd numbers. The multiples of 3 within the set of 1 to 50 are: [ 3, 6, 9, 12, ldots, 48 ] We only need to consider the odd multiples of 3 from this sequence, which fit the form (3k) where (k) is odd: [ 3, 9, 15, 21, 27, 33, 39, 45 ] 4. **Count the odd multiples of 3:** There are (8) numbers in this list: [ {3, 9, 15, 21, 27, 33, 39, 45} ] 5. **Remove the odd multiples of 3 from the remaining 25 odd numbers:** Subtract this count from the previous set of 25 odd numbers: [ 25 - 8 = 17 ] 6. **Conclusion:** Hence, the number of remaining integers in the set (S) after removing multiples of both 2 and 3 is: [ boxed{17} ]

answer:1. **Base Case**: Using the first condition and setting r = 7, we have 7 # 0 = 7 + 1 = 8. 2. **Using the Modified Third Condition**: With r = 5 and s = 3, apply the third condition to find: [(7 # 3) = (5 # 3) + 3 + 2.] We need to find 5 # 3 first. 3. **Recursive Calculation**: - Similarly, by third condition from r = 3 to r = 5: [(5 # 3) = (3 # 3) + 3 + 2,] - From r = 1 to r = 3: [(3 # 3) = (1 # 3) + 3 + 2.] - Base for r = 1: [(1 # 3) = (1 # 0) + 3 + 2 = 8 + 3 + 2 = 13.] - Then (3 # 3) = 13 + 3 + 2 = 18. - Then (5 # 3) = 18 + 3 + 2 = 23. 4. **Final Calculation**: [(7 # 3) = 23 + 3 + 2 = boxed{28}.]