answer:To solve this problem, we use the largest packs available to reach the total of 120 cans while minimizing the total number of packs used. 1. **Using the largest pack (32 cans)**: - Maximum number of 32-can packs without exceeding 120 cans: leftlfloor frac{120}{32} rightrfloor = 3 packs. - Total cans covered by three 32-packs: 3 times 32 = 96 cans. - Remaining cans needed: 120 - 96 = 24 cans. 2. **Using the next largest pack (15 cans)**: - Maximum number of 15-can packs to cover the remaining 24 cans: leftlfloor frac{24}{15} rightrfloor = 1 pack. - Total cans covered by one 15-pack: 15 cans. - Remaining cans needed after using one 15-pack: 24 - 15 = 9 cans. 3. **Using the smallest pack (8 cans)**: - Maximum number of 8-can packs to cover the remaining 9 cans: leftlceil frac{9}{8} rightrceil = 2 packs (because one pack is not enough and we cannot buy a fraction of a pack). - Total cans covered by two 8-packs: 2 times 8 = 16 cans. - This exceeds the 9 cans needed but is the smallest number achievable with full packs. 4. **Total packs used**: - Number of 32-can packs used: 3 - Number of 15-can packs used: 1 - Number of 8-can packs used: 2 - Total number of packs: 3 + 1 + 2 = 6 Thus, the minimum number of packs needed to buy exactly 120 cans of soda is 6. The final answer is boxed{textbf{(B)} 6}

answer:Since persons A, B, and C can each possess the white ball, the events "person A receives the white ball" and "person B receives the white ball" cannot be opposite events. Additionally, the events "person A receives the white ball" and "person B receives the white ball" cannot occur simultaneously; therefore, the relationship between the two events is that they are mutually exclusive. So, let's write this solution step-by-step: Step 1: Recognize that there are three balls of different colors and three individuals, implying that each person will get exactly one ball of a unique color. Step 2: Determine the nature of the events considering person A and person B. The events involving persons A and B each receiving a white ball can be examined for compatibility. Step 3: Analyze the possibility of events occurring simultaneously. If both person A and person B cannot have the white ball at the same time, their respective events are incompatible. Step 4: Conclude the relationship between the events. Since only one person can have the white ball, it must be that if person A has the white ball, person B cannot have it, and vice versa. This means that these events are mutually exclusive. Step 5: Correctly identify the type of events from the multiple-choice options provided. The correct option is that the events are mutually exclusive, which reflects the fact that they cannot both happen simultaneously. Hence, the translated and enhanced solution is that the right answer is: boxed{text{C: Mutually exclusive events}}

answer:Start by rewriting the given equation x^2 + y^2 + 10 = 6x + 12y in standard form. Rearrange it to: [ x^2 - 6x + y^2 - 12y = -10. ] Complete the square for x and y: [ x^2 - 6x = (x-3)^2 - 9, ] [ y^2 - 12y = (y-6)^2 - 36. ] Substitute back into the equation: [ (x-3)^2 - 9 + (y-6)^2 - 36 = -10. ] [ (x-3)^2 + (y-6)^2 - 45 = -10. ] [ (x-3)^2 + (y-6)^2 = 35. ] This is the equation of a circle with center (3,6) and radius sqrt{35}. Therefore, the radius of the cookie is: [ boxed{sqrt{35}}. ]

answer:We start by factoring and simplifying each part of the expression. Notice that x^2 - 4 = (x + 2)(x - 2), x^2 - 2x = x(x - 2), and x^2 - 4x + 4 = (x - 2)^2. Let's rewrite the expression taking these factors into account: begin{align*} (x^{2} - 4)left( frac{x + 2}{x(x - 2)} - frac{x - 1}{(x - 2)^2} right) div frac{x - 4}{x} &= (x + 2)(x - 2) left( frac{x + 2}{x(x - 2)} - frac{x - 1}{(x - 2)^2} right) cdot frac{x}{x - 4} &= (x + 2)(x - 2) cdot left( frac{(x + 2)(x - 2) - (x - 1)x}{x(x - 2)^2} right) cdot frac{x}{x - 4} &= (x + 2)(x - 2) cdot frac{x^2 + 2x - 2x - 4 - x^2 + x}{x(x - 2)^2} cdot frac{x}{x - 4} end{align*} Now we simplify the terms in the numerator of the fraction: (x + 2)(x - 2) cdot frac{x - 4}{x(x - 2)^2} cdot frac{x}{x - 4} Since (x - 4) appears in both the numerator and the denominator, we can cancel them out. The simplified result is: begin{align*} (x + 2)(x - 2) cdot frac{1}{x(x - 2)^2} cdot x &= (x + 2)(x - 2) cdot frac{1}{(x - 2)^2} &= (x + 2) cdot frac{1}{x - 2} &= boxed{frac{x + 2}{x - 2}} end{align*}