answer:The operation # is essentially finding the symmetric difference between the two sets x and y. This means we want to find the elements that are in either x or y but not in both. Let's break it down: - x has 12 integers. - y has 18 integers. - There are 6 integers that are in both x and y. The 6 integers that are in both sets are counted in the totals for both x and y. To find the number of unique elements in x that are not in y, we subtract the 6 shared integers from the total in x: 12 (in x) - 6 (shared with y) = 6 unique to x Similarly, to find the number of unique elements in y that are not in x, we subtract the 6 shared integers from the total in y: 18 (in y) - 6 (shared with x) = 12 unique to y Now, we add the unique elements from both sets to find the total number of elements in x # y: 6 (unique to x) + 12 (unique to y) = 18 Therefore, x # y consists of boxed{18} integers.

answer:First, determine the distance from O (origin) to P: Since P(10,2), [ OP^2 = 10^2 + 2^2 = 100 + 4 = 104 ] Thus, OP = sqrt{104}. This distance matches the radius of the larger circle, so R = sqrt{104}. Assuming QR = 4, the radius of the smaller circle OQ can be calculated as: [ OQ = sqrt{104} - 4 ] Since S(a,a) lies on the smaller circle and on the line y=x, it must satisfy the equation of the smaller circle: [ a^2 + a^2 = (sqrt{104} - 4)^2 ] [ 2a^2 = (sqrt{104} - 4)^2 ] Solving by squaring sqrt{104} - 4: [ 2a^2 = 104 - 8sqrt{104} + 16 ] [ a^2 = 60 - 4sqrt{104} ] Obtaining a: [ a = sqrt{60 - 4sqrt{104}} ] This implies that a = boxed{sqrt{60 - 4sqrt{104}}}.

answer:1. **Calculate the number of potatoes peeled by Homer alone**: Homer peels at a rate of 4 potatoes per minute and peels alone for 6 minutes. Thus, the number of potatoes peeled by Homer in these 6 minutes is: [ 6 text{ minutes} times 4 frac{text{potatoes}}{text{minute}} = 24 text{ potatoes} ] 2. **Determine the number of potatoes remaining**: Initially, there were 60 potatoes. After Homer peeled 24, the number of potatoes left is: [ 60 text{ potatoes} - 24 text{ potatoes} = 36 text{ potatoes} ] 3. **Calculate the combined peeling rate and time taken after Alex joins**: Alex joins Homer, and together they peel at a combined rate of: [ 4 frac{text{potatoes}}{text{minute}} + 6 frac{text{potatoes}}{text{minute}} = 10 frac{text{potatoes}}{text{minute}} ] The time taken to peel the remaining 36 potatoes at this rate is: [ frac{36 text{ potatoes}}{10 frac{text{potatoes}}{text{minute}}} = 3.6 text{ minutes} ] 4. **Calculate the number of potatoes peeled by Alex**: Alex peels at a rate of 6 potatoes per minute. In the 3.6 minutes that they peeled together, Alex peeled: [ 3.6 text{ minutes} times 6 frac{text{potatoes}}{text{minute}} = 21.6 text{ potatoes} ] Thus, Alex peeled a total of 22 potatoes (rounded up since you cannot peel a fraction of a potato). Conclusion: The calculation shows that Alex peeled 22 potatoes. The final answer is boxed{B}

answer:1. Calculating t: Since the probability of picking any teacher (and hence class) is frac{1}{5}, we use: [ t = 60 cdot frac{1}{5} + 30 cdot frac{1}{5} + 20 cdot frac{1}{5} + 5 cdot frac{1}{5} + 5 cdot frac{1}{5} = left(60 + 30 + 20 + 5 + 5right) cdot frac{1}{5} = 120 cdot frac{1}{5} = 24 ] 2. Calculating s: Each student's probability of being in a class is proportional to the class size: [ s = 60 cdot frac{60}{120} + 30 cdot frac{30}{120} + 20 cdot frac{20}{120} + 5 cdot frac{5}{120} + 5 cdot frac{5}{120} ] Simplifying further: [ s = 30 + 7.5 + 3.33 + 0.208 + 0.208 approx 41.25 ] 3. Calculating t - s: [ t - s = 24 - 41.25 = -17.25 ] Thus, the difference between the average number of students per class from a teacher's perspective and from a student's perspective is -17.25. The final answer is boxed{C. -17.25}