💡First, calculate the total number of earrings Barbie bought by multiplying the number of pairs by 2.Next, determine how many earrings Barbie gave to Alissa by finding half of the total earrings.Then, set up an equation where Alissa's total collection is triple the number of earrings she received from Barbie.Finally, solve the equation to find the total number of earrings Alissa has now.

💡First, I need to simplify the complex fraction ( frac{1}{1-i} ). To do this, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is ( 1+i ).[frac{1}{1-i} cdot frac{1+i}{1+i} = frac{1+i}{(1-i)(1+i)} = frac{1+i}{1^2 - i^2} = frac{1+i}{1+1} = frac{1+i}{2} = frac{1}{2} + frac{i}{2}]Next, I'll multiply the simplified fraction by ( i ):[left(frac{1}{2} + frac{i}{2}right) cdot i = frac{1}{2}i + frac{i^2}{2} = frac{1}{2}i - frac{1}{2} = -frac{1}{2} + frac{1}{2}i]Finally, the imaginary part of the resulting complex number is ( frac{1}{2} ).

💡To simplify the expression ((5 - 3i) - (2 + 7i)), I will start by distributing the subtraction across the terms inside the parentheses. This gives me (5 - 3i - 2 - 7i).Next, I will combine the real parts of the expression. The real parts are (5) and (-2), and when combined, they result in (3).Then, I will combine the imaginary parts. The imaginary parts are (-3i) and (-7i), and when combined, they result in (-10i).Finally, by combining the real and imaginary parts, the simplified form of the expression is (3 - 10i).

💡Alright, so I've got these equations here, and I need to figure out the pattern to write the nth equation. Let me take a closer look.First, let's write down the given equations to see if I can spot a pattern:1. (1 + 1) = 2 × 12. (2 + 1)(2 + 2) = 2² × 1 × 33. (3 + 1)(3 + 2)(3 + 3) = 2³ × 1 × 3 × 5Hmm, okay. So for the first equation, n seems to be 1. The left side is just (1 + 1), and the right side is 2 × 1. For the second equation, n is 2, and the left side is the product of (2 + 1) and (2 + 2), which is 3 × 4. The right side is 2 squared times 1 and 3, which is 4 × 3. Wait, that doesn't seem to match because 3 × 4 is 12 and 4 × 3 is also 12. Okay, that works.For the third equation, n is 3. The left side is (3 + 1)(3 + 2)(3 + 3), which is 4 × 5 × 6. That's 120. The right side is 2 cubed times 1 × 3 × 5, which is 8 × 15, and that's also 120. So far, so good.Let me see if I can generalize this. It looks like for each n, the left side is a product of terms starting from (n + 1) up to (n + n). So for n=1, it's just (1 + 1). For n=2, it's (2 + 1)(2 + 2). For n=3, it's (3 + 1)(3 + 2)(3 + 3). So in general, the left side is the product from k=1 to k=n of (n + k).Now, the right side seems to be 2 raised to the power of n multiplied by the product of the first n odd numbers. For n=1, it's 2^1 × 1. For n=2, it's 2^2 × 1 × 3. For n=3, it's 2^3 × 1 × 3 × 5. So, yes, that seems consistent.Wait, but let me check for n=4 to see if this pattern holds. If I follow the pattern, the left side should be (4 + 1)(4 + 2)(4 + 3)(4 + 4) which is 5 × 6 × 7 × 8. Let's calculate that: 5 × 6 is 30, 30 × 7 is 210, 210 × 8 is 1680.The right side should be 2^4 × 1 × 3 × 5 × 7. 2^4 is 16, and 1 × 3 × 5 × 7 is 105. So 16 × 105 is 1680. Yep, that matches. So the pattern holds for n=4.Okay, so it seems like the nth equation is the product from k=1 to k=n of (n + k) equals 2^n multiplied by the product of the first n odd numbers.Let me write that out in mathematical terms. The left side is the product of (n + 1)(n + 2)...(n + n). The right side is 2^n times the product of 1 × 3 × 5 × ... × (2n - 1). So, putting it all together, the nth equation should be:(n + 1)(n + 2)...(n + n) = 2^n × 1 × 3 × 5 × ... × (2n - 1)Let me double-check to make sure I didn't make a mistake. For n=1: (1 + 1) = 2 × 1. That's correct. For n=2: (2 + 1)(2 + 2) = 2² × 1 × 3. 3 × 4 = 12, and 4 × 3 = 12. Correct. For n=3: (3 + 1)(3 + 2)(3 + 3) = 2³ × 1 × 3 × 5. 4 × 5 × 6 = 120, and 8 × 15 = 120. Correct. And for n=4, as I checked earlier, it also works.So, I think I've got the pattern right. The nth equation is the product of (n + 1) up to (n + n) on the left, and on the right, it's 2 to the power of n times the product of the first n odd numbers.

💡Alright, so I have this problem where I need to arrange five athletes, numbered 1 to 5, into five tracks, also numbered 1 to 5. The condition is that exactly two athletes' numbers match their track numbers. I need to figure out how many different arrangements are possible under this condition.Okay, let's break this down. First, I need to understand what it means for exactly two athletes to be in their corresponding tracks. That means out of the five athletes, two will be in the correct track (their number matches the track number), and the remaining three will be in incorrect tracks. So, my task is to count all such possible arrangements.I think this is related to permutations with some restrictions. In combinatorics, when we talk about permutations where certain elements are fixed and others are not, it's called derangements. A derangement is a permutation where none of the objects appear in their original position. But in this case, it's not a complete derangement because exactly two are fixed, and the other three are deranged.So, maybe I can approach this problem by first choosing which two athletes are correctly placed, and then deranging the remaining three. That sounds logical.First, let's figure out how many ways we can choose the two athletes who will be correctly placed. Since there are five athletes, the number of ways to choose two of them is given by the combination formula:C(5, 2) = frac{5!}{2!(5-2)!} = frac{5 times 4}{2 times 1} = 10So, there are 10 ways to choose which two athletes will be in their correct tracks.Now, for the remaining three athletes, they need to be arranged such that none of them are in their correct tracks. This is a derangement of three items. The number of derangements of n items is denoted by !n. For n = 3, the number of derangements is:!3 = 2Wait, how do we calculate derangements? Let me recall. The formula for derangements is:!n = n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + dots + (-1)^n frac{1}{n!}right)So, for n = 3:!3 = 3! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!}right) = 6 left(1 - 1 + frac{1}{2} - frac{1}{6}right) = 6 left(0 + 0.5 - 0.1667right) = 6 times 0.3333 = 2Yes, that's correct. So, there are 2 derangements for three items.Therefore, for each of the 10 ways to choose the two correctly placed athletes, there are 2 ways to arrange the remaining three incorrectly. So, the total number of arrangements is:10 times 2 = 20Wait, but let me make sure I didn't miss anything. Is there another way to think about this?Alternatively, I could think of this as a permutation with exactly two fixed points. The formula for the number of permutations of n elements with exactly k fixed points is given by:C(n, k) times !(n - k)Where C(n, k) is the combination of n things taken k at a time, and !(n - k) is the derangement of the remaining n - k elements.In this case, n = 5 and k = 2, so:C(5, 2) times !3 = 10 times 2 = 20That's the same result as before. So, it seems consistent.Just to double-check, let's consider a smaller case to see if the logic holds. Suppose we have 3 athletes and 3 tracks, and we want exactly 1 athlete to be correctly placed. According to the formula, it should be:C(3, 1) times !2 = 3 times 1 = 3Let's enumerate them:Athletes: 1, 2, 3Tracks: 1, 2, 3Possible permutations with exactly one fixed point:1. Athlete 1 in track 1, athletes 2 and 3 swapped: (1, 3, 2)2. Athlete 2 in track 2, athletes 1 and 3 swapped: (3, 2, 1)3. Athlete 3 in track 3, athletes 1 and 2 swapped: (2, 1, 3)Yes, that's 3 permutations, which matches the formula. So, the formula seems to work for n = 3, k = 1.Therefore, applying the same logic to our original problem with n = 5 and k = 2, we get 20 permutations.I think that's solid. So, the total number of different arrangements is 20.**Final Answer**The total number of different arrangements is boxed{20}.

💡First, I need to identify the side lengths of the triangle. Since the side lengths are three consecutive integers starting from 3, the sides are 3, 4, and 5 units.Next, I should verify if these side lengths can form a valid triangle. According to the triangle inequality theorem, the sum of any two sides must be greater than the third side.Checking the inequalities:- 3 + 4 > 5- 4 + 5 > 3- 5 + 3 > 4All these inequalities hold true, confirming that a triangle with side lengths 3, 4, and 5 is possible.Finally, to find the perimeter, I add up all the side lengths:3 + 4 + 5 = 12 units.Therefore, the smallest possible perimeter of such a triangle is 12 units.