💡First, I need to simplify the expression ((2a + b)^2 - 3a(2a - b)). I'll start by expanding each part separately.Expanding ((2a + b)^2) using the formula ((x + y)^2 = x^2 + 2xy + y^2), I get:[(2a)^2 + 2 cdot 2a cdot b + b^2 = 4a^2 + 4ab + b^2]Next, I'll expand (-3a(2a - b)) by distributing the (-3a) across the terms inside the parentheses:[-3a cdot 2a + (-3a) cdot (-b) = -6a^2 + 3ab]Now, I'll combine the expanded parts:[4a^2 + 4ab + b^2 - 6a^2 + 3ab]Combining like terms:[(4a^2 - 6a^2) + (4ab + 3ab) + b^2 = -2a^2 + 7ab + b^2]With the expression simplified to (-2a^2 + 7ab + b^2), I'll substitute (a = 1) and (b = -2):[-2(1)^2 + 7(1)(-2) + (-2)^2 = -2 - 14 + 4 = -12]Thus, the final evaluated value of the expression is (-12).

💡Alright, so I have this problem where I need to select 4 people out of 6 to visit four different cities: Paris, London, Sydney, and Moscow. Each city is visited by one person, and each person only visits one city. But there's a catch: person A and person B don't want to go to Paris. I need to figure out how many different ways I can assign these four people to the four cities under these conditions.Okay, let's break this down. First, there are 6 people in total, and I need to choose 4 of them. But not just any 4—each person has to go to a specific city, and two of them, A and B, have restrictions about going to Paris. So, I guess the first thing I should figure out is who can go to Paris.Since A and B don't want to go to Paris, that leaves 4 other people who are willing to go there. Let's call them C, D, E, and F for simplicity. So, I have 4 choices for who goes to Paris. Once I've chosen someone for Paris, I need to assign the remaining 3 cities to the other 3 people selected from the remaining 5 people (since one person has already been assigned to Paris).Wait, hold on. If I've already selected 4 people out of 6, and one of them is assigned to Paris, then I have 3 more cities to assign to the remaining 3 people. But actually, I think I need to consider the selection and assignment together. Maybe it's better to think in terms of permutations.Let me recall that when assigning tasks or positions where order matters, permutations are used. In this case, assigning people to specific cities is a permutation problem because each city is distinct, and the order in which people are assigned matters.So, normally, if there were no restrictions, the number of ways to assign 4 people out of 6 to 4 cities would be the number of permutations of 6 people taken 4 at a time, which is calculated as:[ P(6, 4) = 6 times 5 times 4 times 3 = 360 ]But we have restrictions here: A and B can't go to Paris. So, I need to adjust this number to account for those restrictions.Let me think about how to handle the restriction. Since Paris is a specific city, and only 4 people are allowed to go there, I should first choose who goes to Paris and then assign the remaining cities to the other people.So, step 1: Choose someone to go to Paris. As A and B can't go, we have 4 choices: C, D, E, or F.Step 2: After choosing someone for Paris, we have 5 people left (since one person has already been assigned to Paris). From these 5, we need to choose 3 people to assign to the remaining 3 cities: London, Sydney, and Moscow.Wait, but actually, since we are assigning specific cities, it's not just about choosing the people but also assigning them to specific cities. So, after selecting someone for Paris, we need to assign the remaining 3 cities to the remaining 3 people selected from the 5.But hold on, actually, the total number of people to assign is 4, one to each city. So, if we have already selected 4 people out of 6, and one of them is assigned to Paris, the other three are assigned to the other three cities.But perhaps a better way is to think of it as:Total number of ways = (number of ways to choose who goes to Paris) × (number of ways to assign the remaining 3 cities to the remaining 3 people selected from the remaining 5).So, first, choose who goes to Paris: 4 choices (C, D, E, F).Then, from the remaining 5 people (since one has been assigned to Paris), we need to assign 3 cities: London, Sydney, and Moscow. But wait, actually, we need to assign 3 cities to 3 people, which is a permutation of 5 people taken 3 at a time.But wait, no. Because we have already selected 4 people in total, one for each city. So, actually, after choosing who goes to Paris, we need to choose 3 more people from the remaining 5 to assign to the other 3 cities.But hold on, that might not be correct because we are selecting 4 people in total, one for each city. So, perhaps the process is:1. Choose who goes to Paris: 4 choices (C, D, E, F).2. Then, choose 3 more people from the remaining 5 (since A and B are still available, except for the one who went to Paris) to assign to the other 3 cities.But wait, no, because if we choose 3 more people from the remaining 5, that would give us 4 people in total, which is what we need.But actually, once we've chosen someone for Paris, we have 5 people left (including A and B), and we need to assign 3 more cities to 3 of these 5 people.But since each city is distinct, the assignment matters. So, for each choice of who goes to Paris, we have:Number of ways = 4 (choices for Paris) × [number of ways to assign 3 cities to 3 people from the remaining 5].The number of ways to assign 3 cities to 3 people from 5 is the number of permutations of 5 people taken 3 at a time, which is:[ P(5, 3) = 5 times 4 times 3 = 60 ]So, total number of ways would be 4 × 60 = 240.Wait, but let me double-check that.Alternatively, another way to think about it is:Total number of ways without any restrictions is P(6,4) = 360.But we have restrictions: A and B can't go to Paris. So, we need to subtract the number of assignments where A or B is assigned to Paris.So, total without restrictions: 360.Number of assignments where A is assigned to Paris: If A is assigned to Paris, then we have to assign the remaining 3 cities to 3 people from the remaining 5 (since A is already assigned). So, that's P(5,3) = 60.Similarly, number of assignments where B is assigned to Paris: Also 60.But wait, if we subtract both, we might be double-counting the cases where both A and B are assigned to Paris, but since only one person can be assigned to Paris, that's impossible. So, there's no overlap.Therefore, total number of invalid assignments is 60 + 60 = 120.So, total valid assignments = 360 - 120 = 240.Yes, that matches the previous calculation.So, the total number of different selection schemes is 240.Therefore, the answer is B. 240.

💡First, I need to find the greatest common divisor (GCD) of the numbers 256, 180, and 600. To do this, I'll start by determining the prime factorization of each number.For 256, I know it's a power of 2. Specifically, 256 is (2^8).Next, for 180, I'll break it down into its prime factors. Dividing by 2 gives me 90, and dividing by 2 again gives me 45. Then, dividing by 3 gives me 15, and dividing by 3 again gives me 5. Finally, dividing by 5 gives me 1. So, the prime factors of 180 are (2^2 cdot 3^2 cdot 5).For 600, I'll follow a similar process. Dividing by 2 gives me 300, and dividing by 2 again gives me 150. Dividing by 2 once more gives me 75. Then, dividing by 3 gives me 25, and dividing by 5 gives me 5. Finally, dividing by 5 gives me 1. Therefore, the prime factors of 600 are (2^3 cdot 3 cdot 5^2).Now, I'll identify the common prime factors among all three numbers. The common prime factors are 2 and 3. For each common prime factor, I'll take the lowest power that appears in all three factorizations. For 2, the lowest power is (2^2), and for 3, the lowest power is (3^1).Finally, I'll multiply these lowest powers together to find the GCD: (2^2 cdot 3 = 4 cdot 3 = 12).So, the greatest common divisor of 256, 180, and 600 is 12.

💡To determine the average of the five integers ( k, m, r, s, ) and ( t ) with the given conditions, we start by noting that ( t = 20 ) and the median ( r ) is 13. To maximize the median, we minimize the values of ( k ) and ( m ). The smallest positive integers for ( k ) and ( m ) are 1 and 2, respectively. Next, we set ( r = 13 ). Since ( s ) must be greater than ( r ) but less than ( t ), the smallest possible value for ( s ) is 14. Now, we calculate the average:[text{Average} = frac{k + m + r + s + t}{5} = frac{1 + 2 + 13 + 14 + 20}{5} = frac{50}{5} = 10]

💡First, determine how much money Michael gives to his brother by calculating half of 42.Next, add this amount to the brother's original 17 to find the total money he has after receiving the gift.Finally, subtract the 3 spent on candy from the total to find out how much money the brother has left.

💡First, I need to determine how many walnuts the boy squirrel initially gathered. Let's denote this number as ( W ).The boy squirrel adds ( W ) walnuts to the existing 12 in the burrow but drops 1 on the way. This means the total number of walnuts in the burrow after his contribution is ( 12 + W - 1 ).Next, the girl squirrel brings 5 more walnuts to the burrow, increasing the total to ( 12 + W - 1 + 5 ).However, the girl squirrel eats 2 walnuts, so we need to subtract 2 from the total, resulting in ( 12 + W - 1 + 5 - 2 ).According to the problem, there are 20 walnuts left in the burrow after all these actions. Therefore, I can set up the equation:[12 + W - 1 + 5 - 2 = 20]Simplifying the equation:[W + 14 = 20]To find the value of ( W ), I subtract 14 from both sides of the equation:[W = 20 - 14][W = 6]Thus, the boy squirrel initially gathered 6 walnuts.