💡Alright, so I've got this problem here about the economically active population and unemployment rates. Let me try to wrap my head around it step by step. First, the problem says that the number of economically active people in a city will increase by 4% next year. That means both employed and unemployed people together will go up by 4%. Okay, got that. And then it also says that the number of unemployed people will decrease by 9%. Hmm, that's interesting because even though the total economically active population is increasing, the number of unemployed is actually going down. This year, the unemployed made up 5.6% of the economically active population. So, I need to find out what percentage they'll make up next year after these changes. Let me try to break it down. Maybe I should assign some variables to make it clearer. Let's say the current economically active population is E. So, this year, the number of unemployed people is 5.6% of E, which is 0.056E. Next year, the total economically active population is going to increase by 4%. So, the new total will be E plus 4% of E, which is E * 1.04. Now, the number of unemployed is going to decrease by 9%. So, next year's unemployed will be this year's unemployed minus 9% of that. That would be 0.056E minus 0.09 * 0.056E. Let me calculate that: 0.09 * 0.056 is 0.00504, so subtracting that from 0.056 gives 0.05096E. So, next year, the number of unemployed is 0.05096E, and the total economically active population is 1.04E. To find the percentage of unemployed next year, I need to divide the number of unemployed by the total economically active population and then multiply by 100 to get the percentage. So, that would be (0.05096E / 1.04E) * 100. The E cancels out, which is nice because it means the actual number doesn't matter, just the percentages. Now, let's do the division: 0.05096 divided by 1.04. Hmm, 0.05096 divided by 1.04. Let me see, 1.04 goes into 0.05096 how many times? Well, 1.04 times 0.049 is approximately 0.05096. So, that means 0.05096 / 1.04 is roughly 0.049. Multiplying that by 100 gives me 4.9%. So, next year, the unemployed will make up about 4.9% of the economically active population. Wait, let me double-check my calculations to make sure I didn't make a mistake. Starting with the current unemployed: 5.6% of E is 0.056E. Decreasing that by 9% gives 0.056E * 0.91, which is indeed 0.05096E. The total economically active population next year is 1.04E. Dividing 0.05096E by 1.04E gives 0.049, which is 4.9%. Yeah, that seems right. I think I've got it. So, next year, the unemployment rate will drop to 4.9% of the economically active population.

💡Okay, so I have this problem about a cube, ABCD-A₁B₁C₁D₁, and I need to find the number of planes α that pass through four of its vertices. The catch is that the remaining four vertices are all at equal distances from this plane α. The options given are 6, 8, 12, or 16. Hmm, let me try to visualize this.First, I know a cube has 8 vertices. If a plane passes through four of them, the other four are on the opposite side or maybe symmetrically placed. The key here is that these four remaining vertices are all equidistant from the plane. So, I need to figure out how such planes can be arranged.Let me start by thinking about the faces of the cube. Each face is a square, and each face plane passes through four vertices. The remaining four vertices are on the opposite face, which is parallel to the first one. Since the cube is symmetrical, the distance from each of these four vertices to the plane should be the same. So, each of the six faces of the cube satisfies this condition. That gives me six planes right away.But wait, the answer choices go up to 16, so there must be more planes beyond just the faces. Maybe there are planes that cut through the cube diagonally? Let me think about that.I remember that in a cube, there are planes that cut through four vertices diagonally, not just along the edges. For example, imagine a plane that goes through two opposite edges of the cube. These planes are sometimes called "diagonal planes." Each such plane would pass through four vertices, and the other four vertices would be symmetrically placed with respect to this plane.Let me try to count these diagonal planes. In a cube, each space diagonal (a diagonal that goes through the cube from one vertex to the opposite vertex) can be part of such a plane. But how many unique planes are there?I think for each pair of opposite edges, there's a plane that cuts through them. Since a cube has 12 edges, but each plane uses two edges, and each plane is counted multiple times. Wait, maybe it's better to think in terms of the cube's geometry.Each face has two main diagonals, but those are on the face planes. The diagonal planes I'm thinking of are different. They cut through four vertices, not lying on the same face. For example, consider the vertices A, B₁, C, and D₁. These four form a rectangle, and the plane passing through them is a diagonal plane.How many such planes are there? Let me see. For each pair of opposite edges, there's a corresponding diagonal plane. Since the cube has 6 faces, each face has two pairs of opposite edges, but maybe each diagonal plane is shared between two faces.Wait, perhaps it's simpler. I recall that in a cube, there are 6 face planes and 6 "diagonal" planes, each cutting through four vertices diagonally. So, that would give a total of 12 planes. But let me verify this.Each diagonal plane is determined by two opposite edges. Since the cube has 12 edges, but each plane is defined by two edges, and each plane is counted twice (once for each pair of opposite edges). So, the number of diagonal planes would be 12 / 2 = 6. So, that's 6 diagonal planes.Adding the 6 face planes, that gives 12 planes in total. So, the answer should be 12, which is option C.But wait, let me make sure I'm not missing any other planes. Are there any other planes that pass through four vertices and leave the other four equidistant? Maybe planes that aren't aligned with the cube's edges or faces?Hmm, if I consider planes that are not aligned with the cube's edges, would they pass through four vertices and have the other four equidistant? I'm not sure. Maybe, but I think the only planes that satisfy this condition are the face planes and the diagonal planes I mentioned earlier.To confirm, let me think about the cube's structure. Each face plane is straightforward, and the diagonal planes pass through four vertices in a way that the other four are symmetrically opposite. Since the cube is highly symmetrical, these should be the only possibilities.So, 6 face planes plus 6 diagonal planes equal 12 planes. Therefore, the number of such planes α is 12, which corresponds to option C.

💡First, I need to find the length of the unknown parallel side of the trapezium. I know the area of the trapezium is 300 square centimeters, one of the parallel sides is 18 centimeters, and the distance between the two parallel sides is 15 centimeters.I'll use the formula for the area of a trapezium, which is:Area = (1/2) * (sum of the lengths of the parallel sides) * (distance between them)Plugging in the known values:300 = (1/2) * (18 + x) * 15To solve for x, I'll first simplify the equation:300 = (18 + x) * 7.5Next, I'll divide both sides by 7.5 to isolate (18 + x):40 = 18 + xFinally, I'll subtract 18 from both sides to find the value of x:x = 22Therefore, the length of the other parallel side is 22 centimeters.

💡First, I need to determine the value of ( x_0 ) such that ( f(x_0) = 8 ) for the given piecewise function.The function ( f(x) ) is defined differently depending on whether ( x ) is greater than or equal to 2 or less than 2.1. **Case 1: ( x geq 2 )** [ f(x) = x^2 + 2 ] Setting ( f(x_0) = 8 ): [ x_0^2 + 2 = 8 ] Solving for ( x_0 ): [ x_0^2 = 6 implies x_0 = sqrt{6} quad (text{since } x_0 geq 2) ]2. **Case 2: ( x < 2 )** [ f(x) = 2x ] Setting ( f(x_0) = 8 ): [ 2x_0 = 8 implies x_0 = 4 ] However, ( x_0 = 4 ) does not satisfy ( x_0 < 2 ), so this solution is not valid.Therefore, the valid solution is ( x_0 = sqrt{6} ).

💡First, I need to calculate the total cost of the initial grocery items. Adding up the prices of the loaf of bread, three candy bars, box of cereal, two gallons of milk, a dozen eggs, and a pack of cheese gives a subtotal of 35.Next, I'll determine how much money remains after purchasing these initial items by subtracting the subtotal from the total budget of 200, which leaves 165.Then, I'll calculate 15% of the remaining 165 to find out how much is spent on fruits, which is 24.75. Subtracting this amount from 165 leaves 140.25.After that, I'll apply a 5% discount to the 10 must-buy item, resulting in a cost of 9.50. Subtracting this from the remaining 140.25 leaves 130.75.Next, I'll spend 1/5 of the remaining 130.75 on a vegetable mix, which amounts to 26.15. Subtracting this from 130.75 leaves 104.60.Finally, I'll calculate the 7% sales tax on the total cost of the groceries before tax, which is 95.40. The sales tax adds 6.68, making the total cost after tax 102.08. Subtracting this from the remaining 104.60 leaves approximately 2.52.Therefore, after all transactions and applying the sales tax, the remaining amount of money is 2.52.

💡Okay, so I have this problem where I need to find the sum of all three-digit numbers from 500 to 999 where all the digits are distinct. Then, I have to find the remainder when this sum is divided by 1000. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let's understand the range of numbers we're dealing with. The numbers are from 500 to 999. So, all these are three-digit numbers where the hundreds digit is 5, 6, 7, 8, or 9. That gives us 5 possible choices for the hundreds place.Now, the problem specifies that all digits must be distinct. That means once I choose a digit for the hundreds place, I can't use it again in the tens or units place. Similarly, the tens digit can't be the same as the hundreds or units digit, and the same goes for the units digit.So, let's think about how many such numbers there are. For the hundreds digit, we have 5 choices (5-9). For the tens digit, since one digit is already used in the hundreds place, we have 9 remaining choices (0-9 excluding the hundreds digit). Then, for the units digit, we have 8 choices left (0-9 excluding the hundreds and tens digits). So, the total number of such numbers is 5 * 9 * 8 = 360. That's a lot of numbers!But instead of listing all 360 numbers and adding them up, which would take forever, I need a smarter way to calculate the sum. Maybe I can calculate the contribution of each digit place (hundreds, tens, units) separately and then add them up.Let's start with the hundreds place. Each number in our range has a hundreds digit of 5, 6, 7, 8, or 9. Each of these digits will appear in the hundreds place the same number of times. How many times does each hundreds digit appear?Well, for each hundreds digit, there are 9 choices for the tens digit and 8 choices for the units digit, so each hundreds digit appears 9 * 8 = 72 times. Therefore, the total contribution from the hundreds place is:(5 + 6 + 7 + 8 + 9) * 72 * 100Let me compute that. The sum of the hundreds digits is 5 + 6 + 7 + 8 + 9 = 35. So, 35 * 72 = 2520. Then, multiplying by 100 gives 2520 * 100 = 252000.Okay, so the hundreds place contributes 252000 to the total sum.Now, moving on to the tens place. The tens digit can be any digit from 0 to 9 except the hundreds digit. So, for each hundreds digit, there are 9 possible choices for the tens digit. But wait, since the digits must be distinct, once we choose the hundreds and tens digits, the units digit is limited to 8 choices.But how many times does each digit from 0 to 9 appear in the tens place across all numbers?Let's think about it. For each hundreds digit (5-9), each digit from 0-9 (excluding the hundreds digit) can appear in the tens place. So, for each hundreds digit, there are 9 possible tens digits, each appearing the same number of times.But how many times does each tens digit appear? For each hundreds digit, each tens digit appears 8 times (since the units digit can be any of the remaining 8 digits). So, for each hundreds digit, each tens digit appears 8 times.Since there are 5 hundreds digits, each tens digit (0-9, excluding the hundreds digit) appears 5 * 8 = 40 times. Wait, but hold on, the hundreds digit is fixed for each group, so actually, each digit from 0-9 (excluding the hundreds digit) appears 8 times per hundreds digit. So, across all hundreds digits, each digit from 0-9 appears 5 * 8 = 40 times in the tens place.But wait, is that correct? Let's see. For example, take the digit 0. It can appear in the tens place for each hundreds digit (5-9), and for each, it can pair with 8 different units digits. So, 0 appears 5 * 8 = 40 times in the tens place.Similarly, take the digit 5. But wait, 5 is a hundreds digit, so it can't appear in the tens place. So, actually, each digit from 0-9, except for the hundreds digits (5-9), can appear in the tens place. Wait, no, that's not correct. The hundreds digit is fixed, but the tens digit can be any digit except the hundreds digit, including 0.Wait, maybe I need to clarify. For each hundreds digit (5-9), the tens digit can be any digit from 0-9 except the hundreds digit. So, for each hundreds digit, there are 9 possible tens digits, each of which can pair with 8 units digits. Therefore, each tens digit (excluding the hundreds digit) appears 8 times per hundreds digit.But since the hundreds digit varies from 5-9, each digit from 0-9 (excluding 5-9) will appear in the tens place multiple times. Wait, no, actually, digits 5-9 can appear in the tens place as long as they are not the hundreds digit. For example, if the hundreds digit is 5, then the tens digit can be 0-9 except 5, so digits 6-9 can appear in the tens place. Similarly, if the hundreds digit is 6, the tens digit can be 0-9 except 6, so digits 5,7-9 can appear.Therefore, each digit from 0-9 appears in the tens place a certain number of times. Let's compute how many times each digit appears in the tens place.For digits 0-4: Each can appear in the tens place for all 5 hundreds digits (since none of them are the hundreds digit). For each hundreds digit, each of these digits can appear 8 times (since units digit has 8 choices). So, total appearances for digits 0-4 in the tens place: 5 * 8 = 40 times each.For digits 5-9: Each of these digits can only appear in the tens place when the hundreds digit is not themselves. So, for digit 5, it can appear in the tens place when the hundreds digit is 6,7,8,9. That's 4 hundreds digits. For each, digit 5 can appear 8 times. So, total appearances for digit 5: 4 * 8 = 32 times. Similarly, digits 6-9 each can appear in the tens place when the hundreds digit is not themselves. So, each of these digits also appears 4 * 8 = 32 times.Wait, so digits 0-4 appear 40 times each in the tens place, and digits 5-9 appear 32 times each in the tens place.Therefore, the total contribution from the tens place is:Sum of (digit * number of appearances) for each digit.So, for digits 0-4: each appears 40 times, so sum is (0+1+2+3+4) * 40 = 10 * 40 = 400.For digits 5-9: each appears 32 times, so sum is (5+6+7+8+9) * 32 = 35 * 32 = 1120.Therefore, total sum for the tens place is 400 + 1120 = 1520. But wait, this is the sum of the tens digits, but each digit is in the tens place, so we need to multiply by 10. So, total contribution is 1520 * 10 = 15200.Wait, let me double-check that. The sum of the tens digits is 1520, so when we consider their place value, it's 1520 * 10 = 15200.Okay, so the tens place contributes 15200 to the total sum.Now, moving on to the units place. The units digit can be any digit from 0-9 except the hundreds and tens digits. So, similar to the tens place, each digit can appear in the units place a certain number of times.Let's figure out how many times each digit appears in the units place.For digits 0-4: Each can appear in the units place for all 5 hundreds digits, but for each hundreds digit, the tens digit can be 9 choices (excluding the hundreds digit), and the units digit is then 8 choices (excluding hundreds and tens). Wait, no, actually, for each hundreds digit, each units digit (excluding hundreds and tens) can appear.Wait, perhaps a better approach is to realize that for each hundreds digit, each units digit can be any digit except the hundreds and tens digits. So, for each hundreds digit, each units digit (excluding the hundreds digit) can appear a certain number of times.But this might get complicated. Alternatively, since the problem is symmetric for the tens and units places, except that the units place can include 0, just like the tens place.Wait, actually, the logic for the units place is similar to the tens place. For each hundreds digit, the units digit can be any digit except the hundreds and tens digits. So, for each hundreds digit, the units digit can be 8 choices (since two digits are already used: hundreds and tens). But how many times does each digit appear in the units place?Let's think about it. For each hundreds digit, each units digit (excluding the hundreds digit) can appear multiple times, depending on the tens digit.Wait, maybe it's similar to the tens place. For each hundreds digit, each digit from 0-9 (excluding the hundreds digit) can appear in the units place a certain number of times.But actually, for each hundreds digit, each digit (excluding the hundreds digit) can appear in the units place 8 times (since for each units digit, there are 8 possible tens digits). Wait, no, that's not quite right.Wait, for each hundreds digit, the units digit can be any digit except the hundreds digit and the tens digit. So, for each hundreds digit, the units digit can be 8 choices (since two digits are already used: hundreds and tens). But the tens digit itself varies, so for each hundreds digit, each units digit (excluding the hundreds digit) can appear multiple times.Wait, perhaps it's better to think that for each hundreds digit, each units digit (excluding the hundreds digit) can appear 8 times, similar to the tens place. But actually, for each hundreds digit, each units digit can appear 8 times because for each units digit, there are 8 possible tens digits.Wait, no, that's not correct. For each hundreds digit, the units digit is determined after choosing the tens digit. So, for each hundreds digit, the units digit can be any of the remaining 8 digits (excluding hundreds and tens). So, for each hundreds digit, each units digit (excluding the hundreds digit) can appear 8 times, but only if the tens digit is chosen appropriately.Wait, maybe it's similar to the tens place. For each hundreds digit, each digit (excluding the hundreds digit) can appear in the units place 8 times. So, similar to the tens place, each digit from 0-9 (excluding the hundreds digit) appears 8 times in the units place for each hundreds digit.But then, across all hundreds digits, how many times does each digit appear in the units place?For digits 0-4: Each can appear in the units place for all 5 hundreds digits, and for each, 8 times. So, total appearances: 5 * 8 = 40 times each.For digits 5-9: Each can appear in the units place for 4 hundreds digits (since they can't be the hundreds digit themselves), and for each, 8 times. So, total appearances: 4 * 8 = 32 times each.Wait, that seems similar to the tens place. So, the sum for the units place would be the same as the tens place, right?Wait, let's compute it.Sum of units digits:For digits 0-4: each appears 40 times, so sum is (0+1+2+3+4) * 40 = 10 * 40 = 400.For digits 5-9: each appears 32 times, so sum is (5+6+7+8+9) * 32 = 35 * 32 = 1120.Total sum for units digits: 400 + 1120 = 1520.But since the units place is just the units digit, we don't need to multiply by 10. So, the contribution from the units place is 1520.Wait, but hold on, in the tens place, we had to multiply by 10 because it's the tens place, but for the units place, it's just the digit itself. So, yes, the units place contributes 1520.So, putting it all together:Total sum S = hundreds contribution + tens contribution + units contribution = 252000 + 15200 + 1520.Let me compute that:252000 + 15200 = 267200267200 + 1520 = 268720So, the total sum S is 268720.Now, the problem asks for the remainder when S is divided by 1000. So, we need to compute 268720 mod 1000.To find the remainder when divided by 1000, we can look at the last three digits of S. The last three digits of 268720 are 720. Therefore, the remainder is 720.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, hundreds place contribution: 35 * 72 * 100 = 35 * 7200 = 252000. That seems correct.Tens place: sum of digits is 1520, multiplied by 10 is 15200. Units place: sum of digits is 1520. So, total sum is 252000 + 15200 + 1520 = 268720. Yes, that looks right.And 268720 divided by 1000 is 268 with a remainder of 720. So, the remainder is indeed 720.I think that's the correct answer.