💡To find ( g(50) ), we start by analyzing the given functional equation ( g(xy) = xg(y) ). First, we observe that if we set ( y = 1 ), the equation simplifies to:[g(x cdot 1) = xg(1) implies g(x) = xg(1)]This indicates that ( g(x) ) is a linear function of the form ( g(x) = kx ), where ( k = g(1) ).Next, we use the given value ( g(2) = 10 ) to determine the constant ( k ):[g(2) = 2k = 10 implies k = 5]Thus, the function is ( g(x) = 5x ).Finally, we substitute ( x = 50 ) into the function to find ( g(50) ):[g(50) = 5 times 50 = 250]

💡First, I need to expand the given complex multiplication ((1 + i)(1 - bi)). Multiplying the terms, I get:[(1)(1) + (1)(-bi) + (i)(1) + (i)(-bi) = 1 - bi + i - b i^2]Since (i^2 = -1), the equation simplifies to:[1 - bi + i + b = (1 + b) + (1 - b)i]Given that this equals (a), which is a real number, the imaginary part must be zero. Therefore, I set up the following system of equations:[begin{cases}1 + b = a 1 - b = 0end{cases}]Solving the second equation, I find (b = 1). Substituting (b = 1) into the first equation gives (a = 2).Thus, the complex number (a + bi) is (2 + i), which corresponds to the point ((2, 1)) in the complex plane. This point lies in the first quadrant.

💡First, I need to determine the total perimeter of the rectangular plot to understand how much fencing is required.The formula for the perimeter of a rectangle is:[P = 2 times (text{length} + text{width})]Given the length is 90 meters and the width is 50 meters, the perimeter is:[P = 2 times (90 + 50) = 2 times 140 = 280 text{ meters}]Next, I have 28 poles to place around this perimeter. To find the distance between each pole, I divide the total perimeter by the number of poles. However, since the poles are placed around a closed loop, the number of intervals between poles is equal to the number of poles. Therefore, the distance between each pole is:[text{Distance between poles} = frac{280}{28} = 10 text{ meters}]

💡Alright, so I'm looking at this problem with four statements, and I need to figure out which one is correct. Let's take them one by one.**Statement A: A rectangular prism is a cuboid.**Okay, I remember that a rectangular prism has six faces, all of which are rectangles. A cuboid is also a three-dimensional figure with six rectangular faces. So, at first glance, they seem to be the same thing. But wait, sometimes definitions can be tricky. Is there a difference between a rectangular prism and a cuboid? Maybe in some contexts, a cuboid is considered a special type of rectangular prism where all angles are right angles, but I think in general, they're used interchangeably. So, I'm leaning towards this statement being correct.**Statement B: A polyhedron with two parallel faces and all other faces being parallelograms is a prism.**Hmm, prisms are defined as polyhedrons with two congruent, parallel bases and rectangular lateral faces. But this statement says that all other faces are parallelograms, not necessarily rectangles. So, if the lateral faces are parallelograms, does that still make it a prism? I think it depends on whether the lateral faces are rectangles or not. If they're just parallelograms, it might not be a prism but something else, like a parallelepiped. So, maybe this statement isn't entirely correct.**Statement C: A cube can be obtained by cutting off a corner from a simple combination.**This one is a bit confusing. Cutting off a corner of a cube would typically result in a new shape, but would it still be a cube? If you cut off a corner, you're removing part of the cube, so the resulting shape would have more faces or edges, but it wouldn't be a cube anymore. Unless "simple combination" refers to something else, like combining multiple cubes, but the statement doesn't specify. I'm not sure about this one. Maybe it's trying to say that by cutting off a corner from a larger shape, you can get a cube, but that doesn't seem right either.**Statement D: A frustum is the part between a section obtained by cutting a cone and its base.**I remember that a frustum is what remains of a solid (like a cone or pyramid) after cutting off the top with a plane parallel to the base. So, the part between the cut and the base is indeed called a frustum. This seems correct.Okay, let's recap:- **A**: Seems correct, but I'm not entirely sure if there's a subtle difference.- **B**: Might not be correct because prisms have rectangular lateral faces, not just any parallelograms.- **C**: Unclear; cutting off a corner doesn't seem to result in a cube.- **D**: Definitely correct based on my understanding.So, between A and D, I think D is definitely correct. As for A, I'm still a bit unsure, but I think it's correct too. Maybe both A and D are correct? But the question asks for which statement is correct, implying only one. I'll go with D because it's more straightforward.

💡Okay, so I need to figure out how many diagonals a polygon with 150 sides has. Hmm, I remember that polygons have diagonals, which are lines connecting non-adjacent vertices. But I'm a bit fuzzy on the exact formula. Let me think.I think the formula for the number of diagonals in a polygon is something like n(n-3)/2, where n is the number of sides. Wait, why is that? Let me try to break it down.So, if I have a polygon with n sides, each vertex can be connected to n-3 other vertices to form a diagonal. Why n-3? Well, because you can't connect a vertex to itself, and you can't connect it to its two adjacent vertices since those connections are just the sides of the polygon, not diagonals. So, for each vertex, there are n-3 diagonals.But wait, if I just take n(n-3), won't I be counting each diagonal twice? Because each diagonal connects two vertices, so if I count from one vertex and then from the other, I'd be counting the same diagonal twice. That makes sense. So to get the actual number of unique diagonals, I should divide by 2. So the formula becomes n(n-3)/2.Let me test this formula with a simple polygon, like a pentagon, which has 5 sides. Plugging into the formula: 5(5-3)/2 = 5*2/2 = 5. Does a pentagon have 5 diagonals? Yes, I think that's correct. Each vertex connects to two non-adjacent vertices, and since there are 5 vertices, that gives 5 diagonals. Okay, so the formula works for a pentagon.Another test: a hexagon, which has 6 sides. Using the formula: 6(6-3)/2 = 6*3/2 = 9. Does a hexagon have 9 diagonals? Let me visualize a hexagon. Each vertex connects to three non-adjacent vertices, so 6 vertices times 3 gives 18, but since each diagonal is counted twice, we divide by 2, resulting in 9. Yes, that seems right.Alright, so the formula seems solid. Now, applying it to a polygon with 150 sides. Let's plug n = 150 into the formula:Number of diagonals = 150(150 - 3)/2First, calculate 150 - 3, which is 147. Then, multiply 150 by 147. Hmm, 150 times 147. Let me do that step by step.150 * 147: I can break this down into (100 + 50) * 147 = 100*147 + 50*147.100*147 is straightforward: 14,700.50*147: Well, 50 is half of 100, so half of 14,700 is 7,350.Adding them together: 14,700 + 7,350 = 22,050.So, 150*147 = 22,050.Now, divide that by 2 to get the number of diagonals: 22,050 / 2 = 11,025.So, according to my calculation, a polygon with 150 sides has 11,025 diagonals.Let me double-check my steps to make sure I didn't make a mistake. First, I used the formula n(n-3)/2, which I verified with a pentagon and a hexagon. Then, I substituted n = 150, calculated 150 - 3 = 147, multiplied 150 by 147 to get 22,050, and then divided by 2 to get 11,025. That seems correct.Wait, let me make sure I didn't make an arithmetic error when calculating 150*147. Let me do it another way to confirm.150 * 147 can also be calculated as 150*(100 + 40 + 7) = 150*100 + 150*40 + 150*7.150*100 = 15,000150*40 = 6,000150*7 = 1,050Adding them up: 15,000 + 6,000 = 21,000; 21,000 + 1,050 = 22,050. Yep, same result.So, 22,050 divided by 2 is indeed 11,025. Okay, I feel confident about this answer.Looking back at the options provided:(A) 11025(B) 12000(C) 15000(D) 2100011,025 is option A. So, the correct answer should be A.I don't see any mistakes in my reasoning, so I think I'm good.

💡First, I need to determine how many tickets Nathaniel gave away in total. He started with 37 tickets and ended up with 2 tickets. To find the total number of tickets given away, I subtract the remaining tickets from the initial number of tickets:37 tickets - 2 tickets = 35 tickets given away.Next, I know that Nathaniel gives away 5 tickets to each of his best friends. To find out how many best friends he has, I divide the total number of tickets given away by the number of tickets each friend receives:35 tickets ÷ 5 tickets per friend = 7 best friends.Therefore, Nathaniel has 7 best friends.