💡To compute (cos 240^circ), I start by recognizing that (240^circ) is in the third quadrant of the unit circle, where both cosine and sine values are negative.Next, I determine the reference angle by subtracting (180^circ) from (240^circ), which gives (60^circ).I recall that the cosine of (60^circ) is (frac{1}{2}).Since (240^circ) is in the third quadrant, the cosine value will be negative.Therefore, (cos 240^circ = -frac{1}{2}).

💡Alright, so I've got this problem here about folding patterns of squares into a regular tetrahedron. I need to figure out which of the given patterns can't be folded into one. Let me think through this step by step.First off, what's a regular tetrahedron? It's a type of pyramid with four triangular faces, right? Each face is an equilateral triangle, and all the edges are the same length. So, if I'm going to fold a pattern of squares into this shape, the squares must somehow form those triangular faces when folded.Now, the question gives me four different patterns labeled A, B, C, and D. Each pattern is made up of identical squares arranged differently, with lines indicating where the folds can be made. My task is to determine which one can't be folded into a regular tetrahedron.Let's go through each option one by one.**Option A: A cross shape of five squares with a central square and four directly attached to each side.**Okay, so this is like a plus sign made out of squares. There's one central square, and then one square attached to each side of it—up, down, left, and right. That makes five squares in total. Now, can this be folded into a tetrahedron?Well, a tetrahedron has four faces, so having five squares seems a bit confusing. Wait, but maybe the central square becomes the base, and the four surrounding squares fold up to form the triangular faces. Hmm, that might work. Each of the four surrounding squares could fold up along the edges of the central square to form the sides of the tetrahedron. I think this one might actually work. So, A is probably not the answer.**Option B: Four squares making a larger square.**This one is a bit trickier. So, imagine four smaller squares arranged in a bigger square—two squares on the top and two on the bottom, right? Each side of the larger square is made up of two smaller squares.Now, can this be folded into a tetrahedron? Let me visualize it. If I take the larger square and try to fold it, I might fold along the lines that divide the larger square into the four smaller ones. If I fold two opposite sides up, they could form two triangular faces, and the other two sides could fold up to meet them. But wait, a tetrahedron has four triangular faces, not just two. So, maybe I need to fold it differently.Alternatively, perhaps each smaller square becomes a face of the tetrahedron. But wait, a tetrahedron only has four faces, and this pattern has four squares, so that might work. If each square is folded appropriately, they could form the four triangular faces. I'm not entirely sure, but I think this one might also work. So, B might not be the answer either.**Option C: Four squares arranged in a "T" shape.**Alright, a "T" shape. So, that's like three squares in a row, and then one square attached to the center of the top square, making a T. So, in total, four squares.Can this be folded into a tetrahedron? Let me think. If I have a T-shape, I can try folding the arms of the T up to form the sides of the tetrahedron. The central square could be the base, and the three attached squares could fold up to form the triangular faces. But wait, a tetrahedron only has four faces, and here I have four squares. So, maybe this works too. I'm not entirely certain, but it seems plausible. So, C might not be the answer.**Option D: Four squares aligned in a straight row.**Okay, this one is four squares all in a straight line, like a 1x4 rectangle. So, just four squares connected edge-to-edge in a straight line.Can this be folded into a tetrahedron? Hmm, let's see. A tetrahedron has four triangular faces, and each face is connected to the others. If I have four squares in a straight line, how would I fold them to make four triangles?If I try to fold the first square up, then the second one could fold up to meet it, but the third and fourth squares would just be in the way. They wouldn't form triangles but would instead overlap or create some kind of flat shape. It doesn't seem like I can get four distinct triangular faces from a straight line of four squares. There's no way to fold them without overlapping or leaving gaps.So, this one doesn't seem to work. It looks like option D can't be folded into a regular tetrahedron.Wait, let me double-check. Maybe there's a way to fold it that I'm not seeing. If I take the first square and fold the second square up to form a right angle, then the third square could fold up to meet the first one, and the fourth square could fold up to meet the second one. But even then, I don't think it would form four triangular faces. It would probably end up with some overlapping squares or not enough faces to make a tetrahedron.Yeah, I think I'm right. Option D can't be folded into a regular tetrahedron.

💡Alright, so I have this problem about the school buying projectors. There are two parts. Let me try to tackle them one by one.**Problem (1):** Find out how much it costs to buy one A projector and one B projector each.Okay, so the problem says that buying one A projector costs 800 yuan more than buying one B projector. And purchasing 5 A projectors and 4 B projectors in total costs 32,800 yuan. I need to find the individual costs of A and B.Hmm, this sounds like a system of equations problem. Let me define variables:Let’s say the cost of one B projector is ( y ) yuan. Then, the cost of one A projector would be ( y + 800 ) yuan because it's 800 yuan more expensive.Now, the total cost for 5 A projectors and 4 B projectors is 32,800 yuan. So, in terms of ( y ), that would be:( 5(y + 800) + 4y = 32,800 )Let me expand this equation:( 5y + 4000 + 4y = 32,800 )Combine like terms:( 9y + 4000 = 32,800 )Subtract 4000 from both sides:( 9y = 28,800 )Divide both sides by 9:( y = 3,200 )So, the cost of one B projector is 3,200 yuan. Then, the cost of one A projector is ( 3,200 + 800 = 4,000 ) yuan.Wait, let me double-check that. If A is 4,000 and B is 3,200, then 5 A's would be 20,000 and 4 B's would be 12,800. Adding those together gives 32,800, which matches the problem statement. Okay, that seems right.**Problem (2):** The store has a promotion: a 10% discount on one A projector, and a 200 yuan reduction on one B projector. The school needs to purchase a total of 30 projectors, with the total cost not exceeding 97,200 yuan. Find the maximum number of A projectors the school can purchase.Alright, so first, let's understand the promotion:- For A projectors: 10% discount on the original price. The original price of A is 4,000 yuan, so the discounted price would be ( 4,000 times 0.9 = 3,600 ) yuan per A projector. - For B projectors: 200 yuan reduction on the original price. The original price of B is 3,200 yuan, so the discounted price would be ( 3,200 - 200 = 3,000 ) yuan per B projector.Let’s denote the number of A projectors as ( m ) and the number of B projectors as ( n ). The total number of projectors is 30, so:( m + n = 30 )Which means ( n = 30 - m ).The total cost should not exceed 97,200 yuan. So, the cost equation is:( 3,600m + 3,000n leq 97,200 )Substituting ( n = 30 - m ) into the cost equation:( 3,600m + 3,000(30 - m) leq 97,200 )Let me expand this:( 3,600m + 90,000 - 3,000m leq 97,200 )Combine like terms:( 600m + 90,000 leq 97,200 )Subtract 90,000 from both sides:( 600m leq 7,200 )Divide both sides by 600:( m leq 12 )So, the maximum number of A projectors the school can purchase is 12.Wait, let me verify this. If they buy 12 A projectors at 3,600 each, that's ( 12 times 3,600 = 43,200 ) yuan. Then, they buy 18 B projectors at 3,000 each, which is ( 18 times 3,000 = 54,000 ) yuan. Adding those together gives ( 43,200 + 54,000 = 97,200 ) yuan, which is exactly the limit. So, that checks out.But wait, what if they try to buy 13 A projectors? Let's see:13 A projectors would cost ( 13 times 3,600 = 46,800 ) yuan. Then, 17 B projectors would cost ( 17 times 3,000 = 51,000 ) yuan. Total cost would be ( 46,800 + 51,000 = 97,800 ) yuan, which exceeds the 97,200 limit. So, 13 is too many.Therefore, 12 is indeed the maximum number of A projectors they can buy without exceeding the budget.I think that's it. Let me summarize:1. The cost of one A projector is 4,000 yuan, and one B projector is 3,200 yuan.2. With the promotion, the school can buy a maximum of 12 A projectors while staying within the budget.

💡First, determine the cost of a small monkey doll. Since it is 2 cheaper than a large monkey doll, and a large doll costs 6, the small doll costs 4.Let ( X ) represent the total amount of money Dhoni spent.If Dhoni buys only large monkey dolls, the number of dolls he can purchase is ( frac{X}{6} ).If he buys only small monkey dolls, the number of dolls he can purchase is ( frac{X}{4} ).According to the problem, buying small dolls allows him to purchase 25 more dolls than buying large ones. This leads to the equation:[frac{X}{4} - frac{X}{6} = 25]To solve for ( X ), find a common denominator for the fractions, which is 12:[frac{3X}{12} - frac{2X}{12} = 25]Combine the fractions:[frac{X}{12} = 25]Multiply both sides by 12 to find ( X ):[X = 25 times 12 = 300]Therefore, Dhoni spent a total of 300 on the dolls.

💡First, I need to identify all the positive divisors of 24. The divisors are 1, 2, 3, 4, 6, 8, 12, and 24.Next, I need to determine which of these divisors are divisible by 4. Checking each divisor:- 1 is not divisible by 4.- 2 is not divisible by 4.- 3 is not divisible by 4.- 4 is divisible by 4.- 6 is not divisible by 4.- 8 is divisible by 4.- 12 is divisible by 4.- 24 is divisible by 4.The divisors that are divisible by 4 are 4, 8, 12, and 24. Therefore, there are 4 possible values of ( a ) that satisfy the given conditions.

💡To determine the probability of selecting a green ball, I will follow these steps:1. **Identify the probability of choosing each container**: Since there are four containers, each has an equal chance of being selected, which is ( frac{1}{4} ).2. **Calculate the probability of drawing a green ball from each container**: - **Container A**: Contains 5 red and 5 green balls. The probability of drawing a green ball is ( frac{5}{10} = frac{1}{2} ). - **Container B**: Contains 7 red and 3 green balls. The probability of drawing a green ball is ( frac{3}{10} ). - **Container C**: Contains 10 red balls and no green balls. The probability of drawing a green ball is ( frac{0}{10} = 0 ). - **Container D**: Contains 4 red and 6 green balls. The probability of drawing a green ball is ( frac{6}{10} = frac{3}{5} ).3. **Compute the total probability of drawing a green ball**: - Multiply the probability of choosing each container by the probability of drawing a green ball from that container. - Sum these products to get the total probability.4. **Perform the calculations**: [ text{Total probability} = left( frac{1}{4} times frac{1}{2} right) + left( frac{1}{4} times frac{3}{10} right) + left( frac{1}{4} times 0 right) + left( frac{1}{4} times frac{3}{5} right) ] [ = frac{1}{8} + frac{3}{40} + 0 + frac{3}{20} ] [ = frac{1}{8} + frac{3}{40} + frac{6}{40} ] [ = frac{10}{40} + frac{9}{40} ] [ = frac{19}{40} ]5. **Conclusion**: The total probability of selecting a green ball is ( frac{19}{40} ).