answer：First, I need to determine the probability of it not snowing on each of the three days. For the first day, the chance of snow is 1/2, so the chance of no snow is 1 - 1/2 = 1/2. On the second day, the chance of snow is 2/3, so the chance of no snow is 1 - 2/3 = 1/3. On the third day, the chance of snow is 3/4, so the chance of no snow is 1 - 3/4 = 1/4.Next, I calculate the probability of it not snowing on all three days by multiplying the probabilities of no snow each day: (1/2) × (1/3) × (1/4) = 1/24.Finally, to find the probability that it snows at least once during the three days, I subtract the probability of no snow all three days from 1: 1 - 1/24 = 23/24.

answer：Okay, so I need to simplify the expression (cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{8pi}{7}). Hmm, I remember that sometimes when dealing with sums of cosines, especially with angles that are multiples of (pi/7), it can be helpful to use complex numbers or roots of unity. Let me think about how to approach this.First, I recall that the seventh roots of unity are the solutions to the equation (z^7 = 1). These roots can be expressed as (e^{2pi i k/7}) for (k = 0, 1, 2, ..., 6). The real parts of these roots are the cosines of the angles (2pi k/7). So, maybe I can use these roots to express the sum of cosines.Let me denote (omega = e^{2pi i /7}). Then, the seventh roots of unity are (1, omega, omega^2, omega^3, omega^4, omega^5, omega^6). The sum of all these roots is zero because they are the roots of the polynomial (z^7 - 1 = 0), which factors as ((z - 1)(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) = 0). Since (omega neq 1), the sum (1 + omega + omega^2 + omega^3 + omega^4 + omega^5 + omega^6 = 0). Therefore, (omega + omega^2 + omega^3 + omega^4 + omega^5 + omega^6 = -1).Now, the original expression involves (cos frac{2pi}{7}), (cos frac{4pi}{7}), and (cos frac{8pi}{7}). Let me note that (cos frac{8pi}{7}) can be rewritten because cosine is an even function and has a period of (2pi). Specifically, (cos frac{8pi}{7} = cos left( frac{8pi}{7} - 2pi right) = cos left( -frac{6pi}{7} right) = cos frac{6pi}{7}). So, the expression becomes (cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7}).Looking back at the roots of unity, the real parts of (omega), (omega^2), (omega^3), (omega^4), (omega^5), and (omega^6) are (cos frac{2pi}{7}), (cos frac{4pi}{7}), (cos frac{6pi}{7}), (cos frac{8pi}{7}), (cos frac{10pi}{7}), and (cos frac{12pi}{7}), respectively. But since cosine is periodic and even, some of these angles can be simplified:- (cos frac{8pi}{7} = cos left( pi + frac{pi}{7} right) = -cos frac{pi}{7})- (cos frac{10pi}{7} = cos left( pi + frac{3pi}{7} right) = -cos frac{3pi}{7})- (cos frac{12pi}{7} = cos left( 2pi - frac{2pi}{7} right) = cos frac{2pi}{7})Wait, that might complicate things. Maybe I should think differently. Since I have the sum of all the roots except 1 equals -1, and the real parts of the roots are the cosines, maybe I can relate the sum of the cosines to this.Let me denote (x = cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7}). Then, the sum of all the real parts of the roots except 1 is (2x), because each cosine term appears twice in the sum of the roots (since (cos theta = cos (-theta))). Wait, is that correct?Actually, the sum of all the real parts of the roots is the sum of (cos frac{2pi k}{7}) for (k = 0) to (6). But (cos 0 = 1), so the sum is (1 + 2(cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7})). But we know that the sum of all roots is zero, so the sum of their real parts is zero. Therefore:(1 + 2x = 0)So, (2x = -1), which implies (x = -frac{1}{2}).Wait, that seems too straightforward. Let me verify.We have the seventh roots of unity: (1, omega, omega^2, omega^3, omega^4, omega^5, omega^6). The sum of these is zero. The real part of each root is (cos frac{2pi k}{7}) for (k = 0) to (6). So, the sum of the real parts is:(sum_{k=0}^{6} cos frac{2pi k}{7} = 0)But (cos 0 = 1), and the other terms come in pairs: (cos frac{2pi}{7} + cos frac{12pi}{7}), (cos frac{4pi}{7} + cos frac{10pi}{7}), (cos frac{6pi}{7} + cos frac{8pi}{7}). But since (cos theta = cos (2pi - theta)), each pair is equal. So, we have:(1 + 2(cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7}) = 0)Therefore, (2(cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7}) = -1), so the sum is (-frac{1}{2}).But wait, the original expression was (cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{8pi}{7}). Earlier, I noted that (cos frac{8pi}{7} = cos frac{6pi}{7}), so the sum is indeed (cos frac{2pi}{7} + cos frac{4pi}{7} + cos frac{6pi}{7}), which equals (-frac{1}{2}).So, the simplified value is (-frac{1}{2}).I think that makes sense. Let me double-check by considering specific values or perhaps using another method.Alternatively, I could use the identity for the sum of cosines. There's a formula for the sum of cosines of angles in arithmetic progression, but in this case, the angles aren't in arithmetic progression. However, they are related through powers of (omega), which is a root of unity.Another approach might be to use multiple-angle formulas or product-to-sum identities, but that might get complicated. The method using roots of unity seems more straightforward and leads directly to the result.Therefore, I'm confident that the simplified value is (-frac{1}{2}).

answer：Okay, so I have this problem where I need to find the value of λ such that the three vectors a, b, and c lie in the same plane. The vectors are given as:- Vector a = (2, -1, 3)- Vector b = (-1, 4, -2)- Vector c = (1, 3, λ)And the options are A: 1, B: 2, C: 3, D: 4.Hmm, I remember that if three vectors lie in the same plane, they are coplanar. One way to determine if vectors are coplanar is to check if the scalar triple product of the vectors is zero. The scalar triple product is calculated as the determinant of a matrix formed by the vectors. If the determinant is zero, the vectors are coplanar.So, let me recall the formula for the scalar triple product. It's given by:a · (b × c) = 0Which means the dot product of vector a with the cross product of vectors b and c should be zero.First, I need to compute the cross product of vectors b and c. Let me write down the components of vectors b and c:Vector b = (-1, 4, -2)Vector c = (1, 3, λ)The cross product b × c can be calculated using the determinant formula:b × c = |i j k| |-1 4 -2| |1 3 λ|So, expanding this determinant:i * (4 * λ - (-2) * 3) - j * (-1 * λ - (-2) * 1) + k * (-1 * 3 - 4 * 1)Let me compute each component step by step.First, the i component:4 * λ = 4λ(-2) * 3 = -6So, 4λ - (-6) = 4λ + 6Then, the j component:-1 * λ = -λ(-2) * 1 = -2So, -λ - (-2) = -λ + 2But since it's subtracted, it becomes -(-λ + 2) = λ - 2Wait, no, let me double-check that. The formula is:b × c = i*(b2c3 - b3c2) - j*(b1c3 - b3c1) + k*(b1c2 - b2c1)So, for the j component, it's negative of (b1c3 - b3c1). So:b1c3 = (-1)*λ = -λb3c1 = (-2)*1 = -2So, b1c3 - b3c1 = -λ - (-2) = -λ + 2Then, the j component is -(-λ + 2) = λ - 2Okay, that seems correct.Now, the k component:b1c2 = (-1)*3 = -3b2c1 = 4*1 = 4So, b1c2 - b2c1 = -3 - 4 = -7Putting it all together, the cross product b × c is:(4λ + 6, λ - 2, -7)Now, I need to compute the dot product of vector a with this cross product.Vector a = (2, -1, 3)So, the dot product a · (b × c) is:2*(4λ + 6) + (-1)*(λ - 2) + 3*(-7)Let me compute each term:First term: 2*(4λ + 6) = 8λ + 12Second term: (-1)*(λ - 2) = -λ + 2Third term: 3*(-7) = -21Now, adding all these together:8λ + 12 - λ + 2 - 21Combine like terms:(8λ - λ) + (12 + 2 - 21) = 7λ + (14 - 21) = 7λ - 7So, the scalar triple product is 7λ - 7.Since the vectors are coplanar, this must equal zero:7λ - 7 = 0Solving for λ:7λ = 7λ = 1So, λ is equal to 1, which corresponds to option A.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, cross product b × c:i*(4λ - (-6)) = i*(4λ + 6)j component: - [(-1)*λ - (-2)*1] = - [ -λ + 2 ] = λ - 2k component: (-1)*3 - 4*1 = -3 - 4 = -7Dot product with a:2*(4λ + 6) = 8λ + 12-1*(λ - 2) = -λ + 23*(-7) = -21Total: 8λ + 12 - λ + 2 -21 = 7λ -7Set to zero: 7λ -7 =0 => λ=1Yes, that seems correct. So, the answer should be A:1.

answer：Alright, so I've got this problem here about two types of cars, type A and type B, and their capacities to transport goods. The problem has two parts: first, figuring out how much each type of car can carry individually, and second, calculating the total amount of goods a logistics company needs to transport using a certain number of these cars.Okay, let's start with the first part. I'm told that using 2 type A cars and 1 type B car can transport 10 tons of goods at a time. And separately, using 1 type A car and 2 type B cars can transport 11 tons of goods at a time. So, I need to find out how much each type A and type B car can carry on their own.Hmm, this sounds like a system of equations problem. Let me assign variables to make it clearer. Let's say the capacity of one type A car is 'x' tons, and the capacity of one type B car is 'y' tons. So, based on the information given, I can set up two equations:1. 2x + y = 102. x + 2y = 11Alright, so now I have two equations with two variables. I need to solve this system to find the values of x and y.I think the best way to solve this is by substitution or elimination. Let me try substitution first. From the first equation, I can solve for y in terms of x.From equation 1:2x + y = 10So, y = 10 - 2xNow, I can substitute this expression for y into the second equation.From equation 2:x + 2y = 11Substituting y:x + 2(10 - 2x) = 11Let me simplify this:x + 20 - 4x = 11Combine like terms:-3x + 20 = 11Now, subtract 20 from both sides:-3x = -9Divide both sides by -3:x = 3Okay, so x is 3 tons. That means one type A car can carry 3 tons. Now, let's find y using the expression we had earlier.y = 10 - 2xSubstitute x = 3:y = 10 - 2(3)y = 10 - 6y = 4So, y is 4 tons. That means one type B car can carry 4 tons.Let me double-check these values to make sure they satisfy both equations.First equation:2x + y = 102(3) + 4 = 6 + 4 = 10Yes, that works.Second equation:x + 2y = 113 + 2(4) = 3 + 8 = 11That also works.Great, so part one is solved. One type A car can carry 3 tons, and one type B car can carry 4 tons.Now, moving on to part two. A logistics company has a certain amount of goods to transport and plans to rent 6 type A cars and 8 type B cars, with each car fully loaded. I need to find out how many tons of goods the company needs to transport.Alright, so if one type A car carries 3 tons, then 6 type A cars would carry 6 times 3 tons. Similarly, one type B car carries 4 tons, so 8 type B cars would carry 8 times 4 tons.Let me calculate that.Total weight carried by type A cars:6 cars * 3 tons/car = 18 tonsTotal weight carried by type B cars:8 cars * 4 tons/car = 32 tonsNow, to find the total weight, I just add these two amounts together.Total weight = 18 tons + 32 tons = 50 tonsSo, the logistics company needs to transport a total of 50 tons of goods.Let me just recap to make sure I didn't make any mistakes. I found the capacities of each car by solving the system of equations, then multiplied those capacities by the number of cars the company plans to use and summed them up. Everything seems to check out.I think that's it. I've got the answers for both parts of the problem.

answer：To find ( f^{-1}(4) ) for the function ( f(x) = (x-1)^2 ) where ( x leq 1 ), we start by setting ( f(t) = 4 ). This gives us the equation ( (t-1)^2 = 4 ).Solving for ( t ), we take the square root of both sides, resulting in ( t - 1 = pm 2 ). This yields two potential solutions: ( t = 3 ) and ( t = -1 ).However, since the domain of ( f ) is restricted to ( x leq 1 ), we discard ( t = 3 ) as it does not satisfy the condition. Therefore, the valid solution is ( t = -1 ).Thus, ( f^{-1}(4) = -1 ).

answer：Alright, so I have this problem here about sets A, B, and C. Let me try to wrap my head around it step by step. First, let me write down what each set is:- Set A is all real numbers x such that -5 is less than x, and x is less than 5. So, A is the interval (-5, 5).- Set B is all real numbers x such that -7 is less than x, and x is less than some number a. So, B is the interval (-7, a).- Set C is all real numbers x such that some number b is less than x, and x is less than 2. So, C is the interval (b, 2).The problem says that the intersection of A and B is equal to C. So, A ∩ B = C. I need to find the value of a + b.Hmm, okay. Let me recall what the intersection of two sets means. The intersection of A and B, which is A ∩ B, is the set of all elements that are in both A and B. So, in terms of intervals, it's the overlap between (-5, 5) and (-7, a).Let me visualize this on a number line. A starts at -5 and goes up to 5, while B starts at -7 and goes up to a. So, the overlap between A and B would depend on where a is.Since A starts at -5 and B starts at -7, the left endpoint of the intersection should be the larger of -5 and -7. But since -5 is greater than -7, the left endpoint of A ∩ B is -5.Now, the right endpoint of the intersection would be the smaller of 5 and a. So, if a is greater than 5, the intersection would end at 5. If a is less than 5, the intersection would end at a.But the problem says that A ∩ B equals C, which is (b, 2). So, the intersection must be (b, 2). That means the left endpoint of the intersection is b, and the right endpoint is 2.From earlier, I determined that the left endpoint of A ∩ B is -5. Therefore, b must be -5 because the left endpoint of C is b.Now, looking at the right endpoint, the intersection A ∩ B must end at 2 because C ends at 2. That means the smaller of 5 and a must be 2. So, the right endpoint is 2, which is less than 5. Therefore, a must be 2 because if a were greater than 2, the intersection would end at 5, but since it ends at 2, a has to be 2.So, now I have b = -5 and a = 2. Therefore, a + b is 2 + (-5) = -3.Wait, let me double-check that. If a is 2, then set B is (-7, 2). The intersection of A (-5, 5) and B (-7, 2) would indeed be (-5, 2), which is set C. So, yes, that makes sense.I think that's correct. So, the value of a + b is -3.**Final Answer**The value of the real number (a + b) is boxed{-3}.