answer：Okay, so I have this problem with two vectors, a and b. They have an angle of 60 degrees between them. The magnitude of vector a is 4, and the magnitude of vector b is 2. I need to find two things: first, the dot product of a and b, and second, the magnitude of the vector a plus b. Hmm, let me think about how to approach this.Starting with the first part, the dot product. I remember that the dot product of two vectors can be found using the formula involving the magnitudes of the vectors and the cosine of the angle between them. The formula is:a · b = |a| |b| cosθWhere θ is the angle between them. So in this case, |a| is 4, |b| is 2, and θ is 60 degrees. Let me plug those values into the formula.First, I need to calculate the cosine of 60 degrees. I remember that cos(60°) is 0.5 or 1/2. So, substituting the values:a · b = 4 * 2 * (1/2)Let me compute that step by step. 4 multiplied by 2 is 8, and then 8 multiplied by 1/2 is 4. So, the dot product of a and b is 4. That seems straightforward.Now, moving on to the second part: finding the magnitude of the vector a + b. I recall that the magnitude of the sum of two vectors can be found using the formula:|a + b| = sqrt(|a|² + |b|² + 2(a · b))This formula comes from the law of cosines, considering the angle between the vectors. Let me verify that. If I have two vectors a and b, the magnitude of their sum is the square root of the sum of their magnitudes squared plus twice their dot product. Yes, that makes sense because when you add vectors, the resultant vector's magnitude depends on both their lengths and the angle between them.So, I already know |a| is 4, |b| is 2, and from the first part, a · b is 4. Let me plug these into the formula.First, compute |a|², which is 4 squared, so that's 16. Then, |b|² is 2 squared, which is 4. The dot product a · b is 4, so 2 times that is 8. Now, add all these together:16 (from |a|²) + 4 (from |b|²) + 8 (from 2(a · b)) = 28So, |a + b| is the square root of 28. Let me simplify that. The square root of 28 can be broken down into the square root of 4 times 7, which is sqrt(4) times sqrt(7). Since sqrt(4) is 2, that simplifies to 2√7.Wait, let me double-check my calculations to make sure I didn't make a mistake. For the dot product, 4 times 2 is 8, times 0.5 is 4. That seems right. For the magnitude, 4 squared is 16, 2 squared is 4, and 2 times 4 is 8. Adding them up: 16 + 4 is 20, plus 8 is 28. Square root of 28 is indeed 2√7. Okay, that seems correct.I think I've got both parts figured out. The dot product is 4, and the magnitude of a + b is 2√7. I don't see any errors in my reasoning, so I feel confident about these answers.**Final Answer**(1) The dot product is boxed{4}.(2) The magnitude of the sum is boxed{2sqrt{7}}.

answer：Okay, so I have this problem where a school wants to select 8 students from six senior classes for an English speech contest. The catch is that each class has to have at least one student participating. I need to figure out how many different ways the school can allocate these 8 spots. Hmm, let me think about how to approach this.First, I remember that when dealing with problems where you have to distribute items (in this case, spots) into different groups (the classes) with certain conditions, combinatorics is the way to go. Specifically, I think the stars and bars method might be useful here. But wait, stars and bars is usually for distributing identical items into distinct groups, right? In this case, the spots are identical because it doesn't matter which specific spot goes to which class, just the number of students from each class. So, yeah, stars and bars should work.But there's a condition: each class must have at least one student. That means I can't have any class with zero students. So, how does that affect the distribution? I think it means I need to adjust the problem so that each class already has one spot allocated, and then distribute the remaining spots without any restrictions. That makes sense because if each class already has one, I just need to figure out how to distribute the extra spots.So, let me break it down step by step. There are 8 spots in total and 6 classes. If each class must have at least one student, I can start by giving one spot to each class. That uses up 6 spots, leaving me with 8 - 6 = 2 spots left to distribute. Now, these remaining 2 spots can go to any of the 6 classes, and a class can receive more than one additional spot.This is now a problem of distributing 2 identical spots into 6 distinct classes, where a class can receive zero or more spots. This is a classic stars and bars problem. The formula for distributing n identical items into k distinct groups is C(n + k - 1, k - 1), where C is the combination function.Plugging in the numbers, n is 2 (the remaining spots) and k is 6 (the classes). So, the number of ways should be C(2 + 6 - 1, 6 - 1) = C(7, 5). Wait, let me make sure I'm using the formula correctly. The formula is C(n + k - 1, k - 1), so that would be C(2 + 6 - 1, 6 - 1) = C(7, 5). Alternatively, since combinations have symmetry, C(7,5) is the same as C(7,2), which might be easier to compute.Calculating C(7,2): that's 7! / (2! * (7 - 2)!) = (7 * 6) / (2 * 1) = 21. So, there are 21 ways to distribute the remaining 2 spots. Therefore, the total number of ways to allocate the 8 spots with at least one student from each class is 21.Wait, let me double-check. Another way to think about it is that after allocating one spot to each class, we have 2 spots left. These can either go to the same class or different classes. If both go to the same class, there are 6 choices (one for each class). If they go to different classes, we need to choose 2 classes out of 6, which is C(6,2) = 15. So, adding those together, 6 + 15 = 21. Yep, that matches what I got earlier.So, both methods give me the same answer, which is reassuring. I think 21 is the correct number of ways to allocate the spots under the given conditions.**Final Answer**The total number of ways to allocate these 8 spots is boxed{21}.

answer：To determine the value of ( f(1) + f'(1) ), we start by noting that the point ( P(1, f(1)) ) lies on the tangent line ( y = -2x + 10 ). 1. **Finding ( f(1) ):** Substitute ( x = 1 ) into the equation of the tangent line: [ f(1) = -2(1) + 10 = 8 ] 2. **Finding ( f'(1) ):** The slope of the tangent line at ( x = 1 ) is the derivative of the function at that point. The slope of the given tangent line is ( -2 ), so: [ f'(1) = -2 ] 3. **Calculating ( f(1) + f'(1) ):** [ f(1) + f'(1) = 8 + (-2) = 6 ]

answer：Okay, so I have this problem where I need to find the value of 'a' such that the chord AB formed by the intersection of the circle C and the line l has a length of 4. The circle C is given by the equation x² + y² - 2x - 4y + 1 = 0, and the line l is given by x + a y + 1 = 0. Hmm, let me try to figure this out step by step.First, I think it would help to write the equation of the circle in standard form. The general equation of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. So, I need to complete the squares for both x and y terms in the given circle equation.Starting with the given circle equation:x² + y² - 2x - 4y + 1 = 0Let me rearrange the terms:(x² - 2x) + (y² - 4y) + 1 = 0Now, completing the square for x:x² - 2x can be written as (x - 1)² - 1Similarly, for y:y² - 4y can be written as (y - 2)² - 4So substituting these back into the equation:(x - 1)² - 1 + (y - 2)² - 4 + 1 = 0Simplify the constants:-1 - 4 + 1 = -4So the equation becomes:(x - 1)² + (y - 2)² - 4 = 0Which simplifies to:(x - 1)² + (y - 2)² = 4Okay, so the circle has center at (1, 2) and radius 2. That makes sense.Now, the line l is given by x + a y + 1 = 0. I need to find the value of 'a' such that when this line intersects the circle, the chord AB has a length of 4.I remember that the length of a chord in a circle can be related to the distance from the center of the circle to the line. The formula for the length of the chord is 2√(r² - d²), where r is the radius and d is the distance from the center to the line.Given that the length of chord AB is 4, and the radius r is 2, let's plug these into the formula.Length of chord AB = 2√(r² - d²) = 4So, 2√(r² - d²) = 4Divide both sides by 2:√(r² - d²) = 2Square both sides:r² - d² = 4We know that r = 2, so r² = 4.Substitute r² into the equation:4 - d² = 4Subtract 4 from both sides:-d² = 0Multiply both sides by -1:d² = 0Take square root:d = 0So, the distance from the center of the circle to the line l is 0. That means the line passes through the center of the circle.Therefore, the line l must pass through the center (1, 2). Let's substitute x = 1 and y = 2 into the equation of the line to find 'a'.The equation of the line is x + a y + 1 = 0.Substitute x = 1 and y = 2:1 + a*(2) + 1 = 0Simplify:1 + 2a + 1 = 02 + 2a = 0Subtract 2 from both sides:2a = -2Divide both sides by 2:a = -1So, the value of 'a' is -1.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Converted the circle equation to standard form correctly: (x - 1)² + (y - 2)² = 4. That seems right.2. Used the chord length formula: 2√(r² - d²) = 4. Plugged in r = 2, so 2√(4 - d²) = 4. Divided by 2: √(4 - d²) = 2. Squared: 4 - d² = 4. So, d² = 0, hence d = 0. That seems correct.3. Since d = 0, the line passes through the center (1, 2). Plugged into x + a y + 1 = 0: 1 + 2a + 1 = 0. Simplified to 2 + 2a = 0, so a = -1. That seems right.I think I did everything correctly. So, the value of 'a' is indeed -1.**Final Answer**The value of ( a ) is boxed{-1}.

answer：Okay, so I have this problem where a line and a circle intersect at points A and B. The line is given by the equation x + y = 1, and the circle is x² + y² = a. The origin is O, and there's another point C on the circle such that the vector OA plus the vector OB equals the vector OC. I need to find the value of a.First, I should probably visualize this. The circle is centered at the origin with radius √a, and the line x + y = 1 is a straight line that intersects the circle at two points, A and B. Point C is another point on the circle, and the vector from O to C is the sum of the vectors from O to A and O to B. So, in other words, if I add the position vectors of A and B, I get the position vector of C.Let me denote the coordinates of A as (x₁, y₁) and B as (x₂, y₂). Then, the vector OA is (x₁, y₁) and OB is (x₂, y₂). So, OA + OB would be (x₁ + x₂, y₁ + y₂), which should equal the coordinates of point C, (x₀, y₀). Since C is on the circle, it must satisfy the equation x₀² + y₀² = a.So, my goal is to find a such that when I add the vectors OA and OB, the resulting point C lies on the circle. To do this, I probably need to find expressions for x₁ + x₂ and y₁ + y₂.Since points A and B lie on both the line x + y = 1 and the circle x² + y² = a, I can set up a system of equations to solve for their coordinates. Let me substitute y from the line equation into the circle equation.From x + y = 1, we can express y as y = 1 - x. Substituting this into the circle equation:x² + (1 - x)² = aExpanding (1 - x)² gives 1 - 2x + x². So, the equation becomes:x² + 1 - 2x + x² = aCombine like terms:2x² - 2x + 1 = aBring a to the left side:2x² - 2x + (1 - a) = 0This is a quadratic equation in terms of x. Let me write it as:2x² - 2x + (1 - a) = 0To find the solutions for x, I can use the quadratic formula. But before that, since the line intersects the circle at two points, the discriminant of this quadratic must be positive. The discriminant D is given by:D = b² - 4acIn this case, a = 2, b = -2, and c = (1 - a). So,D = (-2)² - 4 * 2 * (1 - a) = 4 - 8(1 - a) = 4 - 8 + 8a = -4 + 8aFor the quadratic to have two real solutions, D must be greater than zero:-4 + 8a > 0Solving for a:8a > 4a > 4/8a > 1/2So, a must be greater than 1/2. That's a useful piece of information.Now, going back to the quadratic equation 2x² - 2x + (1 - a) = 0. The sum of the roots x₁ + x₂ is given by -b/a, which in this case is:x₁ + x₂ = -(-2)/2 = 2/2 = 1So, x₁ + x₂ = 1.Similarly, since y = 1 - x, the sum y₁ + y₂ would be:y₁ + y₂ = (1 - x₁) + (1 - x₂) = 2 - (x₁ + x₂) = 2 - 1 = 1So, y₁ + y₂ = 1.Therefore, the vector OA + OB is (x₁ + x₂, y₁ + y₂) = (1, 1). So, point C is (1, 1).But wait, point C is on the circle x² + y² = a, so substituting (1, 1) into the circle equation:1² + 1² = a1 + 1 = aa = 2So, the value of a is 2.Let me double-check my steps to make sure I didn't make a mistake.1. I substituted y = 1 - x into the circle equation correctly, leading to 2x² - 2x + 1 - a = 0.2. Calculated the discriminant correctly, found that a must be greater than 1/2.3. Found the sum of the roots x₁ + x₂ = 1, and then y₁ + y₂ = 1.4. Concluded that OC is (1, 1), so substituting into the circle equation gives a = 2.Everything seems to check out. So, the answer should be 2.**Final Answer**The value of ( a ) is boxed{2}.

answer：To determine the probability that all four dice will show the number five, I start by understanding that each die has six possible outcomes, with only one outcome being the number five. This means the probability of rolling a five on a single die is 1/6.Since the dice are tossed simultaneously and their outcomes are independent, I can calculate the combined probability by multiplying the probabilities of each individual event. Therefore, for four dice, the probability is (1/6) multiplied by itself four times.Calculating this gives (1/6)^4, which equals 1/1296. Thus, the probability that all four dice will show the number five is 1/1296.