💡Okay, so I have this problem where I need to find the angle between two vectors, a and b. The vectors are given as a = (1, 2) and b = (3, 1). The options are 30°, 45°, 120°, and 135°. Hmm, I remember that to find the angle between two vectors, I can use the dot product formula. Let me try to recall how that works.First, I think I need to find the dot product of vectors a and b. The dot product is calculated by multiplying the corresponding components and then adding them together. So, for a = (1, 2) and b = (3, 1), the dot product should be (1*3) + (2*1). Let me compute that: 1 times 3 is 3, and 2 times 1 is 2. Adding those together, 3 + 2 equals 5. So, the dot product is 5.Next, I need the magnitudes of both vectors. The magnitude of a vector is found by taking the square root of the sum of the squares of its components. For vector a, that would be sqrt(1² + 2²). Calculating that, 1 squared is 1, and 2 squared is 4. Adding them gives 5, so the magnitude of a is sqrt(5). Similarly, for vector b, the magnitude is sqrt(3² + 1²). That's 9 + 1, which is 10, so the magnitude of b is sqrt(10).Now, the formula for the cosine of the angle between two vectors is the dot product divided by the product of their magnitudes. So, cos(theta) = (a · b) / (|a| |b|). Plugging in the numbers I have, that's 5 divided by (sqrt(5) times sqrt(10)). Let me compute the denominator first: sqrt(5) times sqrt(10) is sqrt(5*10) which is sqrt(50). Simplifying sqrt(50), that's 5*sqrt(2). So, the denominator is 5*sqrt(2).Putting it all together, cos(theta) = 5 / (5*sqrt(2)). The 5s cancel out, leaving cos(theta) = 1/sqrt(2). I remember that 1/sqrt(2) is approximately 0.7071, and that corresponds to a 45° angle because cos(45°) is 1/sqrt(2). So, theta must be 45°.Wait, let me double-check my steps to make sure I didn't make a mistake. I calculated the dot product correctly: 1*3 + 2*1 = 5. The magnitudes: sqrt(1+4) = sqrt(5) and sqrt(9+1) = sqrt(10). Then, the cosine of the angle is 5/(sqrt(5)*sqrt(10)) = 5/sqrt(50) = 5/(5*sqrt(2)) = 1/sqrt(2). Yes, that seems right. And 1/sqrt(2) is indeed the cosine of 45°, so the angle is 45°. That makes sense because both vectors are in the first quadrant, and the angle between them isn't too large or too small.I think I got it right. The angle between vectors a and b is 45 degrees, which is option B.

💡Okay, so I have this problem where there are two points, A and B, on the graph of a quadratic function. The quadratic function is given as y = ax² + bx + 2009, and a is not zero. The points A and B have coordinates (x₁, 2012) and (x₂, 2012) respectively. I need to find the value of the quadratic function when x is equal to x₁ + x₂.First, let me try to visualize this. A quadratic function is a parabola, and since a is not zero, it's either opening upwards or downwards. Points A and B are on this parabola, and both have the same y-coordinate, which is 2012. That means these two points are at the same height on the graph, so they must be symmetric with respect to the vertex of the parabola.I remember that for a quadratic function y = ax² + bx + c, the axis of symmetry is at x = -b/(2a). This is the x-coordinate of the vertex. Since points A and B are symmetric around this axis, their x-coordinates, x₁ and x₂, should be equidistant from -b/(2a). So, the average of x₁ and x₂ should be equal to -b/(2a). That is, (x₁ + x₂)/2 = -b/(2a). If I multiply both sides by 2, I get x₁ + x₂ = -b/a.Wait, that seems useful because the problem is asking for the value of the function at x = x₁ + x₂, which is x = -b/a. So, maybe I can substitute x = -b/a into the quadratic function and find the corresponding y-value.Let me write that down:y = a(-b/a)² + b(-b/a) + 2009Let me compute each term step by step.First, (-b/a) squared is (b²)/(a²). So, the first term becomes a*(b²/a²) which simplifies to (a*b²)/a². The a in the numerator cancels with one a in the denominator, leaving b²/a.Next, the second term is b*(-b/a). That's just -b²/a.So, putting it all together:y = (b²/a) - (b²/a) + 2009Hmm, the first two terms cancel each other out. So, y = 0 + 2009, which is 2009.Wait, so the value of the quadratic function at x = x₁ + x₂ is 2009. That seems straightforward, but let me double-check my steps to make sure I didn't make a mistake.Starting again, points A and B are on the parabola y = ax² + bx + 2009 and have the same y-coordinate, 2012. So, plugging in x₁ and x₂ into the equation, we get:2012 = a x₁² + b x₁ + 20092012 = a x₂² + b x₂ + 2009Subtracting 2009 from both sides in both equations:3 = a x₁² + b x₁3 = a x₂² + b x₂So, both x₁ and x₂ satisfy the equation a x² + b x - 3 = 0.Therefore, x₁ and x₂ are the roots of the quadratic equation a x² + b x - 3 = 0.From Vieta's formulas, I know that the sum of the roots x₁ + x₂ is equal to -b/a, which is consistent with what I found earlier.So, x₁ + x₂ = -b/a.Now, evaluating the original quadratic function at x = x₁ + x₂:y = a(-b/a)² + b(-b/a) + 2009Calculating each term:First term: a*(b²/a²) = b²/aSecond term: b*(-b/a) = -b²/aThird term: 2009Adding them up: (b²/a) - (b²/a) + 2009 = 0 + 2009 = 2009So, that confirms my earlier result.Alternatively, I can think about the symmetry of the parabola. The vertex is at x = -b/(2a), and since x₁ and x₂ are symmetric around this point, their average is -b/(2a). Therefore, x₁ + x₂ = -b/a.When I plug x = -b/a into the quadratic function, it's equivalent to plugging in the x-coordinate that is twice the distance from the vertex in the opposite direction. But regardless of that, the calculation shows that the y-value is 2009.I can also consider specific values to test this. Let me pick a simple quadratic function where I know the roots. For example, suppose a = 1, b = 2, and c = 2009. Then the quadratic is y = x² + 2x + 2009.Suppose I set y = 2012:2012 = x² + 2x + 2009Subtract 2009: 3 = x² + 2xSo, x² + 2x - 3 = 0Solving this, x = [-2 ± sqrt(4 + 12)]/2 = [-2 ± sqrt(16)]/2 = [-2 ± 4]/2So, x = (2)/2 = 1 or x = (-6)/2 = -3Thus, x₁ = 1, x₂ = -3, so x₁ + x₂ = -2.Now, plugging x = -2 into y = x² + 2x + 2009:y = (-2)² + 2*(-2) + 2009 = 4 - 4 + 2009 = 0 + 2009 = 2009So, in this specific case, it works out. Therefore, my answer seems correct.Another test case: Let me choose a different a and b. Let’s say a = 2, b = -4, c = 2009.Quadratic function: y = 2x² - 4x + 2009Set y = 2012:2012 = 2x² - 4x + 2009Subtract 2009: 3 = 2x² - 4xSo, 2x² - 4x - 3 = 0Using quadratic formula: x = [4 ± sqrt(16 + 24)]/4 = [4 ± sqrt(40)]/4 = [4 ± 2*sqrt(10)]/4 = [2 ± sqrt(10)]/2So, x₁ = [2 + sqrt(10)]/2, x₂ = [2 - sqrt(10)]/2Sum x₁ + x₂ = [2 + sqrt(10) + 2 - sqrt(10)]/2 = 4/2 = 2Now, plug x = 2 into y = 2x² - 4x + 2009:y = 2*(4) - 4*(2) + 2009 = 8 - 8 + 2009 = 0 + 2009 = 2009Again, the result is 2009. So, this further confirms that the value is indeed 2009.I think I've approached this from multiple angles: algebraically using Vieta's formulas, plugging into the quadratic function, and testing with specific examples. All methods lead to the same conclusion that y = 2009 when x = x₁ + x₂.Therefore, I'm confident that the value of the quadratic function at x = x₁ + x₂ is 2009.**Final Answer**The value of the quadratic function when ( x = x_1 + x_2 ) is boxed{2009}.

💡Alright, so I've got this problem here about a sequence {a_n} that satisfies the relation a₁ + a₂ + … + a_n = 2a₂ for n = 1, 2, 3, and so on. And I need to figure out which of the options A, B, C, or D is true. The options are:A: a₁ < 0 B: a₁ > 0 C: a₁ ≠ a₂ D: a₂ = 0Okay, let's start by understanding what the problem is saying. The sequence {a_n} has a property that the sum of its first n terms equals twice the second term, 2a₂, for every n starting from 1. That seems a bit strange because as n increases, the sum of the terms is always equal to 2a₂. So, no matter how many terms you add, the total is always the same, which is 2a₂. That suggests that after a certain point, the terms might be zero or something like that.Let me try plugging in specific values of n to see what happens.First, let's take n = 1. Then the equation becomes:a₁ = 2a₂So, that tells me that the first term is twice the second term. Okay, so a₁ is related to a₂ in that way.Next, let's take n = 2. Then the equation becomes:a₁ + a₂ = 2a₂Hmm, so if I substitute a₁ from the first equation into this, I get:2a₂ + a₂ = 2a₂Which simplifies to:3a₂ = 2a₂Subtracting 2a₂ from both sides:3a₂ - 2a₂ = 0 a₂ = 0Oh, so a₂ is zero. That's interesting. So, if a₂ is zero, then from the first equation, a₁ = 2a₂, which would mean a₁ is also zero. Wait, but that would make all the terms zero, right? Because if a₁ is zero and a₂ is zero, then for n = 3, the sum would be a₁ + a₂ + a₃ = 2a₂, which is zero. So, a₃ would also have to be zero. Similarly, all subsequent terms would have to be zero to satisfy the condition for any n.But hold on, if all the terms are zero, then a₁ = a₂ = 0, which would mean a₁ = a₂, contradicting option C which says a₁ ≠ a₂. But since a₂ is zero, and a₁ is also zero, they are equal. So, option C would be false.Now, looking back at the options:A: a₁ < 0 B: a₁ > 0 C: a₁ ≠ a₂ D: a₂ = 0From our deductions, a₂ = 0, so option D is true. As for options A and B, since a₁ = 2a₂ and a₂ = 0, a₁ is also zero. So, a₁ is neither less than zero nor greater than zero; it's equal to zero. Therefore, both A and B are false.Option C says a₁ ≠ a₂, but since both are zero, they are equal, so C is also false.So, the only correct option is D: a₂ = 0.Wait, but let me double-check to make sure I didn't make any mistakes. If a₂ is zero, then a₁ is also zero. For n = 3, the sum should be a₁ + a₂ + a₃ = 2a₂, which is zero. So, a₃ must also be zero. Similarly, for any n, all terms must be zero. So, the entire sequence is just zeros. That makes sense because the sum of any number of zeros is zero, which equals 2a₂ since a₂ is zero.Yeah, that seems consistent. So, I think D is definitely the correct answer.

💡Okay, so I have this problem here about reduced proper fractions with a denominator of 100, and I need to find their sum. The options are 20, 30, 35, and 45. Hmm, let me think about how to approach this.First, I remember that a reduced fraction is one where the numerator and denominator have no common divisors other than 1. So, for a fraction with denominator 100, the numerator has to be a number that doesn't share any common factors with 100 except 1. That means I need to find all numbers between 1 and 99 (since it's a proper fraction) that are coprime with 100.I think the first step is to figure out how many such numbers there are. I recall something called Euler's Totient Function, φ(n), which counts the number of integers up to n that are coprime with n. Maybe I can use that here. Let me try to calculate φ(100).To compute φ(100), I need the prime factors of 100. 100 can be factored into 2^2 * 5^2. The formula for Euler's Totient Function is φ(n) = n * (1 - 1/p1) * (1 - 1/p2) * ... where p1, p2, etc., are the distinct prime factors of n. So, applying that here:φ(100) = 100 * (1 - 1/2) * (1 - 1/5) = 100 * (1/2) * (4/5) = 100 * (2/5) = 40.So, there are 40 numbers between 1 and 99 that are coprime with 100. That means there are 40 reduced proper fractions with denominator 100.Now, I need to find the sum of all these fractions. Each fraction is of the form i/100 where i is coprime with 100. So, the sum S is:S = (1/100) + (3/100) + (7/100) + ... + (99/100)Wait, but listing all 40 terms would be tedious. There must be a smarter way to compute this sum without adding each term individually.I remember that in some cases, fractions can be paired in a way that their sum is a whole number. For example, if I take a fraction i/100 and pair it with (100 - i)/100, their sum is (i + (100 - i))/100 = 100/100 = 1. So, each such pair adds up to 1.But does this pairing work for all the fractions? Let me check. If i is coprime with 100, is (100 - i) also coprime with 100? Let's see. Suppose d is a common divisor of (100 - i) and 100. Then d divides 100 and (100 - i). But since d divides 100, it also divides i because d divides (100 - (100 - i)) = i. But since i and 100 are coprime, d must be 1. Therefore, (100 - i) is also coprime with 100. So, yes, each fraction can be paired with another fraction such that their sum is 1.Now, how many such pairs are there? Since there are 40 fractions, pairing them would give 20 pairs, each summing to 1. Therefore, the total sum S would be 20 * 1 = 20.Wait, but let me make sure I didn't make a mistake here. Is there a case where a fraction cannot be paired? For example, if i = 50, then (100 - 50) = 50, but 50 is not coprime with 100. But in our case, since we're only considering i that are coprime with 100, i cannot be 50 because 50 and 100 share a common factor of 50. So, all the i's we're considering are such that i and 100 are coprime, and thus (100 - i) is also coprime with 100. Therefore, every i has a unique pair (100 - i), and there are no unpaired fractions.So, the total number of pairs is indeed 20, and each pair sums to 1, so the total sum is 20.Let me cross-verify this with another approach. The sum of all fractions i/100 where i is coprime with 100 can be written as (1/100) * sum of i where i is coprime with 100. So, if I can find the sum of all such i, then divide by 100, I should get the same result.The sum of all numbers coprime with 100 up to 100 is known to be φ(100) * 100 / 2, but wait, is that correct? Let me think. For numbers coprime with n, their sum is indeed n * φ(n) / 2. So, in this case, it would be 100 * 40 / 2 = 2000. Therefore, the sum of all such i is 2000.Then, the sum of the fractions is 2000 / 100 = 20. So, that's the same result as before. That gives me more confidence that the answer is indeed 20.Wait, but let me make sure I didn't confuse the formula. I think the formula for the sum of numbers coprime to n is n * φ(n) / 2. Let me verify that.Suppose n is a prime number, say 5. Then φ(5) = 4, and the numbers coprime to 5 are 1, 2, 3, 4. Their sum is 1 + 2 + 3 + 4 = 10, which is 5 * 4 / 2 = 10. So, yes, the formula seems to hold.Similarly, for n = 6, φ(6) = 2, numbers coprime to 6 are 1 and 5, their sum is 6, which is 6 * 2 / 2 = 6. So, the formula works here too.Therefore, applying it to n = 100, the sum of numbers coprime to 100 is 100 * 40 / 2 = 2000. Thus, the sum of the fractions is 2000 / 100 = 20.Okay, so both methods give me the same answer, 20. That makes me more confident that 20 is the correct answer.I think I've covered all the steps and cross-verified the result using two different approaches. I don't see any mistakes in my reasoning, so I'm pretty sure the answer is 20.

💡Okay, so I have two systems of equations to solve. Let me start with the first one.**Problem 1:**The system is:1. (3x + 2y = 10)2. (frac{x}{2} - frac{y + 1}{3} = 1)Hmm, I need to solve for x and y. I remember that to solve a system of equations, I can use either substitution or elimination. Let me see which method would be easier here.Looking at the first equation, (3x + 2y = 10), it's already pretty simple. The second equation has fractions, which might be a bit tricky, but I can eliminate the denominators by multiplying through by a common multiple.Let me rewrite the second equation without fractions. The denominators are 2 and 3, so the least common multiple is 6. If I multiply every term by 6, that should clear the fractions.So, multiplying the second equation by 6:[6 times left(frac{x}{2}right) - 6 times left(frac{y + 1}{3}right) = 6 times 1]Simplifying each term:[3x - 2(y + 1) = 6]Expanding the terms inside the parentheses:[3x - 2y - 2 = 6]Now, I'll combine like terms:[3x - 2y = 6 + 2][3x - 2y = 8]So now, my system of equations is:1. (3x + 2y = 10)2. (3x - 2y = 8)Hmm, this looks promising because if I add these two equations together, the y terms will cancel out. Let me try that.Adding equation 1 and equation 2:[(3x + 2y) + (3x - 2y) = 10 + 8]Simplifying:[6x = 18]So, solving for x:[x = frac{18}{6} = 3]Okay, so x is 3. Now, I can plug this value back into one of the original equations to find y. Let's use the first equation because it looks simpler.Substituting x = 3 into (3x + 2y = 10):[3(3) + 2y = 10][9 + 2y = 10]Subtracting 9 from both sides:[2y = 1]Dividing by 2:[y = frac{1}{2}]So, y is 1/2. Let me just check my solution in both equations to make sure I didn't make a mistake.Checking in equation 1:[3(3) + 2left(frac{1}{2}right) = 9 + 1 = 10]Which matches the equation.Checking in equation 2:[frac{3}{2} - frac{frac{1}{2} + 1}{3} = frac{3}{2} - frac{frac{3}{2}}{3} = frac{3}{2} - frac{1}{2} = 1]Which also matches. Okay, so that seems correct.**Problem 2:**Now, moving on to the second system:1. (4x - 5y = 3)2. (frac{x - 2y}{0.4} = 0.6)Again, I need to solve for x and y. Let's see what's the best approach here.The first equation is straightforward, but the second equation has a decimal in the denominator. Maybe I can simplify that first.Starting with the second equation:[frac{x - 2y}{0.4} = 0.6]I can multiply both sides by 0.4 to eliminate the denominator:[x - 2y = 0.6 times 0.4]Calculating the right side:[x - 2y = 0.24]Hmm, now I have:1. (4x - 5y = 3)2. (x - 2y = 0.24)I can use substitution or elimination again. Let me try elimination because the coefficients are manageable.First, let me make the coefficients of x or y the same in both equations. Looking at the coefficients, if I multiply the second equation by 4, the coefficient of x will be 4, matching the first equation.Multiplying the second equation by 4:[4(x - 2y) = 4 times 0.24][4x - 8y = 0.96]Now, the system is:1. (4x - 5y = 3)2. (4x - 8y = 0.96)Now, I can subtract the second equation from the first to eliminate x.Subtracting equation 2 from equation 1:[(4x - 5y) - (4x - 8y) = 3 - 0.96]Simplifying:[4x - 5y - 4x + 8y = 2.04][3y = 2.04]Solving for y:[y = frac{2.04}{3} = 0.68]Okay, so y is 0.68. Now, I can substitute this back into one of the equations to find x. Let's use the second original equation because it's simpler.Substituting y = 0.68 into (x - 2y = 0.24):[x - 2(0.68) = 0.24][x - 1.36 = 0.24]Adding 1.36 to both sides:[x = 0.24 + 1.36][x = 1.6]So, x is 1.6. Let me verify this solution in both equations.Checking in equation 1:[4(1.6) - 5(0.68) = 6.4 - 3.4 = 3]Which matches.Checking in equation 2:[frac{1.6 - 2(0.68)}{0.4} = frac{1.6 - 1.36}{0.4} = frac{0.24}{0.4} = 0.6]Which also matches. Great, that seems correct.So, both systems are solved!

💡Okay, so I have this problem where I need to find the standard equation of a hyperbola C. It shares the same focal length as an ellipse given by the equation x²/9 + y²/4 = 1. Also, one of the hyperbola's asymptotes is x - 2y = 0. Hmm, let me break this down step by step.First, I remember that for an ellipse, the standard form is x²/a² + y²/b² = 1, where a > b. The distance from the center to each focus (c) is calculated using c = √(a² - b²). In this case, a² is 9 and b² is 4, so c = √(9 - 4) = √5. Therefore, the focal length, which is the distance between the two foci, is 2c, so that's 2√5.Now, the hyperbola shares the same focal length, which means its foci are also 2√5 apart. For hyperbolas, the standard form is either x²/A² - y²/B² = 1 or y²/B² - x²/A² = 1, depending on whether it opens horizontally or vertically. The distance to the foci for a hyperbola is also given by c = √(A² + B²). So, in this case, c must be √5 because 2c is 2√5.Next, the problem mentions that one of the asymptotes is x - 2y = 0. I know that for hyperbolas, the asymptotes are lines that the hyperbola approaches but never touches. For a hyperbola of the form x²/A² - y²/B² = 1, the asymptotes are y = ±(B/A)x. Similarly, for a hyperbola of the form y²/B² - x²/A² = 1, the asymptotes are y = ±(B/A)x as well, but in this case, the hyperbola opens vertically.Given the asymptote equation x - 2y = 0, I can rewrite this as y = (1/2)x. So, the slope of the asymptote is 1/2. Comparing this to the standard asymptote equations, if the hyperbola is of the form x²/A² - y²/B² = 1, then B/A should be 1/2. So, B = (1/2)A. Alternatively, if the hyperbola is of the form y²/B² - x²/A² = 1, then B/A is still the slope, so again, B = (1/2)A.Now, I need to find A and B such that c = √(A² + B²) = √5. Let's consider both cases.Case 1: Hyperbola opens horizontally, so equation is x²/A² - y²/B² = 1.We have B = (1/2)A, so let's substitute that into the equation for c:c = √(A² + B²) = √(A² + ( (1/2)A )² ) = √(A² + (1/4)A²) = √( (5/4)A² ) = (√5)/2 * A.We know c = √5, so:(√5)/2 * A = √5Multiply both sides by 2:√5 * A = 2√5Divide both sides by √5:A = 2Then, since B = (1/2)A, B = (1/2)*2 = 1.So, the equation becomes x²/4 - y²/1 = 1, which simplifies to x²/4 - y² = 1.Case 2: Hyperbola opens vertically, so equation is y²/B² - x²/A² = 1.Again, the asymptote slope is 1/2, so B/A = 1/2, meaning B = (1/2)A.Using the same c formula:c = √(A² + B²) = √(A² + ( (1/2)A )² ) = √(A² + (1/4)A²) = √( (5/4)A² ) = (√5)/2 * A.Set this equal to √5:(√5)/2 * A = √5Multiply both sides by 2:√5 * A = 2√5Divide both sides by √5:A = 2Then, B = (1/2)A = 1.So, the equation becomes y²/1 - x²/4 = 1, which simplifies to y² - x²/4 = 1.Therefore, there are two possible equations for the hyperbola: x²/4 - y² = 1 or y² - x²/4 = 1. Looking back at the answer choices, option B is exactly these two equations. So, the correct answer should be B.**Final Answer**boxed{B}