💡Okay, so I have this problem about a parabola in the coordinate system. The equation is given as ( y = ax^2 + bx ) where ( a neq 0 ). There are some inequalities provided: ( 3a + b > 0 ) and ( a + b < 0 ). Points ( (-3, m) ), ( (2, n) ), and ( (4, t) ) lie on this parabola, and I need to figure out the relationship between ( m ), ( n ), and ( t ). The options are A: ( t < n < m ), B: ( m < t < n ), C: ( n < t < m ), and D: ( n < m < t ).Alright, let's start by understanding what the inequalities tell us about the parabola. The first inequality is ( 3a + b > 0 ). Maybe I can rewrite this to make it more meaningful. If I subtract ( a ) from both sides, I get ( 2a + (a + b) > 0 ). But wait, the second inequality is ( a + b < 0 ). So substituting that into the first inequality, I have ( 2a + (text{something negative}) > 0 ). That suggests that ( 2a ) must be positive enough to offset the negative term. So ( 2a > 0 ), which means ( a > 0 ). So the parabola opens upwards because the coefficient of ( x^2 ) is positive.Now, knowing that ( a > 0 ), let's think about the vertex of the parabola. The vertex occurs at ( x = -frac{b}{2a} ). To find where this vertex is, maybe I can use the inequalities given. Let's see, from ( a + b < 0 ), I can write ( b < -a ). And from ( 3a + b > 0 ), I can write ( b > -3a ). So combining these, ( -3a < b < -a ).Substituting ( b ) into the vertex formula, ( x = -frac{b}{2a} ). Since ( b ) is between ( -3a ) and ( -a ), let's plug those in. If ( b = -3a ), then ( x = -frac{-3a}{2a} = frac{3}{2} ). If ( b = -a ), then ( x = -frac{-a}{2a} = frac{1}{2} ). So the vertex is somewhere between ( x = frac{1}{2} ) and ( x = frac{3}{2} ).So the vertex is in the middle of the first and second quadrants, somewhere around there. Since the parabola opens upwards, it has a minimum point at the vertex. Points further away from the vertex on either side will have higher y-values.Now, looking at the points given: ( (-3, m) ), ( (2, n) ), and ( (4, t) ). Let's plot these in our minds. The point ( (-3, m) ) is far to the left, ( (2, n) ) is to the right of the vertex, and ( (4, t) ) is further to the right.Since the parabola opens upwards, the further a point is from the vertex, the higher its y-value. So, let's see:- ( (-3, m) ) is quite far from the vertex, which is around ( x = 1 ). So ( m ) should be quite large.- ( (2, n) ) is closer to the vertex, so ( n ) should be smaller than ( m ).- ( (4, t) ) is even further from the vertex than ( (2, n) ), so ( t ) should be larger than ( n ).Wait, but hold on. If the vertex is at ( x ) between 0.5 and 1.5, then ( x = 2 ) is just a little to the right of the vertex, and ( x = 4 ) is much further to the right. So, the y-values at ( x = 4 ) should be higher than at ( x = 2 ). Similarly, ( x = -3 ) is quite far to the left, so its y-value should be higher than both ( x = 2 ) and ( x = 4 ).So putting it all together: ( m ) is the highest, then ( t ), then ( n ). So the order is ( n < t < m ). That would correspond to option C.Wait, let me double-check. Maybe I should plug in the points into the equation and see.Given ( y = ax^2 + bx ), so:For ( (-3, m) ): ( m = a(-3)^2 + b(-3) = 9a - 3b )For ( (2, n) ): ( n = a(2)^2 + b(2) = 4a + 2b )For ( (4, t) ): ( t = a(4)^2 + b(4) = 16a + 4b )So we have:( m = 9a - 3b )( n = 4a + 2b )( t = 16a + 4b )Now, let's see if we can express these in terms of the inequalities given.We know ( 3a + b > 0 ) and ( a + b < 0 ). Let's denote ( 3a + b = k ) where ( k > 0 ), and ( a + b = l ) where ( l < 0 ).So, from ( 3a + b = k ), we can write ( b = k - 3a ).From ( a + b = l ), substituting ( b ), we get ( a + (k - 3a) = l ), which simplifies to ( -2a + k = l ). So, ( -2a = l - k ), hence ( a = frac{k - l}{2} ).Since ( k > 0 ) and ( l < 0 ), ( k - l ) is positive, so ( a ) is positive, which we already knew.Now, let's express ( m ), ( n ), and ( t ) in terms of ( a ) and ( b ):We have ( m = 9a - 3b ). Substitute ( b = k - 3a ):( m = 9a - 3(k - 3a) = 9a - 3k + 9a = 18a - 3k )Similarly, ( n = 4a + 2b = 4a + 2(k - 3a) = 4a + 2k - 6a = -2a + 2k )And ( t = 16a + 4b = 16a + 4(k - 3a) = 16a + 4k - 12a = 4a + 4k )So now, ( m = 18a - 3k ), ( n = -2a + 2k ), ( t = 4a + 4k )We need to compare ( m ), ( n ), and ( t ). Let's see if we can find relationships between them.First, let's find ( m - n ):( m - n = (18a - 3k) - (-2a + 2k) = 18a - 3k + 2a - 2k = 20a - 5k )Similarly, ( t - n = (4a + 4k) - (-2a + 2k) = 4a + 4k + 2a - 2k = 6a + 2k )And ( m - t = (18a - 3k) - (4a + 4k) = 14a - 7k )Now, let's analyze these differences.First, ( m - n = 20a - 5k ). Since ( a > 0 ) and ( k > 0 ), but we don't know the relation between ( a ) and ( k ). Similarly, ( t - n = 6a + 2k ), which is definitely positive because both ( a ) and ( k ) are positive. So ( t > n ).Next, ( m - t = 14a - 7k ). Let's factor out 7: ( 7(2a - k) ). So, the sign depends on ( 2a - k ). If ( 2a > k ), then ( m > t ); if ( 2a < k ), then ( m < t ).But from earlier, we have ( a = frac{k - l}{2} ). Since ( l < 0 ), ( k - l > k ), so ( a = frac{k - l}{2} > frac{k}{2} ). Therefore, ( 2a > k ). So ( m - t = 7(2a - k) > 0 ), meaning ( m > t ).So, putting it all together:- ( t > n ) because ( t - n > 0 )- ( m > t ) because ( m - t > 0 )Therefore, ( n < t < m ), which is option C.Wait, let me make sure I didn't make a mistake in the algebra.Starting from ( m = 18a - 3k ), ( n = -2a + 2k ), ( t = 4a + 4k ).Calculating ( m - n = 20a - 5k ). Since ( a > 0 ) and ( k > 0 ), but we don't know which is larger. However, earlier we found that ( a > frac{k}{2} ), so ( 2a > k ). Therefore, ( 20a - 5k = 5(4a - k) ). Since ( 4a > 2a > k ), ( 4a - k > 0 ), so ( m - n > 0 ), meaning ( m > n ).Similarly, ( t - n = 6a + 2k > 0 ), so ( t > n ).And ( m - t = 14a - 7k = 7(2a - k) ). Since ( 2a > k ), ( m - t > 0 ), so ( m > t ).Therefore, the order is ( n < t < m ), which is option C.I think that's solid. I don't see any mistakes in the reasoning.

💡First, calculate the total money Cathy received from her parents:- Money from dad: 25- Money from mom: 2 × 25 = 50- Total from parents: 25 + 50 = 75Add the money she initially had:- Total money: 75 + 12 = 87Allocate her budget based on the given percentages:- Food budget: 40% of 87 = 0.40 × 87 = 34.80- Transportation budget: 30% of 87 = 0.30 × 87 = 26.10- Entertainment budget: 10% of 87 = 0.10 × 87 = 8.70- Savings: 20% of 87 = 0.20 × 87 = 17.40Subtract her expenditures:1. Bought a book for 15: - Remaining money: 87 - 15 = 722. Took a cab ride costing 3% of the remaining money: - Cab ride cost: 3% of 72 = 0.03 × 72 = 2.16 - Remaining money: 72 - 2.16 = 69.843. Went out for dinner costing 50% of her food budget: - Dinner cost: 50% of 34.80 = 0.50 × 34.80 = 17.40 - Remaining money: 69.84 - 17.40 = 52.44Therefore, Cathy has 52.44 left in her wallet after all expenditures.

💡Okay, so I have this problem here: If α is an angle in the third quadrant, then the terminal side of α/2 is in the ______ quadrant. Hmm, I need to figure out which quadrant α/2 will be in. Let me think about this step by step.First, I remember that the quadrants are divided based on the angles in standard position. The first quadrant is from 0° to 90°, the second from 90° to 180°, the third from 180° to 270°, and the fourth from 270° to 360°. So, if α is in the third quadrant, that means α is between 180° and 270°, right?But wait, angles can also be more than 360°, like in multiple rotations. So, actually, α could be any angle that's coterminal with an angle between 180° and 270°. So, in general, α can be written as 180° + 360°k < α < 270° + 360°k, where k is any integer. That makes sense because angles repeat every 360°, so adding 360°k just accounts for full rotations.Now, the problem is asking about α/2. So, if I divide the entire inequality by 2, what happens? Let me write that down:180° + 360°k < α < 270° + 360°kDivide all parts by 2:(180° + 360°k)/2 < α/2 < (270° + 360°k)/2Simplify each part:90° + 180°k < α/2 < 135° + 180°kHmm, okay. So, α/2 is between 90° + 180°k and 135° + 180°k. Now, I need to figure out which quadrant this range falls into.Let me consider different values of k to see where α/2 could be.First, let's take k = 0:90° < α/2 < 135°That's between 90° and 135°, which is the second quadrant. So, if k is 0, α/2 is in the second quadrant.What if k = 1:90° + 180°(1) = 270°135° + 180°(1) = 315°So, 270° < α/2 < 315°That's between 270° and 315°, which is the fourth quadrant. So, if k is 1, α/2 is in the fourth quadrant.What about k = -1:90° + 180°(-1) = -90°135° + 180°(-1) = -45°Hmm, negative angles. I need to think about where those would be. Negative angles go clockwise from the positive x-axis. So, -90° is the same as 270°, and -45° is the same as 315°. So, -90° < α/2 < -45° is equivalent to 270° < α/2 < 315°, which again is the fourth quadrant.Wait, so whether k is positive or negative, α/2 can be in either the second or the fourth quadrant. Let me check another value, say k = 2:90° + 180°(2) = 450°135° + 180°(2) = 495°But 450° is more than 360°, so subtracting 360°, that's 90°, and 495° - 360° = 135°. So, 450° < α/2 < 495° is the same as 90° < α/2 < 135°, which is again the second quadrant.Similarly, k = -2:90° + 180°(-2) = -270°135° + 180°(-2) = -225°Again, negative angles. -270° is the same as 90°, and -225° is the same as 135°. So, -270° < α/2 < -225° is equivalent to 90° < α/2 < 135°, which is the second quadrant.Wait, so it seems like depending on whether k is even or odd, α/2 can be in the second or fourth quadrant. Let me think about that.If k is even, say k = 2n, then:90° + 180°(2n) = 90° + 360°n135° + 180°(2n) = 135° + 360°nSo, α/2 is between 90° + 360°n and 135° + 360°n, which is the second quadrant.If k is odd, say k = 2n + 1, then:90° + 180°(2n + 1) = 90° + 360°n + 180° = 270° + 360°n135° + 180°(2n + 1) = 135° + 360°n + 180° = 315° + 360°nSo, α/2 is between 270° + 360°n and 315° + 360°n, which is the fourth quadrant.Therefore, depending on whether k is even or odd, α/2 is in the second or fourth quadrant. So, the terminal side of α/2 is in either the second or fourth quadrant.Let me just visualize this to make sure. If α is in the third quadrant, say 225°, then α/2 is 112.5°, which is in the second quadrant. If α is 270°, then α/2 is 135°, still in the second quadrant. Wait, but 270° is actually on the negative y-axis, not in the third quadrant. The third quadrant goes up to 270°, but doesn't include 270°, right? So, α is strictly less than 270°, so α/2 is strictly less than 135°, but still in the second quadrant.Wait, but earlier when k was 1, we got α/2 in the fourth quadrant. How does that work? Let me take an example where k = 1. So, α is in the third quadrant, which would be 180° + 360°(1) = 540°, but 540° is more than 360°, so subtracting 360°, that's 180°, which is still the third quadrant. Wait, no, 540° - 360° = 180°, which is on the negative x-axis, not in the third quadrant. Hmm, maybe I need to adjust my thinking.Wait, perhaps I'm confusing the values. Let me take a specific example where k = 1. So, α is between 180° + 360°(1) = 540° and 270° + 360°(1) = 630°. But 540° is equivalent to 180°, and 630° is equivalent to 270°, so α is between 180° and 270°, which is the third quadrant. Then, α/2 would be between 270° and 315°, which is the fourth quadrant. So, in this case, α is in the third quadrant, but α/2 is in the fourth quadrant.Wait, so depending on how many full rotations α has, α/2 can be in different quadrants. So, if α is between 180° and 270°, then α/2 is between 90° and 135°, which is the second quadrant. But if α is between 540° and 630°, which is the same as 180° to 270°, then α/2 is between 270° and 315°, which is the fourth quadrant.So, it's not just about the principal value of α, but also about how many full rotations α has. Therefore, depending on whether α is in the third quadrant with k even or odd, α/2 can be in the second or fourth quadrant.Therefore, the answer is that the terminal side of α/2 is in either the second or fourth quadrant.I think that makes sense. Let me just recap:1. α is in the third quadrant: 180° + 360°k < α < 270° + 360°k.2. Divide by 2: 90° + 180°k < α/2 < 135° + 180°k.3. Depending on whether k is even or odd, α/2 falls in the second or fourth quadrant.Yes, that seems consistent. So, the terminal side of α/2 is in the second or fourth quadrant.

💡Alright, so I have this problem here: The line y = kx + 1 is tangent to the curve y = x³ + ax + b at point A(1,3). I need to find the value of 2a + b. The options are A: 2, B: -1, C: 1, D: -2.Hmm, okay. Let me try to break this down step by step. First, I know that if a line is tangent to a curve at a particular point, two conditions must be satisfied. One, the point must lie on both the line and the curve. Two, the slope of the line must be equal to the derivative of the curve at that point. So, I can use these two conditions to set up some equations and solve for a and b.Let me start by writing down what I know. The point A is (1,3). So, plugging x = 1 into both the line and the curve should give me y = 3.For the line: y = kx + 1. Plugging in x = 1, we get y = k*1 + 1 = k + 1. But we know y should be 3 at this point, so:k + 1 = 3That simplifies to k = 3 - 1 = 2. So, k is 2. Okay, that gives me the slope of the tangent line.Now, moving on to the curve: y = x³ + ax + b. Plugging in x = 1 and y = 3, we get:3 = (1)³ + a*(1) + bSimplifying that, 3 = 1 + a + bSo, 1 + a + b = 3Subtracting 1 from both sides, we get:a + b = 2Alright, so that's equation one: a + b = 2.Now, the second condition is about the slopes. The slope of the tangent line is k, which we found to be 2. The slope of the curve at any point x is given by its derivative. Let me find the derivative of the curve.The curve is y = x³ + ax + b. Taking the derivative with respect to x:dy/dx = 3x² + aSo, the slope of the curve at x = 1 is:dy/dx at x=1 = 3*(1)² + a = 3 + aBut we know this slope must be equal to the slope of the tangent line, which is k = 2. So:3 + a = 2Solving for a:a = 2 - 3 = -1Okay, so a is -1. Now, going back to the first equation where a + b = 2, and plugging in a = -1:-1 + b = 2Adding 1 to both sides:b = 2 + 1 = 3So, b is 3. Now, the question asks for the value of 2a + b. Let's compute that:2a + b = 2*(-1) + 3 = -2 + 3 = 1So, 2a + b equals 1. Looking back at the options, that's option C.Wait, let me double-check my steps to make sure I didn't make any mistakes. First, I found k by plugging in the point into the line equation, which gave me k = 2. Then, plugging the point into the curve gave me a + b = 2. Taking the derivative of the curve gave me the slope at x=1 as 3 + a, which equals k=2, so a = -1. Then, substituting a into a + b = 2 gave me b=3. Finally, 2a + b is 2*(-1) + 3 = 1. Yeah, that seems correct.I think I might have confused myself if I tried to do it differently, but this step-by-step approach seems solid. I don't see any errors in my calculations or reasoning. So, I'm confident that the answer is 1, which is option C.**Final Answer**boxed{C}

💡First, I need to understand what it means for two monomials to be of the same type. Monomials are considered of the same type if they have identical variables raised to the same exponents, regardless of their coefficients.Looking at the given monomial (-3ab^{2}), I identify the variables and their exponents:- The exponent of (a) is 1.- The exponent of (b) is 2.Next, I examine each option to see if it matches these exponents:- **Option A: (-3ab^{3})** - The exponent of (a) is 1. - The exponent of (b) is 3. - The exponent of (b) does not match, so this is not the same type.- **Option B: (frac{1}{2}ba^{2})** - The exponent of (a) is 2. - The exponent of (b) is 1. - Neither exponent matches, so this is not the same type.- **Option C: (2ab^{2})** - The exponent of (a) is 1. - The exponent of (b) is 2. - Both exponents match, so this is the same type.- **Option D: (3a^{2}b^{2})** - The exponent of (a) is 2. - The exponent of (b) is 2. - The exponent of (a) does not match, so this is not the same type.After evaluating all options, only **Option C** has the same exponents for both variables as the original monomial. Therefore, it is the correct answer.

💡Okay, so I have this problem here: Point P is moving on the parabola C₁: y² = 4x, and from P, two tangents are drawn to the circle (x - 3)² + y² = 2. I need to find the maximum angle between these two tangents. Hmm, okay, let me try to visualize this.First, the parabola y² = 4x is a standard right-opening parabola with vertex at the origin. Its focus is at (1, 0), and it has a directrix at x = -1. So, point P is somewhere on this parabola, and from P, we can draw two tangents to the given circle.The circle is centered at (3, 0) with a radius of √2. So, it's a relatively small circle located to the right of the parabola's vertex. The tangents from P to the circle will touch the circle at exactly one point each, and the angle between these two tangents is what we're interested in.I remember that for a point outside a circle, the angle between the two tangents can be found using the formula involving the distance from the point to the center of the circle and the radius of the circle. Specifically, if d is the distance from the external point to the center, and r is the radius, then the angle θ between the tangents satisfies sin(θ/2) = r/d. So, θ = 2 arcsin(r/d).In this case, the radius r is √2, and the distance d is the distance from point P to the center of the circle, which is at (3, 0). So, if I can express d in terms of the coordinates of P on the parabola, I can then find θ as a function of P's position and then find its maximum.Since P is on the parabola y² = 4x, I can parametrize P as (t², 2t) for some parameter t. That might make the calculations easier. So, let me set P = (t², 2t).Now, the distance d from P to the center (3, 0) is:d = sqrt[(t² - 3)² + (2t - 0)²] = sqrt[(t² - 3)² + 4t²]Let me expand that:(t² - 3)² = t⁴ - 6t² + 9So, d² = t⁴ - 6t² + 9 + 4t² = t⁴ - 2t² + 9Therefore, d = sqrt(t⁴ - 2t² + 9)Now, the angle θ between the tangents is given by θ = 2 arcsin(r/d) = 2 arcsin(√2 / sqrt(t⁴ - 2t² + 9))To maximize θ, I need to minimize d because as d decreases, arcsin(√2 / d) increases, so θ increases. Therefore, the maximum angle occurs when d is minimized.So, I need to find the minimum value of d, which is sqrt(t⁴ - 2t² + 9). To find the minimum, I can consider the function f(t) = t⁴ - 2t² + 9 and find its minimum.Taking the derivative of f(t):f'(t) = 4t³ - 4tSet f'(t) = 0:4t³ - 4t = 0 => 4t(t² - 1) = 0 => t = 0 or t² = 1 => t = 0, t = 1, t = -1Now, let's evaluate f(t) at these critical points:At t = 0: f(0) = 0 - 0 + 9 = 9At t = 1: f(1) = 1 - 2 + 9 = 8At t = -1: f(-1) = 1 - 2 + 9 = 8So, the minimum value of f(t) is 8, which occurs at t = 1 and t = -1. Therefore, the minimum distance d is sqrt(8) = 2√2.Now, plugging this back into the angle formula:θ = 2 arcsin(√2 / (2√2)) = 2 arcsin(1/2)Since arcsin(1/2) = 30°, so θ = 2 * 30° = 60°Therefore, the maximum angle between the two tangents is 60 degrees.Wait, let me double-check. If d is minimized, then the angle is maximized. So, when d is 2√2, which is the minimum, then the angle is 60°, which is the maximum possible angle. That makes sense because as the point P moves closer to the circle, the angle between the tangents increases.Is there a way to confirm this? Maybe by considering specific points.For example, when t = 1, P is (1, 2). The distance from (1, 2) to (3, 0) is sqrt[(1 - 3)² + (2 - 0)²] = sqrt[4 + 4] = sqrt[8] = 2√2, which matches our earlier calculation. So, the angle is indeed 60°.What if I take another point, say t = 2. Then P is (4, 4). The distance from (4, 4) to (3, 0) is sqrt[(4 - 3)² + (4 - 0)²] = sqrt[1 + 16] = sqrt[17] ≈ 4.123. Then, θ = 2 arcsin(√2 / 4.123) ≈ 2 arcsin(0.333) ≈ 2 * 19.47° ≈ 38.94°, which is less than 60°, confirming that 60° is indeed the maximum.Similarly, if I take t approaching infinity, the distance d would approach infinity, making θ approach 0°, which also makes sense.Therefore, I'm confident that the maximum angle between the two tangents is 60 degrees.**Final Answer**The maximum value of the angle between the two tangents is boxed{60^circ}.