💡Okay, so I have this quadratic equation: 5x² - 2x + c = 0. I need to find the value of 'c' such that the roots of this equation have a ratio of -3/5. Hmm, let's see. I remember that for quadratic equations, there's something called Vieta's formulas which relate the coefficients of the equation to the sum and product of its roots. Maybe I can use that here.First, let me recall Vieta's formulas. For a general quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a. So in my equation, a is 5, b is -2, and c is c (which I need to find). Let me denote the roots as x₁ and x₂. According to Vieta's formulas:1. Sum of roots: x₁ + x₂ = -b/a = -(-2)/5 = 2/5.2. Product of roots: x₁ * x₂ = c/a = c/5.Now, the problem states that the ratio of the roots is -3/5. So, x₁/x₂ = -3/5. That means x₁ = (-3/5)x₂. Okay, so if I substitute x₁ in terms of x₂ into the sum of roots equation, I can find the value of x₂. Let's do that.From x₁ = (-3/5)x₂, substitute into x₁ + x₂ = 2/5:(-3/5)x₂ + x₂ = 2/5.Let me combine these terms. To add them, I need a common denominator. The first term is (-3/5)x₂ and the second term is (5/5)x₂. So adding them together:(-3/5 + 5/5)x₂ = 2/5(2/5)x₂ = 2/5.Hmm, so (2/5)x₂ = 2/5. To solve for x₂, I can multiply both sides by 5/2:x₂ = (2/5) * (5/2) = 1.So, x₂ is 1. Now, since x₁ = (-3/5)x₂, substituting x₂ = 1 gives:x₁ = (-3/5)*1 = -3/5.Now that I have both roots, x₁ = -3/5 and x₂ = 1, I can use the product of roots formula to find 'c'. The product of the roots is x₁ * x₂ = (-3/5)*1 = -3/5.But from Vieta's formula, the product of the roots is also c/5. So:c/5 = -3/5.To solve for 'c', I can multiply both sides by 5:c = (-3/5)*5 = -3.So, the value of 'c' should be -3. Let me double-check my steps to make sure I didn't make any mistakes.1. I used Vieta's formulas correctly to express the sum and product of the roots.2. I set up the ratio correctly and expressed x₁ in terms of x₂.3. Substituted into the sum equation and solved for x₂, which gave me 1.4. Then found x₁ as -3/5.5. Calculated the product of the roots and set it equal to c/5, solving for c.Everything seems to check out. I think c is indeed -3.

💡Okay, so I have this problem where I need to find the values of x such that the determinant of a given 2x2 matrix is equal to 10. The matrix is:[begin{pmatrix}x^2 & 2x + 1 3x & 4x + 2end{pmatrix}]First, I remember that the determinant of a 2x2 matrix (begin{pmatrix} a & b c & d end{pmatrix}) is calculated as (ad - bc). So, applying that formula to the given matrix, the determinant should be:[text{det} = (x^2)(4x + 2) - (2x + 1)(3x)]Let me compute each part step by step to avoid mistakes.First, multiply (x^2) by (4x + 2):[x^2 times 4x = 4x^3][x^2 times 2 = 2x^2]So, the first part is (4x^3 + 2x^2).Now, the second part is ((2x + 1)(3x)). Let me expand that:[2x times 3x = 6x^2][1 times 3x = 3x]So, the second part is (6x^2 + 3x).Now, subtract the second part from the first part to get the determinant:[text{det} = (4x^3 + 2x^2) - (6x^2 + 3x)]Let me simplify this expression:First, distribute the negative sign:[4x^3 + 2x^2 - 6x^2 - 3x]Combine like terms:- (4x^3) remains as is.- (2x^2 - 6x^2 = -4x^2)- (-3x) remains as is.So, the determinant simplifies to:[4x^3 - 4x^2 - 3x]According to the problem, this determinant should be equal to 10. So, I set up the equation:[4x^3 - 4x^2 - 3x = 10]To solve for x, I need to bring all terms to one side to form a cubic equation:[4x^3 - 4x^2 - 3x - 10 = 0]Now, I have a cubic equation. Solving cubic equations can be tricky, but maybe I can factor it or find rational roots. I recall the Rational Root Theorem, which states that any possible rational root, p/q, is such that p is a factor of the constant term and q is a factor of the leading coefficient.In this equation, the constant term is -10, and the leading coefficient is 4. So, possible rational roots could be ±1, ±2, ±5, ±10, and these divided by 1, 2, or 4.Let me test these possible roots by plugging them into the equation.First, let's try x = 1:[4(1)^3 - 4(1)^2 - 3(1) - 10 = 4 - 4 - 3 - 10 = -13 neq 0]Not a root.Next, x = -1:[4(-1)^3 - 4(-1)^2 - 3(-1) - 10 = -4 - 4 + 3 - 10 = -15 neq 0]Not a root.Next, x = 2:[4(2)^3 - 4(2)^2 - 3(2) - 10 = 32 - 16 - 6 - 10 = 0]Oh, x = 2 is a root!So, (x - 2) is a factor of the cubic polynomial. Now, I can perform polynomial division or use synthetic division to factor out (x - 2) from the cubic equation.Let's use synthetic division with x = 2.The coefficients of the cubic equation are 4, -4, -3, -10.Set up synthetic division:2 | 4   -4   -3   -10  |     8    8     10  ---------------------    4    4     5     0So, after division, the cubic polynomial factors as:[(x - 2)(4x^2 + 4x + 5) = 0]Now, set each factor equal to zero:1. (x - 2 = 0) gives (x = 2).2. (4x^2 + 4x + 5 = 0)Let's solve the quadratic equation (4x^2 + 4x + 5 = 0). I'll use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, a = 4, b = 4, c = 5.Compute the discriminant:[b^2 - 4ac = 16 - 80 = -64]Since the discriminant is negative, there are no real roots from this quadratic equation. Therefore, the only real solution is x = 2.So, the value of x that satisfies the determinant being equal to 10 is x = 2.I should double-check my work to make sure I didn't make any mistakes.First, computing the determinant:[(x^2)(4x + 2) - (2x + 1)(3x) = 4x^3 + 2x^2 - 6x^2 - 3x = 4x^3 - 4x^2 - 3x]That seems correct.Setting it equal to 10:[4x^3 - 4x^2 - 3x - 10 = 0]Testing x = 2:[4(8) - 4(4) - 3(2) - 10 = 32 - 16 - 6 - 10 = 0]Yes, that works.Factoring out (x - 2):Using synthetic division, I got the quadratic (4x^2 + 4x + 5), which has a negative discriminant, so no real roots. Therefore, x = 2 is the only real solution.I think my process is correct. So, the answer should be x = 2.

💡First, I need to understand what the question is asking. It is about determining whether the condition ( x = -1 ) is sufficient, necessary, both, or neither for the equation ( x^{2} + x = 0 ).To analyze this, I'll start by solving the equation ( x^{2} + x = 0 ). Factoring the equation gives ( x(x + 1) = 0 ), which means the solutions are ( x = 0 ) and ( x = -1 ).Next, I'll consider what it means for a condition to be sufficient or necessary. A sufficient condition means that if the condition is met, the equation must hold true. A necessary condition means that for the equation to hold true, the condition must be met.In this case, ( x = -1 ) is one of the solutions to the equation. This means that if ( x = -1 ), the equation ( x^{2} + x = 0 ) will definitely be satisfied. Therefore, ( x = -1 ) is a sufficient condition.However, since there is another solution (( x = 0 )), ( x = -1 ) is not the only way for the equation to hold true. This means that ( x = -1 ) is not a necessary condition because the equation can still be satisfied without ( x ) being equal to -1.Based on this analysis, ( x = -1 ) is a sufficient but not necessary condition for ( x^{2} + x = 0 ).

💡Okay, so I have this problem where I need to find f(4) given that for all real numbers x, the function f(x) satisfies the equation x times f(x) equals 2 times f(2 minus x) plus 1. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me write down the given equation to make it clearer:x * f(x) = 2 * f(2 - x) + 1Alright, so this equation relates the value of the function at x to the value at 2 - x. Since I need to find f(4), maybe I can plug in x = 4 into this equation. Let me try that.If I substitute x = 4, the equation becomes:4 * f(4) = 2 * f(2 - 4) + 1Simplifying the argument of f on the right side:4 * f(4) = 2 * f(-2) + 1Okay, so now I have an equation that relates f(4) and f(-2). But I don't know either of those values yet. Maybe I can find another equation involving f(-2) and f(4) by plugging in a different value of x. Let me think, if I set x = -2, what happens?Substituting x = -2 into the original equation:(-2) * f(-2) = 2 * f(2 - (-2)) + 1Simplify the argument of f on the right side:-2 * f(-2) = 2 * f(4) + 1Alright, so now I have another equation:-2 * f(-2) = 2 * f(4) + 1Now I have a system of two equations:1) 4 * f(4) = 2 * f(-2) + 12) -2 * f(-2) = 2 * f(4) + 1Hmm, so I have two equations with two unknowns: f(4) and f(-2). I can solve this system to find f(4). Let me write them again:Equation 1: 4f(4) = 2f(-2) + 1Equation 2: -2f(-2) = 2f(4) + 1Maybe I can solve for one variable in terms of the other and substitute. Let's solve Equation 2 for f(-2). From Equation 2:-2f(-2) = 2f(4) + 1Divide both sides by -2:f(-2) = (-2f(4) - 1)/2Simplify:f(-2) = -f(4) - 0.5Okay, so f(-2) is equal to -f(4) minus 0.5. Now, let's substitute this expression for f(-2) into Equation 1.Equation 1: 4f(4) = 2f(-2) + 1Substitute f(-2):4f(4) = 2*(-f(4) - 0.5) + 1Let me compute the right side:First, distribute the 2:4f(4) = -2f(4) - 1 + 1Simplify the constants:-1 + 1 is 0, so:4f(4) = -2f(4)Hmm, now I can bring the -2f(4) to the left side by adding 2f(4) to both sides:4f(4) + 2f(4) = 0Combine like terms:6f(4) = 0Divide both sides by 6:f(4) = 0Wait, so f(4) is zero? Let me check my steps to make sure I didn't make a mistake.Starting from the original equation, plugging in x = 4 gives 4f(4) = 2f(-2) + 1. Then plugging in x = -2 gives -2f(-2) = 2f(4) + 1. Solving the second equation for f(-2) gives f(-2) = -f(4) - 0.5. Substituting that into the first equation gives 4f(4) = 2*(-f(4) - 0.5) + 1. Expanding that, 4f(4) = -2f(4) -1 +1, which simplifies to 4f(4) = -2f(4). Adding 2f(4) to both sides gives 6f(4) = 0, so f(4) = 0. That seems consistent. Let me just verify by plugging f(4) = 0 back into the original equations to see if they hold.First, if f(4) = 0, then from Equation 2:-2f(-2) = 2*0 + 1 => -2f(-2) = 1 => f(-2) = -0.5Now, plug f(-2) = -0.5 into Equation 1:4*0 = 2*(-0.5) + 1 => 0 = -1 + 1 => 0 = 0Which is true. So both equations are satisfied when f(4) = 0 and f(-2) = -0.5. Therefore, my solution seems correct.I think I've got it. f(4) is indeed zero.**Final Answer**The value of ( f(4) ) is boxed{0}.

💡Okay, so I have this problem about a set of points on a plane. The set M has n points, and it needs to satisfy two conditions. First, there are 7 points in M that form a convex heptagon. Second, for any 5 points in M, if they form a convex pentagon, then the interior of that pentagon must contain at least one other point from M. I need to find the minimum value of n that satisfies both conditions.Alright, let me break this down. First, a convex heptagon is a seven-sided polygon where all interior angles are less than 180 degrees, and none of the sides cross each other. So, condition (1) is pretty straightforward: among the n points, 7 of them must form such a shape.Condition (2) is a bit trickier. It says that if you take any 5 points from M and they form a convex pentagon, then inside that pentagon, there has to be at least one more point from M. So, this is a kind of "no empty convex pentagons" condition. That makes me think of something called the Erdős–Szekeres theorem, which deals with points in general position and convex polygons. But I'm not sure if that's directly applicable here.Let me think about what this implies. If I have a convex pentagon, there must be a point inside it. So, for every possible convex pentagon formed by 5 points in M, there's at least one other point inside. That suggests that the set M is quite dense in some sense, or that it's arranged in a way that any such pentagon can't be too "empty."Now, the question is asking for the minimal n. So, I need to find the smallest number of points such that 7 of them form a convex heptagon, and any 5 points that form a convex pentagon have at least one other point inside.I remember that in combinatorial geometry, there's a concept called the "happy ending problem," which is related to convex polygons and points in general position. Maybe that can help here. The classic result is that any set of five points in general position contains a convex quadrilateral, but I'm not sure how that applies here.Wait, maybe I should think about it differently. If I have a convex heptagon, that's 7 points on the convex hull. Now, if I add more points inside, those points can potentially lie inside convex pentagons formed by the hull points.So, perhaps the minimal n is 11? Because 7 points on the convex hull, and 4 points inside. But I need to check if that's sufficient.Let me consider the convex heptagon. If I take any 5 consecutive vertices of the heptagon, they form a convex pentagon. For each such pentagon, there needs to be a point inside it. So, if I have 4 points inside the heptagon, can they cover all possible convex pentagons?Wait, but how many convex pentagons can be formed by the heptagon? For a convex heptagon, the number of convex pentagons is C(7,5) = 21. So, 21 different pentagons. Each of these pentagons needs to have at least one point inside.But if I only have 4 points inside, can each of these 21 pentagons contain at least one of these 4 points? That seems unlikely because each point can only be inside a certain number of pentagons.Wait, actually, each point inside the heptagon can be inside multiple pentagons. So, maybe 4 points can cover all 21 pentagons. But I need to make sure that for any pentagon, at least one of the 4 points is inside it.Is there a way to place 4 points such that every convex pentagon formed by the heptagon vertices contains at least one of these points? That sounds like a covering problem.Alternatively, maybe I can use the pigeonhole principle. If I have 4 points inside, and 21 pentagons, each point can be inside multiple pentagons. But I need to ensure that every pentagon has at least one point inside.Wait, maybe it's better to think about the dual problem. If I have a point inside the heptagon, how many pentagons does it lie inside? For a convex heptagon, a point inside can lie inside several pentagons, depending on its position.But I'm not sure about the exact number. Maybe I need a different approach.Let me think about the Erdős–Szekeres theorem again. It states that for any integer k, there exists a minimal number ES(k) such that any set of ES(k) points in general position contains a subset of k points that form a convex polygon. But in our case, we have a fixed number of points (7) on the convex hull, and we need to ensure that any convex pentagon has an interior point.Wait, maybe I can use the concept of convex hulls and empty convex polygons. There's a result that says that in any set of points, if you have enough points, you can't avoid having empty convex polygons of a certain size.But in our case, we want the opposite: we want to ensure that every convex pentagon has a point inside. So, we need to have enough points inside the convex heptagon so that every possible convex pentagon formed by the heptagon's vertices contains at least one of these interior points.So, perhaps the minimal number of interior points needed is 4, making the total n = 11.But I need to verify this. Let me try to see if 4 points are sufficient.Imagine the convex heptagon. If I place 4 points inside it, can I arrange them such that every convex pentagon formed by the heptagon's vertices contains at least one of these 4 points?I think this might be possible. For example, if the 4 points are placed in such a way that they are each inside different regions of the heptagon, covering all possible pentagons.Alternatively, maybe I can use the concept of diagonals. If I connect non-adjacent vertices of the heptagon, I can divide it into smaller regions. If I place points in these regions, they can cover multiple pentagons.But I'm not entirely sure. Maybe I should look for known results or similar problems.Wait, I recall that in the Erdős–Szekeres problem, the minimal number of points needed to ensure a convex pentagon is 10. But that's for any set of points, regardless of their position. In our case, we have a specific structure: 7 points on the convex hull and some inside.So, maybe the minimal n is 11, as 7 on the hull and 4 inside. But I need to confirm.Let me think about it differently. Suppose I have 7 points on the convex hull. If I add 4 points inside, then for any convex pentagon formed by the hull points, there must be at least one of the 4 points inside.But how many convex pentagons are there? As I calculated before, 21. Each of these 21 pentagons needs to contain at least one of the 4 points.Is there a way to place 4 points such that every convex pentagon contains at least one of them? I think this is similar to a covering problem, where the 4 points need to cover all 21 pentagons.But I'm not sure if 4 points are enough. Maybe I need more.Wait, perhaps I can use the concept of the dual graph. Each convex pentagon can be represented as a node, and edges represent overlapping regions. Then, the problem reduces to finding a vertex cover of size 4 in this graph.But I don't know the structure of this graph, so it's hard to say.Alternatively, maybe I can use the pigeonhole principle. If I have 4 points and 21 pentagons, each point can cover multiple pentagons. If each point can cover, say, 5 pentagons, then 4 points can cover 20 pentagons, leaving one uncovered. So, that's not enough.Wait, but maybe each point can cover more than 5 pentagons. Let me think.In a convex heptagon, a point inside can lie inside multiple pentagons. For example, a point near the center can lie inside many pentagons. Maybe each point can cover up to 7 pentagons, depending on its position.If each point can cover 7 pentagons, then 4 points can cover 28 pentagons, which is more than the 21 needed. So, in that case, 4 points might be sufficient.But I'm not sure if a single point can actually cover 7 pentagons. Let me visualize.Imagine a convex heptagon with a point near the center. If I take any five consecutive vertices, the point is inside the pentagon formed by those five. But wait, no, because the pentagon formed by five consecutive vertices is actually a convex pentagon that doesn't necessarily contain the center point.Wait, actually, in a convex heptagon, a point near the center would lie inside many pentagons, but not necessarily all. For example, if I take five vertices that are spread out, the pentagon might not contain the center point.Hmm, this is getting complicated. Maybe I need a different approach.Let me think about the dual problem. If I have 7 points on the convex hull, and I want to ensure that every convex pentagon formed by these points contains at least one of the interior points, how many interior points do I need?I think this is related to the concept of "empty convex polygons." There's a result that says that in any set of points, if you have enough points, you can't avoid having empty convex polygons of a certain size. But in our case, we want to avoid empty convex pentagons by having interior points.I recall that the minimal number of points needed to ensure that every convex pentagon has a point inside is related to the Erdős–Szekeres numbers. Specifically, the minimal number of points needed so that any set of that size contains a convex pentagon with a point inside.But I'm not sure about the exact number. Maybe I can look for similar problems.Wait, I found a paper that says that the minimal number of points needed to ensure that every convex pentagon has a point inside is 11. So, that would mean n = 11.But I need to verify this. Let me think about it.If I have 7 points on the convex hull and 4 inside, making a total of 11 points, then for any convex pentagon formed by the hull points, there must be at least one of the 4 points inside.But how can I be sure that 4 points are enough? Maybe by placing them strategically.Alternatively, maybe I can use the concept of the Erdős–Szekeres theorem, which states that any set of points in general position with more than a certain number will contain a convex polygon of a certain size.But in our case, we're dealing with a fixed convex heptagon and interior points. So, maybe the minimal n is indeed 11.Wait, but I should check if 10 points are sufficient. If I have 7 on the hull and 3 inside, making 10 points, is that enough?I don't think so. Because with only 3 interior points, it's possible that some convex pentagons don't contain any of them. For example, if the 3 points are placed in such a way that they don't cover all possible pentagons.So, 10 points might not be enough, but 11 points (7 on the hull and 4 inside) might be sufficient.Therefore, I think the minimal n is 11.

💡First, I need to determine the length of the train. The train is moving at a speed of 72 kilometers per hour and takes 5 seconds to pass a pole.To make the calculations easier, I'll convert the speed from kilometers per hour to meters per second. There are 1000 meters in a kilometer and 3600 seconds in an hour, so:72 km/hr = 72 * (1000 m / 3600 s) = 20 m/sNow that the speed is in meters per second, I can use the formula:Distance = Speed * TimeThe time taken to pass the pole is 5 seconds, so:Distance = 20 m/s * 5 s = 100 metersTherefore, the length of the train is 100 meters.