💡Okay, so I'm trying to solve this geometry problem about triangle ABC. The base AB is fixed in length and position, and the vertex C moves on a circle centered at the midpoint M of AB. I need to figure out what path the centroid (the intersection of the three medians) takes as C moves around that circle.First, let me recall what a centroid is. The centroid of a triangle is the point where the three medians intersect, and it's also the center of mass of the triangle. An important property of the centroid is that it divides each median into a ratio of 2:1, with the longer part being closer to the vertex.Since AB is fixed, its midpoint M is also fixed. The vertex C is moving on a circle centered at M, which means that the distance from C to M is constant. Let's denote this distance as r. So, the length of CM is always r as C moves around the circle.Now, the centroid G of triangle ABC is located along the median CM, and it divides CM in a 2:1 ratio. That means the distance from G to M is two-thirds of the distance from C to M. Since CM is always r, the distance from G to M must be (2/3)r. Wait, so if G is always (2/3)r away from M, and M is fixed, then G must be moving on a circle centered at M with radius (2/3)r. That makes sense because as C moves around the original circle, G just moves around a smaller circle with the same center.Let me double-check this reasoning. If C is moving on a circle, then the centroid, being a fixed ratio along the median from C to M, should also move on a circle. The radius of this new circle would be scaled by the factor of 2/3 compared to the original circle. So, yes, G should trace out a circle.I don't think it's a parabola or an ellipse because those are more complex curves and usually result from different kinds of constraints, like the sum or difference of distances being constant. Here, the constraint is purely on the distance from M, which is a circle. A straight line doesn't make sense either because the centroid isn't moving back and forth along a line but rather around a point.So, putting it all together, as C moves on a circle centered at M, the centroid G moves on a smaller circle centered at the same point M with radius two-thirds of the original circle's radius. Therefore, the correct answer should be a circle.**Final Answer**boxed{textbf{(A)} text{a circle}}

💡Okay, so I have this problem here where I need to find all possible values of 'a' such that the set N is a subset of set M. Let me try to understand what this means.First, let's look at set M. It's defined as all real numbers x that satisfy the equation x² + x - 6 = 0. I remember that solving a quadratic equation can give me the elements of set M. So, I'll start by solving x² + x - 6 = 0.To solve this quadratic equation, I can try factoring. Let me see if I can factor it into two binomials. The equation is x² + x - 6, so I need two numbers that multiply to -6 and add up to 1. Hmm, 3 and -2 work because 3 * (-2) = -6 and 3 + (-2) = 1. So, the equation factors to (x + 3)(x - 2) = 0.Setting each factor equal to zero gives me the solutions:1. x + 3 = 0 ⇒ x = -32. x - 2 = 0 ⇒ x = 2So, set M is {-3, 2}. Got that down.Now, set N is defined as all real numbers x that satisfy the equation ax + 2 = 0, where 'a' is a real number. I need to find all possible 'a' such that every element in N is also in M. In other words, N must be a subset of M.Let me think about what N can be. The equation ax + 2 = 0 is a linear equation, so it can have at most one solution, unless 'a' is zero, in which case the equation becomes 0x + 2 = 0, which simplifies to 2 = 0, which is never true. So, if 'a' is zero, N is the empty set because there's no solution.If 'a' is not zero, then the equation ax + 2 = 0 has exactly one solution, which is x = -2/a. So, N can either be empty or contain a single element.Since N must be a subset of M, which has two elements {-3, 2}, N can be:1. The empty set, which is always a subset of any set.2. A singleton set containing either -3 or 2.So, let's consider these cases.Case 1: N is the empty set.This happens when 'a' is zero because the equation becomes 2 = 0, which has no solution. So, a = 0 is one possible value.Case 2: N contains exactly one element, which must be either -3 or 2.Subcase 2a: N = {-3}This means that x = -3 is a solution to ax + 2 = 0. Plugging x = -3 into the equation:a*(-3) + 2 = 0-3a + 2 = 0-3a = -2a = (-2)/(-3) = 2/3So, a = 2/3 is another possible value.Subcase 2b: N = {2}This means that x = 2 is a solution to ax + 2 = 0. Plugging x = 2 into the equation:a*2 + 2 = 02a + 2 = 02a = -2a = -1So, a = -1 is another possible value.Therefore, the possible values of 'a' are 0, 2/3, and -1.Let me just double-check to make sure I didn't miss anything. If a = 0, N is empty, which is fine. If a = 2/3, then N = {-3}, which is in M. If a = -1, then N = {2}, which is also in M. So, all these cases satisfy N being a subset of M.I think that's all. I don't see any other possibilities because if 'a' is anything else, then the solution x = -2/a would not be in M. For example, if a = 1, then x = -2, which is not in M. So, only when a is 0, 2/3, or -1 does N become a subset of M.**Final Answer**The set of all possible values of ( a ) is boxed{{-1, 0, frac{2}{3}}}.

💡First, I need to determine the common multiples of 36 and 48 that are less than 300. To do this, I'll find the Least Common Multiple (LCM) of 36 and 48.Breaking down the numbers into their prime factors:- 36 = 2² × 3²- 48 = 2⁴ × 3The LCM is found by taking the highest power of each prime number present in the factorizations:- LCM = 2⁴ × 3² = 16 × 9 = 144Next, I'll identify how many multiples of 144 are less than 300:- 144 × 1 = 144- 144 × 2 = 288- 144 × 3 = 432 (which is greater than 300)So, there are 2 multiples of 144 less than 300.Now, I'll calculate the total number of possible pairs of numbers Billy and Bobbi can choose:- Billy has 8 multiples of 36 to choose from.- Bobbi has 6 multiples of 48 to choose from.The total number of possible pairs is 8 × 6 = 48.Finally, to find the probability that they selected the same number, I'll divide the number of common multiples by the total number of possible pairs:- Probability = 2 / 48 = 1/24

💡First, I recognize that the equation of the parabola is given in vertex form, which is ( y = a(x - h)^2 + k ). In this form, the vertex of the parabola is at the point ( (h, k) ).Looking at the given equation ( y = -(x + 2)^2 + 6 ), I can identify the values of ( h ) and ( k ). The term inside the squared expression is ( x + 2 ), which can be rewritten as ( x - (-2) ). This tells me that ( h = -2 ).Next, the constant term outside the squared expression is ( +6 ), which means ( k = 6 ).By substituting these values into the vertex form, I determine that the vertex of the parabola is at the coordinates ( (-2, 6) ).

💡First, I need to establish the relationships between the investments of A, B, and C. It's given that A invests 3 times as much as B. So, if B's investment is ( B ), then A's investment is ( 3B ).Next, it's stated that B invests a fraction of what C invests. Let's denote this fraction as ( F ). Therefore, B's investment can be expressed as ( B = F times C ), where ( C ) is C's investment.The total profit is Rs. 7700, and B's share of the profit is Rs. 1400. The profit is distributed in the ratio of their investments. So, the ratio of B's investment to the total investment is equal to the ratio of B's profit share to the total profit.Setting up the equation for B's share:[frac{B}{A + B + C} times 7700 = 1400]Substituting ( A = 3B ) and ( B = F times C ) into the equation:[frac{F times C}{3F times C + F times C + C} times 7700 = 1400]Simplifying the equation:[frac{F}{4F + 1} times 7700 = 1400]Solving for ( F ):[frac{F}{4F + 1} = frac{14}{77}][77F = 14(4F + 1)][77F = 56F + 14][21F = 14][F = frac{2}{3}]Thus, the ratio of B's investment to C's investment is ( 2:3 ).

💡First, I need to estimate the value of the expression (sqrt{3} times (sqrt{10} - sqrt{3})). To simplify this, I'll distribute (sqrt{3}) across the terms inside the parentheses.This gives me (sqrt{3} times sqrt{10} - sqrt{3} times sqrt{3}), which simplifies to (sqrt{30} - 3).Next, I'll estimate the value of (sqrt{30}). I know that (5^2 = 25) and (6^2 = 36), so (sqrt{30}) must be between 5 and 6.Subtracting 3 from these bounds, I find that (2 < sqrt{30} - 3 < 3). Therefore, the value of the original expression is between 2 and 3.