💡Okay, so I have this problem about a hyperbola, and I need to find the distance from a point P on the hyperbola to the right focus, given that the distance to the left focus is 3. Hmm, let me try to remember what I know about hyperbolas.First, the equation given is (frac{x^2}{4} - frac{y^2}{3} = 1). I think this is a standard form of a hyperbola centered at the origin, opening left and right because the positive term is with (x^2). So, the general form is (frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1), where ((h, k)) is the center. In this case, (h = 0) and (k = 0), so the center is at (0,0).From the equation, (a^2 = 4), so (a = 2). That means the vertices are at ((pm a, 0)), which would be (2,0) and (-2,0). Now, for hyperbolas, I remember that the distance to the foci from the center is given by (c), where (c^2 = a^2 + b^2). Here, (b^2 = 3), so (c^2 = 4 + 3 = 7), which means (c = sqrt{7}). Therefore, the foci are at ((pm sqrt{7}, 0)). So, (F_1) is ((- sqrt{7}, 0)) and (F_2) is ((sqrt{7}, 0)).The problem states that (P) is a point on the hyperbola, and the distance from (P) to (F_1) is 3. I need to find the distance from (P) to (F_2).I recall that one of the defining properties of a hyperbola is the difference of distances from any point on the hyperbola to the two foci. Specifically, for any point (P) on the hyperbola, the absolute difference of the distances to the foci is constant and equal to (2a). So, (|PF_1 - PF_2| = 2a).Given that (a = 2), this means (|PF_1 - PF_2| = 4).We know that (PF_1 = 3), so plugging that into the equation:(|3 - PF_2| = 4)This absolute value equation can have two solutions:1. (3 - PF_2 = 4)2. (3 - PF_2 = -4)Let me solve both:1. (3 - PF_2 = 4) Subtract 3 from both sides: (-PF_2 = 1) Multiply both sides by -1: (PF_2 = -1) Hmm, distance can't be negative, so this solution doesn't make sense in the context.2. (3 - PF_2 = -4) Subtract 3 from both sides: (-PF_2 = -7) Multiply both sides by -1: (PF_2 = 7) That seems reasonable. So, the distance from (P) to (F_2) is 7.Wait, let me double-check. If (PF_1 = 3) and (PF_2 = 7), then the difference is (7 - 3 = 4), which is equal to (2a = 4). That matches the property of the hyperbola. So, it seems correct.But just to be thorough, let me visualize this. The hyperbola opens left and right. The left focus is at ((- sqrt{7}, 0)) and the right focus is at ((sqrt{7}, 0)). Point (P) is somewhere on the hyperbola such that it's closer to the left focus. If the distance to the left focus is 3, then it must be farther from the right focus. The difference in distances is 4, so the distance to the right focus should be 7. That makes sense because the hyperbola is symmetric, and the point can't be too close to both foci.I think I've got it. The distance from (P) to (F_2) is 7.**Final Answer**The distance (PF_2) is boxed{7}.

💡First, I recognize that both circles share the same center, denoted as O.Point X is located two-thirds of the distance from O to P along the segment OP. This means that the length of OX is two-thirds the length of OP.To find the ratio of the areas of the two circles, I need to consider the relationship between their radii. The area of a circle is proportional to the square of its radius.Therefore, the ratio of the areas is the square of the ratio of their radii. Squaring two-thirds gives four-ninths.Thus, the ratio of the area of the circle with radius OX to the area of the circle with radius OP is four-ninths.

💡First, I need to find the total sum of all six numbers: 1856 + 1975 + 2042 + 2071 + 2150 + 2203.Calculating this sum gives me 12297.Next, since the mean of four of these numbers is 2035, the total sum of these four numbers is 4 multiplied by 2035, which equals 8140.To find the sum of the remaining two numbers, I subtract the sum of the four numbers from the total sum: 12297 - 8140 = 4157.Finally, to determine the mean of these two numbers, I divide their sum by 2: 4157 ÷ 2 = 2078.5.

💡Okay, so I have this problem about vectors in a triangle. It says that in triangle ABC, the vector BD is equal to lambda times the vector DC, where lambda is greater than zero. I need to prove that the vector AD is equal to (1/(1+lambda)) times vector AB plus (lambda/(1+lambda)) times vector AC. Hmm, okay, let me try to visualize this.First, I should probably draw triangle ABC. Let me label the points A, B, and C. Now, point D is somewhere on side BC because BD and DC are vectors along BC. The condition given is that BD equals lambda times DC. So, BD = λ DC. That means that D divides BC in the ratio of BD to DC as lambda to one. So, if I think about it, D is closer to B if lambda is greater than one, and closer to C if lambda is less than one.Now, I need to express vector AD in terms of vectors AB and AC. Vector AD goes from point A to point D. Since D is on BC, maybe I can express AD in terms of AB and AC by using some kind of section formula or something like that.Wait, in vector problems, sometimes it's helpful to express vectors in terms of position vectors relative to a common origin. Maybe I can assign position vectors to points A, B, and C, and then express D in terms of these.Let me denote the position vectors of points A, B, and C as vectors a, b, and c respectively. Then, the position vector of point D can be expressed in terms of b and c because D lies on BC.Since BD = λ DC, the ratio of BD to DC is λ:1. So, using the section formula, the position vector of D is given by (c + λ b)/(1 + λ). Wait, is that right? Let me recall the section formula. If a point divides a line segment joining points with position vectors b and c in the ratio m:n, then the position vector of the point is (n b + m c)/(m + n). So, in this case, since BD:DC = λ:1, the position vector of D should be (c + λ b)/(1 + λ). Yeah, that seems correct.So, position vector of D is (c + λ b)/(1 + λ). Now, vector AD is the vector from A to D, which is the position vector of D minus the position vector of A. So, vector AD = (c + λ b)/(1 + λ) - a.But I need to express AD in terms of vectors AB and AC. Let me recall that vector AB is b - a and vector AC is c - a. So, maybe I can rewrite the expression for AD in terms of AB and AC.Let me try that. So, vector AD = (c + λ b)/(1 + λ) - a. Let's write this as (λ b + c)/(1 + λ) - a. Now, I can express this as (λ b)/(1 + λ) + c/(1 + λ) - a.Hmm, maybe I can factor out 1/(1 + λ) from the first two terms. So, vector AD = (λ b + c)/(1 + λ) - a. Let me see if I can write this as a combination of vectors AB and AC.Since vector AB is b - a and vector AC is c - a, perhaps I can express b and c in terms of AB and AC. Let me solve for b and c.From vector AB = b - a, we get b = AB + a. Similarly, from vector AC = c - a, we get c = AC + a.So, substituting back into the expression for vector AD:vector AD = (λ (AB + a) + (AC + a))/(1 + λ) - a.Let me expand this:= (λ AB + λ a + AC + a)/(1 + λ) - a.Combine like terms:= (λ AB + AC + (λ + 1) a)/(1 + λ) - a.Now, split the fraction:= (λ AB)/(1 + λ) + (AC)/(1 + λ) + ((λ + 1) a)/(1 + λ) - a.Simplify each term:= (λ/(1 + λ)) AB + (1/(1 + λ)) AC + a - a.Wait, the last two terms are ((λ + 1) a)/(1 + λ) - a, which simplifies to a - a = 0.So, vector AD = (λ/(1 + λ)) AB + (1/(1 + λ)) AC.But wait, the problem states that vector AD should be (1/(1 + λ)) AB + (λ/(1 + λ)) AC. Hmm, I have the coefficients reversed. Did I make a mistake somewhere?Let me go back. The position vector of D is (c + λ b)/(1 + λ). So, vector AD is (c + λ b)/(1 + λ) - a.Expressed in terms of AB and AC, which are b - a and c - a respectively, so b = AB + a and c = AC + a.Substituting back:vector AD = (AC + a + λ (AB + a))/(1 + λ) - a.Expanding:= (AC + a + λ AB + λ a)/(1 + λ) - a.Combine like terms:= (λ AB + AC + (1 + λ) a)/(1 + λ) - a.Split the fraction:= (λ AB)/(1 + λ) + (AC)/(1 + λ) + ((1 + λ) a)/(1 + λ) - a.Simplify:= (λ/(1 + λ)) AB + (1/(1 + λ)) AC + a - a.Again, the a terms cancel out, leaving:vector AD = (λ/(1 + λ)) AB + (1/(1 + λ)) AC.Wait, that's the same result as before. But the problem states it should be (1/(1 + λ)) AB + (λ/(1 + λ)) AC. So, I must have messed up the ratio somewhere.Let me revisit the section formula. If BD = λ DC, then the ratio BD:DC is λ:1. So, point D divides BC internally in the ratio λ:1. Therefore, the position vector of D should be (λ c + b)/(λ + 1). Wait, is that correct?Wait, no, the section formula for internal division in the ratio m:n is (n b + m c)/(m + n). So, if BD:DC = λ:1, then m = λ and n = 1. So, position vector of D is (1 * b + λ * c)/(λ + 1). So, that's (b + λ c)/(1 + λ). Oh! So, earlier I had (c + λ b)/(1 + λ), which is incorrect. It should be (b + λ c)/(1 + λ).Ah, that was my mistake. I had the coefficients reversed. So, correcting that, position vector of D is (b + λ c)/(1 + λ). Therefore, vector AD = (b + λ c)/(1 + λ) - a.Now, let's express this in terms of AB and AC.Again, b = AB + a and c = AC + a.Substituting back:vector AD = (AB + a + λ (AC + a))/(1 + λ) - a.Expanding:= (AB + a + λ AC + λ a)/(1 + λ) - a.Combine like terms:= (AB + λ AC + (1 + λ) a)/(1 + λ) - a.Split the fraction:= (AB)/(1 + λ) + (λ AC)/(1 + λ) + ((1 + λ) a)/(1 + λ) - a.Simplify each term:= (1/(1 + λ)) AB + (λ/(1 + λ)) AC + a - a.The a terms cancel out, leaving:vector AD = (1/(1 + λ)) AB + (λ/(1 + λ)) AC.There we go! That matches the statement we needed to prove. So, the key mistake was in the initial application of the section formula where I incorrectly assigned the coefficients. Correcting that led to the desired result.So, to summarize, by correctly applying the section formula to find the position vector of D, and then expressing vector AD in terms of AB and AC, we arrive at the required expression for vector AD.

💡First, I need to determine the dimensions of the cistern and the depth of the water. The cistern is 10 meters long, 8 meters wide, and contains water up to a depth of 1.5 meters.Next, I'll calculate the area of the bottom of the cistern. Since the bottom is a rectangle, its area is simply the product of its length and width:Area_bottom = Length × Width = 10 m × 8 m = 80 m²Then, I'll calculate the area of the four walls that are in contact with the water. The two longer walls have a length of 10 meters and a height of 1.5 meters:Area_longer_walls = 2 × (Length × Height) = 2 × (10 m × 1.5 m) = 30 m²The two shorter walls have a width of 8 meters and the same height of 1.5 meters:Area_shorter_walls = 2 × (Width × Height) = 2 × (8 m × 1.5 m) = 24 m²Finally, I'll add the areas of the bottom and the four walls to find the total wet surface area:Total_wet_surface_area = Area_bottom + Area_longer_walls + Area_shorter_wallsTotal_wet_surface_area = 80 m² + 30 m² + 24 m² = 134 m²

💡To determine the probability that three randomly selected sticks can form a triangle, we first calculate the total number of ways to choose three sticks from the eight available. This is given by the combination formula:[binom{8}{3} = 56]Next, we identify the sets of three sticks that satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side. We systematically check each possible combination:1. **If 2 is one of the sticks:** - No valid triangles can be formed because the sum of 2 and any other stick is not greater than the third stick.2. **If 3 is the smallest stick:** - Valid sets: (3, 5, 7), (3, 11, 13), (3, 13, 19)3. **If 5 is the smallest stick:** - Valid sets: (5, 7, 11), (5, 11, 13), (5, 13, 17), (5, 17, 19)4. **If 7 is the smallest stick:** - Valid sets: (7, 11, 13), (7, 11, 17), (7, 13, 17), (7, 13, 19), (7, 17, 19)5. **If 11 is the smallest stick:** - Valid sets: (11, 13, 17), (11, 13, 19), (11, 17, 19)6. **If 13 is the smallest stick:** - Valid sets: (13, 17, 19)Counting all valid sets, we have a total of 18 sets that can form a triangle.Finally, the probability is calculated by dividing the number of valid sets by the total number of possible sets:[frac{18}{56} = frac{9}{28}]