answer：First, I need to determine the scale factor between the actual lighthouse and the model. The actual base holds 18850 liters of air, while the model should hold 0.01885 liters. To find the ratio of the volumes, I divide the actual volume by the model volume:[text{Volume ratio} = frac{18850}{0.01885} = 1000000]Since volume scales with the cube of the scale factor, I take the cube root of the volume ratio to find the linear scale factor:[text{Scale factor} = sqrt[3]{1000000} = 100]Now, I apply this scale factor to the actual height of the lighthouse to find the model's height:[text{Model height} = frac{60 text{ meters}}{100} = 0.6 text{ meters} = 60 text{ centimeters}]Therefore, Ella should make her model lighthouse 60 centimeters tall.

answer：To solve the problem, we start by defining the function ( f(x) = 3^x ).Next, we calculate ( f(x+1) ) by substituting ( x + 1 ) into the function:[f(x+1) = 3^{x+1}]We then find the difference ( f(x+1) - f(x) ):[f(x+1) - f(x) = 3^{x+1} - 3^x]Using the exponent property ( a^{m+n} = a^m cdot a^n ), we rewrite ( 3^{x+1} ) as:[3^{x+1} = 3^x cdot 3]Substituting this back into the expression, we get:[f(x+1) - f(x) = 3^x cdot 3 - 3^x = 3^x(3 - 1) = 3^x cdot 2]Recognizing that ( 3^x cdot 2 = 2f(x) ) since ( f(x) = 3^x ), we conclude:[f(x+1) - f(x) = 2f(x)]Thus, the final answer is (boxed{C}).

answer：First, I recognize that Friend P is walking 20% faster than Friend Q. This means if Friend Q's speed is ( v ) km/h, Friend P's speed is ( 1.20v ) km/h.When they start walking towards each other from opposite ends of the trail, they will meet at some point. At the meeting time, the total distance they have covered together equals the total length of the trail, which I'll denote as ( L ).Friend P has walked 12 kilometers when they meet. Using the formula ( text{distance} = text{speed} times text{time} ), I can express the time taken for them to meet as ( t = frac{12}{1.20v} ).During the same time ( t ), Friend Q has walked ( L - 12 ) kilometers. So, ( L - 12 = v times t ).Substituting the expression for ( t ) from Friend P's equation into Friend Q's equation gives:[L - 12 = v times frac{12}{1.20v}]Simplifying this, the ( v ) cancels out:[L - 12 = frac{12}{1.20} = 10]Finally, solving for ( L ) yields:[L = 10 + 12 = 22]

answer：Alright, so I've got this problem here with two parts, both involving some algebra and inequalities. Let me try to tackle them one by one.Starting with part (i): We have two numbers, a and b, such that a is greater than or equal to b, both are positive, and their sum is 1. So, a + b = 1, and a ≥ b > 0. We need to prove that if m and n are positive integers with m < n, then a^m - a^n is greater than or equal to b^m - b^n, and both of these differences are greater than 0.Hmm, okay. So, first, let's understand what this means. Since a and b are positive and add up to 1, both are less than 1. Also, since a is greater than or equal to b, a is at least 1/2, and b is at most 1/2.Now, when we raise numbers between 0 and 1 to higher powers, they get smaller. So, a^m is greater than a^n when m < n because a is less than 1. Similarly, b^m is greater than b^n for the same reason. So, both a^m - a^n and b^m - b^n are positive, which explains why the second part of the inequality, b^m - b^n > 0, holds.But why is a^m - a^n greater than or equal to b^m - b^n? Let's think about it. Since a is greater than or equal to b, and both are less than 1, their higher powers decrease faster. So, the difference a^m - a^n should be larger than b^m - b^n because a is bigger, so the decrease from a^m to a^n is more significant than the decrease from b^m to b^n.Maybe I can formalize this. Let's consider the function f(x) = x^m - x^n. Since m < n, and x is between 0 and 1, f(x) is positive because x^m > x^n. Also, since the derivative f’(x) = m x^{m-1} - n x^{n-1} is positive for x in (0,1) because m < n and x^{m-1} > x^{n-1} when x < 1. So, f(x) is increasing on (0,1). Since a ≥ b, f(a) ≥ f(b), which means a^m - a^n ≥ b^m - b^n. That makes sense.Okay, so part (i) seems manageable with considering the function f(x) and its properties.Moving on to part (ii): For each positive integer n, we have a quadratic function f_n(x) = x^2 - b^n x - a^n. We need to show that this quadratic has two roots between -1 and 1.Quadratic equations have two roots, and we can use the quadratic formula to find them. The roots are given by [b^n ± sqrt(b^{2n} + 4a^n)] / 2. We need to show that both roots are between -1 and 1.First, let's note that since a and b are positive and less than 1, a^n and b^n are also positive and less than 1. So, b^n is between 0 and 1, and a^n is between 0 and 1 as well.Looking at the quadratic formula, the roots are [b^n ± sqrt(b^{2n} + 4a^n)] / 2. Let's analyze the discriminant first: sqrt(b^{2n} + 4a^n). Since both b^{2n} and 4a^n are positive, the discriminant is positive, so we have two real roots.Now, let's consider the upper root: [b^n + sqrt(b^{2n} + 4a^n)] / 2. We need to show this is less than 1. Let's see:[b^n + sqrt(b^{2n} + 4a^n)] / 2 < 1Multiply both sides by 2:b^n + sqrt(b^{2n} + 4a^n) < 2Since b^n < 1 and sqrt(b^{2n} + 4a^n) < sqrt(1 + 4) = sqrt(5) ≈ 2.236, but wait, that's more than 2. Hmm, that seems problematic. Maybe my approach is wrong.Wait, actually, since a + b = 1, and a ≥ b, a is at least 1/2, so a^n is at least (1/2)^n, which decreases as n increases. Similarly, b^n is at most (1/2)^n.Let me try to bound sqrt(b^{2n} + 4a^n). Since b^{2n} ≤ (1/2)^{2n} and a^n ≥ (1/2)^n, so 4a^n ≥ 4*(1/2)^n.But I'm not sure if that helps. Maybe I need another approach.Alternatively, perhaps I can use the fact that the quadratic f_n(x) = x^2 - b^n x - a^n can be analyzed for its roots by checking the values at x = -1 and x = 1.If f_n(-1) and f_n(1) are both positive or both negative, then the roots might lie outside the interval. But if f_n(-1) and f_n(1) have opposite signs, then there's a root in between. Wait, but we need both roots to be between -1 and 1.Actually, for a quadratic, if f(-1) and f(1) are both positive, and the vertex is below the x-axis, then both roots are between -1 and 1. Alternatively, if f(-1) and f(1) are both negative, and the vertex is above the x-axis, then both roots are between -1 and 1.Let's compute f_n(-1) and f_n(1):f_n(-1) = (-1)^2 - b^n*(-1) - a^n = 1 + b^n - a^nf_n(1) = (1)^2 - b^n*(1) - a^n = 1 - b^n - a^nSince a + b = 1, a^n + b^n < 1 because for n ≥ 1, a^n + b^n ≤ a + b = 1, with equality only when n=1.Wait, actually, for n=1, a + b =1, so f_n(1) = 1 - b - a = 0. So, x=1 is a root when n=1. Hmm, but for n >1, a^n + b^n <1, so f_n(1) =1 - b^n - a^n >0.Similarly, f_n(-1) =1 + b^n - a^n. Since a ≥ b, a^n ≥ b^n, so 1 + b^n - a^n ≤1. But is it positive?We need to check if 1 + b^n - a^n >0. Since a + b =1, and a ≥ b, a ≥1/2, so a^n ≥ (1/2)^n, and b^n ≤ (1/2)^n.So, 1 + b^n - a^n ≥1 - (a^n - b^n). From part (i), we know that a^n - b^n ≤ a^m - b^m for m <n, but not sure if that helps.Wait, let's compute f_n(-1):f_n(-1) =1 + b^n - a^nSince a + b =1, and a ≥ b, a^n ≥ b^n, so 1 + b^n - a^n =1 - (a^n - b^n). From part (i), a^m - a^n ≥ b^m - b^n >0, but m <n.But for m=1, we have a - a^n ≥ b - b^n >0.So, a - a^n ≥ b - b^n.But I'm not sure if that directly helps with f_n(-1).Alternatively, since a ≥ b, and both are positive, a^n ≥ b^n, so 1 + b^n - a^n ≤1.But is it positive? Let's see:Since a + b =1, and a ≥ b, a ≥1/2, so a^n ≥ (1/2)^n.Similarly, b^n ≤ (1/2)^n.So, 1 + b^n - a^n ≥1 - (a^n - b^n). But from part (i), a^n - b^n ≤ a^1 - b^1 = a - b.Since a + b =1, a - b = 2a -1.But a ≥1/2, so 2a -1 ≥0.So, a^n - b^n ≤2a -1.But 2a -1 can be up to 1 (when a=1, b=0, but b>0, so approaching 1).Wait, this seems a bit convoluted. Maybe another approach.Let me consider specific values. Suppose n=1: f_1(x)=x^2 -b x -a. Since a + b=1, f_1(x)=x^2 -b x - (1 - b)=x^2 -b x -1 +b.At x=1: f_1(1)=1 -b -1 +b=0, so x=1 is a root.At x=-1: f_1(-1)=1 +b -1 +b=2b. Since b>0, f_1(-1)=2b>0.So, for n=1, one root is at x=1, and the other root is at x= [b - sqrt(b^2 +4a)] /2. Since a=1 -b, it's [b - sqrt(b^2 +4(1 -b))]/2.Let me compute the other root:x = [b - sqrt(b^2 +4 -4b)] /2 = [b - sqrt(4 -4b +b^2)] /2 = [b - sqrt((2 -b)^2)] /2 = [b - (2 -b)] /2 = [2b -2]/2 = b -1.Since b ≤1/2, b -1 ≤-1/2, so the other root is ≤-1/2, which is greater than -1. So, for n=1, roots are at x=1 and x=b -1, which is between -1 and 1.Wait, but the problem says "two roots that are in between -1 and 1". For n=1, one root is exactly at 1, and the other is at b -1, which is greater than -1. So, it's within the interval.For n=2: f_2(x)=x^2 -b^2 x -a^2.Compute f_2(-1)=1 +b^2 -a^2.Since a + b=1, a=1 -b, so a^2=1 -2b +b^2.Thus, f_2(-1)=1 +b^2 - (1 -2b +b^2)=1 +b^2 -1 +2b -b^2=2b>0.Similarly, f_2(1)=1 -b^2 -a^2=1 -b^2 - (1 -2b +b^2)=1 -b^2 -1 +2b -b^2=2b -2b^2=2b(1 -b).Since b>0 and 1 -b>0, f_2(1)=2b(1 -b)>0.So, both f_n(-1) and f_n(1) are positive. Now, the quadratic opens upwards (coefficient of x^2 is positive), so if f_n(-1) >0 and f_n(1) >0, the roots must be outside the interval (-1,1). But the problem says they are inside. Hmm, contradiction?Wait, no, because for n=2, let's compute the roots:x = [b^2 ± sqrt(b^4 +4a^2)] /2.Since a=1 -b, a^2=1 -2b +b^2.So, sqrt(b^4 +4(1 -2b +b^2))=sqrt(b^4 +4 -8b +4b^2).This seems complicated, but let's pick a specific value for b to test.Let me choose b=1/2, so a=1/2.Then, f_2(x)=x^2 - (1/2)^2 x - (1/2)^2 =x^2 - (1/4)x -1/4.Compute the roots:x = [1/4 ± sqrt(1/16 +1)] /2 = [1/4 ± sqrt(17/16)] /2 = [1/4 ± sqrt(17)/4] /2 = [1 ± sqrt(17)] /8.sqrt(17)≈4.123, so roots are approximately (1 +4.123)/8≈5.123/8≈0.640 and (1 -4.123)/8≈-3.123/8≈-0.390.So, both roots are between -1 and 1. Okay, so for n=2, it works.Wait, but earlier when I thought f_n(-1) and f_n(1) are both positive, I thought the roots would be outside, but in reality, they are inside. So, maybe my earlier reasoning was flawed.Perhaps because the quadratic is positive at both ends, but since it opens upwards, it must dip below the x-axis between -1 and 1, hence having two roots within that interval.Wait, no, if a quadratic is positive at both ends and opens upwards, it might not necessarily have roots in between. It depends on the minimum.Actually, the vertex of the quadratic is at x = b^n /2. Since b^n is between 0 and1, the vertex is at x between 0 and 0.5. The minimum value is f_n(b^n /2) = (b^n /2)^2 - b^n*(b^n /2) -a^n = b^{2n}/4 - b^{2n}/2 -a^n = -b^{2n}/4 -a^n <0.So, the quadratic has a minimum below the x-axis, and since f_n(-1) >0 and f_n(1) >0, it must cross the x-axis twice between -1 and 1.Yes, that makes sense. So, the quadratic is positive at x=-1 and x=1, has a minimum below the x-axis, hence two roots between -1 and 1.Therefore, part (ii) holds.Overall, I think I've got a handle on both parts now.

answer：First, I need to determine the number of 1 coins collected. Since the total number of coins is 275 and there are 148 2 coins, I can subtract the number of 2 coins from the total to find the number of 1 coins.Next, I'll calculate the total value of the 1 coins by multiplying the number of 1 coins by 1.Then, I'll calculate the total value of the 2 coins by multiplying the number of 2 coins by 2.Finally, I'll add the total values of the 1 and 2 coins to find the overall total value of all the coins collected.

answer：First, I need to understand the given equation ((x + 2)^2 = m + 2) and determine the values of (m) that allow the equation to have real roots in terms of (x).I know that for a quadratic equation to have real roots, the discriminant must be non-negative. However, in this case, the equation is already in a squared form, which simplifies the process.Expanding the left side of the equation:[(x + 2)^2 = x^2 + 4x + 4]So, the equation becomes:[x^2 + 4x + 4 = m + 2]Subtracting (m + 2) from both sides to set the equation to zero:[x^2 + 4x + (4 - m - 2) = 0]Simplifying the constant terms:[x^2 + 4x + (2 - m) = 0]Now, this is a standard quadratic equation of the form (ax^2 + bx + c = 0), where:[a = 1, quad b = 4, quad c = 2 - m]The discriminant (D) of a quadratic equation is given by:[D = b^2 - 4ac]Substituting the values of (a), (b), and (c):[D = 4^2 - 4 times 1 times (2 - m) = 16 - 8 + 4m = 8 + 4m]For the equation to have real roots, the discriminant must be greater than or equal to zero:[8 + 4m geq 0]Solving for (m):[4m geq -8 m geq -2]Therefore, any value of (m) that is greater than or equal to (-2) will ensure that the equation has real roots. To provide a specific example, I can choose (m = 1), which satisfies the condition (m geq -2).