answer：First, I need to understand the problem. The travel blogger starts in Indianapolis and makes several stops, with each subsequent journey being 50 miles longer than the previous one. The first leg is 150 miles, the second is 200 miles, the third is 250 miles, and so on. The total distance she plans to travel is 1500 miles.To find out how many stops she can make, I'll model the total distance as the sum of an arithmetic series. The formula for the sum of the first ( n ) terms of an arithmetic series is:[S_n = frac{n}{2} times (2a + (n - 1)d)]Here, ( a = 150 ) miles (the first term), ( d = 50 ) miles (the common difference), and ( S_n = 1500 ) miles (the total distance). Plugging these values into the formula gives:[1500 = frac{n}{2} times (300 + 50(n - 1))]Simplifying this equation:[1500 = frac{n}{2} times (250n + 50)][3000 = 250n^2 + 50n][250n^2 + 50n - 3000 = 0]Dividing the entire equation by 50 to simplify:[5n^2 + n - 60 = 0]This is the quadratic equation that will help determine the number of stops.Next, to find the total time spent traveling, I'll use the average speed of 60 miles per hour. The total time is the total distance divided by the average speed:[text{Total Time} = frac{1500 text{ miles}}{60 text{ mph}} = 25 text{ hours}]

answer：First, I recognize that the bacteria population is modeled by a quadratic function ( B(t) = -2t^2 + 12t + 20 ). Since the coefficient of ( t^2 ) is negative, the parabola opens downward, indicating that the function has a maximum point.To find the time ( t ) at which the maximum population occurs, I'll use the vertex formula for a quadratic function ( at^2 + bt + c ), which is ( t = -frac{b}{2a} ). Here, ( a = -2 ) and ( b = 12 ), so plugging these values into the formula gives ( t = -frac{12}{2(-2)} = 3 ) hours.Next, to determine the time interval during which the bacteria population is at least 30,000, I need to solve the inequality ( -2t^2 + 12t + 20 geq 30 ). Subtracting 30 from both sides results in ( -2t^2 + 12t - 10 geq 0 ).I'll solve the equation ( -2t^2 + 12t - 10 = 0 ) using the quadratic formula ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = -2 ), ( b = 12 ), and ( c = -10 ). Calculating the discriminant gives ( 144 - 80 = 64 ), so the roots are ( t = frac{-12 pm 8}{-4} ), resulting in ( t = 1 ) and ( t = 5 ).Since the quadratic opens downward, the inequality ( -2t^2 + 12t - 10 geq 0 ) holds between the roots. Therefore, the bacteria population is at least 30,000 for ( t ) in the interval [1, 5] hours.

answer：First, I need to determine the maximum height of the projectile described by the quadratic equation ( y = -16x^2 + 160x ). Since this is a quadratic equation in the form ( y = ax^2 + bx + c ), I can find the vertex to identify the maximum point. The vertex occurs at ( x = -frac{b}{2a} ). Plugging in the values from the equation, ( a = -16 ) and ( b = 160 ), I calculate ( x = -frac{160}{2(-16)} = 5 ) seconds. To find the maximum height, I substitute ( x = 5 ) back into the equation: ( y = -16(5)^2 + 160(5) = -400 + 800 = 400 ) feet.Next, I need to solve the inequality ( -16x^2 + 160x > 300 ) to determine the time interval during which the projectile is above 300 feet. I start by rearranging the inequality to ( -16x^2 + 160x - 300 > 0 ). To simplify, I divide the entire inequality by -4, remembering to reverse the inequality sign: ( 4x^2 - 40x + 75 < 0 ). Solving the quadratic equation ( 4x^2 - 40x + 75 = 0 ) using the quadratic formula, I find the roots to be ( x = frac{5}{2} ) and ( x = frac{15}{2} ). Since the coefficient of ( x^2 ) is positive, the quadratic opens upwards, and the inequality ( 4x^2 - 40x + 75 < 0 ) holds between the roots. Therefore, the projectile is above 300 feet between ( 2.5 ) seconds and ( 7.5 ) seconds.

answer：First, I need to define the distance of Route A as a variable. Let's call it ( x ) kilometers.According to the first condition, Route B is twice as long as Route A. So, the distance of Route B is ( 2x ) kilometers.The second condition states that Route C is 50 kilometers longer than Route B. Therefore, the distance of Route C is ( 2x + 50 ) kilometers.The total distance traveled in a day is the sum of the distances of all three routes, which equals 450 kilometers. So, I can write the equation:[x + 2x + (2x + 50) = 450]Combining like terms, the equation simplifies to:[5x + 50 = 450]Subtracting 50 from both sides gives:[5x = 400]Finally, dividing both sides by 5, I find:[x = 80]So, the distance of Route A is 80 kilometers.

answer：Okay, so I have this function ( g(x) = ln(2^x + 1) ), and I need to compute the expression ( g(-4) - g(-3) + g(-2) - g(-1) + g(1) - g(2) + g(3) - g(4) ). Hmm, that seems a bit complicated at first glance, but maybe I can break it down step by step.First, let me write down each term separately so I can see what's going on:1. ( g(-4) = ln(2^{-4} + 1) )2. ( -g(-3) = -ln(2^{-3} + 1) )3. ( g(-2) = ln(2^{-2} + 1) )4. ( -g(-1) = -ln(2^{-1} + 1) )5. ( g(1) = ln(2^{1} + 1) )6. ( -g(2) = -ln(2^{2} + 1) )7. ( g(3) = ln(2^{3} + 1) )8. ( -g(4) = -ln(2^{4} + 1) )So, if I substitute all these back into the expression, it becomes:( ln(2^{-4} + 1) - ln(2^{-3} + 1) + ln(2^{-2} + 1) - ln(2^{-1} + 1) + ln(2^{1} + 1) - ln(2^{2} + 1) + ln(2^{3} + 1) - ln(2^{4} + 1) )Hmm, that's a lot of logarithms. Maybe I can simplify each term individually before combining them. Let's compute each ( g(x) ) term:Starting with the negative exponents:1. ( g(-4) = ln(2^{-4} + 1) = lnleft(frac{1}{16} + 1right) = lnleft(frac{17}{16}right) )2. ( g(-3) = ln(2^{-3} + 1) = lnleft(frac{1}{8} + 1right) = lnleft(frac{9}{8}right) )3. ( g(-2) = ln(2^{-2} + 1) = lnleft(frac{1}{4} + 1right) = lnleft(frac{5}{4}right) )4. ( g(-1) = ln(2^{-1} + 1) = lnleft(frac{1}{2} + 1right) = lnleft(frac{3}{2}right) )Now the positive exponents:5. ( g(1) = ln(2^{1} + 1) = ln(2 + 1) = ln(3) )6. ( g(2) = ln(2^{2} + 1) = ln(4 + 1) = ln(5) )7. ( g(3) = ln(2^{3} + 1) = ln(8 + 1) = ln(9) )8. ( g(4) = ln(2^{4} + 1) = ln(16 + 1) = ln(17) )So substituting these back into the expression, we have:( lnleft(frac{17}{16}right) - lnleft(frac{9}{8}right) + lnleft(frac{5}{4}right) - lnleft(frac{3}{2}right) + ln(3) - ln(5) + ln(9) - ln(17) )Hmm, this looks like a telescoping series or something where terms might cancel out. Let me see if I can rearrange or combine these logarithms.Recall that ( ln(a) - ln(b) = lnleft(frac{a}{b}right) ) and ( ln(a) + ln(b) = ln(ab) ). Maybe I can pair the terms in a way that allows me to combine them.Looking at the expression:1. ( lnleft(frac{17}{16}right) - lnleft(frac{9}{8}right) )2. ( + lnleft(frac{5}{4}right) - lnleft(frac{3}{2}right) )3. ( + ln(3) - ln(5) )4. ( + ln(9) - ln(17) )Let me compute each pair:First pair: ( lnleft(frac{17}{16}right) - lnleft(frac{9}{8}right) = lnleft(frac{17}{16} div frac{9}{8}right) = lnleft(frac{17}{16} times frac{8}{9}right) = lnleft(frac{17 times 8}{16 times 9}right) )Simplify numerator and denominator:17 and 9 don't have common factors, 8 and 16 can be simplified: 8/16 = 1/2So, ( lnleft(frac{17 times 1}{2 times 9}right) = lnleft(frac{17}{18}right) )Second pair: ( lnleft(frac{5}{4}right) - lnleft(frac{3}{2}right) = lnleft(frac{5}{4} div frac{3}{2}right) = lnleft(frac{5}{4} times frac{2}{3}right) = lnleft(frac{10}{12}right) = lnleft(frac{5}{6}right) )Third pair: ( ln(3) - ln(5) = lnleft(frac{3}{5}right) )Fourth pair: ( ln(9) - ln(17) = lnleft(frac{9}{17}right) )So now, substituting back, the entire expression becomes:( lnleft(frac{17}{18}right) + lnleft(frac{5}{6}right) + lnleft(frac{3}{5}right) + lnleft(frac{9}{17}right) )Now, since all these are logarithms being added, I can combine them into a single logarithm by multiplying their arguments:( lnleft(frac{17}{18} times frac{5}{6} times frac{3}{5} times frac{9}{17}right) )Let me compute the product inside the logarithm step by step.First, multiply ( frac{17}{18} times frac{5}{6} ):( frac{17 times 5}{18 times 6} = frac{85}{108} )Next, multiply this by ( frac{3}{5} ):( frac{85}{108} times frac{3}{5} = frac{85 times 3}{108 times 5} = frac{255}{540} )Simplify ( frac{255}{540} ): both are divisible by 15.255 ÷ 15 = 17, 540 ÷ 15 = 36. So, ( frac{17}{36} )Now, multiply this by ( frac{9}{17} ):( frac{17}{36} times frac{9}{17} = frac{17 times 9}{36 times 17} )Here, 17 cancels out, and 9 and 36 can be simplified: 9/36 = 1/4So, ( frac{1}{4} )Therefore, the entire expression inside the logarithm is ( frac{1}{4} ), so the expression becomes:( lnleft(frac{1}{4}right) )Which is equal to ( ln(1) - ln(4) = 0 - ln(4) = -ln(4) )But wait, ( ln(1/4) ) is indeed ( -ln(4) ). Alternatively, ( ln(1/4) = ln(2^{-2}) = -2ln(2) ). So, depending on how the answer is expected, it could be either ( -ln(4) ) or ( -2ln(2) ). But both are correct; it's just a matter of simplification.But let me double-check my steps to make sure I didn't make a mistake.Starting from the expression:( lnleft(frac{17}{18}right) + lnleft(frac{5}{6}right) + lnleft(frac{3}{5}right) + lnleft(frac{9}{17}right) )Multiplying all together:( frac{17}{18} times frac{5}{6} times frac{3}{5} times frac{9}{17} )Let me rearrange the multiplication to see if terms cancel:- 17 in numerator cancels with 17 in denominator.- 5 in numerator cancels with 5 in denominator.- 9 and 18: 9 cancels with 18, leaving 2 in the denominator.- 3 in numerator cancels with 6 in denominator, leaving 2 in the denominator.So, step by step:- 17/18 * 5/6 = (17*5)/(18*6)- Then, *3/5 = (17*5*3)/(18*6*5) = (17*3)/(18*6)- Then, *9/17 = (17*3*9)/(18*6*17) = (3*9)/(18*6)- Simplify numerator and denominator:Numerator: 3*9 = 27Denominator: 18*6 = 108So, 27/108 = 1/4Yes, that's correct. So, the product is indeed 1/4, so the logarithm is ( ln(1/4) = -ln(4) ).Alternatively, since 4 is 2 squared, ( ln(4) = 2ln(2) ), so ( -ln(4) = -2ln(2) ).But the question doesn't specify the form, so either is acceptable, but perhaps ( -ln(4) ) is simpler.Wait, but let me think again. The original expression was a combination of logs with positive and negative signs. Did I handle the signs correctly?Let me recount:Original expression:( g(-4) - g(-3) + g(-2) - g(-1) + g(1) - g(2) + g(3) - g(4) )Which became:( ln(17/16) - ln(9/8) + ln(5/4) - ln(3/2) + ln(3) - ln(5) + ln(9) - ln(17) )Then I grouped them as:[ ( ln(17/16) - ln(9/8) ) ] + [ ( ln(5/4) - ln(3/2) ) ] + [ ( ln(3) - ln(5) ) ] + [ ( ln(9) - ln(17) ) ]Which became:( ln(17/18) + ln(5/6) + ln(3/5) + ln(9/17) )Then multiplied all together to get ( ln(1/4) ).Wait, but when I multiplied all together, it's the product of all four fractions:( (17/18) * (5/6) * (3/5) * (9/17) )Which simplifies to 1/4, as above.So, the total expression is ( ln(1/4) = -ln(4) ).Alternatively, maybe I can think of it another way. Let me see if there's a telescoping pattern.Looking back at the original expression:( g(-4) - g(-3) + g(-2) - g(-1) + g(1) - g(2) + g(3) - g(4) )Notice that the terms are alternating signs, starting with positive for ( g(-4) ), then negative for ( g(-3) ), and so on.Maybe if I pair them differently, I can see a telescoping effect.Let me write the expression as:( [g(-4) - g(-3)] + [g(-2) - g(-1)] + [g(1) - g(2)] + [g(3) - g(4)] )So, each pair is ( g(n) - g(n+1) ) for n = -4, -2, 1, 3.Wait, let's check:First pair: ( g(-4) - g(-3) )Second pair: ( g(-2) - g(-1) )Third pair: ( g(1) - g(2) )Fourth pair: ( g(3) - g(4) )So, each pair is ( g(k) - g(k+1) ) where k is -4, -2, 1, 3.So, each pair is ( ln(2^k + 1) - ln(2^{k+1} + 1) )Which is ( lnleft( frac{2^k + 1}{2^{k+1} + 1} right) )So, for each pair, we have:1. ( lnleft( frac{2^{-4} + 1}{2^{-3} + 1} right) )2. ( lnleft( frac{2^{-2} + 1}{2^{-1} + 1} right) )3. ( lnleft( frac{2^{1} + 1}{2^{2} + 1} right) )4. ( lnleft( frac{2^{3} + 1}{2^{4} + 1} right) )So, the entire expression is the sum of these four logarithms:( lnleft( frac{2^{-4} + 1}{2^{-3} + 1} right) + lnleft( frac{2^{-2} + 1}{2^{-1} + 1} right) + lnleft( frac{2^{1} + 1}{2^{2} + 1} right) + lnleft( frac{2^{3} + 1}{2^{4} + 1} right) )Which can be combined into a single logarithm:( lnleft( frac{2^{-4} + 1}{2^{-3} + 1} times frac{2^{-2} + 1}{2^{-1} + 1} times frac{2^{1} + 1}{2^{2} + 1} times frac{2^{3} + 1}{2^{4} + 1} right) )Let me compute this product step by step.First, compute each fraction:1. ( frac{2^{-4} + 1}{2^{-3} + 1} = frac{frac{1}{16} + 1}{frac{1}{8} + 1} = frac{frac{17}{16}}{frac{9}{8}} = frac{17}{16} times frac{8}{9} = frac{17 times 8}{16 times 9} = frac{136}{144} = frac{17}{18} )2. ( frac{2^{-2} + 1}{2^{-1} + 1} = frac{frac{1}{4} + 1}{frac{1}{2} + 1} = frac{frac{5}{4}}{frac{3}{2}} = frac{5}{4} times frac{2}{3} = frac{10}{12} = frac{5}{6} )3. ( frac{2^{1} + 1}{2^{2} + 1} = frac{2 + 1}{4 + 1} = frac{3}{5} )4. ( frac{2^{3} + 1}{2^{4} + 1} = frac{8 + 1}{16 + 1} = frac{9}{17} )So, the product becomes:( frac{17}{18} times frac{5}{6} times frac{3}{5} times frac{9}{17} )Which is exactly the same as before. So, as before, this product simplifies to ( frac{1}{4} ), so the entire expression is ( lnleft( frac{1}{4} right) = -ln(4) ).So, regardless of how I approach it, whether by combining pairs or looking for telescoping, I end up with the same result.Alternatively, maybe there's a pattern or a property of the function ( g(x) ) that can help simplify this expression without computing each term individually.Let me think about the function ( g(x) = ln(2^x + 1) ). Is there any symmetry or property that could be useful here?Notice that ( 2^{-x} = frac{1}{2^x} ), so ( g(-x) = ln(2^{-x} + 1) = lnleft( frac{1}{2^x} + 1 right) = lnleft( frac{1 + 2^x}{2^x} right) = ln(1 + 2^x) - ln(2^x) = g(x) - x ln(2) )So, ( g(-x) = g(x) - x ln(2) )That's an interesting relation. Maybe I can use this to rewrite some terms.Looking back at the original expression:( g(-4) - g(-3) + g(-2) - g(-1) + g(1) - g(2) + g(3) - g(4) )Let me substitute ( g(-x) = g(x) - x ln(2) ) for each negative term:1. ( g(-4) = g(4) - 4 ln(2) )2. ( g(-3) = g(3) - 3 ln(2) )3. ( g(-2) = g(2) - 2 ln(2) )4. ( g(-1) = g(1) - 1 ln(2) )So, substituting these into the expression:( [g(4) - 4 ln(2)] - [g(3) - 3 ln(2)] + [g(2) - 2 ln(2)] - [g(1) - 1 ln(2)] + g(1) - g(2) + g(3) - g(4) )Let me expand this:= ( g(4) - 4 ln(2) - g(3) + 3 ln(2) + g(2) - 2 ln(2) - g(1) + 1 ln(2) + g(1) - g(2) + g(3) - g(4) )Now, let's combine like terms:- ( g(4) ) and ( -g(4) ) cancel out.- ( -g(3) ) and ( +g(3) ) cancel out.- ( +g(2) ) and ( -g(2) ) cancel out.- ( -g(1) ) and ( +g(1) ) cancel out.Now, the logarithmic terms:- ( -4 ln(2) + 3 ln(2) - 2 ln(2) + 1 ln(2) )Let's compute the coefficients:-4 + 3 = -1-1 -2 = -3-3 +1 = -2So, the total is ( -2 ln(2) )Therefore, the entire expression simplifies to ( -2 ln(2) ), which is the same as ( -ln(4) ) since ( ln(4) = 2 ln(2) ).So, whether I compute each term individually and combine them or use the functional equation ( g(-x) = g(x) - x ln(2) ), I end up with the same result: ( -2 ln(2) ).Therefore, the value of the expression is ( -2 ln(2) ).**Final Answer**The value of the expression is boxed{-2 ln 2}.

answer：Alright, so I have this problem about a parabola that models a rock formation. The equation given is ( y = -2x^2 + 8x ). There are two parts: first, finding the maximum height of the rock formation, and second, figuring out the equation of the parabola when it's scaled down by a factor of 1:100. Let me tackle each part step by step.Starting with the first question: Determine the maximum height of the rock formation along this parabolic curve. Hmm, okay. I remember that for a quadratic equation in the form ( y = ax^2 + bx + c ), the vertex represents either the maximum or minimum point. Since the coefficient of ( x^2 ) here is negative (-2), the parabola opens downward, which means the vertex will be the highest point, or the maximum. So, the maximum height is at the vertex.Now, how do I find the vertex? I recall there are a couple of ways. One is using the vertex formula, which is ( x = -frac{b}{2a} ). Another method is completing the square, but since this equation is already in standard form, maybe the formula is quicker.Let me write down the equation again: ( y = -2x^2 + 8x ). Comparing this to the standard form ( y = ax^2 + bx + c ), I can see that ( a = -2 ), ( b = 8 ), and ( c = 0 ) (since there's no constant term). So, plugging into the vertex formula, ( x = -frac{b}{2a} = -frac{8}{2*(-2)} ).Calculating that: the denominator is 2*(-2) which is -4. So, ( x = -frac{8}{-4} = 2 ). So, the x-coordinate of the vertex is 2. Now, to find the y-coordinate, which will be the maximum height, I substitute x = 2 back into the original equation.So, ( y = -2*(2)^2 + 8*(2) ). Let's compute that step by step. First, ( (2)^2 = 4 ), so ( -2*4 = -8 ). Then, ( 8*2 = 16 ). Adding those together: ( -8 + 16 = 8 ). So, the maximum height is 8 units.Wait, just to make sure I didn't make a mistake. Let me double-check my calculations. The vertex formula: ( x = -b/(2a) ). Plugging in, ( x = -8/(2*(-2)) = -8/-4 = 2 ). That seems right. Then plugging x=2 into the equation: ( y = -2*(4) + 16 = -8 + 16 = 8 ). Yep, that looks correct.Alternatively, I could have completed the square. Let me try that method just to confirm. Starting with ( y = -2x^2 + 8x ). Factor out the coefficient of ( x^2 ) from the first two terms: ( y = -2(x^2 - 4x) ). Now, to complete the square inside the parentheses, I take half of the coefficient of x, which is -4, so half is -2, and square it: (-2)^2 = 4. So, I add and subtract 4 inside the parentheses: ( y = -2[(x^2 - 4x + 4 - 4)] ). Wait, actually, since I factored out a -2, adding 4 inside is like subtracting 8 outside. Let me write it properly.So, ( y = -2[(x^2 - 4x + 4) - 4] ). That becomes ( y = -2[(x - 2)^2 - 4] ). Distribute the -2: ( y = -2(x - 2)^2 + 8 ). So, in vertex form, it's ( y = -2(x - 2)^2 + 8 ). Therefore, the vertex is at (2, 8), confirming the maximum height is 8. Okay, that solidifies my previous answer.Moving on to the second part: Ali wants to create a scale model scaled down by a factor of 1:100. So, I need to write the equation of the parabola for the scaled model. Scaling down by 1:100 means that each unit in the model represents 100 units in real life. So, both the x and y dimensions will be scaled by a factor of 1/100.Wait, how does scaling affect the equation of a parabola? Let me recall. If you scale the graph horizontally by a factor of k, you replace x with x/k. Similarly, scaling vertically by a factor of k, you replace y with y/k. Since the scale factor is 1:100, meaning the model is 1/100th the size, both x and y will be scaled by 1/100.So, to get the scaled equation, I need to replace x with (x)/100 and y with (y)/100 in the original equation. Let me write that out.Original equation: ( y = -2x^2 + 8x ).Scaled equation: ( frac{y}{100} = -2left(frac{x}{100}right)^2 + 8left(frac{x}{100}right) ).Now, let me simplify this equation step by step.First, compute each term:1. ( frac{y}{100} ) remains as is.2. ( -2left(frac{x}{100}right)^2 ) becomes ( -2*frac{x^2}{10000} = -frac{2x^2}{10000} ).3. ( 8left(frac{x}{100}right) ) becomes ( frac{8x}{100} ).So, putting it all together:( frac{y}{100} = -frac{2x^2}{10000} + frac{8x}{100} ).To make this equation look cleaner, I can multiply both sides by 100 to eliminate the denominators:Multiplying each term by 100:Left side: ( y ).Right side: ( -frac{2x^2}{10000} * 100 = -frac{2x^2}{100} ), and ( frac{8x}{100} * 100 = 8x ).So, the equation becomes:( y = -frac{2}{100}x^2 + 8x ).Simplify ( -frac{2}{100} ) to ( -frac{1}{50} ). So, ( y = -frac{1}{50}x^2 + 8x ).Wait, is that correct? Let me double-check my steps.Starting from scaling both x and y by 1/100. So, replacing x with x/100 and y with y/100.Original equation: ( y = -2x^2 + 8x ).After scaling: ( frac{y}{100} = -2left(frac{x}{100}right)^2 + 8left(frac{x}{100}right) ).Calculating each term:- ( left(frac{x}{100}right)^2 = frac{x^2}{10000} ), so multiplied by -2 is ( -frac{2x^2}{10000} ).- ( 8*frac{x}{100} = frac{8x}{100} ).So, equation is ( frac{y}{100} = -frac{2x^2}{10000} + frac{8x}{100} ).Multiplying both sides by 100:Left side: y.Right side: ( -frac{2x^2}{100} + 8x ).Simplify ( -frac{2}{100} ) to ( -frac{1}{50} ), so ( y = -frac{1}{50}x^2 + 8x ).Wait, that seems correct. Alternatively, if I factor out 1/100 from both sides, but I think the way I did it is correct.But hold on, is the scaling factor applied correctly? Because scaling the model down by 1:100, so the model is 1/100 the size, meaning that the coordinates in the model are 1/100 of the original. So, if the original is y = -2x² + 8x, then the model's coordinates (let's say x' and y') relate to the original by x' = x / 100 and y' = y / 100. So, substituting back, x = 100x' and y = 100y'. So, plugging into the original equation:100y' = -2*(100x')² + 8*(100x').Wait, hold on, that seems different. Let me think about this.If the model is scaled down by 1:100, then each unit in the model corresponds to 100 units in real life. So, if the original parabola is in real coordinates (x, y), then the model's coordinates (x', y') are related by x = 100x' and y = 100y'. So, substituting into the original equation:100y' = -2*(100x')² + 8*(100x').Compute each term:- ( (100x')² = 10000x'^2 ), so multiplied by -2 is -20000x'^2.- 8*(100x') = 800x'.So, the equation becomes:100y' = -20000x'^2 + 800x'.Divide both sides by 100 to solve for y':y' = (-20000x'^2 + 800x') / 100 = -200x'^2 + 8x'.So, the scaled equation is ( y' = -200x'^2 + 8x' ).Wait, that's different from what I got earlier. Hmm, so which one is correct?I think my confusion arises from whether we're scaling the graph or scaling the equation. Let me clarify.When scaling a graph, if you scale the axes, the transformation affects the equation inversely. So, if you scale the x-axis by a factor of k, you replace x with x/k in the equation. Similarly, scaling the y-axis by k, replace y with y/k.But in this case, the model is scaled down by 1:100, meaning that each unit in the model is 1/100 of the original. So, to get the model's equation, we need to scale both x and y by 1/100.Alternatively, if we think in terms of transformations, scaling the graph horizontally by 1/100 and vertically by 1/100.So, starting with the original equation ( y = -2x^2 + 8x ), scaling x by 1/100 would replace x with 100x (since x in the model is 1/100 of the original x), and scaling y by 1/100 would replace y with 100y.Wait, no, actually, scaling the graph horizontally by a factor of k is replacing x with x/k. So, if the model is scaled down by 1/100, then each x in the model is 1/100 of the original, so to get the original x, you multiply by 100. So, x_original = 100x_model.Similarly, y_original = 100y_model.So, substituting into the original equation:100y_model = -2*(100x_model)^2 + 8*(100x_model).Which is what I did earlier, leading to y_model = -200x_model^2 + 8x_model.Alternatively, if I consider the scaling as replacing x with x/100 and y with y/100 in the original equation, that would be:(y)/100 = -2*(x/100)^2 + 8*(x/100).Which simplifies to y = -2*(x^2)/100 + 8x/100 * 100.Wait, no, let's do it step by step.Starting with original equation: y = -2x² + 8x.If we scale the model by 1:100, meaning that the model's x and y are 1/100 of the original. So, to express the model's equation, we have:Let x' = x / 100 and y' = y / 100.So, substituting into the original equation:y = -2x² + 8x.Expressed in terms of x' and y':100y' = -2*(100x')² + 8*(100x').Which is the same as:100y' = -2*10000x'^2 + 800x'.Divide both sides by 100:y' = -200x'^2 + 8x'.So, the equation of the scaled model is ( y' = -200x'^2 + 8x' ).But earlier, when I did it the other way, replacing x with x/100 and y with y/100, I got ( y = -frac{1}{50}x^2 + 8x ). But that seems conflicting.Wait, perhaps the confusion is in the direction of scaling. If we are creating a model that is scaled down, then the model's coordinates are smaller. So, if the original is in real units, the model is in model units, which are 1/100th. So, to get the model's equation, we need to express y_model in terms of x_model.So, if we let x_model = x_real / 100 and y_model = y_real / 100, then substituting into the original equation:y_real = -2x_real² + 8x_real.Expressed in model terms:100y_model = -2*(100x_model)^2 + 8*(100x_model).Which is 100y_model = -20000x_model² + 800x_model.Divide both sides by 100:y_model = -200x_model² + 8x_model.So, that's the equation in terms of the model's coordinates.Alternatively, if I think about it as a transformation of the graph, scaling both x and y by 1/100. So, to scale the graph, you replace x with 100x and y with 100y in the original equation.Wait, no, actually, scaling the graph horizontally by a factor of k is replacing x with x/k. So, if the model is scaled down by 1/100, then k = 1/100, so x is replaced with x/(1/100) = 100x. Similarly, y is replaced with y/(1/100) = 100y.So, substituting into the original equation:100y = -2*(100x)^2 + 8*(100x).Which is 100y = -20000x² + 800x.Divide both sides by 100:y = -200x² + 8x.So, that's consistent with the previous result.Therefore, the equation of the scaled model is ( y = -200x^2 + 8x ).Wait, but earlier when I did the substitution by replacing x with x/100 and y with y/100, I got a different result. So, which approach is correct?I think the confusion comes from whether we're scaling the axes or scaling the graph. If we scale the axes by 1/100, meaning each unit on the axes is now 1/100th, then the graph itself is stretched by 100. But in this case, we are scaling the graph down, so the graph is compressed by 1/100.Wait, perhaps another way to think about it is using function transformations. If you have a function f(x), and you scale it horizontally by a factor of k, the new function is f(x/k). Similarly, scaling vertically by k, it's k*f(x).But in this case, scaling down the model by 1:100, so the model is 1/100 the size. So, both horizontally and vertically, it's scaled by 1/100.So, the horizontal scaling would replace x with x/(1/100) = 100x, and vertical scaling would replace y with y/(1/100) = 100y.So, starting from the original function y = -2x² + 8x, scaling both axes by 1/100, the new function becomes:100y = -2*(100x)^2 + 8*(100x).Which simplifies to 100y = -20000x² + 800x, so y = -200x² + 8x.Therefore, the scaled model equation is ( y = -200x^2 + 8x ).So, that seems consistent. Therefore, my initial approach where I replaced x with x/100 and y with y/100 was incorrect because I was scaling the variables instead of scaling the graph.So, the correct equation is ( y = -200x^2 + 8x ).Wait, but let me verify this by considering the vertex. In the original equation, the vertex is at (2, 8). In the scaled model, the vertex should be at (2/100, 8/100) = (0.02, 0.08). Let's plug x = 0.02 into the scaled equation and see if y = 0.08.Compute y = -200*(0.02)^2 + 8*(0.02).First, (0.02)^2 = 0.0004. So, -200*0.0004 = -0.08.Then, 8*0.02 = 0.16.Adding them together: -0.08 + 0.16 = 0.08. Perfect, that matches. So, the vertex is correctly scaled down to (0.02, 0.08).Alternatively, if I had used the equation ( y = -frac{1}{50}x^2 + 8x ), let's see what the vertex would be.Using vertex formula: x = -b/(2a). Here, a = -1/50, b = 8.So, x = -8/(2*(-1/50)) = -8/(-2/50) = (-8)*(-50/2) = (-8)*(-25) = 200.Then, y = -1/50*(200)^2 + 8*(200).Compute that: (200)^2 = 40000. So, -1/50*40000 = -800. 8*200 = 1600. So, y = -800 + 1600 = 800.Wait, that's way off. The vertex in this case would be at (200, 800), which is way larger than the original. That can't be right because the model should be scaled down, not up. So, clearly, this equation is incorrect.Therefore, my initial substitution method was wrong because I misunderstood how scaling affects the equation. The correct approach is to scale the graph, which involves replacing x with 100x and y with 100y, leading to the equation ( y = -200x^2 + 8x ).So, to summarize:1. The maximum height is 8 units.2. The scaled model equation is ( y = -200x^2 + 8x ).I think that's solid now.**Final Answer**1. The maximum height of the rock formation is boxed{8}.2. The equation of the scaled model is boxed{y = -200x^2 + 8x}.