💡Okay, so I have this problem about a sequence where a₁ is 4, and each subsequent term is defined by aₙ₊₁ = 2aₙ - 2n + 1. There are two parts to the problem. The first part asks me to prove that the sequence {aₙ - 2n - 1} is a geometric sequence. The second part introduces another sequence bₙ = tn + 2 and wants me to find the range of t such that bₙ < 2aₙ for all natural numbers n.Starting with part (1). I need to show that {aₙ - 2n - 1} is a geometric sequence. A geometric sequence has the property that each term is a constant multiple of the previous term. So, if I can show that (aₙ₊₁ - 2(n+1) - 1) is a constant multiple of (aₙ - 2n - 1), then I can prove it's geometric.Let me write down the recursive formula again: aₙ₊₁ = 2aₙ - 2n + 1. Maybe I can manipulate this to express aₙ₊₁ - 2(n+1) - 1 in terms of aₙ - 2n - 1.So, let's compute aₙ₊₁ - 2(n+1) - 1:aₙ₊₁ - 2(n+1) - 1 = (2aₙ - 2n + 1) - 2n - 2 - 1Simplify the right-hand side:= 2aₙ - 2n + 1 - 2n - 2 - 1Combine like terms:= 2aₙ - 4n - 2Now, let's factor out a 2:= 2(aₙ - 2n - 1)So, we have aₙ₊₁ - 2(n+1) - 1 = 2(aₙ - 2n - 1). This shows that each term is twice the previous term, which is the definition of a geometric sequence with common ratio 2.Now, let's check the initial term to make sure it's consistent. For n=1, a₁ = 4. So, a₁ - 2(1) - 1 = 4 - 2 - 1 = 1. So, the first term of the new sequence is 1, and each subsequent term is multiplied by 2. Therefore, the sequence {aₙ - 2n - 1} is indeed a geometric sequence with first term 1 and common ratio 2.That takes care of part (1). Now, moving on to part (2). We have bₙ = tn + 2, and we need to find the range of t such that bₙ < 2aₙ for all n in natural numbers.First, let's express aₙ in terms of n. From part (1), we know that {aₙ - 2n - 1} is a geometric sequence with first term 1 and ratio 2. So, the general term of this geometric sequence is:aₙ - 2n - 1 = 1 * 2^{n-1}Therefore, solving for aₙ:aₙ = 2^{n-1} + 2n + 1So, 2aₙ would be:2aₙ = 2*(2^{n-1} + 2n + 1) = 2^{n} + 4n + 2Now, the inequality we need is:bₙ < 2aₙSubstituting bₙ and 2aₙ:tn + 2 < 2^{n} + 4n + 2Subtract 2 from both sides:tn < 2^{n} + 4nDivide both sides by n (assuming n > 0, which it is since n is a natural number):t < (2^{n} + 4n)/nSimplify the right-hand side:t < (2^{n}/n) + 4So, t must be less than (2^{n}/n) + 4 for all n in natural numbers. To find the range of t, we need to find the minimum value of the function f(n) = (2^{n}/n) + 4 over all natural numbers n. The smallest value that f(n) can take will be the upper bound for t.Let me define f(n) = (2^{n}/n) + 4. I need to find the minimum of f(n) for n = 1, 2, 3, ...Let's compute f(n) for the first few natural numbers:For n=1:f(1) = (2^1)/1 + 4 = 2 + 4 = 6For n=2:f(2) = (2^2)/2 + 4 = 4/2 + 4 = 2 + 4 = 6For n=3:f(3) = (2^3)/3 + 4 = 8/3 + 4 ≈ 2.666 + 4 ≈ 6.666For n=4:f(4) = (2^4)/4 + 4 = 16/4 + 4 = 4 + 4 = 8For n=5:f(5) = (2^5)/5 + 4 = 32/5 + 4 = 6.4 + 4 = 10.4For n=6:f(6) = (2^6)/6 + 4 = 64/6 + 4 ≈ 10.666 + 4 ≈ 14.666So, from n=1 onwards, f(n) starts at 6, remains 6 at n=2, then increases to approximately 6.666 at n=3, and continues to increase as n increases. So, the minimum value of f(n) is 6, achieved at both n=1 and n=2.Therefore, for the inequality t < f(n) to hold for all natural numbers n, t must be less than the minimum value of f(n), which is 6.Hence, the range of t is all real numbers less than 6.But let me double-check to make sure I haven't missed anything. Is f(n) always increasing for n ≥ 1? Let's see.Compute f(n+1) - f(n):f(n+1) - f(n) = [2^{n+1}/(n+1) + 4] - [2^{n}/n + 4] = 2^{n+1}/(n+1) - 2^{n}/nFactor out 2^{n}:= 2^{n} [2/(n+1) - 1/n] = 2^{n} [ (2n - (n+1)) / (n(n+1)) ) ] = 2^{n} [ (n - 1) / (n(n+1)) ) ]So, f(n+1) - f(n) = 2^{n} * (n - 1)/(n(n+1))Now, for n=1:f(2) - f(1) = 2^{1}*(1 - 1)/(1*2) = 0, which is consistent with our earlier calculation where f(2) = f(1) = 6.For n=2:f(3) - f(2) = 2^{2}*(2 - 1)/(2*3) = 4*(1)/6 ≈ 0.666 > 0, so f(3) > f(2)For n=3:f(4) - f(3) = 2^{3}*(3 - 1)/(3*4) = 8*(2)/12 ≈ 1.333 > 0, so f(4) > f(3)Similarly, for n ≥ 2, (n - 1) is positive, and since 2^{n} is always positive, the difference f(n+1) - f(n) is positive. Therefore, f(n) is increasing for n ≥ 2.Thus, the minimum value of f(n) occurs at n=1 and n=2, both giving f(n)=6.Therefore, t must be less than 6 to satisfy the inequality for all n.I think that's thorough enough. I considered the function f(n) and its behavior, checked the initial terms, and even derived the difference f(n+1) - f(n) to confirm the increasing nature for n ≥ 2. So, I'm confident that the range of t is all real numbers less than 6.**Final Answer**The range of values for ( t ) is boxed{(-infty, 6)}.

💡First, I'll define the total capacity of the pool as ( x ) cubic meters.The pump removes ( frac{2}{3} ) of the pool's water in 7.5 minutes. Converting 0.15 hours to minutes gives 9 minutes.Next, I'll calculate the pump's rate of removing water. Since it removes ( frac{2}{3}x ) in 7.5 minutes, the rate is:[frac{frac{2}{3}x}{7.5} = frac{2x}{22.5} = frac{4}{15}x text{ m}^3 text{ per minute}]In 9 minutes, the pump removes:[9 times frac{4}{15}x = frac{36}{15}x = 0.8x]The remaining water in the pool is:[x - 0.8x = 0.2x]Given that the remaining water is 25 cubic meters:[0.2x = 25]Solving for ( x ):[x = frac{25}{0.2} = 125]Therefore, the capacity of the pool is 125 cubic meters.

💡Okay, so I have this problem with four propositions, and I need to figure out how many of them have both the original statement and its converse true. The options are A:1, B:2, C:3, D:4. Let me go through each proposition one by one.Starting with proposition 1: "If a > b, then a + b > 0." Hmm, okay. So, if a is greater than b, does that necessarily mean their sum is positive? Let me think of some examples. Suppose a is 3 and b is 2. Then a > b is true, and a + b is 5, which is greater than 0. That works. But what if a and b are negative numbers? Let's say a is -1 and b is -2. Here, a > b because -1 is greater than -2, but a + b is -3, which is less than 0. So in this case, the original statement doesn't hold. Therefore, proposition 1 is not always true. So, the original statement is sometimes false, which means it's not a valid proposition. So, I can eliminate proposition 1.Moving on to proposition 2: "If a ≠ b, then a² ≠ b²." Hmm, okay. So, if two numbers are not equal, does that mean their squares are not equal? Let me think. If a is 2 and b is -2, then a ≠ b because 2 is not equal to -2. But a² is 4 and b² is also 4, so a² equals b². So, in this case, the original statement is false because even though a ≠ b, their squares are equal. So, proposition 2 is not always true either. So, I can eliminate proposition 2 as well.Now, proposition 3: "The points on the angle bisector are equidistant from the two sides of the angle." Okay, so this is a geometry statement. I remember that the angle bisector theorem states that any point on the angle bisector is equidistant from both sides of the angle. So, the original statement seems true. Now, what about its converse? The converse would be: "If a point is equidistant from the two sides of an angle, then it lies on the angle bisector." Is that also true? I think so. Because if a point is equidistant from both sides, it must lie on the bisector. So, both the original statement and its converse are true. So, proposition 3 is good.Finally, proposition 4: "The diagonals of a parallelogram bisect each other." Okay, I remember that in a parallelogram, the diagonals do bisect each other. So, the original statement is true. Now, what about the converse? The converse would be: "If the diagonals of a quadrilateral bisect each other, then it is a parallelogram." Is that true? I think yes. Because if the diagonals bisect each other, then the quadrilateral must be a parallelogram. So, both the original statement and its converse hold. So, proposition 4 is also good.So, summarizing: propositions 3 and 4 have both the original statement and their converse true. Propositions 1 and 2 do not. Therefore, the number of such propositions is 2.**Final Answer**boxed{B}

💡Okay, so I have this function f(x) = log base 2 of (x squared plus a). And I know that when x is 3, f(3) equals 1. I need to find the value of a. Hmm, let me think about how to approach this.First, I remember that a logarithm equation like log_b(y) = x means that b raised to the power of x equals y. So, in this case, log base 2 of something equals 1. That means 2 to the power of 1 should equal that something. So, if I plug in x = 3 into the function, I get log base 2 of (3 squared plus a) equals 1. Let me write that down: log₂(3² + a) = 1. Now, applying the logarithm rule I just remembered, this should mean that 2^1 equals 3² + a. Let me calculate that: 2^1 is 2, and 3 squared is 9. So, 2 equals 9 plus a. Wait, that seems a bit off because 9 is bigger than 2. If I subtract 9 from both sides, I get 2 minus 9 equals a. That would be -7 equals a. So, a is -7. Let me double-check that. If a is -7, then the function becomes log base 2 of (x² - 7). Plugging in x = 3, we get log₂(9 - 7) which is log₂(2). And log base 2 of 2 is indeed 1 because 2^1 = 2. So, that checks out. I think that makes sense. Even though a is negative, it works because when we plug in x = 3, the expression inside the logarithm becomes positive, which is valid because you can't take the logarithm of a negative number. So, everything seems to add up correctly.

💡Okay, so I need to find the distance between the points (3,7) and (3,-4). Hmm, let's see. I remember that to find the distance between two points in a coordinate plane, there's a formula. I think it's something like the square root of the sum of the squares of the differences in each coordinate. Maybe it's the distance formula? Yeah, that sounds right.Wait, let me recall the exact formula. I think it's distance equals the square root of [(x2 minus x1) squared plus (y2 minus y1) squared]. So, if I have two points, (x1, y1) and (x2, y2), I can plug those into the formula. Okay, so in this case, my first point is (3,7), so x1 is 3 and y1 is 7. The second point is (3,-4), so x2 is 3 and y2 is -4.Let me write that down to keep track:x1 = 3, y1 = 7x2 = 3, y2 = -4Now, plugging these into the distance formula:Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2]So, substituting the values:Distance = sqrt[(3 - 3)^2 + (-4 - 7)^2]Let me compute each part step by step. First, calculate (x2 - x1):3 - 3 = 0Okay, so that part is zero. Now, (y2 - y1):-4 - 7 = -11Hmm, so that's negative eleven. Now, I need to square both of these results.First, (0)^2 = 0Second, (-11)^2 = 121So now, I add these two squared differences together:0 + 121 = 121Then, take the square root of that sum:sqrt(121) = 11So, the distance between the two points is 11 units.Wait a minute, let me double-check that. Since both points have the same x-coordinate, they lie on a vertical line. That means the distance between them should just be the difference in their y-coordinates. Let's see:The y-coordinates are 7 and -4. The difference is 7 - (-4) = 7 + 4 = 11. Yep, that matches what I got using the distance formula. So, that makes sense.I guess when the x-coordinates are the same, the horizontal distance is zero, so the distance is just the vertical difference. Similarly, if the y-coordinates were the same, the distance would just be the horizontal difference.Alright, so I think I've got it right. The distance between (3,7) and (3,-4) is 11 units.

💡Alright, so I have this problem where I need to find the largest possible area of triangle ABC. The points A, B, and C are on the parabola y = x² - 3x. Specifically, A is (-2, 0), B is (5, 4), and C is (r, s) where s = r² - 3r, and r is between -2 and 5. First, I remember that the area of a triangle given three points can be found using the shoelace formula. The formula is:Area = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|So, plugging in the coordinates for A, B, and C:A = (-2, 0), so x₁ = -2, y₁ = 0B = (5, 4), so x₂ = 5, y₂ = 4C = (r, s), so x₃ = r, y₃ = s = r² - 3rSubstituting these into the shoelace formula:Area = (1/2) | (-2)(4 - s) + 5(s - 0) + r(0 - 4) |Simplify each term:First term: (-2)(4 - s) = -8 + 2sSecond term: 5(s - 0) = 5sThird term: r(0 - 4) = -4rSo, combining these:Area = (1/2) | (-8 + 2s) + 5s - 4r |Combine like terms:2s + 5s = 7sSo, Area = (1/2) | -8 + 7s - 4r |But since s = r² - 3r, substitute that in:Area = (1/2) | -8 + 7(r² - 3r) - 4r |Let me compute 7(r² - 3r):7r² - 21rSo, now the expression inside the absolute value becomes:-8 + 7r² - 21r - 4rCombine like terms again:-21r - 4r = -25rSo, Area = (1/2) |7r² -25r -8|Wait, but in the initial problem, the user had 7r² -29r -8. Hmm, maybe I made a mistake in substitution.Wait, let me double-check. The shoelace formula is:Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|So, plugging in:x₁(y₂ - y₃) = (-2)(4 - s)x₂(y₃ - y₁) = 5(s - 0) = 5sx₃(y₁ - y₂) = r(0 - 4) = -4rSo, combining:(-2)(4 - s) + 5s -4rWhich is:-8 + 2s + 5s -4rWhich is:-8 + 7s -4rThen, since s = r² - 3r, substitute:-8 + 7(r² - 3r) -4r= -8 +7r² -21r -4r= 7r² -25r -8Wait, so the user had 7r² -29r -8. So, perhaps I made a mistake in the calculation.Wait, let me check the shoelace formula again. Maybe I messed up the signs.Shoelace formula is:Area = (1/2)|x₁y₂ + x₂y₃ + x₃y₁ - x₂y₁ - x₃y₂ - x₁y₃|So, plugging in:x₁y₂ = (-2)(4) = -8x₂y₃ = 5sx₃y₁ = r*0 = 0Minus:x₂y₁ = 5*0 = 0x₃y₂ = r*4 = 4rx₁y₃ = (-2)sSo, putting it all together:Area = (1/2)| (-8 + 5s + 0) - (0 + 4r -2s) |Simplify inside the absolute value:First part: -8 +5sSecond part: - (4r -2s) = -4r +2sSo, combining:-8 +5s -4r +2s= -8 +7s -4rWhich is the same as before. So, substituting s = r² -3r:= -8 +7(r² -3r) -4r= -8 +7r² -21r -4r=7r² -25r -8So, the area is (1/2)|7r² -25r -8|But the user had 7r² -29r -8. So, perhaps the user made a mistake in their calculation.But regardless, I need to proceed with my calculation.So, the area is (1/2)|7r² -25r -8|Now, to maximize the area, we need to maximize |7r² -25r -8| over the interval -2 ≤ r ≤5.Since the quadratic inside the absolute value is 7r² -25r -8, which is a parabola opening upwards (since the coefficient of r² is positive). Therefore, it will have a minimum at its vertex, and the maximum will occur at one of the endpoints of the interval.So, first, let's find the vertex of the quadratic 7r² -25r -8.The vertex occurs at r = -b/(2a) = 25/(2*7) = 25/14 ≈ 1.7857So, the vertex is at r ≈1.7857, which is within the interval [-2,5]. Since the parabola opens upwards, the minimum value is at the vertex, and the maximum will be at one of the endpoints.Therefore, we need to evaluate 7r² -25r -8 at r = -2 and r =5, and see which one gives a larger absolute value.First, at r = -2:7*(-2)^2 -25*(-2) -8 = 7*4 +50 -8 =28 +50 -8=70So, 70.At r =5:7*(5)^2 -25*5 -8=7*25 -125 -8=175 -125 -8=42So, 42.So, at r=-2, the quadratic is 70, at r=5, it's 42.Since 70 >42, the maximum value of the quadratic is 70 at r=-2.But since we have the absolute value, we need to check if the quadratic is positive or negative in the interval.Since the parabola opens upwards, and the vertex is at r≈1.7857, which is a minimum, the quadratic will be positive at both endpoints.Therefore, the maximum of |7r² -25r -8| is 70 at r=-2.Therefore, the maximum area is (1/2)*70=35.Wait, but the user had 7r² -29r -8, which at r=-2 gave 78, leading to area 39.But in my calculation, I have 7r² -25r -8, which at r=-2 gives 70, leading to area 35.So, perhaps I made a mistake in the shoelace formula.Wait, let me double-check the shoelace formula again.Shoelace formula is:Area = (1/2)|x₁y₂ + x₂y₃ + x₃y₁ - x₂y₁ - x₃y₂ - x₁y₃|So, plugging in:x₁ = -2, y₁=0x₂=5, y₂=4x₃=r, y₃=s=r²-3rSo,x₁y₂ = (-2)(4) = -8x₂y₃ =5*s=5(r² -3r)x₃y₁ = r*0=0Minus:x₂y₁=5*0=0x₃y₂=r*4=4rx₁y₃=(-2)*s=(-2)(r² -3r)So, putting it all together:Area = (1/2)| (-8 +5(r² -3r) +0) - (0 +4r -2(r² -3r)) |Simplify inside the absolute value:First part: -8 +5r² -15rSecond part: - (4r -2r² +6r) = - (10r -2r²) = -10r +2r²So, combining:(-8 +5r² -15r) + (-10r +2r²) = -8 +5r² -15r -10r +2r² = -8 +7r² -25rSo, Area = (1/2)|7r² -25r -8|Yes, that's correct. So, the user had 7r² -29r -8, which is different.Wait, perhaps the user made a mistake in their calculation.But regardless, proceeding with my calculation, the maximum area is 35.But let me check again.Wait, perhaps I made a mistake in the signs when expanding the shoelace formula.Let me write it step by step.Compute:First part: x₁y₂ + x₂y₃ + x₃y₁ = (-2)(4) +5s + r*0 = -8 +5s +0 = -8 +5sSecond part: x₂y₁ + x₃y₂ + x₁y₃ =5*0 + r*4 + (-2)s =0 +4r -2sSo, subtracting the second part from the first part:(-8 +5s) - (4r -2s) = -8 +5s -4r +2s = -8 +7s -4rThen, since s = r² -3r, substitute:-8 +7(r² -3r) -4r = -8 +7r² -21r -4r =7r² -25r -8Yes, that's correct. So, the area is (1/2)|7r² -25r -8|So, the maximum occurs at r=-2, giving 70, so area is 35.But the user had 7r² -29r -8, which at r=-2 gave 78, leading to area 39.So, perhaps the user made a mistake in their calculation.Alternatively, perhaps I made a mistake in the shoelace formula.Wait, let me try another approach.Another formula for the area of a triangle given three points is:Area = (1/2)| (x₂ -x₁)(y₃ - y₁) - (x₃ -x₁)(y₂ - y₁) |Let me try this formula.So, points A(-2,0), B(5,4), C(r,s)Compute vectors AB and AC.Vector AB = (5 - (-2), 4 -0) = (7,4)Vector AC = (r - (-2), s -0) = (r+2, s)The area is (1/2)|AB × AC|, which is (1/2)|(7)(s) - (4)(r+2)|So, Area = (1/2)|7s -4r -8|Since s = r² -3r, substitute:Area = (1/2)|7(r² -3r) -4r -8| = (1/2)|7r² -21r -4r -8| = (1/2)|7r² -25r -8|Same result as before.So, the area is (1/2)|7r² -25r -8|So, the maximum occurs at r=-2, giving 70, so area is 35.But wait, let me check the value at r=-2:7*(-2)^2 -25*(-2) -8 =7*4 +50 -8=28+50-8=70Yes, so area is 70/2=35.At r=5:7*25 -25*5 -8=175 -125 -8=42So, area is 21.But wait, the user had 7r² -29r -8, which at r=-2 gave 78, leading to area 39.So, perhaps the user made a mistake in their calculation.Alternatively, perhaps I made a mistake in the shoelace formula.Wait, let me try to compute the area using another method, perhaps the determinant method.The area can also be computed as (1/2)| determinant of the matrix formed by the coordinates |.So, the determinant is:| (-2)(4 - s) +5(s -0) +r(0 -4) |Which is the same as before.So, I think my calculation is correct.Therefore, the maximum area is 35.But wait, let me check the value at the vertex.The vertex is at r=25/14≈1.7857Compute 7r² -25r -8 at r=25/14:7*(25/14)^2 -25*(25/14) -8First, compute (25/14)^2=625/196So, 7*(625/196)=4375/196≈22.3214Then, 25*(25/14)=625/14≈44.6429So, 4375/196 -625/14 -8Convert all to 196 denominator:4375/196 - (625/14)*(14/14)=625*14/196=8750/196Wait, no, 625/14 = (625*14)/196=8750/196Similarly, 8=8*196/196=1568/196So, 4375/196 -8750/196 -1568/196= (4375 -8750 -1568)/196= (4375 -10318)/196= (-5943)/196≈-30.32So, the value at the vertex is approximately -30.32, which is the minimum.Therefore, the maximum occurs at r=-2, giving 70, so area is 35.But wait, the user had 7r² -29r -8, which at r=-2 gave 78, leading to area 39.So, perhaps the user made a mistake in their calculation.Alternatively, perhaps I made a mistake in the shoelace formula.Wait, let me check the user's calculation.The user wrote:Area = (1/2)| (-2)(4) +5(r² -3r) +r(0) -0*5 -4*r - (r² -3r)(-2) |= (1/2)| -8 +5r² -15r +0 -0 -4r +2r² -6r |Wait, that seems different from my calculation.Wait, let's see:The user expanded:(-2)(4) = -85(r² -3r)=5r² -15rr(0)=0Minus:0*5=04*r=4r(r² -3r)(-2)= -2r² +6rSo, putting it all together:-8 +5r² -15r +0 -0 -4r +2r² -6rCombine like terms:5r² +2r²=7r²-15r -4r -6r= -25r-8So, 7r² -25r -8Which is the same as my result.But the user wrote:= (1/2)|7r² -29r -8|So, perhaps the user made a mistake in their calculation.Wait, in their step-by-step, they wrote:= (1/2)|7r² -29r -8|But according to the expansion, it should be 7r² -25r -8.So, the user incorrectly wrote -29r instead of -25r.Therefore, their calculation is wrong.So, the correct expression is 7r² -25r -8.Therefore, the maximum area is 35.But wait, let me check again.Wait, perhaps I made a mistake in the signs.Wait, in the user's calculation, they had:-8 +5r² -15r +0 -0 -4r +2r² -6rWhich is:5r² +2r²=7r²-15r -4r -6r= -25r-8So, 7r² -25r -8Yes, correct.But the user wrote 7r² -29r -8, which is incorrect.Therefore, the user's final result of 39 is wrong.So, the correct maximum area is 35.But wait, let me check the value at r=-2:7*(-2)^2 -25*(-2) -8=28 +50 -8=70So, area is 70/2=35.At r=5:7*25 -25*5 -8=175 -125 -8=42Area=21.So, the maximum area is 35.But wait, perhaps I made a mistake in the shoelace formula.Wait, let me compute the area using another method.Compute the base AB, then find the height from point C.First, compute the length of AB.Points A(-2,0) and B(5,4).Distance AB= sqrt[(5 - (-2))² + (4 -0)²]=sqrt[7² +4²]=sqrt[49+16]=sqrt[65]Then, the area is (1/2)*base*height.To find the height, we need the distance from point C(r,s) to the line AB.First, find the equation of line AB.Slope of AB: m=(4 -0)/(5 - (-2))=4/7So, equation: y -0 = (4/7)(x +2)So, y= (4/7)x +8/7The distance from point C(r,s) to this line is:| (4/7)r - s +8/7 | / sqrt[(4/7)^2 +1^2]Simplify denominator:sqrt[16/49 +1]=sqrt[65/49]=sqrt(65)/7So, distance= | (4r/7 -s +8/7) | / (sqrt(65)/7 )= |4r -7s +8| / sqrt(65)Therefore, area= (1/2)*sqrt(65)*( |4r -7s +8| / sqrt(65)) )= (1/2)|4r -7s +8|But s=r² -3r, so substitute:= (1/2)|4r -7(r² -3r) +8|= (1/2)|4r -7r² +21r +8|= (1/2)| -7r² +25r +8 |Which is the same as (1/2)|7r² -25r -8|, since absolute value makes it positive.So, same result as before.Therefore, the area is (1/2)|7r² -25r -8|Thus, the maximum area is 35.But wait, the user had 7r² -29r -8, leading to area 39.So, perhaps the user made a mistake in their calculation.Therefore, the correct maximum area is 35.But wait, let me check the value at r=-2:7*(-2)^2 -25*(-2) -8=28 +50 -8=70So, area=70/2=35.At r=5:7*25 -25*5 -8=175 -125 -8=42Area=21.So, the maximum area is 35.Therefore, the largest possible area of triangle ABC is 35.