💡First, I'll assign variables to each batsman's score: A for batsman A, B for batsman B, and so on.Since the average score is 36 runs across 5 batsmen, the total runs scored by all batsmen combined is 5 multiplied by 36, which equals 180 runs. This gives me the first equation:A + B + C + D + E = 180Next, I'll translate the given relationships into mathematical equations. D scored 5 more runs than E, so:D = E + 5E scored 8 fewer runs than A, which means:E = A - 8B scored as many runs as D and E combined, so:B = D + EAdditionally, B and C together scored 107 runs:B + C = 107Now, I'll substitute the expressions for D and E into the equation for B:B = (E + 5) + E = 2E + 5Using the equation for B and C, I can express C in terms of E:C = 107 - B = 107 - (2E + 5) = 102 - 2ENext, I'll substitute A, D, B, and C in terms of E into the total runs equation:A + B + C + D + E = (E + 8) + (2E + 5) + (102 - 2E) + (E + 5) + E = 180Combining like terms:5E + 120 = 180Solving for E:5E = 60E = 12Therefore, batsman E scored 12 runs.

💡Okay, so I have this problem where I need to find the locus of points that are the vertices of a family of parabolas given by the equation ( y = a x^2 + z x + c ). Here, ( a ) and ( c ) are fixed positive numbers, and ( z ) can be any real number. Hmm, okay, so I need to figure out what shape these vertices form as ( z ) changes.First, I remember that for a quadratic equation in the form ( y = ax^2 + bx + c ), the vertex can be found using the formula ( x = -frac{b}{2a} ). In this case, our equation is ( y = a x^2 + z x + c ), so comparing it to the standard form, ( b ) is actually ( z ). So, the x-coordinate of the vertex should be ( x_z = -frac{z}{2a} ). That makes sense.Now, to find the corresponding y-coordinate of the vertex, I need to plug this x-value back into the original equation. Let me do that step by step. So, substituting ( x_z = -frac{z}{2a} ) into ( y = a x^2 + z x + c ), we get:( y_z = a left( -frac{z}{2a} right)^2 + z left( -frac{z}{2a} right) + c ).Let me compute each term separately. The first term is ( a left( frac{z^2}{4a^2} right) ), which simplifies to ( frac{z^2}{4a} ). The second term is ( -frac{z^2}{2a} ). Adding these together with the constant term ( c ), we have:( y_z = frac{z^2}{4a} - frac{z^2}{2a} + c ).Combining the like terms ( frac{z^2}{4a} ) and ( -frac{z^2}{2a} ), I get:( y_z = left( frac{1}{4a} - frac{2}{4a} right) z^2 + c = -frac{z^2}{4a} + c ).Hmm, so ( y_z = -frac{z^2}{4a} + c ). But I also know that ( x_z = -frac{z}{2a} ). Maybe I can express ( z ) in terms of ( x_z ) and substitute it back into the equation for ( y_z ).From ( x_z = -frac{z}{2a} ), solving for ( z ) gives ( z = -2a x_z ). Plugging this into ( y_z = -frac{z^2}{4a} + c ), we get:( y_z = -frac{(-2a x_z)^2}{4a} + c = -frac{4a^2 x_z^2}{4a} + c = -a x_z^2 + c ).So, putting it all together, the relationship between ( x_z ) and ( y_z ) is ( y_z = -a x_z^2 + c ). That looks like another parabola, but this time it's opening downward because the coefficient of ( x_z^2 ) is negative.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the vertex formula, I found ( x_z = -frac{z}{2a} ). Then, substituting back into the original equation, I correctly expanded each term and combined them. Then, expressing ( z ) in terms of ( x_z ) and substituting back, I arrived at ( y_z = -a x_z^2 + c ). Yeah, that seems right.So, the locus of the vertices is another parabola given by ( y = -a x^2 + c ). This makes sense because as ( z ) varies, the vertex moves along this downward-opening parabola. The original parabolas all have the same coefficient ( a ) for ( x^2 ), so their vertices form a parabola with the same coefficient but in the opposite direction, hence the negative sign.I think I'm confident with this result. The key was recognizing that the vertex coordinates depend on ( z ), and by expressing ( z ) in terms of ( x_z ), I could eliminate ( z ) and find the relationship purely in terms of ( x_z ) and ( y_z ).**Final Answer**The locus of the vertices is the parabola given by boxed{y = -a x^2 + c}.

💡Okay, so I have this equation here: (5/4)(p + q + r) = 15. And I need to find the average of p, q, and r. Hmm, let's see. First, I remember that the average of three numbers is just the sum of those numbers divided by three. So, if I can find p + q + r, then I can easily find the average by dividing that sum by 3.But right now, the equation has this fraction multiplied by the sum. Maybe I should try to get rid of that fraction to make things simpler. The equation is (5/4)(p + q + r) = 15. So, to isolate p + q + r, I need to do the opposite of multiplying by 5/4, which is multiplying by 4/5. That way, I can cancel out the 5/4 on the left side.Let me write that down: if I multiply both sides of the equation by 4/5, I get (4/5)*(5/4)(p + q + r) = (4/5)*15. On the left side, the 4/5 and 5/4 should cancel each other out, leaving just p + q + r. On the right side, I need to calculate (4/5)*15. Let's see, 15 divided by 5 is 3, and 3 times 4 is 12. So, p + q + r equals 12.Now that I have p + q + r = 12, finding the average is straightforward. The average is (p + q + r)/3, which is 12 divided by 3. 12 divided by 3 is 4. So, the average of p, q, and r is 4.Wait, let me double-check to make sure I didn't make any mistakes. Starting with (5/4)(p + q + r) = 15, multiplying both sides by 4/5 gives p + q + r = 12. Then dividing by 3 gives the average as 4. Yeah, that seems right. I don't think I missed anything here.

💡Alright, so I have this problem here: "Given 40 feet of fencing, what is the greatest possible number of square feet in the area of a rectangular pen enclosed by the fencing?" Hmm, okay. Let me try to figure this out step by step.First, I know that when dealing with fencing and areas, we're usually dealing with perimeters and areas of shapes, probably rectangles in this case since it's specified. So, the fencing is the perimeter, and we need to find the maximum area that can be enclosed with that perimeter.Okay, so the perimeter of a rectangle is calculated by the formula P = 2*(length + width). In this case, the perimeter is 40 feet. So, I can write that as:2*(length + width) = 40If I divide both sides by 2 to simplify, I get:length + width = 20So, the sum of the length and the width of the rectangle must be 20 feet. Now, I need to find the dimensions that will give me the maximum area. The area of a rectangle is given by A = length * width. So, I need to express the area in terms of one variable and then find its maximum.Let me denote the length as 'x'. Then, the width would be 20 - x, since their sum is 20. So, the area A can be expressed as:A = x*(20 - x)Simplifying that, I get:A = 20x - x^2Now, this is a quadratic equation in terms of x, and it's a downward-opening parabola because the coefficient of x^2 is negative. The maximum value of this quadratic equation occurs at the vertex. The vertex of a parabola given by A = ax^2 + bx + c is at x = -b/(2a).In my equation, a = -1 and b = 20. Plugging these into the formula:x = -20/(2*(-1)) = -20/(-2) = 10So, the length that gives the maximum area is 10 feet. Since the width is 20 - x, that would also be 10 feet. Wait, so both length and width are 10 feet? That means the rectangle is actually a square.Okay, so if both sides are 10 feet, the area would be:A = 10*10 = 100 square feetIs that the maximum area? Let me double-check. If I choose different lengths and widths that add up to 20, will the area be less than 100?For example, if length is 15 feet, then width is 5 feet. Area = 15*5 = 75 square feet, which is less than 100.If length is 12 feet, width is 8 feet. Area = 12*8 = 96 square feet, still less than 100.If length is 9 feet, width is 11 feet. Area = 9*11 = 99 square feet, almost 100 but still less.So, it seems like when the rectangle is a square, the area is maximized.But wait, is there a mathematical way to confirm that the vertex gives the maximum area? I think so. Since the coefficient of x^2 is negative, the parabola opens downward, meaning the vertex is indeed the maximum point.Alternatively, I could use calculus to find the maximum. If I take the derivative of the area function with respect to x and set it equal to zero, I should find the critical point which corresponds to the maximum.So, A = 20x - x^2dA/dx = 20 - 2xSetting dA/dx = 0:20 - 2x = 02x = 20x = 10So, again, x = 10 feet, which confirms the earlier result.Therefore, the maximum area is achieved when the rectangle is a square with sides of 10 feet, giving an area of 100 square feet.I think that makes sense. It's interesting that the maximum area is achieved when the shape is a square. I guess symmetry plays a role here. If you have a fixed perimeter, the shape that maximizes the area is a square.But just to be thorough, let me consider another approach. Maybe using the arithmetic mean-geometric mean inequality (AM-GM inequality). It states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean.In this case, the two numbers are length and width, which are positive. So,(length + width)/2 >= sqrt(length*width)We know that length + width = 20, so:20/2 >= sqrt(A)10 >= sqrt(A)Squaring both sides:100 >= ASo, the maximum area A is 100 square feet, achieved when length = width = 10 feet.That's another way to confirm the result. It seems consistent across different methods.I think I've covered multiple ways to approach this problem: algebraic method by completing the square, calculus by taking derivatives, and the AM-GM inequality. All lead to the same conclusion that the maximum area is 100 square feet when the pen is a square with sides of 10 feet.Just to visualize it, if I imagine different rectangles with a perimeter of 40 feet, the square seems to be the most "balanced" shape, not too stretched out in one direction, which probably allows it to enclose the most space.In contrast, if the rectangle becomes more elongated, like a very long and narrow shape, the area decreases because one side becomes much larger while the other becomes much smaller, and the product of the two doesn't compensate for the loss in the other dimension.So, in summary, with 40 feet of fencing, the largest area you can enclose with a rectangular pen is 100 square feet, and this happens when the pen is a square with each side measuring 10 feet.

💡Okay, so I have this problem here about transforming the graph of a cosine function and then finding one of its symmetry centers. Let me try to work through it step by step.First, the original function is f(x) = 2cos(x - π/3) - 1. I know that cosine functions have certain transformations: amplitude, period, phase shift, and vertical shift. In this case, the amplitude is 2, which means the graph will go up to 2 and down to -2 from the midline. The phase shift is π/3 to the right, and the vertical shift is -1, so the midline of the cosine wave is at y = -1.Now, the problem says to translate the graph to the right by π/3 units. Translating a graph to the right by a certain amount affects the phase shift. If I have a function f(x - h), it shifts the graph h units to the right. So, if I already have a phase shift of π/3, shifting it another π/3 to the right would make the new phase shift π/3 + π/3 = 2π/3. So, after the translation, the function becomes f(x) = 2cos(x - 2π/3) - 1.Next, the problem says to shrink all the x-coordinates to half of their original values. Shrinking the x-coordinates by a factor of 1/2 is the same as horizontally compressing the graph by a factor of 2. In terms of the function, this means replacing x with 2x. So, applying this transformation to the translated function, we get g(x) = 2cos(2x - 2π/3) - 1.Let me double-check that. Starting with f(x) = 2cos(x - π/3) - 1, translating right by π/3 gives 2cos(x - 2π/3) - 1. Then, shrinking x-coordinates by half (which is a horizontal compression by factor 2) would replace x with 2x, so 2cos(2x - 2π/3) - 1. Yeah, that seems right.Now, the question is about finding one of the symmetry centers of the graph y = g(x). A symmetry center is a point about which the graph is symmetric. For cosine functions, which are even functions, the graph is symmetric about the vertical line through their maxima and minima. However, after transformations, especially horizontal shifts and compressions, the symmetry centers might change.But wait, symmetry centers are points, not lines. So, for a function that's a cosine wave, which is symmetric about certain vertical lines, but if we're talking about points, maybe it's referring to points of symmetry, like centers of rotational symmetry or something else.Alternatively, perhaps it's referring to points where the function has symmetry about that point, meaning that for any point (a, b) on the graph, the point (2h - a, 2k - b) is also on the graph, where (h, k) is the center of symmetry.Given that, let's see. For a cosine function, which is periodic and symmetric, it's possible that it has multiple centers of symmetry.First, let's analyze the transformed function g(x) = 2cos(2x - 2π/3) - 1. Let's rewrite this as g(x) = 2cos[2(x - π/3)] - 1. So, this is a cosine function with amplitude 2, period π (since the coefficient of x is 2, so period is 2π / 2 = π), phase shift π/3 to the right, and vertical shift -1.So, the midline is y = -1, the amplitude is 2, so it goes from y = -3 to y = 1.Now, the function is periodic with period π, so it repeats every π units. The phase shift is π/3, so the starting point of the cosine wave is shifted to x = π/3.Now, to find a symmetry center, we need to find a point (h, k) such that for any x, g(2h - x) = 2k - g(x). That is, the function is symmetric about the point (h, k).Alternatively, since the function is a cosine function, which is symmetric about its midline, but shifted. The midline is y = -1, so perhaps the vertical line through certain points is a line of symmetry, but the question is about a point.Wait, maybe it's referring to the center of symmetry for the entire graph, which could be the point where the function crosses its midline, or perhaps the midpoint between a maximum and a minimum.Alternatively, considering that the function is a cosine function, which is symmetric about its peaks and troughs, but those are points on the graph, not centers of symmetry.Wait, perhaps the function has a point of symmetry at the midpoint between a maximum and a minimum. Let's think.Given that the function is g(x) = 2cos(2x - 2π/3) - 1, let's find its critical points.First, let's find the maxima and minima.The maxima occur where the cosine function is 1, so:2cos(2x - 2π/3) - 1 = 2*1 - 1 = 1.The minima occur where the cosine function is -1, so:2cos(2x - 2π/3) - 1 = 2*(-1) - 1 = -3.So, the maximum value is 1, and the minimum is -3.The midline is y = -1, so the distance from the midline to the maximum is 2, and to the minimum is also 2, which makes sense with the amplitude.Now, the function is periodic with period π, so it completes a full cycle every π units.Let's find the x-values where the function reaches its maximum and minimum.For maximums:2cos(2x - 2π/3) - 1 = 1So, 2cos(2x - 2π/3) = 2cos(2x - 2π/3) = 1So, 2x - 2π/3 = 2πk, where k is integer.Thus, 2x = 2πk + 2π/3x = πk + π/3Similarly, for minimums:2cos(2x - 2π/3) - 1 = -32cos(2x - 2π/3) = -2cos(2x - 2π/3) = -1So, 2x - 2π/3 = π + 2πk2x = π + 2πk + 2π/32x = (3π + 6πk + 2π)/32x = (5π + 6πk)/3x = (5π + 6πk)/6x = (5π/6) + πkSo, the maxima are at x = πk + π/3, and minima at x = (5π/6) + πk.Now, let's consider the midpoints between a maximum and a minimum.Take k=0:Maximum at x = π/3, y=1Minimum at x = 5π/6, y=-3Midpoint between these two points is:x: (π/3 + 5π/6)/2 = (2π/6 + 5π/6)/2 = (7π/6)/2 = 7π/12y: (1 + (-3))/2 = (-2)/2 = -1So, the midpoint is (7π/12, -1)Similarly, for k=1:Maximum at x = π + π/3 = 4π/3, y=1Minimum at x = 5π/6 + π = 11π/6, y=-3Midpoint:x: (4π/3 + 11π/6)/2 = (8π/6 + 11π/6)/2 = (19π/6)/2 = 19π/12y: same as before, -1So, another midpoint is (19π/12, -1)Similarly, for k=-1:Maximum at x = -π + π/3 = -2π/3, y=1Minimum at x = 5π/6 - π = -π/6, y=-3Midpoint:x: (-2π/3 + (-π/6))/2 = (-4π/6 - π/6)/2 = (-5π/6)/2 = -5π/12y: -1So, midpoint is (-5π/12, -1)So, these midpoints are potential centers of symmetry.Looking at the options given:A: (π/6, 0)B: (π/12, 0)C: (π/6, -1)D: (π/12, -1)None of these are exactly the midpoints I found, which are at (7π/12, -1), (19π/12, -1), (-5π/12, -1), etc.Wait, but 7π/12 is approximately 1.83, and π/12 is approximately 0.26, which is much smaller. So, perhaps I'm missing something.Wait, maybe I need to consider the function's symmetry differently. Let's think about the function g(x) = 2cos(2x - 2π/3) - 1.We can write this as g(x) = 2cos[2(x - π/3)] - 1.So, it's a cosine function with amplitude 2, period π, phase shift π/3, and vertical shift -1.Now, the standard cosine function y = cos(x) has symmetry about the origin if it's shifted appropriately, but in this case, it's a cosine function, which is even, so it's symmetric about the y-axis if not shifted. But here, it's shifted and compressed.Wait, but the question is about a point of symmetry, not a line.Perhaps the function has a point of symmetry at the midpoint of its period.The period is π, so the midpoint between x = 0 and x = π is x = π/2. But let's check if (π/2, -1) is a center of symmetry.Wait, but in our midpoints earlier, we had (7π/12, -1), which is approximately 1.83, which is more than π/2 (which is about 1.57). So, maybe not.Alternatively, perhaps the function has a center of symmetry at the point where it crosses the midline. Let's find where g(x) = -1.So, 2cos(2x - 2π/3) - 1 = -12cos(2x - 2π/3) = 0cos(2x - 2π/3) = 0So, 2x - 2π/3 = π/2 + πk2x = π/2 + 2π/3 + πk2x = (3π/6 + 4π/6) + πk = 7π/6 + πkx = 7π/12 + (π/2)kSo, the function crosses the midline y = -1 at x = 7π/12 + (π/2)k.So, these are points where the function crosses the midline. These could be potential centers of symmetry.Wait, but earlier, when I found midpoints between maxima and minima, I got (7π/12, -1), which is one of these points. So, perhaps each of these points is a center of symmetry.But the options given are (π/6, 0), (π/12, 0), (π/6, -1), (π/12, -1).So, none of these are 7π/12, but perhaps I need to consider the function's symmetry in a different way.Wait, maybe I made a mistake in the transformation. Let me double-check.Original function: f(x) = 2cos(x - π/3) - 1Translate right by π/3: replace x with x - π/3, so f(x - π/3) = 2cos((x - π/3) - π/3) - 1 = 2cos(x - 2π/3) - 1.Then, shrink x-coordinates to half: replace x with 2x, so g(x) = 2cos(2x - 2π/3) - 1.Yes, that seems correct.Alternatively, maybe the problem is referring to a point of symmetry in terms of rotational symmetry, meaning that if you rotate the graph 180 degrees around that point, it maps onto itself.For a cosine function, which is symmetric about its midline, but after transformations, it might have points of rotational symmetry.Alternatively, perhaps the function is symmetric about certain points because of its periodicity.Wait, let's consider the function g(x) = 2cos(2x - 2π/3) - 1.Let me consider if it's symmetric about a point (h, k). For any point (x, y) on the graph, the point (2h - x, 2k - y) should also be on the graph.So, let's suppose that (h, k) is a center of symmetry. Then, for any x, g(2h - x) = 2k - g(x).Let's write that out:2cos[2(2h - x) - 2π/3] - 1 = 2k - [2cos(2x - 2π/3) - 1]Simplify the left side:2cos(4h - 2x - 2π/3) - 1Right side:2k - 2cos(2x - 2π/3) + 1So, set them equal:2cos(4h - 2x - 2π/3) - 1 = 2k - 2cos(2x - 2π/3) + 1Let's rearrange:2cos(4h - 2x - 2π/3) + 2cos(2x - 2π/3) = 2k + 2Divide both sides by 2:cos(4h - 2x - 2π/3) + cos(2x - 2π/3) = k + 1Now, for this equation to hold for all x, the left side must be a constant, independent of x. That is, the sum of these two cosine terms must be constant.Recall that cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]Let me apply this identity:Let A = 4h - 2x - 2π/3Let B = 2x - 2π/3Then,cos A + cos B = 2 cos[(A + B)/2] cos[(A - B)/2]Compute (A + B)/2:(4h - 2x - 2π/3 + 2x - 2π/3)/2 = (4h - 4π/3)/2 = 2h - 2π/3Compute (A - B)/2:(4h - 2x - 2π/3 - (2x - 2π/3))/2 = (4h - 4x)/2 = 2h - 2xSo,cos A + cos B = 2 cos(2h - 2π/3) cos(2h - 2x)Therefore, the equation becomes:2 cos(2h - 2π/3) cos(2h - 2x) = k + 1For this to be true for all x, the coefficient of cos(2h - 2x) must be zero, because otherwise, the left side would vary with x, while the right side is constant.Therefore, 2 cos(2h - 2π/3) = 0Which implies cos(2h - 2π/3) = 0So,2h - 2π/3 = π/2 + πk, where k is integerSolving for h:2h = π/2 + 2π/3 + πkConvert to common denominator:π/2 = 3π/62π/3 = 4π/6So,2h = 3π/6 + 4π/6 + πk = 7π/6 + πkThus,h = 7π/12 + (π/2)kSo, h is of the form 7π/12 + πk/2Now, going back to the equation:2 cos(2h - 2π/3) cos(2h - 2x) = k + 1But we already set 2 cos(2h - 2π/3) = 0, so the left side is 0, which implies k + 1 = 0Therefore, k = -1So, the centers of symmetry are at (h, k) = (7π/12 + πk/2, -1), where k is integer.So, for k=0: (7π/12, -1)For k=1: (7π/12 + π/2, -1) = (7π/12 + 6π/12, -1) = (13π/12, -1)For k=-1: (7π/12 - π/2, -1) = (7π/12 - 6π/12, -1) = (π/12, -1)Ah, so when k=-1, h = π/12, so the center is (π/12, -1), which is option D.Therefore, one of the symmetry centers is (π/12, -1), which is option D.I think that makes sense. So, the answer should be D.

💡Okay, so I have this problem where a polynomial p(x) leaves a remainder of -2 when divided by x - 1, a remainder of 4 when divided by x - 2, and a remainder of 5 when divided by x + 3. I need to find r(7), where r(x) is the remainder when p(x) is divided by (x - 1)(x - 2)(x + 3). Hmm, let me try to figure this out step by step.First, I remember that when a polynomial is divided by a linear divisor like x - a, the remainder is just a constant. But when it's divided by a quadratic or higher degree polynomial, the remainder will be of a lower degree. In this case, since we're dividing by a cubic polynomial (because (x - 1)(x - 2)(x + 3) is a cubic), the remainder r(x) should be a quadratic polynomial, right? So, I can write r(x) as ax² + bx + c, where a, b, and c are constants that I need to find.So, p(x) can be expressed as:p(x) = (x - 1)(x - 2)(x + 3)q(x) + r(x)where q(x) is the quotient polynomial, and r(x) is the remainder, which is ax² + bx + c.Now, since we know the remainders when p(x) is divided by x - 1, x - 2, and x + 3, we can use those to set up equations for a, b, and c. Specifically, by the Remainder Theorem, we know that:p(1) = -2,p(2) = 4,p(-3) = 5.Substituting x = 1 into the expression for p(x), the term with q(x) becomes zero because (1 - 1) is zero, so we have:p(1) = a(1)² + b(1) + c = a + b + c = -2.Similarly, substituting x = 2:p(2) = a(2)² + b(2) + c = 4a + 2b + c = 4.And substituting x = -3:p(-3) = a(-3)² + b(-3) + c = 9a - 3b + c = 5.So now I have a system of three equations:1) a + b + c = -2,2) 4a + 2b + c = 4,3) 9a - 3b + c = 5.I need to solve this system to find the values of a, b, and c. Let me write them down again for clarity:1) a + b + c = -2,2) 4a + 2b + c = 4,3) 9a - 3b + c = 5.Hmm, maybe I can subtract equation 1 from equation 2 to eliminate c. Let's try that:Equation 2 - Equation 1:(4a + 2b + c) - (a + b + c) = 4 - (-2),Which simplifies to:3a + b = 6.Similarly, subtract equation 2 from equation 3 to eliminate c again:Equation 3 - Equation 2:(9a - 3b + c) - (4a + 2b + c) = 5 - 4,Which simplifies to:5a - 5b = 1.So now I have two equations:4) 3a + b = 6,5) 5a - 5b = 1.Let me see if I can solve these two equations for a and b. Maybe I can solve equation 4 for b and substitute into equation 5.From equation 4:b = 6 - 3a.Substitute this into equation 5:5a - 5(6 - 3a) = 1,5a - 30 + 15a = 1,20a - 30 = 1,20a = 31,a = 31/20.Okay, so a is 31/20. Now, substitute a back into equation 4 to find b:3*(31/20) + b = 6,93/20 + b = 6,b = 6 - 93/20,Convert 6 to 120/20,b = 120/20 - 93/20 = 27/20.Wait, that doesn't seem right. Let me check my calculations again. From equation 4:3a + b = 6,a = 31/20,So 3*(31/20) = 93/20,So 93/20 + b = 6,Convert 6 to 120/20,So b = 120/20 - 93/20 = 27/20.Hmm, okay, so b is 27/20. Now, let's find c using equation 1:a + b + c = -2,31/20 + 27/20 + c = -2,(31 + 27)/20 + c = -2,58/20 + c = -2,Simplify 58/20 to 29/10,29/10 + c = -2,c = -2 - 29/10,Convert -2 to -20/10,c = -20/10 - 29/10 = -49/10.Wait, that doesn't match the previous result. Earlier, I thought c was -51/10, but now I'm getting -49/10. Let me check my steps again.Wait, in equation 1:a + b + c = -2,a = 31/20,b = 27/20,So 31/20 + 27/20 = 58/20 = 29/10,So 29/10 + c = -2,c = -2 - 29/10,Convert -2 to -20/10,So c = -20/10 - 29/10 = -49/10.Hmm, so c is -49/10. Wait, but earlier I thought it was -51/10. Maybe I made a mistake earlier. Let me check the equations again.Wait, perhaps I made a mistake in solving equation 5. Let me go back to equation 5:5a - 5b = 1,We had a = 31/20,So 5*(31/20) - 5b = 1,155/20 - 5b = 1,Simplify 155/20 to 31/4,31/4 - 5b = 1,Subtract 31/4 from both sides:-5b = 1 - 31/4,Convert 1 to 4/4,-5b = 4/4 - 31/4 = -27/4,So b = (-27/4)/(-5) = 27/(4*5) = 27/20.Okay, so b is indeed 27/20. Then, plugging back into equation 1:a + b + c = -2,31/20 + 27/20 + c = -2,58/20 + c = -2,29/10 + c = -2,c = -2 - 29/10 = -20/10 - 29/10 = -49/10.So c is -49/10. Wait, but in the initial problem, when I thought through it, I had c as -51/10. Maybe I made a mistake in the initial thought process. Let me check.Wait, in the initial thought process, I had:From equation 1: a + b + c = -2,a = 31/20, b = 29/20,So 31/20 + 29/20 = 60/20 = 3,So 3 + c = -2,c = -5.Wait, that's different. Hmm, so there's inconsistency here. Let me see where I went wrong.Wait, in the initial thought process, I had a = 31/20, b = 29/20, which would make a + b = 60/20 = 3, so c = -5. But in the current calculation, I have a = 31/20, b = 27/20, so a + b = 58/20 = 29/10, so c = -49/10. So which one is correct?Wait, perhaps I made a mistake in the initial thought process when solving for b. Let me check again.From equation 4: 3a + b = 6,a = 31/20,So 3*(31/20) = 93/20,So 93/20 + b = 6,Convert 6 to 120/20,So b = 120/20 - 93/20 = 27/20.So b is 27/20, not 29/20. So in the initial thought process, I must have made a mistake when solving for b.So, correct values are a = 31/20, b = 27/20, c = -49/10.Wait, but let me check if these values satisfy all three equations.Equation 1: a + b + c = 31/20 + 27/20 + (-49/10),Convert all to 20 denominators:31/20 + 27/20 - 98/20 = (31 + 27 - 98)/20 = (-40)/20 = -2. Correct.Equation 2: 4a + 2b + c = 4*(31/20) + 2*(27/20) + (-49/10),Calculate each term:4*(31/20) = 124/20 = 31/5,2*(27/20) = 54/20 = 27/10,-49/10 remains as is.Convert all to 10 denominators:31/5 = 62/10,27/10 remains,-49/10 remains.So total: 62/10 + 27/10 - 49/10 = (62 + 27 - 49)/10 = 40/10 = 4. Correct.Equation 3: 9a - 3b + c = 9*(31/20) - 3*(27/20) + (-49/10),Calculate each term:9*(31/20) = 279/20,-3*(27/20) = -81/20,-49/10 = -98/20.So total: 279/20 - 81/20 - 98/20 = (279 - 81 - 98)/20 = (279 - 179)/20 = 100/20 = 5. Correct.Okay, so the correct values are a = 31/20, b = 27/20, c = -49/10.So, r(x) = (31/20)x² + (27/20)x - 49/10.Now, I need to find r(7). Let's compute that.First, compute each term:(31/20)*(7)² + (27/20)*(7) - 49/10.Calculate 7² = 49,So first term: (31/20)*49 = (31*49)/20.Calculate 31*49:31*50 = 1550,Subtract 31: 1550 - 31 = 1519,So first term: 1519/20.Second term: (27/20)*7 = 189/20.Third term: -49/10 = -98/20.Now, add them all together:1519/20 + 189/20 - 98/20 = (1519 + 189 - 98)/20.Calculate numerator:1519 + 189 = 1708,1708 - 98 = 1610.So, 1610/20 = 80.5.Wait, 1610 divided by 20 is 80.5? Wait, 20*80 = 1600, so 1610 - 1600 = 10, so 80 + 10/20 = 80.5. Yes.But 80.5 is equal to 161/2, but in fraction form, 1610/20 simplifies to 161/2, which is 80.5.Wait, but in the initial thought process, I had r(7) = 77. That must have been a mistake. So, the correct value is 80.5, which is 161/2.Wait, but let me double-check my calculations because 1610/20 is indeed 80.5. Hmm, but maybe I made a mistake in the coefficients.Wait, let me recompute r(7):r(7) = (31/20)*(49) + (27/20)*(7) - 49/10.Compute each term:31/20 * 49: 31*49 = 1519, so 1519/20.27/20 *7: 27*7=189, so 189/20.-49/10: which is -98/20.So, adding them up:1519/20 + 189/20 - 98/20 = (1519 + 189 - 98)/20.Calculate 1519 + 189: 1519 + 189 = 1708.Then, 1708 - 98 = 1610.So, 1610/20 = 80.5.Hmm, so r(7) is 80.5, which is 161/2.Wait, but in the initial thought process, I had 77. That must have been incorrect. So, the correct answer is 80.5, or 161/2.But let me check if I made a mistake in the coefficients. Wait, in the initial thought process, I had a = 31/20, b = 29/20, c = -51/10, which led to r(7) = 77. But in reality, after correctly solving, I have a = 31/20, b = 27/20, c = -49/10, which gives r(7) = 80.5.Wait, so perhaps in the initial thought process, I made a mistake in solving for b, leading to incorrect coefficients and thus an incorrect r(7). So, the correct answer should be 80.5, which is 161/2.But let me confirm once more.Given a = 31/20, b = 27/20, c = -49/10.Compute r(7):r(7) = (31/20)*(49) + (27/20)*(7) + (-49/10).Compute each term:31/20 * 49: 31*49 = 1519, so 1519/20.27/20 *7: 27*7=189, so 189/20.-49/10: which is -490/100 or -49/10.Wait, but to add them together, let's convert all to 20 denominators:1519/20 + 189/20 - 98/20 = (1519 + 189 - 98)/20 = 1610/20 = 80.5.Yes, that's correct. So, r(7) is 80.5, which is 161/2.Wait, but 80.5 is equal to 161/2, which is correct.Hmm, so in the initial thought process, I must have made a mistake in solving for b, leading to incorrect coefficients. Therefore, the correct answer is 80.5, which is 161/2.But wait, 161/2 is 80.5, but the problem might expect an integer. Let me check if I made a mistake in calculations.Wait, let me recompute the system of equations.We had:1) a + b + c = -2,2) 4a + 2b + c = 4,3) 9a - 3b + c = 5.Subtracting equation 1 from equation 2:(4a + 2b + c) - (a + b + c) = 4 - (-2),Which is 3a + b = 6. Correct.Subtracting equation 2 from equation 3:(9a - 3b + c) - (4a + 2b + c) = 5 - 4,Which is 5a - 5b = 1. Correct.So, 3a + b = 6,5a - 5b = 1.Let me solve these again.From 3a + b = 6,b = 6 - 3a.Substitute into 5a -5b =1,5a -5*(6 - 3a) =1,5a -30 +15a =1,20a -30 =1,20a=31,a=31/20.Then, b=6 -3*(31/20)=6 -93/20=120/20 -93/20=27/20.So, a=31/20, b=27/20.Then, from equation1,a + b + c = -2,31/20 +27/20 +c= -2,58/20 +c= -2,29/10 +c= -2,c= -2 -29/10= -20/10 -29/10= -49/10.So, correct.Therefore, r(x)= (31/20)x² + (27/20)x -49/10.Thus, r(7)= (31/20)*49 + (27/20)*7 -49/10.Compute each term:31*49=1519, so 1519/20=75.95,27*7=189, so 189/20=9.45,-49/10= -4.9.So, 75.95 +9.45=85.4,85.4 -4.9=80.5.So, 80.5 is correct, which is 161/2.Wait, but 161/2 is 80.5, which is correct.So, the initial thought process had an error in solving for b, leading to incorrect coefficients and thus an incorrect r(7). The correct answer is 80.5, which is 161/2.But wait, the problem asks for r(7). So, 80.5 is correct, but maybe we can write it as a fraction.80.5 is equal to 161/2.So, the final answer is 161/2, which is 80.5.But let me check if I can simplify it further or if I made any mistake in the calculations.Wait, 31/20 is 1.55,27/20 is 1.35,-49/10 is -4.9.So, r(7)=1.55*(49) +1.35*(7) -4.9.Compute 1.55*49:1.55*50=77.5,Minus 1.55=77.5-1.55=75.95.1.35*7=9.45.So, 75.95 +9.45=85.4.85.4 -4.9=80.5.Yes, correct.So, the correct answer is 80.5, which is 161/2.But in the initial thought process, I had 77, which was incorrect due to a mistake in solving for b.Therefore, the correct value of r(7) is 80.5, or 161/2.Wait, but 161/2 is an improper fraction, but it's correct.Alternatively, as a mixed number, it's 80 1/2.But since the problem doesn't specify the form, either is acceptable, but likely as an improper fraction, 161/2.Wait, but let me check once more if I made any mistake in the system of equations.From the three equations:1) a + b + c = -2,2) 4a + 2b + c = 4,3) 9a - 3b + c = 5.We subtracted 1 from 2 to get 3a + b =6,Subtracted 2 from 3 to get 5a -5b=1.Solved for a=31/20, b=27/20, c=-49/10.Plugging back into all three equations, they satisfy.Thus, r(x)= (31/20)x² + (27/20)x -49/10.Thus, r(7)=80.5=161/2.So, the final answer is 161/2, which is 80.5.But wait, in the initial thought process, I had 77, which was wrong. So, the correct answer is 161/2.Wait, but let me check if the problem expects an integer. Maybe I made a mistake in the calculations.Wait, let me compute r(7) again:r(7)= (31/20)*(49) + (27/20)*(7) -49/10.Compute each term:31*49=1519, so 1519/20=75.95,27*7=189, so 189/20=9.45,-49/10= -4.9.Adding them up:75.95 +9.45=85.4,85.4 -4.9=80.5.Yes, correct.So, the answer is 80.5, which is 161/2.But wait, 161/2 is 80.5, which is correct.Therefore, the correct answer is 161/2.But in the initial thought process, I had 77, which was incorrect. So, the correct answer is 161/2.Wait, but 161/2 is 80.5, which is correct.So, the final answer is 161/2, which can be written as boxed{dfrac{161}{2}}.But let me check if I can simplify it further or if I made any mistake in the calculations.Wait, 31/20 is 1.55,27/20 is 1.35,-49/10 is -4.9.So, r(7)=1.55*(49) +1.35*(7) -4.9.Compute 1.55*49:1.55*50=77.5,Minus 1.55=77.5-1.55=75.95.1.35*7=9.45.So, 75.95 +9.45=85.4.85.4 -4.9=80.5.Yes, correct.So, the correct answer is 80.5, which is 161/2.Therefore, the final answer is boxed{dfrac{161}{2}}.