💡Okay, so I have this inequality: (x^2 - (a^2 + a)x + a^3 > 0). The solution set is given as ({x | x < a^2 text{ or } x > a}). I need to find the range of the real number (a). Hmm, let me think about how to approach this.First, I remember that quadratic inequalities can be solved by finding their roots and then determining the intervals where the quadratic expression is positive or negative. Since the solution set is given as (x < a^2) or (x > a), it suggests that the quadratic opens upwards (since the solution is outside the roots) and that (a^2) and (a) are the roots of the corresponding equation (x^2 - (a^2 + a)x + a^3 = 0).Let me verify that (a^2) and (a) are indeed the roots. If I plug (x = a^2) into the quadratic equation:[(a^2)^2 - (a^2 + a)(a^2) + a^3 = a^4 - a^4 - a^3 + a^3 = 0]Yes, that works. Now, plugging (x = a):[a^2 - (a^2 + a)a + a^3 = a^2 - a^3 - a^2 + a^3 = 0]That also works. So, (a^2) and (a) are the roots.Since the quadratic opens upwards (the coefficient of (x^2) is positive), the expression (x^2 - (a^2 + a)x + a^3) will be greater than zero when (x) is less than the smaller root or greater than the larger root. Therefore, for the solution set to be (x < a^2) or (x > a), it must be that (a^2 < a). Wait, is that necessarily true?Hold on, actually, the order of the roots depends on the values of (a). So, if (a^2 < a), then the smaller root is (a^2) and the larger root is (a). But if (a^2 > a), then the smaller root would be (a) and the larger root would be (a^2). So, the given solution set implies that (a^2 < a), right? Because the solution is (x < a^2) or (x > a), meaning (a^2) is the smaller root and (a) is the larger root.So, let's write that inequality:[a^2 < a]Subtracting (a^2) from both sides:[0 < a - a^2][0 < a(1 - a)]So, the product (a(1 - a)) is positive. This happens when both factors are positive or both are negative.Case 1: Both (a > 0) and (1 - a > 0). That implies (a > 0) and (a < 1). So, (0 < a < 1).Case 2: Both (a < 0) and (1 - a < 0). That implies (a < 0) and (a > 1). But this is impossible because (a) cannot be both less than 0 and greater than 1 at the same time.Therefore, the only valid case is (0 < a < 1).But wait, I should also consider the case when (a^2 = a). That is, when (a = 0) or (a = 1). If (a = 0), then the quadratic becomes (x^2 > 0), which is true for all (x neq 0). But the given solution set is (x < 0) or (x > 0), which is actually all real numbers except 0. So, that's consistent.Similarly, if (a = 1), the quadratic becomes (x^2 - (1 + 1)x + 1 = x^2 - 2x + 1 = (x - 1)^2 > 0). This is true for all (x neq 1), which is (x < 1) or (x > 1). So, that's also consistent with the given solution set.Therefore, including (a = 0) and (a = 1), the range of (a) is (0 leq a leq 1).But just to make sure, I should check if the quadratic actually has real roots for these values of (a). The discriminant of the quadratic equation (x^2 - (a^2 + a)x + a^3 = 0) is:[Delta = (a^2 + a)^2 - 4 times 1 times a^3][= a^4 + 2a^3 + a^2 - 4a^3][= a^4 - 2a^3 + a^2][= a^2(a^2 - 2a + 1)][= a^2(a - 1)^2]Since (Delta = a^2(a - 1)^2), which is always non-negative for all real (a). So, the quadratic always has real roots, which are (a^2) and (a). Therefore, the discriminant condition doesn't impose any additional restrictions on (a).So, putting it all together, the range of (a) is all real numbers from 0 to 1, inclusive. That is, (0 leq a leq 1).**Final Answer**The range of the real number (a) is boxed{[0, 1]}.

💡First, let's denote the total number of eggs Tricia ordered as ( E ).She used ( frac{1}{4} ) of the eggs for making crepes, so the remaining eggs after making crepes are:[E - frac{1}{4}E = frac{3}{4}E]Next, she used ( frac{2}{3} ) of the remaining eggs for making cupcakes. The number of eggs used for cupcakes is:[frac{2}{3} times frac{3}{4}E = frac{1}{2}E]After making cupcakes, the number of eggs left is:[frac{3}{4}E - frac{1}{2}E = frac{1}{4}E]According to the problem, she has 9 eggs left:[frac{1}{4}E = 9]Solving for ( E ):[E = 9 times 4 = 36]Therefore, Tricia ordered (boxed{36}) eggs in total.

💡Alright, so I've got this problem about a regular 20-gon. It's asking for the number of lines of symmetry, which is L, and the smallest positive angle for rotational symmetry, which is R. Then, I need to find the sum of L and R. The answer choices are from 36 to 44, so I guess it's a matter of figuring out L and R correctly and then adding them up.First, let me recall what a regular 20-gon is. A regular polygon has all sides equal and all angles equal. So, a regular 20-gon would have 20 equal sides and 20 equal angles. That should mean it has a lot of symmetry, both in terms of lines of symmetry and rotational symmetry.Starting with lines of symmetry, L. For regular polygons, the number of lines of symmetry is equal to the number of sides. So, a regular triangle has 3 lines of symmetry, a square has 4, and so on. Therefore, a regular 20-gon should have 20 lines of symmetry. That seems straightforward. So, L = 20.Next, the smallest positive angle for rotational symmetry, R. Rotational symmetry means that when you rotate the shape by a certain angle around its center, it looks the same as before. The smallest such angle is called the order of rotational symmetry. For regular polygons, this angle is calculated by dividing 360 degrees by the number of sides. So, for a square, it's 360/4 = 90 degrees, and for a regular triangle, it's 360/3 = 120 degrees.Applying that to a regular 20-gon, the smallest angle R should be 360 divided by 20. Let me calculate that: 360 ÷ 20 = 18 degrees. So, R = 18 degrees.Now, to find L + R, I just add these two numbers together. L is 20 and R is 18, so 20 + 18 = 38.Looking back at the answer choices, 38 is option B. So, I think the answer is B) 38.Wait, just to make sure I didn't make a mistake. Let me double-check. Lines of symmetry in a regular polygon are equal to the number of sides, so 20 makes sense. For rotational symmetry, the formula is 360/n, which for n=20 is 18 degrees. Adding them together, 20 + 18 is indeed 38. Hmm, that seems right.I don't think I made any calculation errors here. The logic follows from the properties of regular polygons. So, yeah, I'm confident that the answer is 38.**Final Answer**boxed{B}

💡Alright, so I've got this problem about an ellipse with foci at (-1, 0) and (1, 0). The equation of the ellipse is given as (x²/a²) + (y²/3) = 1, where a > 0. I need to find the equation of the ellipse first, and then solve some geometry problems involving points A and B on specific lines and certain conditions.Starting with part (I): Find the equation of the ellipse C.Hmm, okay. I remember that for an ellipse, the standard form is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis and b is the semi-minor axis. The distance between the center and each focus is c, and the relationship between a, b, and c is a² = b² + c². Given the foci are at (-1, 0) and (1, 0), the center of the ellipse is at the midpoint, which is (0, 0). The distance from the center to each focus is c = 1. The given equation has y²/3, so b² = 3. Therefore, using the relationship a² = b² + c², I can plug in the known values.So, a² = 3 + 1² = 4. Therefore, a² = 4, so a = 2. Thus, the equation of the ellipse is (x²/4) + (y²/3) = 1. That seems straightforward.Moving on to part (II): Points A and B are located on the lines x = -2 and x = 2 respectively, and AF₁ is perpendicular to BF₁. First, let me visualize this. Points A and B are on vertical lines x = -2 and x = 2. F₁ is at (-1, 0). So, AF₁ is the line segment from A(-2, y₁) to F₁(-1, 0), and BF₁ is the line segment from B(2, y₂) to F₁(-1, 0). These two segments are perpendicular.I need to find the coordinates of A and B such that AF₁ is perpendicular to BF₁. Then, for part (i), when triangle ABF₁ is isosceles, find its area. For part (ii), find the minimum value of the sum of distances from F₁ and F₂ to the line AB.Starting with part (II)(i): When triangle ABF₁ is an isosceles triangle, find the area.First, let's denote points A(-2, m) and B(2, n). So, A is on x = -2, so x-coordinate is fixed at -2, and y-coordinate is m. Similarly, B is on x = 2, so x-coordinate is 2, y-coordinate is n.Given that AF₁ is perpendicular to BF₁. Let's find the vectors AF₁ and BF₁.Vector AF₁ is from A(-2, m) to F₁(-1, 0). So, the vector is (-1 - (-2), 0 - m) = (1, -m).Vector BF₁ is from B(2, n) to F₁(-1, 0). So, the vector is (-1 - 2, 0 - n) = (-3, -n).Since AF₁ is perpendicular to BF₁, their dot product should be zero.So, (1)(-3) + (-m)(-n) = 0 => -3 + mn = 0 => mn = 3.So, the product of m and n is 3.Now, triangle ABF₁ is isosceles. So, either two sides are equal. The sides are AF₁, BF₁, and AB.We need to consider the cases where either AF₁ = BF₁, AF₁ = AB, or BF₁ = AB.But since A is on x = -2 and B is on x = 2, and F₁ is at (-1, 0), the distances AF₁ and BF₁ might be equal, or AB might be equal to one of them.Let me compute the lengths.First, AF₁: distance from A(-2, m) to F₁(-1, 0). Using distance formula:AF₁ = sqrt[(-1 - (-2))² + (0 - m)²] = sqrt[(1)² + (-m)²] = sqrt(1 + m²).Similarly, BF₁: distance from B(2, n) to F₁(-1, 0):BF₁ = sqrt[(-1 - 2)² + (0 - n)²] = sqrt[(-3)² + (-n)²] = sqrt(9 + n²).AB: distance from A(-2, m) to B(2, n):AB = sqrt[(2 - (-2))² + (n - m)²] = sqrt[(4)² + (n - m)²] = sqrt(16 + (n - m)²).So, triangle ABF₁ is isosceles. So, either AF₁ = BF₁, AF₁ = AB, or BF₁ = AB.Case 1: AF₁ = BF₁.So, sqrt(1 + m²) = sqrt(9 + n²). Squaring both sides: 1 + m² = 9 + n² => m² - n² = 8.But we already have mn = 3 from the perpendicular condition.So, we have two equations:1. m² - n² = 82. mn = 3We can solve this system for m and n.From equation 2: n = 3/m. Substitute into equation 1:m² - (9/m²) = 8.Multiply both sides by m²: m⁴ - 9 = 8m² => m⁴ - 8m² - 9 = 0.Let me set u = m², so the equation becomes u² - 8u - 9 = 0.Solving quadratic: u = [8 ± sqrt(64 + 36)] / 2 = [8 ± sqrt(100)] / 2 = [8 ± 10]/2.So, u = (8 + 10)/2 = 18/2 = 9, or u = (8 - 10)/2 = (-2)/2 = -1.Since u = m², it can't be negative. So, u = 9 => m² = 9 => m = ±3.Thus, m = 3 or m = -3.If m = 3, then n = 3/m = 3/3 = 1.If m = -3, then n = 3/(-3) = -1.So, we have two possibilities:1. A(-2, 3) and B(2, 1)2. A(-2, -3) and B(2, -1)Now, let's check if in these cases, triangle ABF₁ is isosceles.Compute AF₁, BF₁, AB.First case: A(-2, 3), B(2, 1), F₁(-1, 0).Compute AF₁: sqrt(1 + 9) = sqrt(10).BF₁: sqrt(9 + 1) = sqrt(10). So, AF₁ = BF₁, which makes triangle ABF₁ isosceles.AB: sqrt(16 + (1 - 3)^2) = sqrt(16 + 4) = sqrt(20) = 2*sqrt(5).So, sides are sqrt(10), sqrt(10), 2*sqrt(5). So, yes, it's isosceles.Second case: A(-2, -3), B(2, -1), F₁(-1, 0).Compute AF₁: sqrt(1 + 9) = sqrt(10).BF₁: sqrt(9 + 1) = sqrt(10). So, same as above.AB: sqrt(16 + (-1 - (-3))²) = sqrt(16 + (2)^2) = sqrt(16 + 4) = sqrt(20) = 2*sqrt(5).Same as above, so triangle is isosceles.So, both cases are valid.Now, we need to find the area of triangle ABF₁.Since AF₁ and BF₁ are equal, and the triangle is isosceles, the area can be calculated in a couple of ways.One way is to use coordinates and the shoelace formula.Let me use coordinates for the first case: A(-2, 3), B(2, 1), F₁(-1, 0).Using shoelace formula:Arrange the points: A(-2, 3), B(2, 1), F₁(-1, 0), back to A(-2, 3).Compute the sum of x_i y_{i+1}:(-2)(1) + (2)(0) + (-1)(3) = (-2) + 0 + (-3) = -5Compute the sum of y_i x_{i+1}:(3)(2) + (1)(-1) + (0)(-2) = 6 + (-1) + 0 = 5Area = (1/2)|sum1 - sum2| = (1/2)|-5 - 5| = (1/2)|-10| = 5.Similarly, for the second case: A(-2, -3), B(2, -1), F₁(-1, 0).Shoelace formula:Points: A(-2, -3), B(2, -1), F₁(-1, 0), back to A(-2, -3).Sum of x_i y_{i+1}:(-2)(-1) + (2)(0) + (-1)(-3) = 2 + 0 + 3 = 5Sum of y_i x_{i+1}:(-3)(2) + (-1)(-1) + (0)(-2) = (-6) + 1 + 0 = -5Area = (1/2)|5 - (-5)| = (1/2)|10| = 5.So, in both cases, the area is 5.Alternatively, since AF₁ and BF₁ are both sqrt(10), and the angle between them is 90 degrees because AF₁ is perpendicular to BF₁. Wait, no, actually, AF₁ is perpendicular to BF₁, but in the triangle ABF₁, the sides AF₁ and BF₁ are equal and the angle between them is 90 degrees? Wait, no, because AF₁ and BF₁ are perpendicular, but in the triangle, the sides AF₁ and BF₁ are equal, and the angle at F₁ is 90 degrees. So, the area can also be calculated as (1/2)*AF₁*BF₁, since it's a right triangle.But wait, AF₁ and BF₁ are equal and perpendicular, so the area is (1/2)*(sqrt(10))*(sqrt(10)) = (1/2)*10 = 5. Yep, same result.So, the area is 5.Moving on to part (II)(ii): Find the minimum value of the sum of distances from points F₁ and F₂ to line AB.So, we need to find the minimum of d₁ + d₂, where d₁ is the distance from F₁ to AB, and d₂ is the distance from F₂ to AB.First, let's find the equation of line AB.Points A(-2, m) and B(2, n). So, the slope of AB is (n - m)/(2 - (-2)) = (n - m)/4.So, the equation of AB can be written as y - m = [(n - m)/4](x + 2).Simplify:y = [(n - m)/4]x + [(n - m)/4]*2 + m= [(n - m)/4]x + (n - m)/2 + m= [(n - m)/4]x + (n - m + 2m)/2= [(n - m)/4]x + (n + m)/2So, the equation is y = [(n - m)/4]x + (n + m)/2.We can write this in standard form: ax + by + c = 0.Multiply both sides by 4:4y = (n - m)x + 2(n + m)Bring all terms to left:(n - m)x - 4y + 2(n + m) = 0.So, the equation of AB is (n - m)x - 4y + 2(n + m) = 0.Now, the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²).So, distance from F₁(-1, 0) to AB:d₁ = |(n - m)(-1) - 4*0 + 2(n + m)| / sqrt((n - m)² + (-4)²)= | - (n - m) + 2(n + m) | / sqrt((n - m)² + 16)Simplify numerator:- n + m + 2n + 2m = ( -n + 2n ) + ( m + 2m ) = n + 3mSo, d₁ = |n + 3m| / sqrt((n - m)² + 16)Similarly, distance from F₂(1, 0) to AB:d₂ = |(n - m)(1) - 4*0 + 2(n + m)| / sqrt((n - m)² + 16)= |(n - m) + 2(n + m)| / sqrt((n - m)² + 16)Simplify numerator:n - m + 2n + 2m = 3n + mSo, d₂ = |3n + m| / sqrt((n - m)² + 16)Therefore, the sum d₁ + d₂ is:[ |n + 3m| + |3n + m| ] / sqrt((n - m)² + 16)But we know from earlier that mn = 3.So, we have the constraint mn = 3.We need to minimize d₁ + d₂.Hmm, this seems a bit complicated. Maybe we can express everything in terms of one variable.Let me denote m as a variable, then n = 3/m.So, let's substitute n = 3/m into the expressions.First, compute d₁ + d₂:Numerator: |(3/m) + 3m| + |3*(3/m) + m| = |3/m + 3m| + |9/m + m|Denominator: sqrt( (3/m - m)² + 16 )Let me compute numerator:First term: |3/m + 3m| = 3|m + 1/m|Second term: |9/m + m| = |m + 9/m|So, numerator = 3|m + 1/m| + |m + 9/m|Denominator: sqrt( (3/m - m)² + 16 ) = sqrt( ( (3 - m²)/m )² + 16 ) = sqrt( (9 - 6m² + m⁴)/m² + 16 )= sqrt( (9 - 6m² + m⁴ + 16m²)/m² ) = sqrt( (m⁴ + 10m² + 9)/m² ) = sqrt( (m² + 1)(m² + 9) ) / |m|But since m can be positive or negative, but let's assume m > 0 for simplicity, as the problem is symmetric.So, denominator = sqrt( (m² + 1)(m² + 9) ) / m.So, putting it all together:d₁ + d₂ = [3|m + 1/m| + |m + 9/m| ] / [ sqrt( (m² + 1)(m² + 9) ) / m ]= [3(m + 1/m) + (m + 9/m)] * m / sqrt( (m² + 1)(m² + 9) )Since m > 0, we can drop the absolute values.So, numerator inside the brackets:3(m + 1/m) + (m + 9/m) = 3m + 3/m + m + 9/m = 4m + 12/mMultiply by m:4m² + 12Denominator:sqrt( (m² + 1)(m² + 9) )So, d₁ + d₂ = (4m² + 12) / sqrt( (m² + 1)(m² + 9) )Let me denote t = m², so t > 0.Then, d₁ + d₂ = (4t + 12) / sqrt( (t + 1)(t + 9) )Simplify numerator and denominator:Numerator: 4(t + 3)Denominator: sqrt( (t + 1)(t + 9) ) = sqrt(t² + 10t + 9)So, d₁ + d₂ = 4(t + 3) / sqrt(t² + 10t + 9)Let me denote f(t) = 4(t + 3) / sqrt(t² + 10t + 9)We need to find the minimum of f(t) for t > 0.To find the minimum, let's compute the derivative of f(t) with respect to t and set it to zero.First, let me write f(t) as 4(t + 3)(t² + 10t + 9)^(-1/2)Compute f'(t):Using product rule:f'(t) = 4[ (1)(t² + 10t + 9)^(-1/2) + (t + 3)(-1/2)(t² + 10t + 9)^(-3/2)(2t + 10) ]Simplify:= 4[ (t² + 10t + 9)^(-1/2) - (t + 3)(t + 5)(t² + 10t + 9)^(-3/2) ]Factor out (t² + 10t + 9)^(-3/2):= 4(t² + 10t + 9)^(-3/2) [ (t² + 10t + 9) - (t + 3)(t + 5) ]Compute the expression inside the brackets:(t² + 10t + 9) - (t + 3)(t + 5) = t² + 10t + 9 - (t² + 8t + 15) = t² + 10t + 9 - t² - 8t - 15 = 2t - 6So, f'(t) = 4(t² + 10t + 9)^(-3/2)(2t - 6)Set f'(t) = 0:4(t² + 10t + 9)^(-3/2)(2t - 6) = 0Since 4 ≠ 0 and (t² + 10t + 9)^(-3/2) ≠ 0 for t > 0, we have 2t - 6 = 0 => t = 3.So, critical point at t = 3.Now, check if this is a minimum.Compute second derivative or test intervals.For t < 3, say t = 2, f'(t) = 4*(...)*(2*2 - 6) = 4*(...)*(-2) < 0For t > 3, say t = 4, f'(t) = 4*(...)*(2*4 - 6) = 4*(...)*(2) > 0So, function decreases before t=3 and increases after t=3, so t=3 is a minimum.Thus, the minimum occurs at t = 3.Compute f(3):f(3) = 4(3 + 3)/sqrt(9 + 30 + 9) = 4*6 / sqrt(48) = 24 / (4*sqrt(3)) = 6 / sqrt(3) = 2*sqrt(3)So, the minimum value of d₁ + d₂ is 2*sqrt(3).Therefore, the minimum sum of distances is 2√3.**Final Answer**(I) The equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(II)(i) The area of triangle ( triangle ABF_1 ) is boxed{5}.(II)(ii) The minimum value of the sum of distances is boxed{2sqrt{3}}.

💡Alright, so I'm trying to figure out whether there are more "humped" five-digit numbers or "valleyed" five-digit numbers. Let me start by understanding what these terms mean.A five-digit number is called "humped" if its middle digit is larger than all the other digits. So, for example, if we have a number like 12345, the middle digit is 3. If 3 is larger than 1, 2, 4, and 5, then it's humped. Wait, but in this case, 3 isn't larger than 4 and 5, so 12345 isn't humped. Maybe 12934 would be humped because 9 is larger than 1, 2, 3, and 4.On the other hand, a five-digit number is called "valleyed" if its middle digit is smaller than all the other digits. Using the same example, 12345 isn't valleyed because 3 isn't smaller than 1, 2, 4, and 5. But a number like 32123 might be valleyed because 1 is smaller than 3, 2, 2, and 3.Okay, so I need to figure out which type is more common. Maybe I can think about how many possibilities there are for each case.Let's consider the middle digit first. For a five-digit number, the middle digit is the third one. So, positions are: first, second, third (middle), fourth, fifth.For a number to be humped, the middle digit must be greater than the first, second, fourth, and fifth digits. Similarly, for it to be valleyed, the middle digit must be smaller than those four digits.I wonder if there's a symmetry here. If I consider all possible five-digit numbers, maybe the number of humped and valleyed numbers are the same? But I'm not sure because the middle digit has to be strictly greater or strictly smaller, not just greater than or equal to or something like that.Wait, maybe there's a way to pair each humped number with a valleyed number. If I take a humped number and subtract each digit from 9, would that give me a valleyed number? Let's see.Suppose I have a humped number, say 12934. If I subtract each digit from 9, I get 87065. Is 87065 a valleyed number? Let's check: the middle digit is 0, and the other digits are 8, 7, 6, and 5. Is 0 smaller than all of them? Yes, 0 is smaller than 5, 6, 7, and 8. So, 87065 is valleyed.That seems to work. So, if I take any humped number and subtract each digit from 9, I get a valleyed number. This suggests that there's a one-to-one correspondence between humped and valleyed numbers.But wait, are there any numbers that are both humped and valleyed? That would mean the middle digit is both greater than and smaller than all other digits, which is impossible. So, no overlap there.Also, are there any numbers that are neither humped nor valleyed? Definitely, because the middle digit could be equal to some other digits or neither greater nor smaller than all.So, if there's a one-to-one correspondence between humped and valleyed numbers, does that mean they are equally common? But the problem is asking which is more common, so maybe I'm missing something.Oh, maybe there are some numbers that can't be paired this way. For example, if a number starts with 9, subtracting each digit from 9 might lead to issues because the first digit can't be 0. Wait, but in five-digit numbers, the first digit can't be 0 anyway. So, if I have a number like 91234, subtracting each digit from 9 would give me 08765, but that's not a valid five-digit number because it starts with 0. So, maybe this pairing doesn't work for numbers starting with 9.Hmm, that complicates things. So, numbers starting with 9 can't be paired with valleyed numbers in the same way because their complements would start with 0, which isn't allowed. Therefore, there might be more valleyed numbers because the numbers starting with 9 can't be paired with valleyed numbers, but valleyed numbers can still exist without being paired with humped numbers.Alternatively, maybe the numbers starting with 9 are humped, and their complements would start with 0, which aren't valid, so those humped numbers don't have corresponding valleyed numbers. Therefore, there are more valleyed numbers because they can exist without being paired with humped numbers, but humped numbers are limited by the pairing.Wait, but I'm not sure if all valleyed numbers can be obtained by subtracting digits from 9 of humped numbers. Maybe some valleyed numbers can't be obtained this way, especially if their complements start with 0.This is getting a bit confusing. Maybe I should think about the total number of five-digit numbers and see how many are humped and how many are valleyed.The total number of five-digit numbers is from 10000 to 99999, which is 90000 numbers.Now, to count the number of humped numbers, I need to consider that the middle digit is the largest. So, for each possible middle digit from 1 to 9, I need to count how many numbers have that digit as the middle one and all other digits less than it.Similarly, for valleyed numbers, the middle digit is the smallest, so for each possible middle digit from 0 to 8, I need to count how many numbers have that digit as the middle one and all other digits greater than it.This seems complicated, but maybe I can find a pattern or use combinatorics.For humped numbers:- Choose the middle digit d (from 1 to 9).- The first digit can be from 1 to d-1 (since it can't be 0 and must be less than d).- The second, fourth, and fifth digits can be from 0 to d-1.So, for each d, the number of humped numbers is:- First digit: (d-1) choices- Second digit: d choices- Fourth digit: d choices- Fifth digit: d choicesSo, total for each d: (d-1) * d^3Summing over d from 1 to 9:Total humped = Σ (d-1)*d^3 from d=1 to 9Similarly, for valleyed numbers:- Choose the middle digit d (from 0 to 8).- The first digit can be from d+1 to 9 (since it can't be 0 and must be greater than d).- The second, fourth, and fifth digits can be from d+1 to 9.So, for each d, the number of valleyed numbers is:- First digit: (9 - d) choices- Second digit: (9 - d) choices- Fourth digit: (9 - d) choices- Fifth digit: (9 - d) choicesWait, that doesn't seem right. For valleyed numbers, the first digit must be greater than d, but d can be 0 to 8, so for d=0, first digit can be 1-9, which is 9 choices. For d=1, first digit can be 2-9, which is 8 choices, and so on.But the second, fourth, and fifth digits also need to be greater than d, so they have (9 - d) choices each.Therefore, for each d, the number of valleyed numbers is:- First digit: (9 - d) choices- Second digit: (9 - d) choices- Fourth digit: (9 - d) choices- Fifth digit: (9 - d) choicesSo, total for each d: (9 - d)^4Summing over d from 0 to 8:Total valleyed = Σ (9 - d)^4 from d=0 to 8Wait, but when d=0, (9 - 0)^4 = 9^4 = 6561When d=1, (9 - 1)^4 = 8^4 = 4096And so on, down to d=8, which is 1^4 = 1So, total valleyed = 6561 + 4096 + 2401 + 1296 + 625 + 256 + 81 + 16 + 1Let me calculate that:6561 + 4096 = 1065710657 + 2401 = 1305813058 + 1296 = 1435414354 + 625 = 1497914979 + 256 = 1523515235 + 81 = 1531615316 + 16 = 1533215332 + 1 = 15333So, total valleyed numbers = 15,333Now, for humped numbers, let's calculate:For each d from 1 to 9:Total humped += (d-1)*d^3Let's compute each term:d=1: (1-1)*1^3 = 0d=2: (2-1)*2^3 = 1*8 = 8d=3: (3-1)*3^3 = 2*27 = 54d=4: (4-1)*4^3 = 3*64 = 192d=5: (5-1)*5^3 = 4*125 = 500d=6: (6-1)*6^3 = 5*216 = 1080d=7: (7-1)*7^3 = 6*343 = 2058d=8: (8-1)*8^3 = 7*512 = 3584d=9: (9-1)*9^3 = 8*729 = 5832Now, sum these up:d=1: 0d=2: 8d=3: 54 → total so far: 62d=4: 192 → total: 254d=5: 500 → total: 754d=6: 1080 → total: 1834d=7: 2058 → total: 3892d=8: 3584 → total: 7476d=9: 5832 → total: 13308So, total humped numbers = 13,308Comparing the two totals:- Valleyed: 15,333- Humped: 13,308So, valleyed numbers are more common than humped numbers.Wait, but earlier I thought about the pairing where humped numbers correspond to valleyed numbers by subtracting digits from 9, but that didn't account for numbers starting with 9. It seems that the combinatorial approach shows that valleyed numbers are indeed more common.I think this makes sense because for valleyed numbers, the middle digit can be 0, which allows for more combinations, whereas for humped numbers, the middle digit has to be at least 1, and the first digit can't be 0, which limits the possibilities.So, in conclusion, valleyed numbers are more common than humped numbers.

💡First, I need to simplify the given complex number ( z = i cdot (2 + i) ).Multiplying ( i ) by ( 2 ) gives ( 2i ), and multiplying ( i ) by ( i ) gives ( i^2 ), which is equal to (-1).So, ( z = 2i + (-1) = -1 + 2i ).Next, I find the conjugate of ( z ), denoted as ( overline{z} ). The conjugate of a complex number ( a + bi ) is ( a - bi ).Therefore, ( overline{z} = -1 - 2i ).In the complex plane, the real part is the x-coordinate and the imaginary part is the y-coordinate. So, the point corresponding to ( overline{z} ) is ( (-1, -2) ).Both the real and imaginary parts are negative, which places the point in the third quadrant of the complex plane.