💡First, define variables for the number of bicycles ((b)), tricycles ((t)), and unicycles ((u)).Set up the equations based on the total number of children and wheels:[b + t + u = 10][2b + 3t + u = 26]Express (u) from the first equation:[u = 10 - b - t]Substitute (u) into the second equation:[2b + 3t + (10 - b - t) = 26]Simplify the equation:[b + 2t = 16]Express (b) in terms of (t):[b = 16 - 2t]Substitute (b) back into the first equation:[16 - 2t + t + u = 10 implies u = t - 6]Since the number of unicycles cannot be negative, solve for (t):[t - 6 geq 0 implies t geq 6]Check the possible values of (t) to satisfy all conditions. The only valid solution is:[t = 6]

💡First, I need to understand the custom operation ( A * B ), which is defined as ( frac{A + B}{2} ). This means that whenever I see ( A * B ), I should replace it with the average of ( A ) and ( B ).Next, I'll evaluate the expression ( (7 * 9) * 4 ) step by step.1. **Evaluate ( 7 * 9 ):** [ 7 * 9 = frac{7 + 9}{2} = frac{16}{2} = 8 ] Here, I added 7 and 9 to get 16, then divided by 2 to find the average, which is 8.2. **Use the result to evaluate ( (7 * 9) * 4 ):** [ (7 * 9) * 4 = 8 * 4 = frac{8 + 4}{2} = frac{12}{2} = 6 ] I substituted the result from the first step (which is 8) into the expression, then added 8 and 4 to get 12, and finally divided by 2 to find the average, resulting in 6.Therefore, the final answer is ( 6 ).

💡Okay, so I have this problem where I need to find the value of ((x+1)(y+1)) given three equations involving (x) and (y). Let me try to figure this out step by step.First, the equations given are:1. (xy = frac{1}{9})2. (x(y+1) = frac{7}{9})3. (y(x+1) = frac{5}{18})I need to find ((x+1)(y+1)). Hmm, I remember that expanding ((x+1)(y+1)) gives (xy + x + y + 1). So if I can find the values of (x) and (y), I can plug them into this expression.Looking at the second equation, (x(y+1) = frac{7}{9}). I can expand this to (xy + x = frac{7}{9}). Since I know from the first equation that (xy = frac{1}{9}), I can substitute that in:(frac{1}{9} + x = frac{7}{9})Subtracting (frac{1}{9}) from both sides gives:(x = frac{7}{9} - frac{1}{9} = frac{6}{9} = frac{2}{3})Okay, so (x = frac{2}{3}). Now, let's look at the third equation: (y(x+1) = frac{5}{18}). Expanding this gives (xy + y = frac{5}{18}). Again, substituting (xy = frac{1}{9}):(frac{1}{9} + y = frac{5}{18})Subtracting (frac{1}{9}) from both sides. Wait, (frac{1}{9}) is equal to (frac{2}{18}), so:(y = frac{5}{18} - frac{2}{18} = frac{3}{18} = frac{1}{6})So, (y = frac{1}{6}). Now that I have both (x) and (y), I can compute ((x+1)(y+1)).Let's compute (x + 1) and (y + 1):(x + 1 = frac{2}{3} + 1 = frac{2}{3} + frac{3}{3} = frac{5}{3})(y + 1 = frac{1}{6} + 1 = frac{1}{6} + frac{6}{6} = frac{7}{6})Now, multiply these two results together:(frac{5}{3} times frac{7}{6} = frac{5 times 7}{3 times 6} = frac{35}{18})So, ((x+1)(y+1) = frac{35}{18}).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. I used the second equation to find (x) by substituting (xy = frac{1}{9}). That gave me (x = frac{2}{3}).2. Then, I used the third equation to find (y) by substituting (xy = frac{1}{9}). That gave me (y = frac{1}{6}).3. Then, I expanded ((x+1)(y+1)) to (xy + x + y + 1), but since I already had (x) and (y), I just added 1 to each and multiplied.Wait, another way to think about it is that since I already know (x) and (y), I can directly compute ((x+1)(y+1)) without expanding. Let me confirm:((x+1)(y+1) = left(frac{2}{3} + 1right)left(frac{1}{6} + 1right) = left(frac{5}{3}right)left(frac{7}{6}right) = frac{35}{18})Yes, that seems correct. Alternatively, if I had expanded it as (xy + x + y + 1), I would have:(frac{1}{9} + frac{2}{3} + frac{1}{6} + 1)Let me compute that:Convert all to eighteenths to add them easily:(frac{2}{18} + frac{12}{18} + frac{3}{18} + frac{18}{18} = frac{2 + 12 + 3 + 18}{18} = frac{35}{18})Same result. So, that confirms it.I think I did everything correctly. I found (x) and (y) using substitution from the given equations, then used those values to compute the desired expression. Both methods of expanding gave me the same answer, so I feel confident that (frac{35}{18}) is correct.**Final Answer**boxed{E}

💡Alright, so I have this problem here: Given that (a) and (b) are real numbers, a sufficient but not necessary condition for (frac{1}{a} < frac{1}{b}) to hold is ( ). The options are A: (b < a < 0), B: (a < b), C: (b(a - b) > 0), and D: (a > b).Okay, let's break this down. First, I need to understand what it means for a condition to be sufficient but not necessary. A sufficient condition means that if the condition is met, then the statement (frac{1}{a} < frac{1}{b}) must be true. However, it's not necessary, meaning there might be other scenarios where (frac{1}{a} < frac{1}{b}) is true even if this condition isn't met.So, I need to check each option to see if it guarantees (frac{1}{a} < frac{1}{b}), but also ensure that there are cases where (frac{1}{a} < frac{1}{b}) is true without that condition being satisfied.Let's start with option A: (b < a < 0). This means both (a) and (b) are negative numbers, and (b) is less than (a). Since both are negative, their reciprocals will also be negative. Remember, when you take reciprocals of negative numbers, the inequality flips. So, if (b < a < 0), then (frac{1}{b} > frac{1}{a}), which means (frac{1}{a} < frac{1}{b}). So, this condition does guarantee the inequality. But is it necessary? Let's see. Suppose (a) is positive and (b) is negative. Then (frac{1}{a}) is positive and (frac{1}{b}) is negative, so (frac{1}{a} > frac{1}{b}). That doesn't help. What if both are positive? If (a > b > 0), then (frac{1}{a} < frac{1}{b}). So, in that case, the condition (b < a < 0) isn't met, but the inequality still holds. Therefore, (b < a < 0) is sufficient but not necessary. Hmm, that seems like a good candidate.Moving on to option B: (a < b). This is a straightforward inequality. If (a < b), does that guarantee (frac{1}{a} < frac{1}{b})? Well, it depends on the signs of (a) and (b). If both are positive, then yes, (a < b) implies (frac{1}{a} > frac{1}{b}), which is the opposite of what we want. If both are negative, then (a < b) would mean that (a) is more negative than (b), so taking reciprocals would flip the inequality again, giving (frac{1}{a} < frac{1}{b}). But if (a) is negative and (b) is positive, then (frac{1}{a}) is negative and (frac{1}{b}) is positive, so (frac{1}{a} < frac{1}{b}). So, (a < b) can lead to (frac{1}{a} < frac{1}{b}) in some cases but not all. Therefore, it's not a sufficient condition because it doesn't always guarantee the inequality.Option C: (b(a - b) > 0). Let's parse this. If (b(a - b) > 0), then either both (b) and (a - b) are positive, or both are negative. So, case 1: (b > 0) and (a - b > 0), which implies (a > b > 0). In this case, (frac{1}{a} < frac{1}{b}) because reciprocals reverse the inequality for positive numbers. Case 2: (b < 0) and (a - b < 0), which implies (a < b < 0). Here, since both are negative, (a < b) means (frac{1}{a} > frac{1}{b}), which is the opposite of what we want. So, in this case, (b(a - b) > 0) can lead to both (frac{1}{a} < frac{1}{b}) and (frac{1}{a} > frac{1}{b}), depending on the signs. Therefore, it's not a guaranteed condition for (frac{1}{a} < frac{1}{b}).Option D: (a > b). Similar to option B, this depends on the signs. If both are positive, (a > b) implies (frac{1}{a} < frac{1}{b}). If both are negative, (a > b) means (a) is less negative than (b), so (frac{1}{a} > frac{1}{b}). If (a) is positive and (b) is negative, then (frac{1}{a} < frac{1}{b}) because (frac{1}{a}) is positive and (frac{1}{b}) is negative. So, (a > b) can lead to (frac{1}{a} < frac{1}{b}) in some cases but not all. Therefore, it's not a sufficient condition.Wait, but in option A, (b < a < 0), we saw that it's sufficient but not necessary because there are other cases where (frac{1}{a} < frac{1}{b}) holds, like when both are positive and (a > b). So, A seems to be the correct answer.But let me double-check. If (b < a < 0), then both are negative, and (b) is less than (a), meaning (b) is more negative. So, taking reciprocals, since they are both negative, the one with the smaller absolute value has a larger reciprocal. So, (frac{1}{b} > frac{1}{a}), which means (frac{1}{a} < frac{1}{b}). That's correct.Is there a case where (frac{1}{a} < frac{1}{b}) is true without (b < a < 0)? Yes, for example, if (a = 2) and (b = 1), both positive, then (frac{1}{2} < frac{1}{1}), but (b < a < 0) isn't satisfied because they're positive. So, A is sufficient but not necessary.Options B and D are not sufficient because they don't always lead to the inequality, and option C can lead to both the inequality and its opposite, so it's not sufficient either.Therefore, the answer should be A.

💡Okay, so I have this problem here: Given a point P(x, y) on the ellipse x² + 4y² = 4, I need to find the maximum value of the expression (3/4)x² + 2x - y². The options are A: 8, B: 7, C: 2, D: -1. Hmm, let me try to figure this out step by step.First, I know that the equation x² + 4y² = 4 is an ellipse. To get a better understanding, maybe I should write it in the standard form. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1. So if I divide both sides of the given equation by 4, I get (x²/4) + (y²/1) = 1. That means a² = 4 and b² = 1, so a = 2 and b = 1. Therefore, the ellipse has a major axis of length 2a = 4 along the x-axis and a minor axis of length 2b = 2 along the y-axis.Now, the point P(x, y) lies on this ellipse, so it must satisfy x² + 4y² = 4. I need to maximize the expression (3/4)x² + 2x - y². Let me write that down:Expression to maximize: (3/4)x² + 2x - y².Since P is on the ellipse, I can express y² in terms of x². From the ellipse equation:x² + 4y² = 4 => 4y² = 4 - x² => y² = (4 - x²)/4 => y² = 1 - (x²)/4.So, y² is equal to 1 - (x²)/4. Let me substitute this into the expression:Expression becomes: (3/4)x² + 2x - [1 - (x²)/4].Let me simplify this step by step. First, distribute the negative sign:= (3/4)x² + 2x - 1 + (x²)/4.Now, combine like terms. The terms with x² are (3/4)x² and (1/4)x². Adding them together:(3/4 + 1/4)x² = (4/4)x² = x².So, the expression simplifies to:x² + 2x - 1.Alright, so now the expression to maximize is x² + 2x - 1. That's a quadratic in terms of x. Since it's a quadratic, I can find its maximum or minimum by completing the square or using calculus. But since we're dealing with a closed interval (because x is bounded by the ellipse), I can also evaluate the expression at the endpoints and critical points.Wait, let me think. The ellipse x² + 4y² = 4 implies that x ranges from -2 to 2 because when y = 0, x² = 4, so x = ±2. So x ∈ [-2, 2]. Therefore, the domain of x is from -2 to 2.So, the expression x² + 2x - 1 is a quadratic function, and since the coefficient of x² is positive (1), it opens upwards, meaning it has a minimum point, not a maximum. But we're looking for the maximum value on the interval [-2, 2]. So, the maximum must occur at one of the endpoints of the interval.Therefore, I should evaluate the expression at x = -2 and x = 2 and see which one gives the higher value.Let me compute the expression at x = -2:f(-2) = (-2)² + 2*(-2) - 1 = 4 - 4 - 1 = -1.Now, at x = 2:f(2) = (2)² + 2*(2) - 1 = 4 + 4 - 1 = 7.So, f(-2) = -1 and f(2) = 7. Therefore, the maximum value is 7.Wait, but just to be thorough, even though the quadratic opens upwards, maybe there's a point inside the interval where the function is higher? Let me check the vertex of the parabola.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). In this case, a = 1, b = 2, so x = -2/(2*1) = -1.So, the vertex is at x = -1. Let me compute f(-1):f(-1) = (-1)² + 2*(-1) - 1 = 1 - 2 - 1 = -2.So, the minimum value is -2 at x = -1, which is within the interval. But since we're looking for the maximum, the maximum occurs at x = 2, which is 7.Therefore, the maximum value is 7, which corresponds to option B.Wait, but let me double-check my substitution earlier. I substituted y² = 1 - (x²)/4 into the expression (3/4)x² + 2x - y², which gave me x² + 2x - 1. Let me verify that:(3/4)x² + 2x - y² = (3/4)x² + 2x - [1 - (x²)/4] = (3/4)x² + 2x - 1 + (x²)/4 = (3/4 + 1/4)x² + 2x - 1 = x² + 2x - 1.Yes, that seems correct. So, the expression simplifies correctly.Alternatively, I could use calculus to find the maximum. Let's try that approach as a verification.Let me consider the expression f(x) = x² + 2x - 1. To find its extrema on the interval [-2, 2], I can take the derivative and set it equal to zero.f'(x) = 2x + 2.Set f'(x) = 0:2x + 2 = 0 => 2x = -2 => x = -1.So, the critical point is at x = -1, which we already found. Evaluating f at x = -1 gives the minimum, as we saw. Therefore, the maximum must be at one of the endpoints, which are x = -2 and x = 2.As calculated earlier, f(-2) = -1 and f(2) = 7. So, the maximum is indeed 7.Another way to approach this problem is by using Lagrange multipliers, which is a method for finding the extrema of a function subject to equality constraints. Let me try that as well.Let me denote the function to maximize as f(x, y) = (3/4)x² + 2x - y², and the constraint is g(x, y) = x² + 4y² - 4 = 0.The method of Lagrange multipliers tells us that at the extrema, the gradient of f is proportional to the gradient of g. That is, ∇f = λ∇g, where λ is the Lagrange multiplier.Compute the gradients:∇f = (df/dx, df/dy) = ( (3/2)x + 2, -2y )∇g = (dg/dx, dg/dy) = (2x, 8y )So, setting ∇f = λ∇g:( (3/2)x + 2, -2y ) = λ(2x, 8y )This gives us two equations:1. (3/2)x + 2 = 2λx 2. -2y = 8λyFrom equation 2: -2y = 8λy.Assuming y ≠ 0, we can divide both sides by y:-2 = 8λ => λ = -2/8 = -1/4.If y = 0, then from the constraint x² + 4y² = 4, we get x² = 4, so x = ±2. Let's check these points later.Now, substitute λ = -1/4 into equation 1:(3/2)x + 2 = 2*(-1/4)x => (3/2)x + 2 = (-1/2)x => (3/2)x + (1/2)x = -2 => (4/2)x = -2 => 2x = -2 => x = -1.So, x = -1. Then, from the constraint x² + 4y² = 4:(-1)² + 4y² = 4 => 1 + 4y² = 4 => 4y² = 3 => y² = 3/4 => y = ±√(3)/2.So, the critical points are (-1, √3/2) and (-1, -√3/2). Let's evaluate f at these points:f(-1, √3/2) = (3/4)(-1)² + 2*(-1) - (√3/2)² = (3/4)(1) - 2 - (3/4) = 3/4 - 2 - 3/4 = (3/4 - 3/4) - 2 = 0 - 2 = -2.Similarly, f(-1, -√3/2) will also be -2 because y² is the same.Now, let's check the points where y = 0, which are x = ±2.At x = 2, y = 0:f(2, 0) = (3/4)(4) + 2*(2) - 0 = 3 + 4 = 7.At x = -2, y = 0:f(-2, 0) = (3/4)(4) + 2*(-2) - 0 = 3 - 4 = -1.So, the function f(x, y) attains the value -2 at (-1, ±√3/2), -1 at (-2, 0), and 7 at (2, 0). Therefore, the maximum value is 7, which occurs at (2, 0).This confirms the earlier result obtained by substitution and calculus.Alternatively, I could parameterize the ellipse and then maximize the expression. Let me try that approach as well for thoroughness.Parametrizing the ellipse: Since the standard parametric equations for an ellipse are x = a cosθ and y = b sinθ, where a and b are the semi-major and semi-minor axes, respectively.From the ellipse equation x² + 4y² = 4, we have a² = 4 and b² = 1, so a = 2 and b = 1.Therefore, the parametric equations are:x = 2 cosθ y = sinθ.Now, substitute these into the expression f(x, y) = (3/4)x² + 2x - y².Compute each term:x² = (2 cosθ)² = 4 cos²θ y² = (sinθ)² = sin²θ.So, f(x, y) becomes:(3/4)(4 cos²θ) + 2*(2 cosθ) - sin²θ = 3 cos²θ + 4 cosθ - sin²θ.Now, let's simplify this expression. Recall that sin²θ = 1 - cos²θ, so substitute that:= 3 cos²θ + 4 cosθ - (1 - cos²θ) = 3 cos²θ + 4 cosθ - 1 + cos²θ = (3 cos²θ + cos²θ) + 4 cosθ - 1 = 4 cos²θ + 4 cosθ - 1.So, f(θ) = 4 cos²θ + 4 cosθ - 1.Now, let me write this as a function of cosθ. Let me set u = cosθ, where u ∈ [-1, 1] because cosine ranges between -1 and 1.So, f(u) = 4u² + 4u - 1.Now, I need to find the maximum of f(u) = 4u² + 4u - 1 for u ∈ [-1, 1].This is a quadratic function in u. The coefficient of u² is 4, which is positive, so the parabola opens upwards, meaning it has a minimum at its vertex and the maximum occurs at one of the endpoints of the interval.The vertex of this parabola is at u = -b/(2a) = -4/(2*4) = -4/8 = -0.5.So, the vertex is at u = -0.5. Let's compute f(-0.5):f(-0.5) = 4*(-0.5)² + 4*(-0.5) - 1 = 4*(0.25) - 2 - 1 = 1 - 2 - 1 = -2.So, the minimum value is -2 at u = -0.5, which corresponds to θ where cosθ = -0.5, i.e., θ = 120° or 240°, which aligns with the earlier critical points.Now, let's evaluate f(u) at the endpoints u = -1 and u = 1.At u = -1:f(-1) = 4*(-1)² + 4*(-1) - 1 = 4*1 - 4 - 1 = 4 - 4 - 1 = -1.At u = 1:f(1) = 4*(1)² + 4*(1) - 1 = 4 + 4 - 1 = 7.So, the maximum value of f(u) is 7 at u = 1, which corresponds to cosθ = 1, meaning θ = 0°, so x = 2*1 = 2, y = sin0 = 0. Therefore, the point is (2, 0), as found earlier.This method also confirms that the maximum value is 7.Another approach could be using the method of completing the square on the quadratic expression in x. Let me try that as well.We had earlier simplified the expression to f(x) = x² + 2x - 1.Completing the square:f(x) = x² + 2x - 1 = (x² + 2x) - 1 = (x² + 2x + 1) - 1 - 1 = (x + 1)² - 2.So, f(x) = (x + 1)² - 2.Since (x + 1)² is always non-negative, the minimum value occurs when (x + 1)² is minimized, which is 0 when x = -1, giving f(x) = -2. The maximum value occurs when (x + 1)² is maximized on the interval x ∈ [-2, 2].The expression (x + 1)² is a parabola opening upwards, so it attains its maximum at the endpoints of the interval. Let's compute (x + 1)² at x = -2 and x = 2.At x = -2:(x + 1)² = (-2 + 1)² = (-1)² = 1 So, f(x) = 1 - 2 = -1.At x = 2:(x + 1)² = (2 + 1)² = 3² = 9 So, f(x) = 9 - 2 = 7.Therefore, the maximum value is 7 at x = 2.This again confirms the result.I think I've approached this problem from multiple angles: substitution, calculus, Lagrange multipliers, parametrization, and completing the square. All methods lead to the conclusion that the maximum value is 7, achieved at the point (2, 0) on the ellipse.Just to ensure there's no oversight, let me consider if there's any other point on the ellipse where the expression could be larger than 7. Since all methods consistently give 7 as the maximum, and the function doesn't have any other critical points outside of what's already considered, I can be confident that 7 is indeed the correct answer.Therefore, the answer is B: 7.

💡First, I need to determine the number of pens Alex and Jane have after 4 weeks.For Alex:- Let ( A ) be the initial number of pens.- Each week, Alex's pens multiply by ( X ).- After 4 weeks, Alex's pens will be ( A times X^4 ).For Jane:- Let ( B ) be the initial number of pens.- Each week, Jane's pens multiply by ( Y ).- After 4 weeks, Jane's pens will be ( B times Y^4 ).Given that Alex ends up with ( M ) pens and Jane with ( N ) pens after 4 weeks, I can set up the following equations:[M = A times X^4][N = B times Y^4]To find out how many more pens Alex has than Jane, I subtract the number of pens Jane has from the number of pens Alex has:[text{Difference} = M - N]Therefore, the number of additional pens Alex has compared to Jane after 4 weeks is ( M - N ).