💡Okay, so I've got this geometry problem here, and I need to find the value of x + y. Let me try to visualize the figure based on the Asymptote code provided. It seems like there's a rectangle or some kind of figure with points A, B, and C at the top, and angles x and y marked at specific points. The given angles are m∠A = 28°, m∠B = 74°, and m∠C = 26°. First, I need to understand the structure of the figure. The Asymptote code draws a rectangle from (0,5) to (0,0) to (15,0) to (15,5), so that's a rectangle with width 15 and height 5. Then, there are some internal lines: from A (0,5) to (2,2), then to B (5,5), then to (12,-2), and finally to C (15,5). So, it's creating a polygon or a series of triangles within the rectangle. Angles x and y are marked at (2.5,2.5) and (12,-2), which are points along the lines drawn inside the rectangle. The figure also has some small right angle symbols at the bottom left and bottom right corners, indicating that those are right angles. So, let me try to break this down. There are two angles, x and y, which are part of some triangles inside the rectangle. The given angles at A, B, and C are 28°, 74°, and 26°, respectively. I need to figure out how these angles relate to x and y and then find their sum.Starting with angle A, which is 28°. Since A is at the top-left corner of the rectangle, and the figure is drawn from A to (2,2), which is somewhere inside the rectangle. Then from (2,2) to B at (5,5), which is another point on the top side of the rectangle. Then from B to (12,-2), which is below the rectangle, and then to C at (15,5), the top-right corner.So, this creates a polygon that goes from A to (2,2) to B to (12,-2) to C and back to A, forming a five-sided figure. So, it's a pentagon with vertices at A, (2,2), B, (12,-2), and C.Given that, I can consider the angles at each vertex of this pentagon. The angles at A, B, and C are given as 28°, 74°, and 26°, respectively. The other two angles are at (2,2) and (12,-2), which are labeled as x° and y°. So, the pentagon has angles 28°, x°, 74°, y°, and 26°. Wait, but in a pentagon, the sum of the interior angles is (5-2)*180° = 540°. So, if I can find the sum of all the angles, I can set up an equation to solve for x + y.But hold on, the angles at (2,2) and (12,-2) might not be the interior angles of the pentagon because the figure is drawn with lines crossing outside the rectangle. So, maybe I need to be careful here.Looking back at the Asymptote code, the figure is drawn as (0,5)--(2,2)--(5,5)--(12,-2)--(15,5). So, from A (0,5) to (2,2), which is inside the rectangle, then to B (5,5), then to (12,-2), which is outside the rectangle below the bottom side, then to C (15,5). So, the figure is a polygon that goes from A to (2,2) to B to (12,-2) to C and back to A.Therefore, the angles at (2,2) and (12,-2) are indeed part of this pentagon. So, the sum of the interior angles of this pentagon is 540°, as I thought earlier. So, if I can find the measures of the other three angles, which are given as 28°, 74°, and 26°, I can set up an equation.But wait, the angles at A, B, and C are given. So, the pentagon has angles at A: 28°, at (2,2): x°, at B: 74°, at (12,-2): y°, and at C: 26°. So, adding those up: 28 + x + 74 + y + 26 = 540.Simplifying that: 28 + 74 + 26 = 128, so 128 + x + y = 540. Therefore, x + y = 540 - 128 = 412°. Wait, that can't be right because angles in a polygon can't exceed 180°, and 412° is way too high. So, I must have made a mistake here.Hmm, maybe the angles at A, B, and C are not the interior angles of the pentagon but rather some other angles. Let me think again.Looking at the figure, angle A is at (0,5), which is a corner of the rectangle. The line from A goes to (2,2), which is inside the rectangle. So, the angle at A in the pentagon is formed by the sides from A to (2,2) and from A to (0,0). But wait, the rectangle has a right angle at A, which is 90°, but the angle given is 28°, which is much smaller. So, perhaps the angle at A in the pentagon is 28°, which is different from the rectangle's angle.Similarly, at point B (5,5), which is another corner on the top side of the rectangle, the angle given is 74°, which is also different from the rectangle's angle, which would be 90°. So, maybe the pentagon's angles at A, B, and C are 28°, 74°, and 26°, respectively, and the other two angles are x and y.But then, as I calculated earlier, 28 + 74 + 26 + x + y = 540, so x + y = 412°, which is impossible because each angle in a convex pentagon must be less than 180°, and their sum is 540°. So, 412° for x + y would mean each is over 200°, which is not possible. So, I must have misunderstood the figure.Wait, maybe the angles at A, B, and C are not the interior angles of the pentagon but some exterior angles or angles formed by the intersecting lines. Let me try to analyze the figure more carefully.Looking at the Asymptote code, the figure is drawn as follows:- The rectangle is from (0,5) to (0,0) to (15,0) to (15,5) to (0,5).- Then, internal lines are drawn from (0,5) to (2,2), then to (5,5), then to (12,-2), then to (15,5).So, the figure is a polygon with vertices at (0,5), (2,2), (5,5), (12,-2), (15,5), and back to (0,5). So, it's a pentagon with those five vertices.At each vertex, there is an angle. The angles at (0,5), (5,5), and (15,5) are given as 28°, 74°, and 26°, respectively. The other two angles are at (2,2) and (12,-2), which are labeled as x° and y°.So, the pentagon has angles at five points: A (28°), (2,2) (x°), B (74°), (12,-2) (y°), and C (26°). So, the sum of these angles should be 540°, as it's a pentagon.Therefore, 28 + x + 74 + y + 26 = 540.Adding the known angles: 28 + 74 + 26 = 128.So, 128 + x + y = 540.Therefore, x + y = 540 - 128 = 412°.But as I thought earlier, this is impossible because in a convex pentagon, each interior angle must be less than 180°, and their sum is 540°. So, x + y = 412° would mean that each of x and y is greater than 200°, which is not possible in a convex polygon.This suggests that my initial assumption is wrong. Maybe the angles at A, B, and C are not the interior angles of the pentagon but some other angles formed by the intersecting lines.Let me try to think differently. Perhaps the angles at A, B, and C are angles formed by the intersection of the internal lines with the sides of the rectangle.Looking at point A (0,5), which is connected to (2,2). So, the line from A to (2,2) forms an angle with the top side of the rectangle. Similarly, at point B (5,5), the line from (2,2) to B forms an angle with the top side, and at point C (15,5), the line from (12,-2) to C forms an angle with the top side.So, maybe the given angles at A, B, and C are the angles between the internal lines and the top side of the rectangle. So, these are angles formed by the intersection of the internal lines with the top side.In that case, these angles are not the interior angles of the pentagon but rather some other angles. So, I need to find a way to relate these given angles to the angles x and y.Let me try to analyze the triangles formed by these internal lines.First, looking at point A (0,5). The line from A goes to (2,2). So, this line forms an angle of 28° with the top side of the rectangle. Since the top side is horizontal, the angle between the line from A to (2,2) and the horizontal is 28°. Similarly, at point B (5,5), the line from (2,2) to B forms an angle of 74° with the top side, and at point C (15,5), the line from (12,-2) to C forms an angle of 26° with the top side.So, these angles are the angles between the internal lines and the top side of the rectangle. Therefore, they are angles formed by the intersection of the internal lines with the top side.Given that, I can consider the triangles formed by these internal lines.First, let's consider the triangle formed by points A, (2,2), and (0,0). Wait, but (0,0) is the bottom-left corner of the rectangle. Is that connected? Looking back at the Asymptote code, it draws the rectangle and then the internal lines. So, the internal lines are from A to (2,2), then to B, then to (12,-2), then to C.So, the figure is a polygon with vertices at A, (2,2), B, (12,-2), C, and back to A. So, it's a pentagon, but the lines from A to (2,2) and from B to (12,-2) are internal lines within the rectangle.So, perhaps I can consider the triangles formed by these internal lines and the sides of the rectangle.Let me try to analyze the triangle at the top-left corner. From A (0,5) to (2,2), which is inside the rectangle, and then to (0,0), the bottom-left corner. So, triangle A-(2,2)-(0,0). Similarly, at the top-right corner, from C (15,5) to (12,-2), which is below the rectangle, and then to (15,0), the bottom-right corner. So, triangle C-(12,-2)-(15,0).In these triangles, the angles at A and C are given as 28° and 26°, respectively. So, in triangle A-(2,2)-(0,0), the angle at A is 28°, and in triangle C-(12,-2)-(15,0), the angle at C is 26°.Similarly, at point B (5,5), the line from (2,2) to B forms an angle of 74° with the top side. So, in triangle (2,2)-B-(5,0), the angle at B is 74°.Wait, but (5,0) is not drawn in the figure. Maybe I need to consider the triangle formed by (2,2), B, and some other point. Alternatively, perhaps the angle at B is part of another triangle.Alternatively, maybe I should consider the angles formed at the intersection points.Wait, perhaps I can use the concept of corresponding angles or alternate interior angles due to the transversal lines crossing the horizontal sides of the rectangle.Given that, the lines from A to (2,2) and from C to (12,-2) are transversals crossing the horizontal top and bottom sides of the rectangle. Since the top and bottom sides are parallel, the angles formed by these transversals can be related.So, at point A, the angle between the top side and the line to (2,2) is 28°, which is the angle of incidence. Similarly, at point C, the angle between the top side and the line to (12,-2) is 26°. Since the top and bottom sides are parallel, the angles formed by the transversals on the bottom side should be equal to the angles on the top side. So, the angle between the line from (2,2) to (0,0) and the bottom side at (0,0) should be equal to 28°, and the angle between the line from (12,-2) to (15,0) and the bottom side at (15,0) should be equal to 26°.Wait, but in the Asymptote code, the figure is drawn with lines from A to (2,2), then to B, then to (12,-2), then to C. So, the lines from (2,2) to B and from B to (12,-2) are also part of the figure.So, perhaps I can consider the angles at (2,2) and (12,-2) as part of triangles formed with the sides of the rectangle.Let me try to analyze the triangle at (2,2). The lines from A to (2,2) and from (2,2) to B form two sides of a triangle, with the base being from A to B along the top side of the rectangle.Wait, but A is at (0,5) and B is at (5,5), so the distance between A and B is 5 units along the top side. The line from A to (2,2) is a diagonal, and the line from (2,2) to B is another diagonal.So, triangle A-(2,2)-B is a triangle with sides from A to (2,2), (2,2) to B, and A to B. The angle at A is 28°, and the angle at B is 74°, so the third angle at (2,2) can be found using the fact that the sum of angles in a triangle is 180°.So, angle at (2,2) = 180° - 28° - 74° = 78°. So, x°, which is the angle at (2,2), is 78°.Wait, but in the figure, x is labeled at (2.5,2.5), which is along the line from A to (2,2). So, maybe x is not the angle at (2,2) but some other angle.Wait, perhaps I need to consider the angle between the line from (2,2) to B and the line from (2,2) to (0,0). Since (0,0) is the bottom-left corner, and the line from (2,2) to (0,0) is part of the rectangle's side.Wait, no, the line from (2,2) to (0,0) is not drawn in the figure. The figure only draws from A to (2,2), then to B, then to (12,-2), then to C.So, perhaps the angle x is the angle between the line from A to (2,2) and the line from (2,2) to B. So, in triangle A-(2,2)-B, the angle at (2,2) is x°, which we calculated as 78°.Similarly, at point (12,-2), the angle y is the angle between the line from B to (12,-2) and the line from (12,-2) to C. So, in triangle B-(12,-2)-C, the angle at (12,-2) is y°.Given that, let's analyze triangle B-(12,-2)-C. The angle at B is 74°, and the angle at C is 26°, so the angle at (12,-2) would be 180° - 74° - 26° = 80°. So, y° is 80°.Wait, but if x is 78° and y is 80°, then x + y = 78 + 80 = 158°. But I'm not sure if this is correct because the figure might involve more complex relationships.Alternatively, perhaps the angles x and y are exterior angles related to the triangles we've considered.Wait, let's think about the lines from (2,2) to B and from B to (12,-2). The angle at B is given as 74°, which is the angle between the top side and the line from B to (12,-2). So, in triangle B-(12,-2)-C, the angle at B is 74°, and the angle at C is 26°, so the angle at (12,-2) is 80°, as I calculated earlier.Similarly, in triangle A-(2,2)-B, the angle at A is 28°, the angle at B is 74°, so the angle at (2,2) is 78°, which is x°.Therefore, x + y = 78° + 80° = 158°.But wait, earlier I thought that the sum of the angles in the pentagon should be 540°, but that led to an impossible result. So, maybe the pentagon's angles are not just the given angles and x and y, but also include other angles.Alternatively, perhaps the figure is not a simple pentagon but a more complex polygon with intersecting lines, creating multiple triangles and angles.Wait, looking back at the Asymptote code, the figure is drawn as follows:- The rectangle is drawn first.- Then, internal lines are drawn from A to (2,2), then to B, then to (12,-2), then to C.So, the figure is a polygon with vertices at A, (2,2), B, (12,-2), C, and back to A. So, it's a pentagon, but the lines from (2,2) to B and from B to (12,-2) are part of this pentagon.Therefore, the angles at (2,2) and (12,-2) are indeed part of the pentagon. So, the sum of the interior angles of the pentagon is 540°, as I thought earlier.Given that, the angles at A, B, and C are 28°, 74°, and 26°, respectively, and the angles at (2,2) and (12,-2) are x and y. So, 28 + 74 + 26 + x + y = 540.Adding the known angles: 28 + 74 + 26 = 128.So, 128 + x + y = 540.Therefore, x + y = 540 - 128 = 412°.But as I thought earlier, this is impossible because in a convex pentagon, each interior angle must be less than 180°, and their sum is 540°. So, x + y = 412° would mean that each of x and y is greater than 200°, which is not possible in a convex polygon.This suggests that the pentagon is not convex but concave, meaning some of its interior angles are greater than 180°. However, even in a concave polygon, the sum of the interior angles remains the same, but individual angles can be reflex angles (greater than 180°).But in this case, the figure is drawn within a rectangle, and the points (2,2) and (12,-2) are inside and outside the rectangle, respectively. So, the pentagon might be concave at (12,-2), where y is located.But even so, if y is a reflex angle, it would be greater than 180°, and x could be a regular angle. But let's see.If y is a reflex angle, then y = 360° - y', where y' is the acute angle at that vertex. Similarly, x could be a regular angle.But I'm not sure if that's the case. Alternatively, perhaps the angles at A, B, and C are not the interior angles of the pentagon but some other angles.Wait, maybe the angles at A, B, and C are the angles between the internal lines and the top side of the rectangle, which are given as 28°, 74°, and 26°, respectively. So, these are not the interior angles of the pentagon but the angles formed by the internal lines with the top side.In that case, the interior angles of the pentagon at A, B, and C would be different.So, perhaps I need to find the interior angles of the pentagon at A, B, and C, which are different from the given angles.Given that, let's consider the lines from A to (2,2) and from C to (12,-2). These lines form angles of 28° and 26° with the top side of the rectangle, respectively.Since the top side is horizontal, the slope of the line from A to (2,2) can be calculated. The change in y is from 5 to 2, which is -3, and the change in x is from 0 to 2, which is +2. So, the slope is -3/2, which corresponds to an angle of arctangent(3/2) below the horizontal. Since the angle given is 28°, which is the angle between the line and the horizontal, that makes sense.Similarly, the line from C (15,5) to (12,-2) has a change in y of -7 and a change in x of -3, so the slope is 7/3, which corresponds to an angle of arctangent(7/3) below the horizontal. The angle given is 26°, which is the angle between the line and the horizontal.So, these angles are the angles between the internal lines and the top side of the rectangle.Given that, perhaps the interior angles of the pentagon at A, B, and C are supplementary to these given angles.Wait, at point A, the internal line makes an angle of 28° with the top side. The interior angle of the pentagon at A would be the angle between the internal line and the side of the pentagon going from A to (2,2). Similarly, at point C, the internal line makes an angle of 26° with the top side, and the interior angle of the pentagon at C would be the angle between the internal line and the side going from C to (12,-2).But I'm not sure if that's the case. Alternatively, perhaps the interior angles at A, B, and C are the given angles, and the lines form those angles with the top side.Wait, maybe I can consider the angles formed by the internal lines with the sides of the rectangle and use that to find the interior angles of the pentagon.At point A, the internal line makes an angle of 28° with the top side. Since the top side is horizontal, the internal line is going downward to the right at 28°. Similarly, at point C, the internal line makes an angle of 26° with the top side, going downward to the left.At point B, the internal line from (2,2) to B makes an angle of 74° with the top side. So, the internal line is going upward to the right at 74° from the top side.Wait, but the top side is horizontal, so an angle of 74° above the horizontal would mean the line is going upward to the right. However, in the figure, the line from (2,2) to B is going from (2,2) to (5,5), which is upward to the right, so that makes sense.Given that, perhaps the interior angles of the pentagon at A, B, and C are related to these angles.Wait, perhaps the interior angles at A, B, and C are the angles between the internal lines and the sides of the pentagon.At point A, the internal line makes an angle of 28° with the top side. The interior angle of the pentagon at A would be the angle between the internal line and the side going from A to (2,2). Since the internal line is going downward to the right at 28°, the interior angle would be 180° - 28° = 152°.Similarly, at point C, the internal line makes an angle of 26° with the top side, going downward to the left. So, the interior angle at C would be 180° - 26° = 154°.At point B, the internal line makes an angle of 74° with the top side, going upward to the right. So, the interior angle at B would be 180° - 74° = 106°.Wait, but the given angles at A, B, and C are 28°, 74°, and 26°, respectively. So, if the interior angles of the pentagon at these points are 152°, 106°, and 154°, then the sum of the interior angles would be 152 + 106 + 154 + x + y = 540.Adding those: 152 + 106 = 258, 258 + 154 = 412. So, 412 + x + y = 540, which means x + y = 128°.That seems more reasonable because 128° is less than 540°, and individual angles could be, say, 64° each, which is possible.But wait, how did I get the interior angles at A, B, and C as 152°, 106°, and 154°? Let me verify that.At point A, the internal line makes an angle of 28° with the top side. Since the pentagon is going from A to (2,2), the interior angle at A is the angle between the top side and the side going to (2,2). Since the internal line is going downward to the right at 28°, the interior angle would be 180° - 28° = 152°.Similarly, at point C, the internal line makes an angle of 26° with the top side, going downward to the left. So, the interior angle at C is 180° - 26° = 154°.At point B, the internal line makes an angle of 74° with the top side, going upward to the right. So, the interior angle at B is 180° - 74° = 106°.Therefore, the interior angles at A, B, and C are 152°, 106°, and 154°, respectively.Adding these up: 152 + 106 + 154 = 412°.Since the sum of the interior angles of a pentagon is 540°, the remaining two angles at (2,2) and (12,-2) must add up to 540 - 412 = 128°. Therefore, x + y = 128°.So, the value of x + y is 128°.

💡Okay, so I need to figure out how many ways there are to choose two items from a set of ( n ) items. Hmm, let me think about this step by step.First, I know that when we're talking about choosing items without considering the order, it's called a combination. So, choosing two items from ( n ) is a combination problem. I remember there's a formula for combinations, something like ( C(n, k) ), where ( n ) is the total number of items, and ( k ) is the number of items we want to choose.But wait, I'm not entirely sure about the exact formula. Let me try to recall. I think it involves factorials. Yeah, something like ( C(n, k) = frac{n!}{k!(n - k)!} ). So, if I plug in ( k = 2 ), it should give me the number of ways to choose two items from ( n ).Let me write that down:[C(n, 2) = frac{n!}{2!(n - 2)!}]Okay, now I need to simplify this expression. I know that ( n! ) means ( n times (n - 1) times (n - 2)! ). So, I can rewrite ( n! ) as ( n times (n - 1) times (n - 2)! ). That should help cancel out the ( (n - 2)! ) in the denominator.So, substituting back in:[C(n, 2) = frac{n times (n - 1) times (n - 2)!}{2! times (n - 2)!}]Now, I can cancel out the ( (n - 2)! ) terms:[C(n, 2) = frac{n times (n - 1)}{2!}]And since ( 2! = 2 times 1 = 2 ), this simplifies to:[C(n, 2) = frac{n times (n - 1)}{2}]So, the number of ways to choose two items from ( n ) items is ( frac{n(n - 1)}{2} ).Let me check if this makes sense with an example. Suppose ( n = 5 ). Then, according to the formula, the number of ways should be:[C(5, 2) = frac{5 times 4}{2} = frac{20}{2} = 10]And I remember from problem 56 that ( C(5, 2) = 10 ), which matches. So, this seems correct.Another way to think about it is by considering each item and how many ways it can pair with others. For the first item, there are ( n - 1 ) items it can pair with. For the second item, there are ( n - 2 ) new items it can pair with, and so on. But this approach counts each pair twice (once for each order), so we need to divide by 2 to correct for that overcounting.So, that's another way to arrive at the same formula:[frac{n(n - 1)}{2}]Yeah, that makes sense. It's like counting all possible ordered pairs and then adjusting for the fact that order doesn't matter in combinations.I think I've got it. The formula ( frac{n(n - 1)}{2} ) gives the number of ways to choose two items from ( n ) items without considering the order.

💡Okay, so I have this problem about complex numbers. It says that a complex number z is four-presentable if there's another complex number w with absolute value 4 such that z equals w minus 1 over w. Then, U is the set of all such z, and it forms a closed curve. I need to find the area inside this curve.Hmm, let me start by understanding what it means for z to be four-presentable. So, z = w - 1/w, and |w| = 4. That means w is a complex number lying on a circle with radius 4 centered at the origin in the complex plane.I remember that for complex numbers, if |w| = 4, then the reciprocal 1/w has absolute value 1/4. So, 1/w is like scaling w by 1/|w|², which in this case is 1/16. So, 1/w is a complex number with absolute value 1/4.So, z is the difference between w and 1/w. Since both w and 1/w are complex numbers, their difference will trace out some curve in the complex plane. The set U is all such z, so it's the image of the circle |w| = 4 under the transformation f(w) = w - 1/w.I think this kind of transformation might result in an ellipse or some kind of conic section. Let me recall that when you have a function like w + 1/w, it can parametrize an ellipse. But here it's w - 1/w, so maybe similar.Let me try to parametrize w. Since |w| = 4, I can write w in polar form as w = 4e^{iθ}, where θ ranges from 0 to 2π. Then, 1/w is (1/4)e^{-iθ}, because 1/w = overline{w}/|w|², which is (4e^{-iθ})/16 = (1/4)e^{-iθ}.So, substituting into z, we have:z = w - 1/w = 4e^{iθ} - (1/4)e^{-iθ}Let me write this in terms of sine and cosine to see the real and imaginary parts.First, e^{iθ} = cosθ + i sinθ, and e^{-iθ} = cosθ - i sinθ.So, substituting:z = 4(cosθ + i sinθ) - (1/4)(cosθ - i sinθ)Let me distribute the constants:z = 4cosθ + 4i sinθ - (1/4)cosθ + (1/4)i sinθNow, combine like terms:Real part: 4cosθ - (1/4)cosθ = (4 - 1/4)cosθ = (16/4 - 1/4)cosθ = (15/4)cosθImaginary part: 4i sinθ + (1/4)i sinθ = (4 + 1/4)i sinθ = (16/4 + 1/4)i sinθ = (17/4)i sinθSo, z can be written as:z = (15/4)cosθ + i(17/4)sinθThis looks like the parametric equation of an ellipse. The standard parametric form of an ellipse is:x = a cosθy = b sinθWhere a is the semi-major axis and b is the semi-minor axis. Comparing, we have:x = (15/4)cosθy = (17/4)sinθWait, but in the standard form, x is associated with the semi-major axis if a > b, or semi-minor if a < b. Here, 15/4 is 3.75 and 17/4 is 4.25. So, 17/4 is larger. So, the semi-major axis is 17/4 along the y-axis, and the semi-minor axis is 15/4 along the x-axis.But in the standard ellipse, the major axis is usually along the x-axis. So, in this case, since the coefficient of sinθ is larger, the major axis is along the y-axis. So, the ellipse is elongated vertically.But regardless, the area of an ellipse is πab, where a and b are the semi-major and semi-minor axes. So, in this case, a = 17/4 and b = 15/4.Therefore, the area should be π*(17/4)*(15/4) = π*(255/16). So, 255/16 π.Wait, let me double-check my steps.1. Expressed w as 4e^{iθ}, correct.2. Calculated 1/w as (1/4)e^{-iθ}, correct.3. Substituted into z = w - 1/w, correct.4. Expanded using Euler's formula, correct.5. Combined like terms: Real part 15/4 cosθ, Imaginary part 17/4 sinθ, correct.6. Identified as parametric equations of an ellipse with semi-axes 15/4 and 17/4, correct.7. Calculated area as π*(15/4)*(17/4) = 255/16 π, correct.So, I think that's the answer. The area inside U is 255/16 π.**Final Answer**The area inside ( U ) is (boxed{dfrac{255}{16}pi}).

💡First, I need to determine the percentage increase from 88 to 132. To do this, I'll start by calculating the difference between the two numbers.Next, I'll divide the difference by the original number to find the ratio of the increase.Finally, I'll convert this ratio into a percentage by multiplying by 100 to get the desired percentage increase.

💡First, I need to understand the given ratio of purple to orange to yellow candies, which is 2:4:5. This means that for every 2 purple candies, there are 4 orange candies and 5 yellow candies.Next, I'll determine the total number of parts in the ratio by adding them together: 2 + 4 + 5 = 11 parts.Since there are 40 yellow candies and yellow candies make up 5 parts of the total ratio, I can set up the equation 5 parts = 40 candies to find the value of one part.Solving for one part: 1 part = 40 candies / 5 = 8 candies.Now, I'll calculate the number of purple candies by multiplying the number of parts for purple by the value of one part: 2 parts * 8 candies/part = 16 purple candies.Similarly, I'll find the number of orange candies: 4 parts * 8 candies/part = 32 orange candies.Finally, I'll add up all the candies to find the total number: 16 purple + 32 orange + 40 yellow = 88 candies in total.

💡Okay, so I need to prove by contradiction that among the natural numbers a, b, and c, exactly one is even. And I have to figure out which assumption is required for this proof. The options are A, B, C, or D.First, let me recall what proof by contradiction means. It involves assuming the opposite of what you want to prove and then showing that this assumption leads to a contradiction, which means the original statement must be true.So, the original proposition is: "Among the natural numbers a, b, c, exactly one is even." To prove this by contradiction, I need to assume the opposite. The opposite of "exactly one is even" would be that either none are even or more than one is even.Let me look at the options:A: a, b, c are all odd.B: a, b, c are all even.C: a, b, c are all odd or at least two are even.D: a, b, c at least two are even.Hmm, option A is assuming all are odd, which would mean none are even. That's part of the opposite of the original statement because if none are even, then it's not true that exactly one is even. But is that the only assumption needed?Option B assumes all are even, which would mean all three are even. That's also part of the opposite because if all are even, then more than one is even. But again, is assuming all are even sufficient?Option C says either all are odd or at least two are even. That seems to cover both possibilities of the opposite: either none are even (all odd) or more than one is even (at least two). So this seems comprehensive.Option D assumes at least two are even, which covers the case where more than one is even, but it doesn't cover the case where none are even. So it's only part of the opposite.Therefore, to fully assume the opposite of "exactly one is even," I need to consider both possibilities: none are even or more than one is even. That's exactly what option C states. So, option C is the correct assumption needed for the proof by contradiction.Let me double-check. If I assume option C, which is that either all are odd or at least two are even, and then show that this leads to a contradiction, then the original statement must be true. Yes, that makes sense.So, I think the answer is option C.