💡Okay, so I have this problem about rhombuses. Let me try to visualize it first. There's a rhombus ABCD with each side equal to 13. On side AB, another rhombus BAFC is constructed outside ABCD, and AF is parallel to the diagonal BD of ABCD. The area of BAFC is given as 65, and I need to find the area of ABCD.Hmm, rhombuses have all sides equal, and their diagonals bisect each other at right angles. So, in ABCD, the diagonals AC and BD intersect at point H, which is the midpoint of both diagonals. Since AF is parallel to BD, and BAFC is a rhombus, maybe there's some similarity or proportionality I can use here.Let me try to draw this out mentally. Rhombus ABCD has sides AB, BC, CD, DA each of length 13. On AB, another rhombus BAFC is constructed outside ABCD. So, BAFC shares the side AB with ABCD but is built outward. AF is one of the sides of BAFC and is parallel to BD.Since BAFC is a rhombus, all its sides are equal, so BA = AF = FC = CB = 13. Wait, no, BAFC is a rhombus, so BA = AF = FC = CB? Wait, BA is a side of ABCD, which is 13, so AF must also be 13. But AF is parallel to BD, which is a diagonal of ABCD.I think I need to find some relationship between the diagonals of ABCD and BAFC. Maybe the area of BAFC can help me find the length of AF or something else.The area of a rhombus is given by (diagonal1 * diagonal2)/2. But in BAFC, if I can find the lengths of the diagonals, I can relate them to ABCD.Wait, but BAFC is a rhombus with side length 13 and area 65. So, maybe I can find the height of BAFC? Because area is also base times height. If the base is 13, then height would be 65/13 = 5. So, the height of BAFC is 5.Since AF is parallel to BD, and the height of BAFC is 5, maybe this height relates to the diagonals of ABCD. Let me think.In rhombus ABCD, the diagonals AC and BD intersect at H, which is the midpoint. So, AC is split into two equal parts, AH and HC, and BD is split into BH and HD. Since AF is parallel to BD, and BAFC is a rhombus, maybe the height of BAFC corresponds to half of AC or something like that.Wait, if AF is parallel to BD, and the height from F to AB in BAFC is 5, maybe that's related to the length of AC. Because in ABCD, the height from D to AB would be related to the area of ABCD, which I need to find.Let me try to formalize this. In BAFC, the height is 5, which is the distance from F to AB. Since AF is parallel to BD, and AF is a side of BAFC, maybe the height of BAFC is half the length of AC in ABCD.Wait, if AF is parallel to BD, and since AF is a side of BAFC, which is a rhombus, then AF is equal in length to AB, which is 13. But BD is a diagonal, not a side. So, maybe the direction of AF is the same as BD, but their lengths are different.Alternatively, maybe the triangles formed by the diagonals are similar.Let me think about the coordinates. Maybe assigning coordinates can help. Let me place point A at (0,0). Since ABCD is a rhombus, I can let point B be at (a,0), point C at (a + b, c), and point D at (b, c). But this might get complicated.Alternatively, since all sides are 13, maybe I can use vectors or trigonometry.Wait, another approach: since AF is parallel to BD, and BAFC is a rhombus, the angle between BA and AF is the same as the angle between BA and BD. Hmm, not sure.Wait, maybe the triangles involved are similar. If AF is parallel to BD, then triangle ABF is similar to triangle ABD or something like that.Wait, no, AF is a side of BAFC, which is a rhombus, so AF is equal to AB, which is 13. BD is a diagonal, which is longer than the sides.Wait, maybe the ratio of the areas can help. The area of BAFC is 65, which is half of 130, but not sure.Wait, let me recall that in a rhombus, the area can also be expressed as base times height. So, for BAFC, area is 65, which is base AB (13) times height from F to AB, which is 5. So, the height is 5.Since AF is parallel to BD, and the height from F to AB is 5, maybe this height corresponds to half the length of AC in ABCD. Because in ABCD, the height from D to AB would be related to the area of ABCD.Wait, let's denote the diagonals AC and BD of ABCD. Let AC = d1 and BD = d2. The area of ABCD is (d1*d2)/2.In BAFC, the height is 5, which might be related to d1. Since AF is parallel to BD, and the height is 5, maybe this is half of AC? So, AC = 10.Wait, if AC is 10, then in ABCD, the diagonals are AC = 10 and BD = d2. Then, the area would be (10*d2)/2 = 5*d2.But I need to find d2. How?In rhombus ABCD, the diagonals satisfy the relationship: (d1/2)^2 + (d2/2)^2 = side^2.So, (10/2)^2 + (d2/2)^2 = 13^2.Which is 25 + (d2/2)^2 = 169.So, (d2/2)^2 = 144, so d2/2 = 12, so d2 = 24.Therefore, the area of ABCD is (10*24)/2 = 120.Wait, that seems straightforward. Let me check if this makes sense.So, in BAFC, the height is 5, which is half of AC, so AC = 10. Then, using the Pythagorean theorem in ABCD, we find BD = 24. Therefore, area is 120.Yes, that seems correct. I think that's the answer.**Final Answer**The area of rhombus (ABCD) is boxed{120}.

💡Okay, so I have this problem where I need to find the real values of x and y such that the complex numbers z₁ and z₂ are opposites. Let me try to understand what that means. First, z₁ is given as x²(1 - 3i) - 10. I need to expand this. So, multiplying x² by (1 - 3i) gives me x² - 3x²i. Then subtracting 10, so z₁ becomes (x² - 10) - 3x²i. That makes sense.Next, z₂ is y(y - i). Let me expand that too. Multiplying y by (y - i) gives y² - yi. So, z₂ is y² - yi.Now, the problem says that z₁ and z₂ are opposites. I think that means z₁ = -z₂. So, if I set (x² - 10) - 3x²i equal to -(y² - yi), that should give me the equations I need to solve.Let me write that out: (x² - 10) - 3x²i = -y² + yi. Now, I can separate this into real and imaginary parts. The real part on the left is (x² - 10), and the imaginary part is -3x²i. On the right, the real part is -y², and the imaginary part is yi.So, setting the real parts equal: x² - 10 = -y². And setting the imaginary parts equal: -3x² = y.Hmm, so from the imaginary part, I get y = -3x². That seems straightforward. Now, I can substitute this into the real part equation.Substituting y = -3x² into x² - 10 = -y² gives me x² - 10 = -(-3x²)². Let me compute that. (-3x²) squared is 9x⁴, so the equation becomes x² - 10 = -9x⁴.Let me rearrange this: 9x⁴ + x² - 10 = 0. Hmm, that's a quartic equation, but maybe I can make a substitution to make it quadratic. Let me set u = x², so the equation becomes 9u² + u - 10 = 0.Now, solving for u: 9u² + u - 10 = 0. Using the quadratic formula, u = [-1 ± sqrt(1 + 360)] / (2*9). The discriminant is 1 + 360 = 361, whose square root is 19. So, u = [-1 ± 19]/18.Calculating the two possibilities: First, u = (-1 + 19)/18 = 18/18 = 1.Second, u = (-1 - 19)/18 = -20/18 = -10/9.But u = x², and x² can't be negative, so u = 1 is the only valid solution. Therefore, x² = 1, which means x = ±1.Now, substituting back to find y: y = -3x². Since x² is 1, y = -3*1 = -3.So, the real values are x = 1 or x = -1, and y = -3 in both cases.Wait, let me double-check. If x = 1, then z₁ = 1²(1 - 3i) - 10 = (1 - 3i) - 10 = -9 - 3i. And z₂ = y(y - i) = (-3)(-3 - i) = 9 + 3i. So, z₁ = -9 - 3i and z₂ = 9 + 3i, which are indeed opposites.Similarly, if x = -1, z₁ = (-1)²(1 - 3i) - 10 = 1 - 3i - 10 = -9 - 3i. And z₂ is still (-3)(-3 - i) = 9 + 3i. So, same result.Okay, that seems correct. So, the solutions are x = 1, y = -3 and x = -1, y = -3.Now, moving on to the next part, which is illustrating the sets of points for the given conditions. Let me see:10.12: -1 ≤ Im z < 2.10.13: |z| ≤ 2.10.14: |z + i| = 1.5.I need to visualize these on the complex plane.Starting with 10.12: -1 ≤ Im z < 2. This is the set of all complex numbers z = x + yi where the imaginary part y is between -1 and 2. So, on the complex plane, this would be the region between the horizontal lines y = -1 and y = 2, including y = -1 but not including y = 2.Next, 10.13: |z| ≤ 2. This is the closed disk centered at the origin with radius 2. So, all points inside and on the circle of radius 2.Then, 10.14: |z + i| = 1.5. This is the set of points z such that the distance from z to -i is 1.5. So, it's a circle centered at (0, -1) with radius 1.5.I should probably sketch these regions. For 10.12, it's a horizontal strip between y = -1 and y = 2. For 10.13, it's a filled circle with radius 2. For 10.14, it's a circle centered at (0, -1) with radius 1.5.I think that's it. I need to make sure I've interpreted the conditions correctly. For 10.12, it's the imaginary part between -1 and 2, so vertical strip? Wait, no, in the complex plane, the imaginary part is the y-axis, so it's horizontal lines. So, yes, between y = -1 and y = 2.And |z| ≤ 2 is the disk centered at the origin with radius 2. And |z + i| = 1.5 is a circle shifted down by 1 unit on the imaginary axis, with radius 1.5.I think that's accurate. I should maybe check if there are any intersections or specific points, but the problem just asks to illustrate the sets, so I think describing them as above is sufficient.Wait, but the user mentioned "sets of points for which the following conditions are met (10.12-10.16)", but only 10.12, 10.13, and 10.14 are given. Maybe 10.15 and 10.16 are not provided, so I'll focus on these three.So, summarizing:10.12: Horizontal strip between y = -1 and y = 2.10.13: Closed disk of radius 2 centered at origin.10.14: Circle centered at (0, -1) with radius 1.5.I think that's all. I should probably draw these on the complex plane to visualize, but since I'm just writing, I'll describe them as above.Wait, let me make sure I didn't mix up real and imaginary parts. For 10.12, it's Im z, which is the y-coordinate, so yes, horizontal lines. For |z|, it's the modulus, so distance from origin. For |z + i|, it's distance from -i, which is (0, -1). Correct.Okay, I think I've got it.

💡First, I need to find the original number that, when halved, results in 9.Let’s denote the original number as ( x ).According to the problem, half of ( x ) is 9. This can be written as:[frac{x}{2} = 9]To find ( x ), I will multiply both sides of the equation by 2:[x = 9 times 2][x = 18]Therefore, the original number is 18.

💡First, I need to find all the positive factors of 480. These are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 80, 96, 120, 160, 240, and 480.Next, I'll identify which of these factors are less than 60. The factors of 480 that are smaller than 60 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, and 48.Then, I'll list the factors of 90, which are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.Now, I'll find the common factors between the two lists. The common factors are 1, 2, 3, 5, 6, 10, 15, and 30.Finally, I'll determine the greatest common factor from these that is less than 60. The greatest common factor is 30.

💡First, determine the probability of a single student being absent and the probability of being present. Each student has a ( frac{1}{20} ) chance of being absent and a ( frac{19}{20} ) chance of being present.Next, identify the number of ways exactly one student can be absent out of three. This can happen in three different ways:1. The first student is absent, and the other two are present.2. The second student is absent, and the other two are present.3. The third student is absent, and the other two are present.For each scenario, calculate the probability:[frac{1}{20} times frac{19}{20} times frac{19}{20} = frac{361}{8000}]Since these scenarios are mutually exclusive, add their probabilities together:[3 times frac{361}{8000} = frac{1083}{8000} = 0.135375]Finally, convert the probability to a percentage and round to the nearest tenth:[0.135375 times 100 = 13.5375% approx 13.5%]

💡First, identify the three consecutive even numbers. Since ( m ) is the smallest, the next two numbers are ( m + 2 ) and ( m + 4 ).Next, calculate the sum of these three numbers:[m + (m + 2) + (m + 4)]Combine like terms:[m + m + 2 + m + 4 = 3m + 6]Therefore, the sum of the three consecutive even numbers is ( 3m + 6 ).