💡First, I will add the numbers in base 12 by starting from the rightmost digits.In the units column, adding 6 and 3 gives 9, which is less than 12, so there is no carry-over.Next, in the tens column, adding 5 and 7 results in 12. Since 12 is equal to 10 in base 12, I will write down 0 and carry over 1 to the next column.Moving to the hundreds column, I add 9, 2, and the carried over 1, which totals 12. Again, this is equal to 10 in base 12, so I will write down 0 and carry over another 1.Finally, in the thousands column, I add the carried over 1, resulting in 1.Putting it all together, the sum of 956₁₂ and 273₁₂ is 1009₁₂.

💡First, I need to understand what it means to project one vector onto another. The projection of vector **a** onto vector **b** essentially finds how much of **a** lies in the direction of **b**.To calculate this, I'll start by finding the dot product of vectors **a** and **b**. The dot product is calculated by multiplying the corresponding components of the vectors and then summing those products.Next, I'll determine the magnitude of vector **b**. The magnitude is found by taking the square root of the sum of the squares of its components.With both the dot product and the magnitude of **b**, I can now use the projection formula. This formula divides the dot product by the magnitude of **b** to give the scalar projection of **a** onto **b**.Finally, I'll simplify the expression to get the final result.

💡To determine the ratio of the area of the smaller rectangle to the area of the original rectangle, let's analyze the problem step by step.1. **Original Rectangle Dimensions**: - Let the original rectangle have a length ( l ) and a width ( w ). - The area of the original rectangle is ( A = l times w ).2. **Midpoints and Smaller Rectangle**: - The midpoints of each side of the original rectangle are connected to form a new, smaller rectangle. - Connecting midpoints divides each side into two equal segments, each of length ( frac{l}{2} ) and ( frac{w}{2} ).3. **Dimensions of the Smaller Rectangle**: - The length of the smaller rectangle is ( frac{l}{2} ). - The width of the smaller rectangle is ( frac{w}{2} ).4. **Area of the Smaller Rectangle**: - The area of the smaller rectangle is ( frac{l}{2} times frac{w}{2} = frac{l times w}{4} ).5. **Ratio of Areas**: - The ratio of the area of the smaller rectangle to the area of the original rectangle is: [ frac{frac{l times w}{4}}{l times w} = frac{1}{4} ]Therefore, the ratio of the area of the smaller rectangle to the area of the original rectangle is (boxed{frac{1}{4}}).

💡Alright, so I have this problem about an ellipse, and I need to find its equation and then determine something about a line intersecting it. Let me try to break it down step by step.First, part (I) asks for the equation of the ellipse. I know the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b > 0). The eccentricity (e) is given as (frac{1}{2}). I remember that the eccentricity of an ellipse is related to (a) and (c) (the distance from the center to a focus) by the formula (e = frac{c}{a}). So, since (e = frac{1}{2}), that means (c = frac{a}{2}).Also, I recall that for an ellipse, (c^2 = a^2 - b^2). Plugging in (c = frac{a}{2}), we get:[left(frac{a}{2}right)^2 = a^2 - b^2 frac{a^2}{4} = a^2 - b^2 b^2 = a^2 - frac{a^2}{4} b^2 = frac{3a^2}{4} b = frac{sqrt{3}a}{2}]So, now I know (b) in terms of (a).Next, the problem mentions point (A) is a vertex on the short axis. Since the ellipse is centered at the origin (I assume because the equation is in standard form), the vertices on the short axis are at ((0, b)) and ((0, -b)). So, point (A) is either ((0, b)) or ((0, -b)). I think it's just ((0, b)) because they mention it as a vertex on the short axis without specifying, but I'll keep that in mind.Point (F) is the left focus. Since the ellipse is centered at the origin, the foci are at ((pm c, 0)). So, the left focus (F) is at ((-c, 0)), which is ((- frac{a}{2}, 0)).Point (B) is on the x-axis, and (AB) is perpendicular to (AF). So, I need to find the coordinates of point (B).Let me write down the coordinates:- (A = (0, b) = left(0, frac{sqrt{3}a}{2}right))- (F = left(-frac{a}{2}, 0right))- (B = (x, 0)) for some (x) on the x-axis.Since (AB) is perpendicular to (AF), their slopes should multiply to (-1).First, let's find the slope of (AF). The slope (m_{AF}) is:[m_{AF} = frac{b - 0}{0 - (-frac{a}{2})} = frac{frac{sqrt{3}a}{2}}{frac{a}{2}} = sqrt{3}]So, the slope of (AF) is (sqrt{3}). Therefore, the slope of (AB) should be the negative reciprocal, which is (-frac{1}{sqrt{3}}).Now, let's find the equation of line (AB). It passes through point (A) and has slope (-frac{1}{sqrt{3}}). Using point-slope form:[y - b = -frac{1}{sqrt{3}}(x - 0) y = -frac{1}{sqrt{3}}x + b]Since point (B) is on the x-axis, its y-coordinate is 0. So, plugging (y = 0) into the equation:[0 = -frac{1}{sqrt{3}}x + b frac{1}{sqrt{3}}x = b x = b sqrt{3}]Substituting (b = frac{sqrt{3}a}{2}):[x = frac{sqrt{3}a}{2} times sqrt{3} = frac{3a}{2}]So, point (B) is at (left(frac{3a}{2}, 0right)).Now, the circle (C) determined by points (A), (B), and (F) is tangent to the line (x + sqrt{3}y + 3 = 0). I need to find the equation of this circle and use the tangency condition to solve for (a).First, let's find the equation of the circle passing through (A), (B), and (F). To find the equation of a circle passing through three points, I can use the general equation of a circle:[x^2 + y^2 + Dx + Ey + F = 0]Plugging in the coordinates of (A), (B), and (F):1. For (A(0, b)):[0 + b^2 + 0 + Eb + F = 0 b^2 + Eb + F = 0 quad (1)]2. For (Bleft(frac{3a}{2}, 0right)):[left(frac{3a}{2}right)^2 + 0 + Dleft(frac{3a}{2}right) + 0 + F = 0 frac{9a^2}{4} + frac{3aD}{2} + F = 0 quad (2)]3. For (Fleft(-frac{a}{2}, 0right)):[left(-frac{a}{2}right)^2 + 0 + Dleft(-frac{a}{2}right) + 0 + F = 0 frac{a^2}{4} - frac{aD}{2} + F = 0 quad (3)]Now, I have three equations:1. (b^2 + Eb + F = 0)2. (frac{9a^2}{4} + frac{3aD}{2} + F = 0)3. (frac{a^2}{4} - frac{aD}{2} + F = 0)Let me subtract equation (3) from equation (2):[left(frac{9a^2}{4} + frac{3aD}{2} + Fright) - left(frac{a^2}{4} - frac{aD}{2} + Fright) = 0 - 0 frac{8a^2}{4} + frac{4aD}{2} = 0 2a^2 + 2aD = 0 2a(a + D) = 0]Since (a neq 0), this implies (a + D = 0), so (D = -a).Now, plug (D = -a) into equation (3):[frac{a^2}{4} - frac{a(-a)}{2} + F = 0 frac{a^2}{4} + frac{a^2}{2} + F = 0 frac{3a^2}{4} + F = 0 F = -frac{3a^2}{4}]Now, plug (F = -frac{3a^2}{4}) into equation (1):[b^2 + Eb - frac{3a^2}{4} = 0]But we know (b^2 = frac{3a^2}{4}), so:[frac{3a^2}{4} + Eb - frac{3a^2}{4} = 0 Eb = 0]Since (b neq 0), this implies (E = 0).So, the equation of the circle is:[x^2 + y^2 - a x + 0 cdot y - frac{3a^2}{4} = 0 x^2 + y^2 - a x - frac{3a^2}{4} = 0]I can rewrite this in standard form by completing the square for the x-terms:[x^2 - a x + y^2 = frac{3a^2}{4} (x^2 - a x + frac{a^2}{4}) + y^2 = frac{3a^2}{4} + frac{a^2}{4} (x - frac{a}{2})^2 + y^2 = a^2]So, the circle has center (left(frac{a}{2}, 0right)) and radius (a).Now, this circle is tangent to the line (x + sqrt{3} y + 3 = 0). The condition for a line to be tangent to a circle is that the distance from the center of the circle to the line equals the radius.The distance (d) from center (left(frac{a}{2}, 0right)) to the line (x + sqrt{3} y + 3 = 0) is:[d = frac{| frac{a}{2} + sqrt{3} cdot 0 + 3 |}{sqrt{1^2 + (sqrt{3})^2}} = frac{| frac{a}{2} + 3 |}{2}]Since the circle is tangent to the line, this distance must equal the radius (a):[frac{| frac{a}{2} + 3 |}{2} = a | frac{a}{2} + 3 | = 2a]This absolute value equation can split into two cases:1. (frac{a}{2} + 3 = 2a)2. (-left(frac{a}{2} + 3right) = 2a)Let's solve each case:**Case 1:**[frac{a}{2} + 3 = 2a 3 = 2a - frac{a}{2} 3 = frac{3a}{2} a = 2]**Case 2:**[- frac{a}{2} - 3 = 2a -3 = 2a + frac{a}{2} -3 = frac{5a}{2} a = -frac{6}{5}]But since (a > 0), we discard the negative solution. So, (a = 2).Now, knowing (a = 2), we can find (b):[b = frac{sqrt{3}a}{2} = frac{sqrt{3} times 2}{2} = sqrt{3}]Therefore, the equation of the ellipse is:[frac{x^2}{4} + frac{y^2}{3} = 1]So, that's part (I) done.Now, moving on to part (II). This seems more complicated. Let me read it again:"Determine whether there exists a line (l) passing through (F) with slope (k) ((k neq 0)) that intersects the ellipse at points (M) and (N), with (P) being the midpoint of segment (MN). Let (O) be the center of the ellipse, and ray (OP) intersects the ellipse at point (Q). If (overrightarrow{OM} + overrightarrow{ON} = overrightarrow{OQ}), find the value of (k) if it exists, otherwise explain why."Alright, so we need to find if such a line (l) exists with slope (k) passing through (F), intersecting the ellipse at (M) and (N), such that the vector sum of (overrightarrow{OM}) and (overrightarrow{ON}) equals (overrightarrow{OQ}), where (Q) is the intersection of ray (OP) with the ellipse.First, let's note the coordinates:- The ellipse is (frac{x^2}{4} + frac{y^2}{3} = 1), so (a = 2), (b = sqrt{3}).- The center (O) is at ((0, 0)).- The left focus (F) is at ((-c, 0)), where (c = frac{a}{2} = 1). So, (F = (-1, 0)).So, line (l) passes through (F(-1, 0)) with slope (k). Its equation is:[y = k(x + 1)]This line intersects the ellipse at points (M) and (N). Let me find the points of intersection by substituting (y = k(x + 1)) into the ellipse equation.Substituting:[frac{x^2}{4} + frac{[k(x + 1)]^2}{3} = 1]Expanding:[frac{x^2}{4} + frac{k^2(x^2 + 2x + 1)}{3} = 1]Multiply both sides by 12 to eliminate denominators:[3x^2 + 4k^2(x^2 + 2x + 1) = 12 3x^2 + 4k^2x^2 + 8k^2x + 4k^2 = 12 (3 + 4k^2)x^2 + 8k^2x + (4k^2 - 12) = 0]So, we have a quadratic in (x):[(3 + 4k^2)x^2 + 8k^2x + (4k^2 - 12) = 0]Let me denote this as (Ax^2 + Bx + C = 0), where:- (A = 3 + 4k^2)- (B = 8k^2)- (C = 4k^2 - 12)Let the roots be (x_1) and (x_2), corresponding to points (M) and (N). Then, by Vieta's formulas:- (x_1 + x_2 = -frac{B}{A} = -frac{8k^2}{3 + 4k^2})- (x_1 x_2 = frac{C}{A} = frac{4k^2 - 12}{3 + 4k^2})The midpoint (P) of (MN) has coordinates:[x_P = frac{x_1 + x_2}{2} = -frac{4k^2}{3 + 4k^2}][y_P = k(x_P + 1) = kleft(-frac{4k^2}{3 + 4k^2} + 1right) = kleft(frac{-4k^2 + 3 + 4k^2}{3 + 4k^2}right) = frac{3k}{3 + 4k^2}]So, (P) is at (left(-frac{4k^2}{3 + 4k^2}, frac{3k}{3 + 4k^2}right)).Now, ray (OP) intersects the ellipse again at point (Q). Since (O) is the origin, the ray (OP) is just the line from the origin through (P), extended to intersect the ellipse again.Parametric equations for ray (OP) can be written as:[x = t cdot x_P y = t cdot y_P]Where (t) is a parameter. When (t = 1), we reach point (P). To find point (Q), we need to find the other intersection with the ellipse. Since (P) is inside the ellipse (as it's the midpoint of a chord), (Q) will be the other intersection point when extending beyond (P).But since (OP) is a line through the origin, and (P) is on it, (Q) can be found by scaling the coordinates of (P) appropriately.Alternatively, since (Q) is on the ellipse, plugging the parametric equations into the ellipse equation:[frac{(t x_P)^2}{4} + frac{(t y_P)^2}{3} = 1]Solving for (t):[t^2 left( frac{x_P^2}{4} + frac{y_P^2}{3} right) = 1]Let me compute (frac{x_P^2}{4} + frac{y_P^2}{3}):[frac{left(-frac{4k^2}{3 + 4k^2}right)^2}{4} + frac{left(frac{3k}{3 + 4k^2}right)^2}{3} = frac{frac{16k^4}{(3 + 4k^2)^2}}{4} + frac{frac{9k^2}{(3 + 4k^2)^2}}{3} = frac{4k^4}{(3 + 4k^2)^2} + frac{3k^2}{(3 + 4k^2)^2} = frac{4k^4 + 3k^2}{(3 + 4k^2)^2} = frac{k^2(4k^2 + 3)}{(3 + 4k^2)^2} = frac{k^2}{3 + 4k^2}]So, the equation becomes:[t^2 cdot frac{k^2}{3 + 4k^2} = 1 t^2 = frac{3 + 4k^2}{k^2} t = pm sqrt{frac{3 + 4k^2}{k^2}} = pm frac{sqrt{3 + 4k^2}}{k}]Since (P) is at (t = 1), the other intersection point (Q) is at (t = -frac{sqrt{3 + 4k^2}}{k}). Therefore, the coordinates of (Q) are:[x_Q = -frac{sqrt{3 + 4k^2}}{k} cdot x_P = -frac{sqrt{3 + 4k^2}}{k} cdot left(-frac{4k^2}{3 + 4k^2}right) = frac{4k^2 sqrt{3 + 4k^2}}{k(3 + 4k^2)} = frac{4k sqrt{3 + 4k^2}}{3 + 4k^2}][y_Q = -frac{sqrt{3 + 4k^2}}{k} cdot y_P = -frac{sqrt{3 + 4k^2}}{k} cdot frac{3k}{3 + 4k^2} = -frac{3 sqrt{3 + 4k^2}}{3 + 4k^2}]Simplify (x_Q):[x_Q = frac{4k sqrt{3 + 4k^2}}{3 + 4k^2} = frac{4k}{sqrt{3 + 4k^2}}]Similarly, (y_Q):[y_Q = -frac{3 sqrt{3 + 4k^2}}{3 + 4k^2} = -frac{3}{sqrt{3 + 4k^2}}]So, (Q) is at (left( frac{4k}{sqrt{3 + 4k^2}}, -frac{3}{sqrt{3 + 4k^2}} right)).Now, the condition given is (overrightarrow{OM} + overrightarrow{ON} = overrightarrow{OQ}).Let me express vectors (overrightarrow{OM}) and (overrightarrow{ON}):- (overrightarrow{OM} = (x_1, y_1))- (overrightarrow{ON} = (x_2, y_2))So, their sum is:[overrightarrow{OM} + overrightarrow{ON} = (x_1 + x_2, y_1 + y_2)]From earlier, (x_1 + x_2 = -frac{8k^2}{3 + 4k^2}). What about (y_1 + y_2)?Since both (M) and (N) lie on line (l), (y = k(x + 1)). So:[y_1 = k(x_1 + 1) y_2 = k(x_2 + 1)]Therefore:[y_1 + y_2 = k(x_1 + x_2 + 2) = kleft(-frac{8k^2}{3 + 4k^2} + 2right) = kleft( frac{-8k^2 + 6 + 8k^2}{3 + 4k^2} right) = k cdot frac{6}{3 + 4k^2} = frac{6k}{3 + 4k^2}]So, (overrightarrow{OM} + overrightarrow{ON} = left( -frac{8k^2}{3 + 4k^2}, frac{6k}{3 + 4k^2} right)).On the other hand, (overrightarrow{OQ} = (x_Q, y_Q) = left( frac{4k}{sqrt{3 + 4k^2}}, -frac{3}{sqrt{3 + 4k^2}} right)).So, setting (overrightarrow{OM} + overrightarrow{ON} = overrightarrow{OQ}):[left( -frac{8k^2}{3 + 4k^2}, frac{6k}{3 + 4k^2} right) = left( frac{4k}{sqrt{3 + 4k^2}}, -frac{3}{sqrt{3 + 4k^2}} right)]This gives us two equations:1. ( -frac{8k^2}{3 + 4k^2} = frac{4k}{sqrt{3 + 4k^2}} )2. ( frac{6k}{3 + 4k^2} = -frac{3}{sqrt{3 + 4k^2}} )Let me solve equation 2 first because it might be simpler.Equation 2:[frac{6k}{3 + 4k^2} = -frac{3}{sqrt{3 + 4k^2}}]Multiply both sides by (sqrt{3 + 4k^2}):[frac{6k}{sqrt{3 + 4k^2}} = -3]Divide both sides by 3:[frac{2k}{sqrt{3 + 4k^2}} = -1]Multiply both sides by (sqrt{3 + 4k^2}):[2k = -sqrt{3 + 4k^2}]Square both sides to eliminate the square root:[(2k)^2 = (-sqrt{3 + 4k^2})^2 4k^2 = 3 + 4k^2 4k^2 - 4k^2 = 3 0 = 3]Wait, that's impossible. 0 = 3 is a contradiction. Hmm, so equation 2 leads to a contradiction. That suggests that there is no such (k) that satisfies the condition. But let me check my steps to make sure I didn't make a mistake.Looking back at equation 2:[frac{6k}{3 + 4k^2} = -frac{3}{sqrt{3 + 4k^2}}]Multiply both sides by (sqrt{3 + 4k^2}):[frac{6k}{sqrt{3 + 4k^2}} = -3]Yes, that's correct. Then dividing by 3:[frac{2k}{sqrt{3 + 4k^2}} = -1]Yes, correct. Then, multiplying both sides by (sqrt{3 + 4k^2}):[2k = -sqrt{3 + 4k^2}]Yes, correct. Then squaring both sides:[4k^2 = 3 + 4k^2]Which simplifies to 0 = 3, which is impossible. So, equation 2 has no solution.Therefore, the condition (overrightarrow{OM} + overrightarrow{ON} = overrightarrow{OQ}) cannot be satisfied for any real (k). So, such a line (l) does not exist.But just to be thorough, let me check equation 1 as well, in case I made a mistake in interpreting the vectors.Equation 1:[-frac{8k^2}{3 + 4k^2} = frac{4k}{sqrt{3 + 4k^2}}]Multiply both sides by (sqrt{3 + 4k^2}):[-frac{8k^2}{sqrt{3 + 4k^2}} = 4k]Divide both sides by 4:[-frac{2k^2}{sqrt{3 + 4k^2}} = k]Multiply both sides by (sqrt{3 + 4k^2}):[-2k^2 = k sqrt{3 + 4k^2}]Square both sides:[4k^4 = k^2 (3 + 4k^2) 4k^4 = 3k^2 + 4k^4 4k^4 - 4k^4 = 3k^2 0 = 3k^2]Which implies (k = 0). But the problem states (k neq 0), so this is also a contradiction.Therefore, both equations lead to contradictions, confirming that no such (k) exists.So, the conclusion is that there does not exist a line (l) with the given properties.**Final Answer**Ⅰ. The equation of the ellipse is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).Ⅱ. Such a line (l) does not exist, so the value of (k) is (boxed{text{不存在}}).

💡First, assume there are 100 birds in the reserve for simplicity.1. **Calculate the number of hawks:** - 30% of the birds are hawks. - Hawks = 30 birds.2. **Determine the number of non-hawks:** - Non-hawks = Total birds - Hawks = 100 - 30 = 70 birds.3. **Find the number of paddyfield-warblers:** - 40% of non-hawks are paddyfield-warblers. - Paddyfield-warblers = 0.4 * 70 = 28 birds.4. **Calculate the number of kingfishers:** - There are 25% as many kingfishers as paddyfield-warblers. - Kingfishers = 0.25 * 28 = 7 birds.5. **Determine the number of white storks:** - The number of white storks is 10% more than the number of kingfishers. - White storks = 7 + (0.10 * 7) = 7 + 0.7 = 7.7 birds. - Since we can't have a fraction of a bird, round to 8 white storks.6. **Sum the numbers of hawks, paddyfield-warblers, kingfishers, and white storks:** - Total = 30 (Hawks) + 28 (Paddyfield-warblers) + 7 (Kingfishers) + 8 (White storks) = 73 birds.7. **Find the remaining birds:** - Remaining birds = Total birds - Total accounted for = 100 - 73 = 27 birds.8. **Calculate the percentage of remaining birds:** - Percentage = (27 / 100) * 100% = 27%.Therefore, 27% of the birds are not hawks, paddyfield-warblers, kingfishers, or white storks.

💡Alright, so I have this problem here: Let ζ be a nonreal root of z⁴ = 1. Find the number of ordered pairs (a, b) of integers such that |aζ + b| = 1.Hmm, okay. Let me try to break this down step by step. First, I need to understand what ζ is. The equation z⁴ = 1 is a quartic equation, and its roots are the fourth roots of unity. I remember that the fourth roots of unity are 1, -1, i, and -i. So, the nonreal roots must be i and -i. That makes sense because i and -i are purely imaginary numbers, so they aren't real.So, ζ is either i or -i. Since the problem doesn't specify which one, I guess it doesn't matter because they are essentially the same in terms of their properties for this problem. Maybe I can just pick one, say ζ = i, and work with that. If I get a solution, it should hold for ζ = -i as well because they are complex conjugates.Now, the main equation I need to solve is |aζ + b| = 1, where a and b are integers. Let me write that out with ζ = i: |ai + b| = 1. I need to find all integer pairs (a, b) such that the absolute value (or modulus) of ai + b is equal to 1.I recall that for a complex number x + yi, the modulus is √(x² + y²). So, in this case, ai + b is a complex number where the real part is b and the imaginary part is a. Therefore, the modulus |ai + b| is √(b² + a²). So, the equation becomes √(b² + a²) = 1.If I square both sides to eliminate the square root, I get b² + a² = 1. So, now the problem reduces to finding all integer solutions (a, b) such that a² + b² = 1.Alright, so I need to find all integer pairs where the sum of their squares is 1. Let's think about the possible integer values for a and b. Since a² and b² are both non-negative, each of them can be 0 or 1 because 1 is the maximum value their sum can take.So, let's list all possible combinations:1. If a = 0, then b² = 1. So, b can be 1 or -1.2. If a = 1, then b² = 0. So, b must be 0.3. If a = -1, then b² = 0. So, b must be 0.4. If a = 2, then a² = 4, which is already greater than 1, so that's not possible. Similarly, any a with absolute value greater than 1 won't work because their square will exceed 1.So, the possible integer pairs (a, b) are:- (0, 1)- (0, -1)- (1, 0)- (-1, 0)Let me double-check each of these to make sure they satisfy the original equation |ai + b| = 1.1. For (0, 1): |0*i + 1| = |1| = 1. That works.2. For (0, -1): |0*i - 1| = |-1| = 1. That also works.3. For (1, 0): |1*i + 0| = |i| = 1. Yep, that's correct.4. For (-1, 0): |-1*i + 0| = |-i| = 1. That works too.Are there any other integer pairs that could satisfy a² + b² = 1? Let me think. If a and b are integers, their squares can only be 0 or 1 because 2² = 4 is already too big. So, the only possibilities are when one of them is 0 and the other is ±1.Therefore, I don't think there are any other solutions besides the four pairs I listed. So, the number of ordered pairs (a, b) is 4.Wait, just to make sure I didn't miss anything, let me consider if a or b could be fractions or something. But no, the problem specifies that a and b are integers, so fractions aren't allowed. So, yeah, only the four pairs work.I think that's it. So, the answer should be 4 ordered pairs.**Final Answer**The number of ordered pairs is boxed{4}.