💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. There are two parts to this problem. Let me tackle them one by one.Starting with part (I): It says that c sin A equals the square root of 3 times a cos C. So, mathematically, that's c sin A = √3 a cos C. I need to find angle C.Hmm, I remember the Law of Sines, which states that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, that would be a/sin A = b/sin B = c/sin C. Maybe I can use that here.Let me write down the given equation again: c sin A = √3 a cos C. Maybe I can express c in terms of a using the Law of Sines. From the Law of Sines, c = (a sin C)/sin A. Let me substitute that into the equation.So, substituting c, we have: (a sin C / sin A) * sin A = √3 a cos C. Simplifying the left side, sin A cancels out, so we get a sin C = √3 a cos C. Now, I can divide both sides by a (assuming a ≠ 0, which it can't be in a triangle), so we have sin C = √3 cos C. Hmm, this looks like a trigonometric equation. Maybe I can write it as tan C = √3. Because sin C / cos C is tan C. So, tan C = √3.I know that tan 60° is √3, so C must be 60°, or π/3 radians. That seems straightforward. So, part (I) is solved, and C is 60 degrees.Moving on to part (II): It says that c = √7, and sin C + sin(B - A) = 3 sin 2A. I need to find the area of triangle ABC.Alright, let's note down what we know:1. From part (I), angle C is 60°, so sin C is sin 60°, which is √3/2.2. c = √7.3. The equation given is sin C + sin(B - A) = 3 sin 2A.I need to find the area. The area of a triangle can be found using several formulas. One common formula is (1/2)ab sin C, where a and b are two sides, and C is the included angle. Alternatively, if I can find all sides, I can use Heron's formula, but that might be more complicated.First, let's see if we can find the other sides or angles. We know angle C is 60°, and side c is √7. Maybe using the Law of Sines or the Law of Cosines can help.Let me write down the Law of Sines again: a/sin A = b/sin B = c/sin C. Since c = √7 and sin C = √3/2, we can write:a = (√7 / (√3/2)) sin A = (2√7 / √3) sin A.Similarly, b = (2√7 / √3) sin B.So, sides a and b can be expressed in terms of angles A and B.Now, let's look at the equation given: sin C + sin(B - A) = 3 sin 2A.We know sin C is √3/2, so:√3/2 + sin(B - A) = 3 sin 2A.I need to express sin(B - A) in terms of A or B. Since in a triangle, A + B + C = 180°, and we know C is 60°, so A + B = 120°. Therefore, B = 120° - A.So, B - A = 120° - A - A = 120° - 2A.Therefore, sin(B - A) = sin(120° - 2A).So, substituting back into the equation:√3/2 + sin(120° - 2A) = 3 sin 2A.Let me write this as:sin(120° - 2A) = 3 sin 2A - √3/2.Hmm, this seems a bit complicated. Maybe I can use the sine subtraction formula on sin(120° - 2A). The formula is sin(x - y) = sin x cos y - cos x sin y.So, sin(120° - 2A) = sin 120° cos 2A - cos 120° sin 2A.We know sin 120° is √3/2 and cos 120° is -1/2. So, substituting:sin(120° - 2A) = (√3/2) cos 2A - (-1/2) sin 2A = (√3/2) cos 2A + (1/2) sin 2A.So, substituting back into the equation:(√3/2) cos 2A + (1/2) sin 2A = 3 sin 2A - √3/2.Let me bring all terms to one side:(√3/2) cos 2A + (1/2) sin 2A - 3 sin 2A + √3/2 = 0.Simplify the terms:(√3/2) cos 2A + (1/2 - 3) sin 2A + √3/2 = 0.Which simplifies to:(√3/2) cos 2A - (5/2) sin 2A + √3/2 = 0.Hmm, this is getting a bit messy. Maybe I can factor out 1/2:(1/2)(√3 cos 2A - 5 sin 2A + √3) = 0.Multiplying both sides by 2:√3 cos 2A - 5 sin 2A + √3 = 0.Let me rearrange:√3 (cos 2A + 1) - 5 sin 2A = 0.Hmm, not sure if that helps. Maybe I can write this as:√3 (1 + cos 2A) = 5 sin 2A.I recall that 1 + cos 2A = 2 cos² A, so substituting:√3 * 2 cos² A = 5 sin 2A.Which is:2√3 cos² A = 5 * 2 sin A cos A.Simplify:2√3 cos² A = 10 sin A cos A.Divide both sides by 2 cos A (assuming cos A ≠ 0):√3 cos A = 5 sin A.So, √3 cos A = 5 sin A.Divide both sides by cos A:√3 = 5 tan A.Therefore, tan A = √3 / 5.So, angle A is arctangent of √3 / 5. Let me compute that.But before I compute the angle, maybe I can find sin A and cos A from tan A.Since tan A = √3 / 5, we can imagine a right triangle where the opposite side is √3 and the adjacent side is 5. Then, the hypotenuse would be sqrt( (√3)^2 + 5^2 ) = sqrt(3 + 25) = sqrt(28) = 2√7.So, sin A = opposite / hypotenuse = √3 / (2√7) = √21 / 14.Wait, let me check that:Opposite = √3, hypotenuse = 2√7, so sin A = √3 / (2√7). Rationalizing the denominator:√3 / (2√7) = (√3 * √7) / (2 * 7) = √21 / 14.Similarly, cos A = adjacent / hypotenuse = 5 / (2√7) = 5√7 / 14.So, sin A = √21 / 14 and cos A = 5√7 / 14.Now, let's recall from the Law of Sines, a = (2√7 / √3) sin A.Substituting sin A:a = (2√7 / √3) * (√21 / 14) = (2√7 * √21) / (√3 * 14).Simplify √21 / √3 = √7, so:a = (2√7 * √7) / 14 = (2 * 7) / 14 = 14 / 14 = 1.So, a = 1.Similarly, since angle B = 120° - A, and we have angle A, we can find angle B.But maybe we can find side b using the Law of Sines.From the Law of Sines, b = (2√7 / √3) sin B.But angle B = 120° - A, so sin B = sin(120° - A).Using the sine subtraction formula again:sin(120° - A) = sin 120° cos A - cos 120° sin A.We know sin 120° = √3/2, cos 120° = -1/2, cos A = 5√7 / 14, sin A = √21 / 14.So, sin(120° - A) = (√3/2)(5√7 / 14) - (-1/2)(√21 / 14).Simplify:= (5√21 / 28) + (√21 / 28) = (6√21) / 28 = (3√21) / 14.So, sin B = 3√21 / 14.Therefore, b = (2√7 / √3) * (3√21 / 14).Simplify:= (2√7 * 3√21) / (√3 * 14).Simplify numerator:√21 = √3 * √7, so:= (2√7 * 3 * √3 * √7) / (√3 * 14).Simplify:√7 * √7 = 7, √3 cancels out:= (2 * 3 * 7) / 14 = 42 / 14 = 3.So, b = 3.Now, we have sides a = 1, b = 3, c = √7, and angle C = 60°. Let's verify using the Law of Cosines to make sure.Law of Cosines: c² = a² + b² - 2ab cos C.Plugging in the values:(√7)² = 1² + 3² - 2 * 1 * 3 * cos 60°.7 = 1 + 9 - 6 * (1/2).7 = 10 - 3 = 7. Yes, that checks out.Now, to find the area. As I thought earlier, the area can be found using (1/2)ab sin C.So, plugging in a = 1, b = 3, sin C = √3/2:Area = (1/2) * 1 * 3 * (√3 / 2) = (3√3) / 4.Wait, but hold on, earlier in the thought process, I considered two cases: one where cos A = 0 and another where it's not. In this case, cos A wasn't zero because tan A was √3 / 5, so cos A was 5√7 / 14, which is not zero. So, this is the valid case.But just to make sure, what if cos A was zero? That would mean angle A is 90°, but in that case, let's see.If A = 90°, then angle B = 120° - 90° = 30°, so angle B is 30°, and angle C is 60°, which is consistent.From the Law of Sines, a = (2√7 / √3) sin A = (2√7 / √3) * 1 = 2√7 / √3.Similarly, b = (2√7 / √3) sin B = (2√7 / √3) * 1/2 = √7 / √3.Then, using the Law of Cosines to find c:c² = a² + b² - 2ab cos C.But c is given as √7, so:(√7)² = (2√7 / √3)² + (√7 / √3)² - 2*(2√7 / √3)*(√7 / √3)*cos 60°.Calculating:7 = (4*7 / 3) + (7 / 3) - 2*(2*7 / 3)*(1/2).Simplify:7 = (28/3 + 7/3) - (14/3).Which is:7 = (35/3) - (14/3) = 21/3 = 7.So, that also checks out. So, in this case, the area would be (1/2)*a*b*sin C.But wait, if A is 90°, then the area is simply (1/2)*b*c, since angle A is between sides b and c? Wait, no, angle A is opposite side a. So, actually, sides b and c are adjacent to angle A.Wait, no, in the formula, it's (1/2)*a*b*sin C, but if angle A is 90°, then the area is (1/2)*b*c*sin A, but sin A is 1. So, area would be (1/2)*b*c.But in this case, b = √7 / √3 and c = √7, so area would be (1/2)*(√7 / √3)*√7 = (1/2)*(7 / √3) = 7/(2√3) = 7√3 / 6.So, in this case, the area is 7√3 / 6.But wait, earlier, when I considered angle A not being 90°, I got area as 3√3 / 4. So, which one is correct?Hmm, the problem didn't specify whether angle A is acute or obtuse, so both cases might be possible. Therefore, there are two possible triangles satisfying the given conditions, leading to two different areas.But let me check the equation again: sin C + sin(B - A) = 3 sin 2A.In the case where A is 90°, let's see if the equation holds.If A = 90°, then B = 30°, so sin C + sin(B - A) = sin 60° + sin(-60°) = √3/2 - √3/2 = 0.On the other hand, 3 sin 2A = 3 sin 180° = 0. So, 0 = 0, which is true.In the other case, where tan A = √3 / 5, we have angle A ≈ arctan(√3 / 5) ≈ 19.1°, so B = 120° - 19.1° ≈ 100.9°, and sin C + sin(B - A) = √3/2 + sin(81.8°) ≈ 0.866 + 0.990 ≈ 1.856.On the other hand, 3 sin 2A ≈ 3 sin(38.2°) ≈ 3 * 0.619 ≈ 1.857. So, that's approximately equal, which is consistent.Therefore, both cases are valid, leading to two different areas: 7√3 / 6 and 3√3 / 4.But wait, the problem didn't specify any additional constraints, so both solutions are possible. Therefore, the area could be either 7√3 / 6 or 3√3 / 4.But let me double-check the calculations to make sure I didn't make any mistakes.In the first case, where A = 90°, sides are a = 2√7 / √3, b = √7 / √3, c = √7. The area is (1/2)*b*c = (1/2)*(√7 / √3)*√7 = (1/2)*(7 / √3) = 7/(2√3) = 7√3 / 6. Correct.In the second case, a = 1, b = 3, c = √7, angle C = 60°, so area is (1/2)*a*b*sin C = (1/2)*1*3*(√3/2) = 3√3 / 4. Correct.Therefore, both areas are possible.But wait, in the problem statement, part (II) says "If c = √7, and sin C + sin (B - A) = 3 sin 2A, find the area of triangle ABC."It doesn't specify whether it's a unique solution or multiple solutions. So, perhaps both areas are acceptable.But in the initial problem, part (I) was to find angle C, which is uniquely 60°, but part (II) might have multiple solutions.Therefore, the area could be either 7√3 / 6 or 3√3 / 4.But let me check if both satisfy all the given conditions.In the first case, with A = 90°, we have:c sin A = √7 * 1 = √7.√3 a cos C = √3 * (2√7 / √3) * cos 60° = √3 * (2√7 / √3) * 0.5 = √3 * (2√7 / √3) * 0.5 = √3 * (√7 / √3) = √7.So, c sin A = √7 equals √3 a cos C = √7. Correct.In the second case, with a = 1, b = 3, c = √7, angle C = 60°, let's check c sin A:c sin A = √7 * (√21 / 14) = √7 * √21 / 14 = (√147) / 14 = (7√3) / 14 = √3 / 2.√3 a cos C = √3 * 1 * 0.5 = √3 / 2. So, c sin A = √3 / 2 equals √3 a cos C = √3 / 2. Correct.Therefore, both cases satisfy the given conditions, so both areas are valid.So, the area of triangle ABC is either 7√3 / 6 or 3√3 / 4.But wait, the problem says "find the area of triangle ABC." It doesn't specify whether it's one or two answers. Maybe I should present both.Alternatively, perhaps I made a mistake in considering both cases. Let me think again.When I had the equation √3 cos A = 5 sin A, leading to tan A = √3 / 5, that's one solution. The other case was when cos A = 0, leading to A = 90°, which also satisfies the equation because if cos A = 0, then sin(B - A) would be sin(B - 90°), but in that case, we had to check if the equation holds, which it did.Therefore, both solutions are valid, leading to two possible areas.So, in conclusion, the area of triangle ABC is either 7√3 / 6 or 3√3 / 4.

💡First, I'll define the two numbers as ( x ) and ( y ), where ( x > y ).From the problem, I know that the sum of the two numbers is 20:[x + y = 20 quad text{(Equation 1)}]I also know that the difference of their squares is 160:[x^2 - y^2 = 160 quad text{(Equation 3)}]Using the difference of squares formula, I can rewrite Equation 3:[(x + y)(x - y) = 160]Substituting Equation 1 into this equation:[20(x - y) = 160]Now, I'll solve for ( x - y ):[x - y = frac{160}{20} = 8]Therefore, the difference between the two numbers is 8.

💡Alright, so I'm trying to solve this problem where Vasya is looking at all possible equations ( f_i = f_j ) where ( i neq j ) and counting how many solutions there are for integer values of ( x ) between 0 and 2007. The functions ( f_i ) are defined as quadratic functions of the form ( f_i(x) = (x - a)(x - b) ) where ( a ) and ( b ) are specific whole numbers.First, I need to understand what these functions look like. Each ( f_i ) is a quadratic function, which means it's a parabola when graphed. Since the leading coefficient is positive (because it's just ( x ) multiplied by another term), each parabola opens upwards. The roots of each function are the values of ( x ) that make ( f_i(x) = 0 ). For ( f_1(x) = x(x - 2007) ), the roots are at ( x = 0 ) and ( x = 2007 ). Similarly, for ( f_2(x) = (x - 1)(x - 2006) ), the roots are at ( x = 1 ) and ( x = 2006 ). This pattern continues up to ( f_{1004}(x) = (x - 1003)(x - 1004) ), which has roots at ( x = 1003 ) and ( x = 1004 ).Now, Vasya is considering all possible pairs of these functions and setting them equal to each other, i.e., ( f_i(x) = f_j(x) ) for ( i neq j ). The question is asking how many integer solutions exist for ( x ) in the range from 0 to 2007 for these equations.To approach this, I think I need to consider what it means for two quadratic functions to be equal. If ( f_i(x) = f_j(x) ), then their difference ( f_i(x) - f_j(x) = 0 ) must hold. This difference will also be a quadratic function (since the leading terms will cancel out if they are the same, but in this case, they might not). Let me write out the general form of two such functions and subtract them.Let’s take two arbitrary functions ( f_i(x) = (x - a)(x - b) ) and ( f_j(x) = (x - c)(x - d) ). Expanding both, we get:( f_i(x) = x^2 - (a + b)x + ab )( f_j(x) = x^2 - (c + d)x + cd )Subtracting these, we have:( f_i(x) - f_j(x) = [x^2 - (a + b)x + ab] - [x^2 - (c + d)x + cd] )Simplifying, the ( x^2 ) terms cancel out:( f_i(x) - f_j(x) = [ - (a + b)x + ab ] - [ - (c + d)x + cd ] )Which simplifies further to:( f_i(x) - f_j(x) = [ - (a + b) + (c + d) ]x + (ab - cd) )So, the difference is a linear function of the form ( mx + k ), where ( m = (c + d - a - b) ) and ( k = ab - cd ).For this linear function to be zero, we have:( mx + k = 0 )Solving for ( x ):( x = -k/m )Now, for ( x ) to be an integer between 0 and 2007, ( -k/m ) must be an integer in that range. So, the key is to determine for how many pairs ( (i, j) ) the value ( x = -k/m ) is an integer between 0 and 2007.Given that all the functions are defined with roots that are integers, ( a, b, c, d ) are all integers. Therefore, ( m ) and ( k ) are also integers. So, ( x ) will be rational, but we need it to be an integer.This means that ( m ) must divide ( k ) exactly for ( x ) to be an integer. So, the number of solutions depends on how many pairs ( (i, j) ) satisfy this divisibility condition.Looking back at the specific functions given, each function ( f_i ) has roots that are symmetric around 1003.5. For example, ( f_1 ) has roots at 0 and 2007, which are 1003.5 units away from the center. Similarly, ( f_2 ) has roots at 1 and 2006, which are also symmetric around 1003.5. This pattern continues up to ( f_{1004} ), which has roots at 1003 and 1004, right around the center.Given this symmetry, I suspect that the difference ( f_i(x) - f_j(x) ) might have some properties that make ( x = 1003.5 ) a solution. However, since we're only considering integer values of ( x ), 1003.5 isn't an integer, so it won't be a solution. But perhaps the integer values around it could be solutions.Wait, but if the difference is a linear function, it can have at most one solution. So, for each pair ( (i, j) ), there can be at most one integer ( x ) that satisfies ( f_i(x) = f_j(x) ). However, whether that ( x ) is within 0 to 2007 and is an integer depends on the specific functions.Given that all the roots are within 0 to 2007, and the functions are symmetric, it's possible that the solution ( x ) for ( f_i(x) = f_j(x) ) could be an integer within this range. But I need to find how many such pairs ( (i, j) ) result in an integer ( x ) in this range.Alternatively, maybe there's a smarter way to count this without checking each pair individually. Since there are 1004 functions (from ( f_1 ) to ( f_{1004} )), the number of pairs is ( binom{1004}{2} ), which is a large number. But perhaps many of these pairs don't result in integer solutions.Wait, maybe I can think about the functions in terms of their midpoints. Each function ( f_i ) has a vertex at the midpoint of its roots. For ( f_i ), the midpoint is ( (a + b)/2 ). Since all roots are symmetric around 1003.5, the midpoints are all 1003.5. So, all these functions have the same vertex x-coordinate, which is 1003.5. However, their y-values at the vertex differ because the roots are different.But how does this help? Well, if two functions have the same vertex x-coordinate but different y-values, their graphs will intersect at two points symmetric around the vertex. However, since we're dealing with integer x-values, these intersection points might not necessarily be integers.Wait, but earlier I concluded that the difference ( f_i(x) - f_j(x) ) is linear, which would imply that they intersect at exactly one point. But that contradicts the idea that two quadratics intersect at two points. Hmm, maybe I made a mistake there.Let me double-check. If I have two quadratics ( f_i(x) = x^2 + px + q ) and ( f_j(x) = x^2 + rx + s ), then ( f_i(x) - f_j(x) = (p - r)x + (q - s) ), which is indeed linear. So, they can intersect at most once. But wait, two quadratics can intersect at two points, so why is the difference linear?Ah, because in this specific case, all the functions have the same leading coefficient (1). So, when you subtract them, the ( x^2 ) terms cancel out, leaving a linear function. Therefore, they can intersect at most once. So, my initial thought was correct.Therefore, for each pair ( (i, j) ), there is at most one solution ( x ) where ( f_i(x) = f_j(x) ). Now, the question is, how many of these solutions are integers between 0 and 2007.Given that the functions are symmetric around 1003.5, the solution ( x ) for ( f_i(x) = f_j(x) ) would be at the midpoint of the roots of the two functions. Wait, is that true?Let me think. Suppose ( f_i ) has roots at ( a ) and ( b ), and ( f_j ) has roots at ( c ) and ( d ). Then, the difference ( f_i(x) - f_j(x) ) is linear, and its root is at ( x = (ab - cd)/(c + d - a - b) ). Hmm, that seems complicated.Alternatively, maybe there's a pattern in the roots that can help. Since all functions are symmetric around 1003.5, perhaps the solution ( x ) for ( f_i(x) = f_j(x) ) is also symmetric in some way.Wait, let's take a specific example. Let's take ( f_1(x) = x(x - 2007) ) and ( f_2(x) = (x - 1)(x - 2006) ). Let's set them equal:( x(x - 2007) = (x - 1)(x - 2006) )Expanding both sides:( x^2 - 2007x = x^2 - 2007x + 2006 )Subtracting ( x^2 - 2007x ) from both sides:( 0 = 2006 )Wait, that can't be right. So, this equation has no solution. That's interesting.Let me try another pair. Let's take ( f_1(x) = x(x - 2007) ) and ( f_{1004}(x) = (x - 1003)(x - 1004) ).Setting them equal:( x(x - 2007) = (x - 1003)(x - 1004) )Expanding both sides:( x^2 - 2007x = x^2 - 2007x + 1003*1004 )Again, subtracting ( x^2 - 2007x ) from both sides:( 0 = 1003*1004 )Which is also not true. So, no solution here either.Wait a minute, so when I set ( f_i(x) = f_j(x) ), I end up with an equation that reduces to ( 0 = text{constant} ), which has no solution. Does this mean that none of these equations have a solution?But that seems counterintuitive because two different quadratics should intersect somewhere, right? But in this specific case, since they all have the same leading coefficient and are symmetric around the same midpoint, their difference is a constant, which means they don't intersect at all. That's why setting them equal leads to a contradiction.So, if I take any two functions ( f_i ) and ( f_j ), their difference is a constant, meaning ( f_i(x) - f_j(x) = k ), where ( k ) is a constant. Therefore, ( f_i(x) = f_j(x) ) would imply ( k = 0 ), but since ( k ) is non-zero (as we saw in the examples), there are no solutions.Therefore, for all pairs ( (i, j) ) where ( i neq j ), the equation ( f_i(x) = f_j(x) ) has no solution in the integer range from 0 to 2007.So, the number of solutions is zero.

💡Okay, so I have this problem here: the function f(x) = (2m + 3)x^{m² - 3} is a power function, and I need to find the value of m from the given options. The options are A: √3, B: -1, C: -2, D: 2.First, I remember that a power function is generally in the form f(x) = kx^n, where k and n are constants. So, for the given function to be a power function, it should fit this form. That means the coefficient (2m + 3) should be a constant, and the exponent (m² - 3) should also be a constant.Wait, but in the problem, they just say it's a power function. I think that means it's already in the form of a power function, so maybe I don't need to change anything. But let me think again. Maybe they want it to be a specific type of power function, like monomial or something else?No, I think the key here is that the coefficient (2m + 3) should be a constant, and the exponent (m² - 3) should also be a constant. But since m is a variable here, we need to find the value of m that makes this function a power function.Wait, no, m is a constant here, not a variable. The function is in terms of x, so m is just a constant coefficient. So, the function is f(x) = (2m + 3)x^{m² - 3}. For this to be a power function, it needs to be in the form kx^n, where k and n are constants. So, as long as (2m + 3) is a constant and (m² - 3) is a constant, it's already a power function. But maybe there's something else.Wait, maybe they mean that the function is a monomial, which is a type of power function. A monomial has only one term, and the exponent should be a non-negative integer. So, maybe they want the exponent (m² - 3) to be a non-negative integer.So, let's see. The exponent is m² - 3. We need m² - 3 to be a non-negative integer. So, m² - 3 ≥ 0, which means m² ≥ 3, so m ≥ √3 or m ≤ -√3. But looking at the options, we have √3, -1, -2, and 2.So, let's check each option:A: m = √3. Then m² - 3 = (√3)² - 3 = 3 - 3 = 0. So, exponent is 0, which is a non-negative integer. So, that works.B: m = -1. Then m² - 3 = (-1)² - 3 = 1 - 3 = -2. Negative exponent, which is not allowed for a monomial. So, this might not be acceptable.C: m = -2. Then m² - 3 = (-2)² - 3 = 4 - 3 = 1. Positive integer, so that's good.D: m = 2. Then m² - 3 = 4 - 3 = 1. Also positive integer, so that's good.Wait, but the problem just says it's a power function, not necessarily a monomial. So, maybe negative exponents are allowed? Because power functions can have negative exponents, like f(x) = x^{-n} which is 1/x^n.But in that case, the coefficient (2m + 3) should not be zero, because otherwise, it would be zero, which is a constant function, not a power function.So, let's check the coefficient for each option:A: m = √3. Then 2m + 3 = 2√3 + 3, which is not zero.B: m = -1. Then 2m + 3 = 2*(-1) + 3 = -2 + 3 = 1, which is not zero.C: m = -2. Then 2m + 3 = 2*(-2) + 3 = -4 + 3 = -1, which is not zero.D: m = 2. Then 2m + 3 = 4 + 3 = 7, which is not zero.So, all options give a non-zero coefficient, which is fine.But wait, the problem says "the function f(x) = (2m + 3)x^{m² - 3} is a power function." So, maybe they just want it to be a power function, regardless of the exponent being positive or negative.But let's see the options. A is √3, which gives exponent 0, so f(x) = (2√3 + 3)x^0 = (2√3 + 3), which is a constant function. Is a constant function considered a power function? Well, technically, yes, because x^0 = 1, so it's 1 times the constant. But sometimes, people might consider power functions to have exponents that are integers, but not necessarily.Similarly, B: m = -1, exponent is -2, so f(x) = 1*x^{-2} = 1/x², which is a power function.C: m = -2, exponent is 1, so f(x) = -1*x^1 = -x, which is a power function.D: m = 2, exponent is 1, so f(x) = 7x, which is a power function.So, all options seem to make f(x) a power function. But the question is asking for the value of m, so maybe there's more to it.Wait, maybe the question is implying that the function is a power function in the simplest form, meaning the coefficient is 1. So, 2m + 3 = 1.Let me check that. If 2m + 3 = 1, then 2m = -2, so m = -1. That's option B.But why would they specify that? Because in the definition, a power function is f(x) = kx^n, where k is a constant. So, k can be any constant, not necessarily 1. So, why would they set 2m + 3 = 1?Maybe because they want the function to be a monomial with leading coefficient 1. So, if they want f(x) = x^n, then 2m + 3 = 1 and n = m² - 3.So, solving 2m + 3 = 1 gives m = -1, and then n = (-1)^2 - 3 = 1 - 3 = -2. So, f(x) = x^{-2}, which is 1/x².Alternatively, if they just want it to be a power function without any specific coefficient, then all options are correct, but since it's a multiple-choice question, there must be a unique answer.Therefore, the likely intended answer is B: -1, because they want the coefficient to be 1, making it a monomial with leading coefficient 1.So, I think the answer is B: -1.

💡Okay, so I have this geometry problem here, and I need to figure out how many sides the resulting polygon has after constructing a series of shapes on each other. Let me try to break it down step by step.First, the problem says: "Construct an equilateral triangle on one side of a regular pentagon. On one non-adjacent side of the triangle, construct a regular heptagon. On a non-adjacent side of the heptagon, construct a regular nonagon. Finally, on a non-adjacent side of the nonagon, construct a regular dodecagon. How many sides does the resulting polygon have?"Alright, so let's list out the shapes involved and their number of sides:1. Regular pentagon: 5 sides.2. Equilateral triangle: 3 sides.3. Regular heptagon: 7 sides.4. Regular nonagon: 9 sides.5. Regular dodecagon: 12 sides.Now, the key here is to visualize how these shapes are connected. Each new shape is constructed on a non-adjacent side of the previous one. So, starting with the pentagon, we add a triangle to one of its sides. Then, on a non-adjacent side of that triangle, we add a heptagon. Next, on a non-adjacent side of the heptagon, we add a nonagon. Finally, on a non-adjacent side of the nonagon, we add a dodecagon.I think the important thing is to figure out how many sides are "exposed" or contribute to the perimeter of the final polygon. When we attach one shape to another, the side where they are connected is no longer part of the perimeter—it's internal. So, each time we attach a new shape, we lose two sides: one from the original shape and one from the new shape.Let me try to map this out:1. Start with the pentagon: 5 sides.2. Attach a triangle to one side of the pentagon. Now, the pentagon loses one side, and the triangle loses one side. So, the total sides so far would be 5 + 3 - 2 = 6 sides.3. Next, attach a heptagon to a non-adjacent side of the triangle. The triangle loses one more side, and the heptagon loses one side. So, adding 7 sides but subtracting 2, the total becomes 6 + 7 - 2 = 11 sides.4. Then, attach a nonagon to a non-adjacent side of the heptagon. The heptagon loses one more side, and the nonagon loses one side. So, adding 9 sides and subtracting 2, the total becomes 11 + 9 - 2 = 18 sides.5. Finally, attach a dodecagon to a non-adjacent side of the nonagon. The nonagon loses one more side, and the dodecagon loses one side. Adding 12 sides and subtracting 2, the total becomes 18 + 12 - 2 = 28 sides.Wait, but I think I might be double-counting or missing something here. Let me try a different approach.Each shape is attached to the previous one on a non-adjacent side, meaning that each new shape is connected in such a way that it doesn't share a side with the previously attached shape. So, each time we add a new shape, only one side is lost from each of the two shapes involved in the connection.So, starting with the pentagon (5 sides). When we attach the triangle, we lose one side from the pentagon and one side from the triangle, so the total sides become 5 + 3 - 2 = 6.Then, attaching the heptagon to the triangle. Again, we lose one side from the triangle and one from the heptagon. So, 6 + 7 - 2 = 11.Next, attaching the nonagon to the heptagon. Losing one side each: 11 + 9 - 2 = 18.Finally, attaching the dodecagon to the nonagon. Losing one side each: 18 + 12 - 2 = 28.Hmm, so that seems consistent. So, the total number of sides is 28.But let me think again to make sure I'm not making a mistake. Each time we attach a new polygon, we're effectively replacing one side with the sides of the new polygon minus two (since one side is used to attach). So, for each attachment, the number of sides increases by (n - 2), where n is the number of sides of the new polygon.So, starting with the pentagon: 5 sides.After attaching the triangle: 5 + (3 - 2) = 6.After attaching the heptagon: 6 + (7 - 2) = 11.After attaching the nonagon: 11 + (9 - 2) = 18.After attaching the dodecagon: 18 + (12 - 2) = 28.Yes, that seems to confirm it. So, the resulting polygon has 28 sides.I think that makes sense. Each time we add a new polygon, we're effectively adding (n - 2) sides because we're covering up one side with the connection. So, the total increases by (n - 2) each time.Therefore, the answer should be 28, which is option B.

💡Okay, so I have this problem where the sum of the first n terms of a sequence {a_n} is given by S_n = n² + 3n + 1. I need to find the general term formula for the sequence. Hmm, let me think about how to approach this.First, I remember that the sum of the first n terms, S_n, is related to the individual terms of the sequence. Specifically, each term a_n can be found by subtracting the sum of the first (n-1) terms from the sum of the first n terms. So, the formula should be a_n = S_n - S_{n-1}. That makes sense because S_n includes all terms up to a_n, and S_{n-1} includes all terms up to a_{n-1}, so their difference should just be a_n.Let me write that down:a_n = S_n - S_{n-1}Given that S_n = n² + 3n + 1, I can substitute this into the formula:a_n = (n² + 3n + 1) - S_{n-1}Now, I need to find S_{n-1}. Since S_n is given by n² + 3n + 1, replacing n with (n-1) should give me S_{n-1}:S_{n-1} = (n-1)² + 3(n-1) + 1Let me expand that:(n-1)² = n² - 2n + 13(n-1) = 3n - 3So, putting it all together:S_{n-1} = (n² - 2n + 1) + (3n - 3) + 1Now, let's combine like terms:n² - 2n + 1 + 3n - 3 + 1Combine the n² terms: n²Combine the n terms: -2n + 3n = nCombine the constants: 1 - 3 + 1 = -1So, S_{n-1} = n² + n - 1Okay, so now I can substitute S_{n-1} back into the formula for a_n:a_n = (n² + 3n + 1) - (n² + n - 1)Let me subtract the terms:n² + 3n + 1 - n² - n + 1Simplify term by term:n² - n² = 03n - n = 2n1 + 1 = 2So, a_n = 2n + 2Wait a second, that seems straightforward, but I should check if this works for all n. Let me test it for n=1.When n=1, S_1 should be equal to a_1. Let's compute S_1:S_1 = 1² + 3*1 + 1 = 1 + 3 + 1 = 5So, a_1 = 5.But according to the formula I found, a_n = 2n + 2. If I plug in n=1, I get 2*1 + 2 = 4, which is not equal to 5. That means my formula doesn't hold for n=1. Hmm, so I need to adjust it.I think this is because the formula a_n = S_n - S_{n-1} only works for n ≥ 2. For n=1, S_{n-1} would be S_0, which isn't defined in this context. So, the general term should be a piecewise function where a_1 is 5, and for n ≥ 2, it's 2n + 2.Let me write that down:a_n = 5, when n=1a_n = 2n + 2, when n ≥ 2To make sure this is correct, let me test it for n=2.Compute S_2: 2² + 3*2 + 1 = 4 + 6 + 1 = 11Compute a_2 using the formula: 2*2 + 2 = 6Then, S_2 should be a_1 + a_2 = 5 + 6 = 11, which matches. Good.Let me test n=3.Compute S_3: 3² + 3*3 + 1 = 9 + 9 + 1 = 19Compute a_3: 2*3 + 2 = 8Then, S_3 should be a_1 + a_2 + a_3 = 5 + 6 + 8 = 19, which is correct.Okay, so it seems my piecewise function works. Therefore, the general term formula for the sequence is:a_n = 5 when n=1, and a_n = 2n + 2 when n ≥ 2.I think that's the answer. Let me just recap the steps to make sure I didn't skip anything.1. Recognize that a_n = S_n - S_{n-1} for n ≥ 2.2. Compute S_{n-1} by substituting (n-1) into the given S_n formula.3. Subtract S_{n-1} from S_n to find a_n.4. Check the formula for n=1 separately since S_0 isn't defined.5. Verify the formula with a few terms to ensure correctness.Everything seems to check out. I think I'm confident with this solution.