💡Okay, so I have this equation: ( x^2 + y^2 - 2x + 6y + m = 0 ). It says it represents a circle, and I need to find the range of the real number ( m ). The options are A: ( m > 10 ), B: ( m geq 10 ), C: ( m leq 10 ), D: ( m < 10 ). Hmm, I remember that the general equation of a circle is ( x^2 + y^2 + Dx + Ey + F = 0 ), and for it to represent a real circle, the radius must be a positive real number. That means the radius squared has to be positive, right?So, let me recall the formula for the radius in terms of the coefficients. The radius ( r ) is given by ( r = sqrt{left(frac{D}{2}right)^2 + left(frac{E}{2}right)^2 - F} ). For the radius to be real and positive, the expression under the square root, which is ( left(frac{D}{2}right)^2 + left(frac{E}{2}right)^2 - F ), must be greater than zero. In the given equation, comparing it to the general form, I can see that ( D = -2 ), ( E = 6 ), and ( F = m ). So, plugging these into the radius formula, we get:( r = sqrt{left(frac{-2}{2}right)^2 + left(frac{6}{2}right)^2 - m} )Simplifying that:( r = sqrt{(-1)^2 + (3)^2 - m} )Which is:( r = sqrt{1 + 9 - m} )So, ( r = sqrt{10 - m} )For the radius to be a real positive number, the expression inside the square root must be positive. So, ( 10 - m > 0 ). Solving for ( m ), we get:( 10 - m > 0 )Subtracting 10 from both sides:( -m > -10 )Multiplying both sides by -1 (and remembering to flip the inequality sign):( m < 10 )So, the range of ( m ) is all real numbers less than 10. Looking back at the options, that's option D: ( m < 10 ).Wait, let me double-check. If ( m = 10 ), then the radius would be zero, which would mean it's just a single point, not a circle. So, ( m ) has to be strictly less than 10, not less than or equal to. That confirms that D is the correct answer.I think that's it. I don't see any mistakes in my reasoning. So, yeah, the answer should be D.**Final Answer**boxed{D}

💡Okay, so I have this problem where there's a sequence defined by b₁ = b₂ = b₃ = 2, and for every positive integer n, b_{n+3} = b_{n+2} + b_{n+1} + b_n. They give me the values of b₂₅, b₂₆, and b₂₇, which are 2,555,757; 4,700,770; and 8,651,555 respectively. I need to find the remainder when the sum of the first 25 terms, from b₁ to b₂₅, is divided by 1000.Hmm, okay, so first, I need to figure out how to compute the sum S = b₁ + b₂ + ... + b₂₅. Since the sequence is defined recursively, maybe there's a clever way to express the sum in terms of later terms?Let me think about the recurrence relation: b_{n+3} = b_{n+2} + b_{n+1} + b_n. If I sum this equation over n from 1 to 24, maybe something will telescope or simplify.So, let's write out the equations for n = 1 to n = 24:For n=1: b₄ = b₃ + b₂ + b₁For n=2: b₅ = b₄ + b₃ + b₂...For n=24: b₂₇ = b₂₆ + b₂₅ + b₂₄If I add all these equations together, the left side will be the sum from b₄ to b₂₇. The right side will be the sum of each triplet (b₃ + b₂ + b₁), (b₄ + b₃ + b₂), ..., up to (b₂₆ + b₂₅ + b₂₄).Let me denote S = b₁ + b₂ + ... + b₂₅. Then, the sum of the left side is (b₄ + b₅ + ... + b₂₇). The sum of the right side is 3*(b₁ + b₂ + ... + b₂₆) minus some terms because each term from b₁ to b₂₆ is added multiple times.Wait, maybe another approach. Let's denote S = b₁ + b₂ + ... + b₂₅. Then, the sum of the left side is (b₄ + b₅ + ... + b₂₇) = (S - b₁ - b₂ - b₃) + b₂₆ + b₂₇. Because S includes up to b₂₅, so subtracting b₁, b₂, b₃ gives us b₄ to b₂₅, and then adding b₂₆ and b₂₇ gives us up to b₂₇.On the right side, each equation contributes b_{n+2} + b_{n+1} + b_n. So, when we sum from n=1 to 24, we get the sum of b_{n+2} + b_{n+1} + b_n for n=1 to 24.Let me compute that sum:Sum_{n=1 to 24} [b_{n+2} + b_{n+1} + b_n] = Sum_{n=1 to 24} b_n + Sum_{n=1 to 24} b_{n+1} + Sum_{n=1 to 24} b_{n+2}Which is equal to:Sum_{n=1 to 24} b_n + Sum_{n=2 to 25} b_n + Sum_{n=3 to 26} b_nSo, combining these:= (Sum_{n=1 to 24} b_n) + (Sum_{n=2 to 25} b_n) + (Sum_{n=3 to 26} b_n)= (b₁ + b₂ + ... + b₂₄) + (b₂ + b₃ + ... + b₂₅) + (b₃ + b₄ + ... + b₂₆)Now, let's express this in terms of S:= (S - b₂₅) + (S - b₁ - b₂₅) + (S - b₁ - b₂ - b₂₅ + b₂₆)Wait, maybe that's getting complicated. Alternatively, let's note that:Sum_{n=1 to 24} b_n = S - b₂₅Sum_{n=2 to 25} b_n = S - b₁Sum_{n=3 to 26} b_n = (S - b₁ - b₂) + b₂₆So, putting it all together:Sum_{n=1 to 24} [b_{n+2} + b_{n+1} + b_n] = (S - b₂₅) + (S - b₁) + (S - b₁ - b₂ + b₂₆)Simplify:= 3S - 2b₁ - b₂ - b₂₅ + b₂₆But we also have that the left side of the summed equations is (S - b₁ - b₂ - b₃) + b₂₆ + b₂₇So:(S - b₁ - b₂ - b₃) + b₂₆ + b₂₇ = 3S - 2b₁ - b₂ - b₂₅ + b₂₆Simplify both sides:Left side: S - b₁ - b₂ - b₃ + b₂₆ + b₂₇Right side: 3S - 2b₁ - b₂ - b₂₅ + b₂₆Subtract b₂₆ from both sides:Left: S - b₁ - b₂ - b₃ + b₂₇Right: 3S - 2b₁ - b₂ - b₂₅Now, let's move all terms to one side:S - b₁ - b₂ - b₃ + b₂₇ = 3S - 2b₁ - b₂ - b₂₅Bring all terms to left:S - b₁ - b₂ - b₃ + b₂₇ - 3S + 2b₁ + b₂ + b₂₅ = 0Simplify:(-2S) + ( -b₁ - b₂ - b₃ + 2b₁ + b₂ ) + b₂₇ + b₂₅ = 0Which is:-2S + (b₁ - b₃) + b₂₅ + b₂₇ = 0So:-2S + (2 - 2) + b₂₅ + b₂₇ = 0Wait, b₁ = 2, b₂ = 2, b₃ = 2, so b₁ - b₃ = 0.Thus:-2S + b₂₅ + b₂₇ = 0So:2S = b₂₅ + b₂₇Therefore:S = (b₂₅ + b₂₇)/2Given that b₂₅ = 2,555,757 and b₂₇ = 8,651,555, so:S = (2,555,757 + 8,651,555)/2 = (11,207,312)/2 = 5,603,656Now, we need to find 5,603,656 mod 1000, which is the remainder when divided by 1000.To find this, we can look at the last three digits of 5,603,656, which are 656.So, the remainder is 656.**Final Answer**The remainder when the sum is divided by 1000 is boxed{656}.

💡Okay, so I have this equation: ( x - 2log_{10} frac{1}{sqrt{x}} = 3 ). I need to find an interval where I can use the bisection method to approximate the solution. The options are (0,1), (1,2), (2,3), and (3,4). Hmm, let me think about how to approach this.First, I remember that the bisection method requires a continuous function that changes sign over the interval. So, I should probably define a function ( f(x) ) based on the equation and then evaluate it at the endpoints of each interval to see where it changes sign.Let me rewrite the equation as ( f(x) = x - 2log_{10} frac{1}{sqrt{x}} - 3 ). That way, finding the root of ( f(x) ) will give me the solution to the original equation.Now, I need to simplify ( log_{10} frac{1}{sqrt{x}} ). I know that ( frac{1}{sqrt{x}} ) is the same as ( x^{-1/2} ), so the logarithm becomes ( log_{10} x^{-1/2} ). Using logarithm properties, this is equal to ( -frac{1}{2} log_{10} x ). So, substituting back into ( f(x) ), I get:( f(x) = x - 2 times left( -frac{1}{2} log_{10} x right) - 3 )Simplifying that, the 2 and the -1/2 cancel out, so:( f(x) = x + log_{10} x - 3 )Alright, that's simpler. Now, I need to evaluate ( f(x) ) at the endpoints of each interval to see where it changes sign.Let's start with interval A: (0,1). But wait, ( x ) is in the denominator inside the logarithm, so ( x ) must be positive. Also, as ( x ) approaches 0 from the right, ( log_{10} x ) goes to negative infinity, so ( f(x) ) would approach negative infinity. At ( x = 1 ), ( log_{10} 1 = 0 ), so ( f(1) = 1 + 0 - 3 = -2 ). So, both ends of interval A are negative. That means there's no sign change, so the bisection method won't work here.Moving on to interval B: (1,2). Let's compute ( f(1) ) and ( f(2) ). We already know ( f(1) = -2 ). Now, ( f(2) = 2 + log_{10} 2 - 3 ). ( log_{10} 2 ) is approximately 0.3010, so:( f(2) ≈ 2 + 0.3010 - 3 = -0.6990 )Still negative. So, both ends of interval B are negative as well. No sign change, so bisection method isn't applicable here either.Next, interval C: (2,3). Let's compute ( f(2) ) and ( f(3) ). We have ( f(2) ≈ -0.6990 ) as before. Now, ( f(3) = 3 + log_{10} 3 - 3 ). ( log_{10} 3 ) is approximately 0.4771, so:( f(3) ≈ 3 + 0.4771 - 3 = 0.4771 )Okay, so ( f(2) ≈ -0.6990 ) and ( f(3) ≈ 0.4771 ). Here, the function changes from negative to positive. That means there's a root in this interval. So, interval C is a good candidate for the bisection method.Just to be thorough, let's check interval D: (3,4). We already know ( f(3) ≈ 0.4771 ). Now, ( f(4) = 4 + log_{10} 4 - 3 ). ( log_{10} 4 ) is approximately 0.6020, so:( f(4) ≈ 4 + 0.6020 - 3 = 1.6020 )Both ( f(3) ) and ( f(4) ) are positive, so no sign change here either. Therefore, interval D isn't suitable.So, the only interval where ( f(x) ) changes sign is (2,3). That means the solution lies in this interval, and we can use the bisection method here to approximate it.**Final Answer**The interval is boxed{C}.

💡Okay, so I have this triangle ABC, and I'm given that side c equals a times the cosine of angle B. I need to find the measure of angle A. Hmm, let me think about how to approach this.First, I remember that in any triangle, the Law of Cosines relates the sides and angles. It says that for any triangle, c² = a² + b² - 2ab cos C. Wait, but in this case, I have c = a cos B. Maybe I can use that to substitute into the Law of Cosines.Let me write down what I know:c = a cos B.So, if I square both sides, I get c² = a² cos² B.Now, using the Law of Cosines for angle B, which would be:b² = a² + c² - 2ac cos B.But wait, I can substitute c² from the earlier equation into this. So, replacing c² with a² cos² B, I get:b² = a² + a² cos² B - 2ac cos B.Hmm, that seems a bit complicated. Maybe there's another way. Let me think about the Law of Sines. It says that a/sin A = b/sin B = c/sin C.Since I know c = a cos B, maybe I can relate that to the Law of Sines. Let's see:From the Law of Sines, c = a (sin C / sin A).But I also have c = a cos B. So, setting these equal:a cos B = a (sin C / sin A).Dividing both sides by a, I get:cos B = sin C / sin A.Hmm, interesting. I also know that in a triangle, the sum of the angles is π radians, or 180 degrees. So, A + B + C = π.Maybe I can express C in terms of A and B. So, C = π - A - B.Then, sin C = sin(π - A - B) = sin(A + B), since sin(π - x) = sin x.So, sin C = sin(A + B).But from earlier, cos B = sin C / sin A, so substituting sin C:cos B = sin(A + B) / sin A.Let me expand sin(A + B) using the sine addition formula:sin(A + B) = sin A cos B + cos A sin B.So, substituting back in:cos B = [sin A cos B + cos A sin B] / sin A.Let me simplify the right side:[sin A cos B + cos A sin B] / sin A = cos B + (cos A sin B)/sin A.So, now I have:cos B = cos B + (cos A sin B)/sin A.Subtracting cos B from both sides:0 = (cos A sin B)/sin A.So, (cos A sin B)/sin A = 0.For this fraction to be zero, either cos A = 0 or sin B = 0.But in a triangle, angles are between 0 and π, so sin B = 0 would mean B = 0 or π, which isn't possible in a triangle. Therefore, cos A must be zero.If cos A = 0, then angle A must be π/2 radians, or 90 degrees.So, angle A is 90 degrees. That makes triangle ABC a right-angled triangle at A.Okay, that answers the first part. Now, moving on to the second part: If sin C = 1/3, find the value of cos(π + B).First, let's recall that cos(π + B) is equal to -cos B, because cosine is negative in the second quadrant and cos(π + θ) = -cos θ.So, cos(π + B) = -cos B.Therefore, I need to find cos B and then take its negative.Since I know sin C = 1/3, and from the first part, angle A is 90 degrees, so angle C = 90 degrees - angle B.Wait, no. Actually, since A is 90 degrees, the sum of angles B and C must be 90 degrees. So, C = 90 degrees - B.Therefore, sin C = sin(90 degrees - B) = cos B.Because sin(90 - x) = cos x.So, sin C = cos B.But we are given that sin C = 1/3, so cos B = 1/3.Therefore, cos(π + B) = -cos B = -1/3.So, the value of cos(π + B) is -1/3.Let me double-check my steps to make sure I didn't make a mistake.1. Given c = a cos B.2. Used Law of Sines and Law of Cosines to find that angle A is 90 degrees.3. For the second part, used the fact that A is 90 degrees, so C = 90 - B.4. Therefore, sin C = cos B.5. Given sin C = 1/3, so cos B = 1/3.6. Then, cos(π + B) = -cos B = -1/3.Everything seems to check out. So, I think my answers are correct.

💡Okay, so I have this function here: y equals (x cubed plus 9x squared plus 26x plus 24) divided by (x plus 3). The problem says I can simplify this into a quadratic function, which is y equals Ax squared plus Bx plus C, and it's defined everywhere except at x equals D. Then I need to find the sum of A, B, C, and D.Alright, let me start by understanding what this means. The original function is a rational function, meaning it's a polynomial divided by another polynomial. The denominator here is (x + 3), which is a linear polynomial. The numerator is a cubic polynomial. So, I think the idea is to perform polynomial division to simplify this expression.Polynomial division, right? So, if I divide the numerator by the denominator, I should get a quadratic polynomial as the quotient, and maybe a remainder. But since the problem says it simplifies into y equals Ax squared plus Bx plus C, I guess the division should come out evenly without any remainder. That makes sense because otherwise, there would be a remainder term, and the function wouldn't be a simple quadratic.So, let me set up the division. The numerator is x cubed plus 9x squared plus 26x plus 24, and the denominator is x plus 3. I can use either long division or synthetic division. I think synthetic division might be quicker here because the denominator is linear.To use synthetic division, I need to find the root that makes the denominator zero. So, x plus 3 equals zero when x equals -3. That means I'll use -3 in my synthetic division setup.Alright, let's write down the coefficients of the numerator: 1 (for x cubed), 9 (for x squared), 26 (for x), and 24 (constant term). I'll set up synthetic division with these coefficients and the root -3.So, I bring down the 1. Then I multiply 1 by -3, which gives me -3. I add that to the next coefficient, 9, giving me 6. Then I multiply 6 by -3, which is -18, and add that to 26, giving me 8. Then I multiply 8 by -3, which is -24, and add that to 24, giving me 0. Perfect, no remainder.So, the coefficients of the quotient are 1, 6, and 8, which means the quadratic polynomial is x squared plus 6x plus 8. So, A is 1, B is 6, and C is 8.Now, the function is undefined when the denominator is zero, which is when x plus 3 equals zero, so x equals -3. Therefore, D is -3.So, now I need to find the sum of A, B, C, and D. That would be 1 plus 6 plus 8 plus (-3). Let me calculate that: 1 plus 6 is 7, plus 8 is 15, minus 3 is 12.Wait, let me double-check my synthetic division to make sure I didn't make a mistake. Starting with 1, multiplying by -3 gives -3, adding to 9 gives 6. Then 6 times -3 is -18, adding to 26 gives 8. Then 8 times -3 is -24, adding to 24 gives 0. Yep, that looks correct. So the quadratic is indeed x squared plus 6x plus 8.And the point where the function is undefined is at x equals -3, so D is -3. Adding them up: 1 + 6 + 8 + (-3) equals 12. That seems right.I don't think I made any mistakes here. The process was straightforward: perform synthetic division to simplify the rational function into a quadratic, identify the coefficients, note the point of discontinuity, and sum them up. Everything checks out.

💡First, I need to determine the coordinates of point ( C ) using the given vector ( overrightarrow{AC} = (-4, -3) ) and the coordinates of point ( A(0, 1) ).Since ( overrightarrow{AC} = (x - 0, y - 1) = (x, y - 1) ), I can set up the following equations:[begin{cases}x = -4 y - 1 = -3end{cases}]Solving these equations, I find ( x = -4 ) and ( y = -2 ). Therefore, the coordinates of point ( C ) are ( (-4, -2) ).Next, to find the vector ( overrightarrow{BC} ), I use the coordinates of points ( B(3, 2) ) and ( C(-4, -2) ):[overrightarrow{BC} = (-4 - 3, -2 - 2) = (-7, -4)]Thus, the coordinates of vector ( overrightarrow{BC} ) are ( (-7, -4) ).