answer：First, I need to find the vector (overrightarrow{a} - overrightarrow{c}). Given that (overrightarrow{a} = (3, 1)) and (overrightarrow{c} = (t, 2)), I subtract the corresponding components:[overrightarrow{a} - overrightarrow{c} = (3 - t, 1 - 2) = (3 - t, -1)]Next, since ((overrightarrow{a} - overrightarrow{c})) is perpendicular to (overrightarrow{b}), their dot product must be zero. The dot product of two vectors ((x_1, y_1)) and ((x_2, y_2)) is calculated as:[(x_1 cdot x_2) + (y_1 cdot y_2)]Applying this to our vectors:[(3 - t) cdot 1 + (-1) cdot 3 = 0]Simplifying the equation:[3 - t - 3 = 0][-t = 0]Solving for (t):[t = 0]

answer：First, I need to find the greatest common factor (GCF) of 2750 and 9450. To do this, I'll start by performing the prime factorization of each number.For 2750, I notice that it's divisible by 10, so I can break it down as 2750 = 275 × 10. Further breaking down 275, it's 5² × 11, and 10 is 2 × 5. Combining these, 2750 becomes 2 × 5³ × 11.Next, for 9450, it's also divisible by 10, so I can express it as 9450 = 945 × 10. Breaking down 945, it's 3³ × 5 × 7, and 10 is again 2 × 5. Combining these, 9450 becomes 2 × 3³ × 5² × 7.Now, to find the GCF, I'll identify the common prime factors and take the lowest powers of those factors. Both numbers have 2 and 5 as common prime factors. The lowest power of 2 is 2¹, and the lowest power of 5 is 5².Multiplying these together, the GCF is 2 × 5² = 2 × 25 = 50.

answer：Okay, so I have this equation to solve: [(x - 1)^4 + (x - 1) = 0.] Hmm, it looks a bit complicated, but maybe I can simplify it somehow. Let me think... Maybe I can factor out something common from both terms. I notice that both terms have a ((x - 1)) in them, so perhaps I can factor that out. Let me try that.Factoring out ((x - 1)), the equation becomes:[(x - 1)left[(x - 1)^3 + 1right] = 0.]Alright, so now I have a product of two factors equal to zero. That means either one of the factors is zero. So, I can set each factor equal to zero and solve for (x).First, let's solve (x - 1 = 0). That seems straightforward. If (x - 1 = 0), then (x = 1). Okay, so one solution is (x = 1). That was easy enough.Now, the second factor is ((x - 1)^3 + 1 = 0). Let me write that down separately:[(x - 1)^3 + 1 = 0.]Hmm, so I can rearrange this to:[(x - 1)^3 = -1.]Alright, so I need to find all the real and complex numbers (x) such that when I subtract 1 and cube the result, I get (-1). I remember that solving equations like (y^3 = a) can sometimes have multiple solutions, especially when dealing with complex numbers. So, maybe I should consider both real and complex solutions here.First, let me think about the real solution. If I have (y^3 = -1), then the real solution is straightforward: (y = -1), because ((-1)^3 = -1). So, substituting back, (y = x - 1), so (x - 1 = -1), which gives (x = 0). So, another solution is (x = 0).But wait, I think there are more solutions because when dealing with complex numbers, a cubic equation like this should have three roots. So, besides the real root (x = 0), there should be two more complex roots. Let me try to find those.I remember that in complex numbers, the equation (y^3 = -1) can be written in polar form. The complex number (-1) can be represented as (1 cdot e^{ipi}) because it's on the negative real axis with magnitude 1 and angle (pi). So, to find all the cube roots of (-1), I can use the formula for the nth roots of a complex number. The general formula for the k-th root is:[ y = r^{1/n} cdot e^{i(theta + 2pi k)/n} ]where (r) is the magnitude, (theta) is the angle, and (k = 0, 1, ..., n-1).In this case, (n = 3), (r = 1), and (theta = pi). So, plugging in, we get:[ y = 1^{1/3} cdot e^{i(pi + 2pi k)/3} ]for (k = 0, 1, 2).Simplifying, since (1^{1/3} = 1), we have:[ y = e^{i(pi + 2pi k)/3} ]for (k = 0, 1, 2).Let me compute each of these for (k = 0, 1, 2):1. For (k = 0):[ y = e^{ipi/3} ]This is equal to (cos(pi/3) + isin(pi/3)), which is ( frac{1}{2} + ifrac{sqrt{3}}{2} ).2. For (k = 1):[ y = e^{i(pi + 2pi)/3} = e^{ipi} ]This is equal to (cos(pi) + isin(pi)), which is (-1 + i0), so just (-1).3. For (k = 2):[ y = e^{i(5pi/3)} ]This is equal to (cos(5pi/3) + isin(5pi/3)), which is ( frac{1}{2} - ifrac{sqrt{3}}{2} ).So, the three cube roots of (-1) are:1. ( y = -1 )2. ( y = frac{1}{2} + ifrac{sqrt{3}}{2} )3. ( y = frac{1}{2} - ifrac{sqrt{3}}{2} )Now, remembering that (y = x - 1), we can solve for (x) in each case:1. For (y = -1):[ x - 1 = -1 implies x = 0 ]We already found this solution earlier.2. For (y = frac{1}{2} + ifrac{sqrt{3}}{2}):[ x - 1 = frac{1}{2} + ifrac{sqrt{3}}{2} implies x = 1 + frac{1}{2} + ifrac{sqrt{3}}{2} = frac{3}{2} + ifrac{sqrt{3}}{2} ]3. For (y = frac{1}{2} - ifrac{sqrt{3}}{2}):[ x - 1 = frac{1}{2} - ifrac{sqrt{3}}{2} implies x = 1 + frac{1}{2} - ifrac{sqrt{3}}{2} = frac{3}{2} - ifrac{sqrt{3}}{2} ]So, the three solutions from the second factor are:- (x = 0)- (x = frac{3}{2} + ifrac{sqrt{3}}{2})- (x = frac{3}{2} - ifrac{sqrt{3}}{2})But wait, I already had (x = 1) from the first factor. So, combining all the solutions, I have four solutions in total:1. (x = 1)2. (x = 0)3. (x = frac{3}{2} + ifrac{sqrt{3}}{2})4. (x = frac{3}{2} - ifrac{sqrt{3}}{2})Let me double-check if these solutions satisfy the original equation.First, checking (x = 1):[(1 - 1)^4 + (1 - 1) = 0 + 0 = 0]Yes, that works.Next, checking (x = 0):[(0 - 1)^4 + (0 - 1) = (-1)^4 + (-1) = 1 - 1 = 0]That also works.Now, checking the complex solutions. Let's take (x = frac{3}{2} + ifrac{sqrt{3}}{2}):First, compute (x - 1):[left(frac{3}{2} + ifrac{sqrt{3}}{2}right) - 1 = frac{1}{2} + ifrac{sqrt{3}}{2}]Now, compute ((x - 1)^4):Let me compute ((x - 1)^3) first since I know it's equal to (-1). Wait, from earlier, we had ((x - 1)^3 = -1), so ((x - 1)^4 = (x - 1)^3 cdot (x - 1) = (-1) cdot (x - 1)).So, ((x - 1)^4 = - (x - 1)). Then, the original equation becomes:[(x - 1)^4 + (x - 1) = - (x - 1) + (x - 1) = 0]Which is indeed zero. So, that works.Similarly, for (x = frac{3}{2} - ifrac{sqrt{3}}{2}), the same logic applies because it's the conjugate of the previous solution, and the equation has real coefficients, so if one complex solution works, its conjugate should also work.Therefore, all four solutions satisfy the original equation.Wait a second, but the original equation is a quartic (degree 4) equation, so it should have four roots in total, counting multiplicities, according to the Fundamental Theorem of Algebra. I have found four solutions: one real at (x = 1), another real at (x = 0), and two complex conjugates. That seems correct.But just to make sure, let me expand the original equation and see if it's indeed a quartic.Starting with:[(x - 1)^4 + (x - 1) = 0]Expanding ((x - 1)^4):[(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1]So, the equation becomes:[x^4 - 4x^3 + 6x^2 - 4x + 1 + x - 1 = 0]Simplify:[x^4 - 4x^3 + 6x^2 - 3x = 0]Factor out an (x):[x(x^3 - 4x^2 + 6x - 3) = 0]So, one solution is (x = 0), and the other solutions come from solving (x^3 - 4x^2 + 6x - 3 = 0).Wait, but earlier, I had solutions at (x = 1), (x = 0), and two complex numbers. But here, factoring gives (x = 0) and a cubic equation. Let me check if (x = 1) is a root of the cubic equation.Plugging (x = 1) into (x^3 - 4x^2 + 6x - 3):[1 - 4 + 6 - 3 = 0]Yes, it is zero. So, (x = 1) is a root of the cubic. Therefore, I can factor out ((x - 1)) from the cubic.Let me perform polynomial division or use synthetic division to factor it.Using synthetic division for (x = 1):Coefficients: 1 | -4 | 6 | -3Bring down the 1.Multiply by 1: 1Add to next coefficient: -4 + 1 = -3Multiply by 1: -3Add to next coefficient: 6 + (-3) = 3Multiply by 1: 3Add to last coefficient: -3 + 3 = 0So, the cubic factors as:[(x - 1)(x^2 - 3x + 3) = 0]Therefore, the equation becomes:[x(x - 1)(x^2 - 3x + 3) = 0]So, the solutions are:1. (x = 0)2. (x = 1)3. Solutions to (x^2 - 3x + 3 = 0)Let me solve (x^2 - 3x + 3 = 0) using the quadratic formula:[x = frac{3 pm sqrt{9 - 12}}{2} = frac{3 pm sqrt{-3}}{2} = frac{3 pm isqrt{3}}{2}]Which simplifies to:[x = frac{3}{2} pm frac{sqrt{3}}{2}i]So, the solutions are:1. (x = 0)2. (x = 1)3. (x = frac{3}{2} + frac{sqrt{3}}{2}i)4. (x = frac{3}{2} - frac{sqrt{3}}{2}i)This matches the solutions I found earlier. So, everything checks out.I think I've covered all the steps, factoring, solving each factor, checking the solutions, and even expanding the original equation to confirm the number of solutions. It seems consistent throughout. I don't think I missed anything here. So, the four solutions are correct.**Final Answer**The solutions are (boxed{1}), (boxed{0}), (boxed{frac{3}{2} + frac{sqrt{3}}{2}i}), and (boxed{frac{3}{2} - frac{sqrt{3}}{2}i}).But wait, the original problem didn't specify whether to present all solutions or just the real ones. Since it's a quartic equation, and in the initial solving, I considered complex solutions, it's appropriate to present all four solutions. However, sometimes problems might only ask for real solutions, but since it wasn't specified, I think providing all solutions is correct.Alternatively, if the problem expects only real solutions, then the solutions would be (x = 1) and (x = 0). But given that complex solutions were found and the equation is a quartic, I think including all four is the right approach.So, to present them neatly, I can write them as:(boxed{1}), (boxed{0}), (boxed{frac{3}{2} + frac{sqrt{3}}{2}i}), and (boxed{frac{3}{2} - frac{sqrt{3}}{2}i}).But perhaps the problem expects them in a single box, separated by commas. Let me check the initial instruction. It says "put your final answer within boxed{}". So, maybe I should list all solutions inside a single box.So, combining them, the solutions are:(boxed{0}), (boxed{1}), (boxed{frac{3}{2} + frac{sqrt{3}}{2}i}), and (boxed{frac{3}{2} - frac{sqrt{3}}{2}i}).But actually, in the previous assistant's answer, they presented all solutions in a single box separated by commas. So, maybe I should do the same for consistency.Thus, the final answer is:(boxed{0, 1, frac{3}{2} + frac{sqrt{3}}{2}i, frac{3}{2} - frac{sqrt{3}}{2}i})But wait, in the initial problem, the equation is [(x - 1)^4 + (x - 1) = 0.] When I factored it, I got ((x - 1)((x - 1)^3 + 1) = 0), leading to solutions (x = 1) and the roots of ((x - 1)^3 = -1). So, in that case, the solutions are (x = 1), and (x = 0) (from the real root), and the two complex roots.But when I expanded the equation, I got (x(x - 1)(x^2 - 3x + 3) = 0), leading to solutions (x = 0), (x = 1), and the roots of (x^2 - 3x + 3 = 0), which are (frac{3}{2} pm frac{sqrt{3}}{2}i). So, same solutions.Therefore, the four solutions are correct.Just to make sure, let me plug in one of the complex solutions into the original equation.Take (x = frac{3}{2} + frac{sqrt{3}}{2}i).Compute (x - 1 = frac{1}{2} + frac{sqrt{3}}{2}i).Compute ((x - 1)^4):First, compute ((x - 1)^2):[left(frac{1}{2} + frac{sqrt{3}}{2}iright)^2 = left(frac{1}{2}right)^2 + 2 cdot frac{1}{2} cdot frac{sqrt{3}}{2}i + left(frac{sqrt{3}}{2}iright)^2 = frac{1}{4} + frac{sqrt{3}}{2}i - frac{3}{4} = -frac{1}{2} + frac{sqrt{3}}{2}i]Then, compute ((x - 1)^4 = left((x - 1)^2right)^2 = left(-frac{1}{2} + frac{sqrt{3}}{2}iright)^2):[left(-frac{1}{2}right)^2 + 2 cdot -frac{1}{2} cdot frac{sqrt{3}}{2}i + left(frac{sqrt{3}}{2}iright)^2 = frac{1}{4} - frac{sqrt{3}}{2}i - frac{3}{4} = -frac{1}{2} - frac{sqrt{3}}{2}i]Now, compute ((x - 1)^4 + (x - 1)):[-frac{1}{2} - frac{sqrt{3}}{2}i + frac{1}{2} + frac{sqrt{3}}{2}i = 0]Yes, it works.Similarly, for (x = frac{3}{2} - frac{sqrt{3}}{2}i), the same steps would apply, and it would also satisfy the equation.Therefore, all four solutions are indeed correct.I think I've thoroughly checked each step and confirmed that all solutions satisfy the original equation. So, I'm confident that these are the correct solutions.**Final Answer**The solutions are (boxed{0}), (boxed{1}), (boxed{frac{3}{2} + frac{sqrt{3}}{2}i}), and (boxed{frac{3}{2} - frac{sqrt{3}}{2}i}).But since the problem asks to put the final answer within a single box, I should present all solutions in one box, separated by commas.(boxed{0, 1, frac{3}{2} + frac{sqrt{3}}{2}i, frac{3}{2} - frac{sqrt{3}}{2}i})Wait, actually, looking back at the initial problem, it's written as:[(x - 1)^4 + (x - 1) = 0.]And the assistant's previous answer was:(boxed{1, 0, 1 + frac{sqrt{3}}{2}i, 1 - frac{sqrt{3}}{2}i})But in my solving, I got (x = frac{3}{2} pm frac{sqrt{3}}{2}i), which is equivalent to (1 + frac{1}{2} pm frac{sqrt{3}}{2}i), so that's the same as (1 + frac{sqrt{3}}{2}i) and (1 - frac{sqrt{3}}{2}i). So, actually, my solutions can be written as (1 + frac{sqrt{3}}{2}i) and (1 - frac{sqrt{3}}{2}i), which matches the previous assistant's answer.So, perhaps it's better to write the complex solutions as (1 pm frac{sqrt{3}}{2}i) instead of (frac{3}{2} pm frac{sqrt{3}}{2}i). Let me adjust that.Since (x = 1 + y), where (y = frac{1}{2} pm frac{sqrt{3}}{2}i), so (x = 1 + frac{1}{2} pm frac{sqrt{3}}{2}i = frac{3}{2} pm frac{sqrt{3}}{2}i). Alternatively, it's also correct to write it as (1 + frac{sqrt{3}}{2}i) and (1 - frac{sqrt{3}}{2}i), but actually, that would be incorrect because (1 + frac{sqrt{3}}{2}i) is not equal to (frac{3}{2} + frac{sqrt{3}}{2}i). Wait, no, actually, (1 + frac{sqrt{3}}{2}i) is different from (frac{3}{2} + frac{sqrt{3}}{2}i). So, perhaps I made a mistake in interpreting the solutions.Wait, let's clarify. When I solved for (y = x - 1), I found (y = -1), (y = frac{1}{2} + frac{sqrt{3}}{2}i), and (y = frac{1}{2} - frac{sqrt{3}}{2}i). Therefore, (x = y + 1), so:1. (x = -1 + 1 = 0)2. (x = frac{1}{2} + frac{sqrt{3}}{2}i + 1 = frac{3}{2} + frac{sqrt{3}}{2}i)3. (x = frac{1}{2} - frac{sqrt{3}}{2}i + 1 = frac{3}{2} - frac{sqrt{3}}{2}i)So, the complex solutions are indeed (frac{3}{2} pm frac{sqrt{3}}{2}i), not (1 pm frac{sqrt{3}}{2}i). Therefore, the previous assistant's answer had a mistake in writing the complex solutions as (1 pm frac{sqrt{3}}{2}i), whereas the correct solutions are (frac{3}{2} pm frac{sqrt{3}}{2}i).Therefore, my final answer should be:(boxed{0}), (boxed{1}), (boxed{frac{3}{2} + frac{sqrt{3}}{2}i}), and (boxed{frac{3}{2} - frac{sqrt{3}}{2}i}).But again, since the problem asks for the final answer in a single box, I should present all solutions together.(boxed{0, 1, frac{3}{2} + frac{sqrt{3}}{2}i, frac{3}{2} - frac{sqrt{3}}{2}i})Yes, that seems correct. I think I've clarified any confusion about the form of the complex solutions. So, the final answer is as above.

answer：To find the value of (cos alpha) given the point (P(-4, 3)) on the terminal side of angle (alpha), follow these steps:1. **Identify the coordinates**: - The point (P(-4, 3)) has coordinates (x = -4) and (y = 3).2. **Calculate the radius (r)**: - Use the distance formula to find the radius: [ r = sqrt{x^2 + y^2} = sqrt{(-4)^2 + 3^2} = sqrt{16 + 9} = sqrt{25} = 5 ]3. **Determine (cos alpha)**: - The cosine of an angle in standard position is given by: [ cos alpha = frac{x}{r} ] - Substitute the known values: [ cos alpha = frac{-4}{5} = -frac{4}{5} ]4. **Select the correct option**: - The value of (cos alpha) is (-frac{4}{5}), which corresponds to option C.

answer：Alright, so I have this problem here where I need to calculate the value of the expression:frac{1}{1+x+xy} + frac{1}{1+y+yz} + frac{1}{1+z+zx}given that ( xyz = 1 ). Hmm, okay. Let me try to figure this out step by step.First, I notice that each term in the expression has a similar structure: each is a fraction where the denominator is a sum of 1, a variable, and the product of that variable with another. The variables involved are x, y, and z, and they multiply together to give 1. Maybe that condition will help simplify things.Let me write down the expression again to get a clear view:frac{1}{1+x+xy} + frac{1}{1+y+yz} + frac{1}{1+z+zx}Given ( xyz = 1 ), perhaps I can express one variable in terms of the others. For example, if ( xyz = 1 ), then ( z = frac{1}{xy} ). Maybe substituting z in terms of x and y will help simplify the expression.Let me try that. So, substituting ( z = frac{1}{xy} ) into the third term:frac{1}{1 + z + zx} = frac{1}{1 + frac{1}{xy} + frac{x}{xy}} = frac{1}{1 + frac{1}{xy} + frac{1}{y}}Hmm, that seems a bit messy. Maybe there's another approach.Looking back at the expression, I notice that each denominator has a similar pattern. Maybe I can manipulate each term to have a common denominator or find a relationship between them.Let me consider the first term:frac{1}{1 + x + xy}I can factor out an x from the last two terms in the denominator:frac{1}{1 + x(1 + y)}Not sure if that helps yet. Let me look at the second term:frac{1}{1 + y + yz}Similarly, I can factor out a y:frac{1}{1 + y(1 + z)}And the third term:frac{1}{1 + z + zx}Factor out a z:frac{1}{1 + z(1 + x)}Still, not seeing a clear path. Maybe I can use the condition ( xyz = 1 ) to relate the terms.Let me think about multiplying numerator and denominator by something to make the denominators similar. For example, in the second term, if I multiply numerator and denominator by x, I get:frac{x}{x(1 + y + yz)} = frac{x}{x + xy + xyz}But since ( xyz = 1 ), this simplifies to:frac{x}{x + xy + 1}Interesting, now the denominator is the same as the denominator of the first term! So, the second term becomes:frac{x}{1 + x + xy}Similarly, let's try the third term. If I multiply numerator and denominator by xy:frac{xy}{xy(1 + z + zx)} = frac{xy}{xy + xyz + xyzx}Again, using ( xyz = 1 ), this becomes:frac{xy}{xy + 1 + x}So, the third term simplifies to:frac{xy}{1 + x + xy}Now, let's rewrite the entire expression with these simplifications:frac{1}{1 + x + xy} + frac{x}{1 + x + xy} + frac{xy}{1 + x + xy}Now, all three terms have the same denominator, which is (1 + x + xy). So, I can combine them:frac{1 + x + xy}{1 + x + xy}Which simplifies to:1Wow, okay, so the entire expression simplifies to 1. That makes sense because when we manipulated each term using the condition ( xyz = 1 ), we were able to express each fraction with the same denominator and then sum them up to get 1.I think that's the answer. Let me just double-check my steps to make sure I didn't make any mistakes.1. I started by trying to substitute z in terms of x and y, but that didn't seem helpful.2. Then, I tried factoring each denominator, which also didn't lead me anywhere immediately.3. Next, I decided to manipulate the second and third terms by multiplying numerator and denominator by x and xy, respectively.4. Using the condition ( xyz = 1 ), I was able to simplify the denominators to match the first term's denominator.5. Finally, adding all three fractions together gave me 1.Everything seems to check out. I don't see any errors in my reasoning, so I'm confident that the value of the expression is indeed 1.

answer：To solve the equation ((x - i)i = y + 2i) where (x) and (y) are real numbers, we start by expanding the left side of the equation. First, distribute (i) across ((x - i)):[(x - i)i = xi - i^2]Since (i^2 = -1), this simplifies to:[xi - (-1) = xi + 1]So, the equation becomes:[xi + 1 = y + 2i]Next, we equate the real and imaginary parts from both sides of the equation. - The real part on the left side is (1), and on the right side, it is (y). Therefore:[1 = y]- The imaginary part on the left side is (x), and on the right side, it is (2). Therefore:[x = 2]With (x = 2) and (y = 1), we can form the complex number (x + yi):[x + yi = 2 + 1i = 2 + i]Thus, the correct answer is (2 + i), which corresponds to option B.