answer：Okay, so I have this complex number expression to calculate: (frac{(1 - i)^3}{i}). Hmm, complex numbers can be tricky, but I think I can handle this step by step. Let me break it down.First, I need to expand the numerator, which is ((1 - i)^3). I remember that when you have something raised to the third power, you can think of it as multiplying the binomial by itself three times. But maybe there's a shortcut or a formula I can use to make this easier. I recall the binomial theorem, which might help here. The binomial theorem says that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k). So, applying that to ((1 - i)^3), where (a = 1), (b = -i), and (n = 3).Let me write that out:[(1 - i)^3 = binom{3}{0}1^3(-i)^0 + binom{3}{1}1^2(-i)^1 + binom{3}{2}1^1(-i)^2 + binom{3}{3}1^0(-i)^3]Calculating each term:1. (binom{3}{0}1^3(-i)^0 = 1 times 1 times 1 = 1)2. (binom{3}{1}1^2(-i)^1 = 3 times 1 times (-i) = -3i)3. (binom{3}{2}1^1(-i)^2 = 3 times 1 times (i^2) = 3 times (-1) = -3)4. (binom{3}{3}1^0(-i)^3 = 1 times 1 times (-i)^3 = (-i)^3)Wait, what's ((-i)^3)? I know that (i^3 = -i), so ((-i)^3 = -i^3 = -(-i) = i). So the last term is (i).Putting all the terms together:[1 - 3i - 3 + i = (1 - 3) + (-3i + i) = -2 - 2i]So, the numerator simplifies to (-2 - 2i).Now, the expression becomes:[frac{-2 - 2i}{i}]Hmm, dividing by (i) can be a bit confusing. I remember that dividing by (i) is the same as multiplying by (-i), because (1/i = -i). Let me verify that:[frac{1}{i} = frac{1}{i} times frac{i}{i} = frac{i}{i^2} = frac{i}{-1} = -i]Yes, that works. So, if I multiply both the numerator and the denominator by (-i), I can eliminate the (i) in the denominator.Let's do that:[frac{-2 - 2i}{i} times frac{-i}{-i} = frac{(-2 - 2i)(-i)}{i times -i}]First, calculate the denominator:[i times -i = -i^2 = -(-1) = 1]So, the denominator becomes 1, which is nice because it simplifies things.Now, the numerator:[(-2 - 2i)(-i) = (-2)(-i) + (-2i)(-i) = 2i + 2i^2]Simplify each term:1. (2i) stays as it is.2. (2i^2 = 2(-1) = -2)So, combining these:[2i - 2 = -2 + 2i]Therefore, the entire expression simplifies to (-2 + 2i).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. Expanded ((1 - i)^3) using the binomial theorem and got (-2 - 2i). That seems right.2. Then, divided by (i), which I converted to multiplying by (-i).3. Multiplied (-2 - 2i) by (-i) and got (-2 + 2i).Everything checks out. I think that's the correct answer.

answer：Okay, so I have this problem here about vectors. Let me read it again: Given vectors a equals (2, -1) and b equals (x, 1). If 2a plus b is collinear with b, find the value of x. Hmm, okay. I need to figure out what x is.First, I remember that when two vectors are collinear, they point in the same or exactly opposite directions. That means one is a scalar multiple of the other. So, if 2a + b is collinear with b, then 2a + b must be equal to some scalar multiplied by b. Let me write that down.So, 2a + b = k * b, where k is some scalar. That makes sense because if they're collinear, one is just a scaled version of the other. Now, let me compute 2a + b. Vector a is (2, -1), so 2a would be 2*(2, -1) which is (4, -2). Vector b is (x, 1), so adding them together, 2a + b is (4 + x, -2 + 1). Simplifying that, it's (4 + x, -1).Alright, so 2a + b is (4 + x, -1). And this should be equal to k times b, which is k*(x, 1) = (kx, k). So, setting the components equal, we have:4 + x = kx (from the x-component)and-1 = k (from the y-component)Wait, that's interesting. From the y-component, we can directly solve for k. It says -1 = k. So, k is -1. That simplifies things. Now, plug that back into the equation for the x-component.So, 4 + x = (-1)*x. Let me write that out: 4 + x = -x. Now, I can solve for x. Let's bring all the x terms to one side. So, I'll add x to both sides: 4 + x + x = 0. That simplifies to 4 + 2x = 0. Then, subtract 4 from both sides: 2x = -4. Finally, divide both sides by 2: x = -2.Wait, so x is -2? Let me double-check that. If x is -2, then vector b is (-2, 1). Then, 2a + b would be 2*(2, -1) + (-2, 1) = (4, -2) + (-2, 1) = (2, -1). Hmm, so 2a + b is (2, -1). And vector b is (-2, 1). Are these collinear?Well, for two vectors to be collinear, one must be a scalar multiple of the other. Let's see: If I take (2, -1) and multiply it by -1, I get (-2, 1), which is exactly vector b. So yes, (2, -1) is -1 times b. That means they are indeed collinear. So, x = -2 seems correct.Just to make sure I didn't make any mistakes in my calculations. Let's go through it again. 2a is (4, -2). Adding b, which is (x, 1), gives (4 + x, -2 + 1) = (4 + x, -1). Since 2a + b is collinear with b, there exists a scalar k such that (4 + x, -1) = k*(x, 1). That gives two equations: 4 + x = kx and -1 = k. Substituting k = -1 into the first equation: 4 + x = -x. Adding x to both sides: 4 + 2x = 0. Then, 2x = -4, so x = -2. Yep, that checks out.I think I've got it right. So, the value of x is -2.**Final Answer**The value of ( x ) is boxed{-2}.

answer：Okay, so I have this problem here: Hayley had 25 meatballs on her plate. Kirsten stole some of her meatballs. Now she has 11 meatballs on her plate. How many meatballs did Kirsten steal?Alright, let's break this down. First, Hayley starts with 25 meatballs. That's the total number she had initially. Then, Kirsten comes along and steals some of them. After that, Hayley is left with 11 meatballs. So, the question is asking how many meatballs Kirsten stole.Hmm, so I think this is a subtraction problem. If Hayley had 25 and now has 11, the difference between those two numbers should tell me how many were stolen. Let me write that down:25 (original number) minus 11 (remaining) equals the number stolen.So, 25 - 11 equals... let me do the math. 25 minus 10 is 15, and then minus 1 more is 14. So, 14 meatballs were stolen.Wait, does that make sense? If Hayley had 25 and lost 14, she would have 11 left. Yes, that adds up. So, Kirsten must have stolen 14 meatballs.I don't think there's anything more to it. It's a straightforward subtraction problem. No need for more complicated calculations or anything. Just subtract the remaining number from the original number to find out how many were taken.So, the answer should be 14 meatballs stolen by Kirsten.

answer：Okay, so I have this problem here about propositions p and q, and I need to figure out which of the given options is true. Let me try to break it down step by step.First, let me understand what each proposition is saying.Proposition p: If a > b, then a + c > b + c.Hmm, this seems like a basic property of inequalities. I remember that if you add the same number to both sides of an inequality, the direction of the inequality doesn't change. So, if a is greater than b, adding c to both sides should preserve that inequality. So, a + c should indeed be greater than b + c. That makes sense. So, proposition p is true.Now, proposition q: If a > b > 0, then ac > bc.Alright, this one is about multiplying both sides of an inequality by a positive number. I recall that when you multiply both sides of an inequality by a positive number, the direction of the inequality remains the same. So, if a is greater than b and both are positive, then multiplying both by c (assuming c is positive) should keep the inequality the same. Wait, hold on, the problem says a > b > 0, but it doesn't specify whether c is positive or negative. Hmm, that's a bit confusing.Wait, no, actually, in proposition q, it's given that a > b > 0, so a and b are both positive. But c is just a real number, so it could be positive or negative. If c is positive, then yes, ac > bc. But if c is negative, then multiplying both sides by c would reverse the inequality, making ac < bc. So, does that mean proposition q is not always true?Wait, but the proposition says "If a > b > 0, then ac > bc." It doesn't specify anything about c. So, if c is positive, it's true, but if c is negative, it's false. Since c can be any real number, including negative, proposition q isn't necessarily always true. So, proposition q is not universally true.Wait, but hold on, in the original problem statement, it just says "a, b, c are real numbers." So, c can be positive or negative. Therefore, proposition q is only true if c is positive, but since c can be negative, the proposition isn't always true. So, proposition q is false.Wait, but in the initial problem, it's stated as a proposition, so maybe I'm misinterpreting it. Maybe in the context of the problem, c is positive? But the problem doesn't specify that. It just says real numbers. So, I think I have to consider c as any real number, positive or negative.Therefore, proposition q is not necessarily true because if c is negative, ac would be less than bc, which would make the conclusion false. So, proposition q is false.Wait, but hold on, let me double-check. If a > b > 0, and c is positive, then ac > bc. If c is negative, then ac < bc. So, if c is positive, the conclusion is true; if c is negative, the conclusion is false. Since c can be either, the proposition q isn't always true. Therefore, q is false.So, summarizing:- Proposition p is true.- Proposition q is false.Now, let's look at the options:A: (¬p) ∨ qSince p is true, ¬p is false. q is false. So, (¬p) ∨ q is false ∨ false, which is false.B: p ∧ qp is true, q is false. So, true ∧ false is false.C: (¬p) ∧ (¬q)¬p is false, ¬q is true (since q is false). So, false ∧ true is false.D: (¬p) ∨ (¬q)¬p is false, ¬q is true. So, false ∨ true is true.Wait, so D is true?But hold on, I thought p was true and q was false. So, ¬p is false, ¬q is true. So, D is (¬p) ∨ (¬q) which is false ∨ true, which is true.But in my initial thought, I thought q was true, but now I'm reconsidering. Wait, maybe I made a mistake earlier.Let me go back to proposition q. It says: If a > b > 0, then ac > bc.Given that a, b, c are real numbers. So, if c is positive, then ac > bc. If c is negative, ac < bc. If c is zero, then ac = bc.So, the conclusion ac > bc is only true when c is positive. Since c can be any real number, including negative and zero, the proposition q is not always true. Therefore, q is false.So, going back, p is true, q is false.Therefore, the options:A: (¬p) ∨ q = false ∨ false = falseB: p ∧ q = true ∧ false = falseC: (¬p) ∧ (¬q) = false ∧ true = falseD: (¬p) ∨ (¬q) = false ∨ true = trueSo, D is true.Wait, but in my initial thought, I thought q was true, but now I realize q is false because c can be negative. So, the correct answer is D.But wait, in the initial problem, the user wrote:"Given that a, b, c are real numbers. Consider the proposition p: If a > b, then a+c > b+c; and proposition q: If a > b > 0, then ac > bc. Among the following propositions, the true one is ( ). A: (¬p)∨q B: p∧q C: (¬p)∧(¬q) D: (¬p)∨(¬q)"And the initial assistant response was:"Given that a, b, c are real numbers. Consider the proposition p: If a > b, then a+c > b+c is a true statement, consequently, ¬p is false. The proposition q: If a > b > 0, then ac > bc is also a true statement. Therefore, (¬p)∨q is true because q is true and the disjunction of a false statement with a true statement is true. p∧q is true because both p and q are true and the conjunction of true statements is true. (¬p)∧(¬q) is false because although ¬q is false, ¬p is also false and the conjunction of false statements is false. (¬p)∨(¬q) is false because both ¬p and ¬q are false, and the disjunction of false statements is false. To find the correct answer, we rely on the truth values of the compound propositions. The correct answer is found by recognizing that both initial propositions p and q are indeed true given the operations allowed with inequalities. Thus, the only option consistent with two true propositions is p∧q, which means both propositions must hold. Therefore, the true proposition among the options given is: boxed{B: p∧q}"Wait, so the initial assistant thought both p and q are true, but I'm now thinking q is false because c can be negative. So, there's a discrepancy here.Let me double-check proposition q again. It says: If a > b > 0, then ac > bc.Given that a, b, c are real numbers. So, if c is positive, then ac > bc. If c is negative, ac < bc. If c is zero, ac = bc.Therefore, the conclusion ac > bc is only true when c is positive. Since c can be any real number, including negative, the implication is not always true. Therefore, proposition q is false.So, the initial assistant was incorrect in assuming q is true. Therefore, the correct answer should be D: (¬p) ∨ (¬q), which is false ∨ true = true.Wait, but in the initial problem, the user wrote that a, b, c are real numbers, so c can be negative. Therefore, q is not necessarily true.So, to summarize:p is true.q is false.Therefore:A: (¬p) ∨ q = false ∨ false = falseB: p ∧ q = true ∧ false = falseC: (¬p) ∧ (¬q) = false ∧ true = falseD: (¬p) ∨ (¬q) = false ∨ true = trueTherefore, the correct answer is D.But the initial assistant thought q was true, leading to answer B. So, I think the initial assistant made a mistake by assuming q is always true, but in reality, q is false because c can be negative.So, the correct answer should be D.

answer：Alright, so I have this quadratic equation: (x^2 - 2a(x - 1) - 1 = 0). I need to find all the values of (a) such that the sum of the roots is equal to the sum of the squares of the roots. Hmm, okay, let me break this down step by step.First, I remember that for any quadratic equation of the form (x^2 + bx + c = 0), the sum of the roots is (-b) and the product of the roots is (c). So, maybe I can rewrite the given equation in the standard form to identify (b) and (c).Let me expand the given equation:(x^2 - 2a(x - 1) - 1 = 0)Expanding the middle term:(x^2 - 2a x + 2a - 1 = 0)So, now it looks like (x^2 - 2a x + (2a - 1) = 0). Comparing this with the standard quadratic equation (x^2 + bx + c = 0), I can see that:- The coefficient (b) is (-2a)- The constant term (c) is (2a - 1)Therefore, using Vieta's formulas:- Sum of the roots, (x_1 + x_2 = -b = 2a)- Product of the roots, (x_1 x_2 = c = 2a - 1)Now, the problem states that the sum of the roots is equal to the sum of the squares of the roots. So, mathematically, that means:(x_1 + x_2 = x_1^2 + x_2^2)I need to express (x_1^2 + x_2^2) in terms of (x_1 + x_2) and (x_1 x_2) because I already have expressions for those in terms of (a). I recall the identity:(x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2)Let me substitute the known values into this identity:(x_1^2 + x_2^2 = (2a)^2 - 2(2a - 1))Calculating this:(x_1^2 + x_2^2 = 4a^2 - 4a + 2)According to the problem, this should be equal to the sum of the roots, which is (2a). So, setting them equal:(2a = 4a^2 - 4a + 2)Now, I need to solve this equation for (a). Let me rearrange the terms to form a quadratic equation:(4a^2 - 4a + 2 - 2a = 0)Combining like terms:(4a^2 - 6a + 2 = 0)So, the quadratic equation is (4a^2 - 6a + 2 = 0). To solve for (a), I can use the quadratic formula:(a = frac{-b pm sqrt{b^2 - 4ac}}{2a})Wait, hold on, that might be confusing because the variable is also (a). Let me rewrite the quadratic equation with different coefficients to avoid confusion. Let me denote:(A = 4), (B = -6), (C = 2)So, the quadratic formula becomes:(a = frac{-B pm sqrt{B^2 - 4AC}}{2A})Plugging in the values:(a = frac{-(-6) pm sqrt{(-6)^2 - 4 times 4 times 2}}{2 times 4})Simplifying:(a = frac{6 pm sqrt{36 - 32}}{8})Calculating the discriminant:(sqrt{36 - 32} = sqrt{4} = 2)So, substituting back:(a = frac{6 pm 2}{8})This gives two solutions:1. (a = frac{6 + 2}{8} = frac{8}{8} = 1)2. (a = frac{6 - 2}{8} = frac{4}{8} = frac{1}{2})So, the values of (a) that satisfy the condition are (1) and (frac{1}{2}).Let me just double-check my steps to make sure I didn't make any mistakes.1. Expanded the original equation correctly: (x^2 - 2a x + 2a - 1 = 0).2. Identified (b = -2a) and (c = 2a - 1) correctly.3. Applied Vieta's formulas correctly: sum is (2a), product is (2a - 1).4. Used the identity (x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2) correctly.5. Substituted the values into the identity and simplified correctly: (4a^2 - 4a + 2).6. Set the sum equal to the sum of squares: (2a = 4a^2 - 4a + 2).7. Rearranged correctly to form the quadratic equation: (4a^2 - 6a + 2 = 0).8. Applied the quadratic formula correctly, even with the confusion of variable names, and arrived at solutions (1) and (frac{1}{2}).Everything seems to check out. So, I'm confident that the values of (a) are (1) and (frac{1}{2}).**Final Answer**The values of (a) are (boxed{1}) and (boxed{frac{1}{2}}).

answer：First, I need to determine the total expenditure from January to June. Since the average expenditure is Rs. 4200 over 6 months, the total expenditure is 6 multiplied by 4200, which equals Rs. 25200.Next, I calculate the total expenditure from February to July. With an average of Rs. 4250 over 6 months, the total expenditure is 6 multiplied by 4250, resulting in Rs. 25500.To find the amount spent in January, I subtract the total expenditure from January to June from the total expenditure from February to July. This gives me 25500 minus 25200, which equals 300.Since the expenditure in July is Rs. 1500, I set up the equation 300 equals J minus 1500, where J is the amount spent in January. Solving for J, I add 1500 to both sides of the equation, resulting in J equals 1800.Therefore, Amithab spent Rs. 1800 in January.