💡First, simplify the complex fraction by multiplying both numerator and denominator by the conjugate of the denominator:z = frac{5}{i(i+2)} cdot frac{-i(i-2)}{-i(i-2)} = frac{-5i(i-2)}{(i+2)(-i)(i-2)} = frac{-5i(i-2)}{-i^3+2i^2+2i} = frac{-5i(i-2)}{-2i+2+2i} = frac{-5i(i-2)}{2}Now, expand and simplify the expression:z = frac{-5i(i-2)}{2} = frac{-5i^2+10i}{2} = frac{5-10i}{2} = frac{5}{2} - 5iThe imaginary part of ( z ) is ( -5i ). However, the given answer choices suggest that the problem may be asking for the coefficient of the imaginary part, which is ( -5 ). None of the answer choices match this result, so let's double-check our work.Since ( i^2 = -1 ), we can rewrite the given complex number as:z = frac{5}{i(i+2)} = frac{5}{-1 + 2i}Now, we can use the formula to find the imaginary part of a complex number in the form ( a + bi ):z = frac{5}{-1 + 2i} = frac{5(-1 - 2i)}{(-1 + 2i)(-1 - 2i)} = frac{-5 - 10i}{1 + 4} = frac{-5 - 10i}{5} = -1 - 2iThe imaginary part of ( z ) is ( -2i ), so the coefficient of the imaginary part is ( -2 ).Answer: (boxed{A: -2})

💡Okay, so I have this problem here, and I need to figure out whether the condition that the quadratic equation ( x^2 - mx + n = 0 ) has two positive roots is a sufficient, necessary, both, or neither condition for the curve ( mx^2 + ny^2 = 1 ) to be an ellipse. The options are A, B, C, or D. Let me try to break this down step by step.First, I remember that for a quadratic equation ( ax^2 + bx + c = 0 ) to have two positive roots, certain conditions must be met. Specifically, the discriminant must be non-negative (so that there are real roots), the sum of the roots must be positive, and the product of the roots must be positive. Given the equation ( x^2 - mx + n = 0 ), let me write down these conditions. The discriminant ( D ) is ( m^2 - 4n ). For two real roots, ( D ) must be greater than or equal to zero. So, ( m^2 - 4n geq 0 ). Next, the sum of the roots is given by ( -b/a ). In this case, ( a = 1 ) and ( b = -m ), so the sum is ( m ). Since we want two positive roots, the sum must be positive, which means ( m > 0 ).Then, the product of the roots is ( c/a ), which here is ( n ). For the roots to be positive, the product must also be positive, so ( n > 0 ).So, summarizing the conditions for the quadratic equation to have two positive roots:1. ( m^2 - 4n geq 0 )2. ( m > 0 )3. ( n > 0 )Now, moving on to the second part of the problem: the curve ( mx^2 + ny^2 = 1 ) being an ellipse. I recall that the general form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a ) and ( b ) are positive real numbers. Comparing this to the given equation, ( mx^2 + ny^2 = 1 ), we can rewrite it as ( frac{x^2}{1/m} + frac{y^2}{1/n} = 1 ). For this to be an ellipse, both denominators must be positive. That means ( m > 0 ) and ( n > 0 ). Additionally, if ( m = n ), the equation would represent a circle, which is a special case of an ellipse. So, technically, even if ( m = n ), it's still an ellipse. Wait, but I should double-check that. Is a circle considered an ellipse? Yes, in mathematics, a circle is a special case of an ellipse where the major and minor axes are equal. So, if ( m = n ), the equation becomes ( mx^2 + my^2 = 1 ), which is a circle. So, it's still an ellipse.Therefore, the conditions for ( mx^2 + ny^2 = 1 ) to be an ellipse are:1. ( m > 0 )2. ( n > 0 )There's no requirement for ( m ) and ( n ) to be different. So, even if ( m = n ), it's still an ellipse (a circle). Now, let's compare the two sets of conditions. From the quadratic equation, we have:1. ( m^2 - 4n geq 0 )2. ( m > 0 )3. ( n > 0 )From the ellipse condition, we have:1. ( m > 0 )2. ( n > 0 )So, the quadratic equation imposes an additional condition ( m^2 - 4n geq 0 ) that isn't required for the ellipse. Let me think about what this means. If the quadratic equation has two positive roots, then we know ( m > 0 ), ( n > 0 ), and ( m^2 geq 4n ). This would certainly satisfy the conditions for the ellipse because ( m > 0 ) and ( n > 0 ). So, having two positive roots is a sufficient condition for the ellipse because it guarantees that ( m ) and ( n ) are positive. But is it a necessary condition? That is, does the ellipse condition imply that the quadratic equation must have two positive roots? Well, the ellipse only requires ( m > 0 ) and ( n > 0 ). It doesn't say anything about the discriminant ( m^2 - 4n ). So, it's possible for ( m ) and ( n ) to be positive without ( m^2 - 4n geq 0 ). For example, if ( m = 1 ) and ( n = 1 ), then ( m^2 - 4n = 1 - 4 = -3 < 0 ), so the quadratic equation would have no real roots. But ( mx^2 + ny^2 = 1 ) would still be an ellipse (in this case, a circle). Therefore, the condition that the quadratic has two positive roots is sufficient for the ellipse but not necessary because the ellipse can exist even when the quadratic doesn't have two positive roots. Wait, hold on. Let me make sure. If the quadratic doesn't have two positive roots, does that mean the ellipse condition fails? No, because the ellipse only needs ( m > 0 ) and ( n > 0 ). The quadratic's discriminant doesn't affect the ellipse's existence. So, the ellipse can exist even if the quadratic doesn't have two positive roots, meaning the quadratic's condition isn't necessary for the ellipse. So, in summary:- If the quadratic has two positive roots, then the ellipse exists. So, it's sufficient.- However, the ellipse can exist without the quadratic having two positive roots. So, it's not necessary.Therefore, the correct answer should be A: A sufficient but not necessary condition.But wait, let me double-check. The quadratic having two positive roots gives more information than just ( m > 0 ) and ( n > 0 ). It also gives ( m^2 geq 4n ). So, if someone tells me that the quadratic has two positive roots, I know not only that ( m ) and ( n ) are positive but also that ( m ) is at least twice the square root of ( n ). But for the ellipse, I don't need that extra condition. So, having two positive roots is a stronger condition than needed for the ellipse. Therefore, it's sufficient because it guarantees the ellipse, but it's not necessary because the ellipse can exist without that extra condition.Yes, that makes sense. So, the answer should be A.**Final Answer**boxed{A}

💡First, I need to find the vector (overrightarrow{BC}) given the vectors (overrightarrow{BA} = (1, 2)) and (overrightarrow{CA} = (4, 5)).I recall that vector subtraction can be used to find (overrightarrow{BC}) by subtracting (overrightarrow{CA}) from (overrightarrow{BA}).So, I subtract the corresponding components of the vectors:[overrightarrow{BC} = overrightarrow{BA} - overrightarrow{CA} = (1 - 4, 2 - 5) = (-3, -3)]This means the vector (overrightarrow{BC}) is ((-3, -3)), which corresponds to option B.

💡First, I need to determine how many runs have already been scored in the first 10 overs. The run rate during this period was 3.5 runs per over.Next, I'll calculate the total runs scored in those 10 overs by multiplying the run rate by the number of overs:3.5 runs/over × 10 overs = 35 runs.Now, I'll find out how many runs are still needed to reach the target of 320 runs by subtracting the runs already scored:320 runs - 35 runs = 285 runs.With 40 overs remaining, I'll calculate the required run rate by dividing the runs needed by the number of overs left:285 runs / 40 overs = 7.125 runs per over.Therefore, the team needs to maintain a run rate of 7.125 runs per over in the remaining 40 overs to achieve the target of 320 runs.

💡Okay, so Mrs. Hilt is baking bread, and I need to figure out how many loaves she's making. Let me see... The problem says she needs 5 cups of flour in total, and each loaf requires 2.5 cups of flour. Hmm, so I guess I need to find out how many loaves she can make with 5 cups if each loaf takes 2.5 cups.Alright, so if one loaf needs 2.5 cups, then two loaves would need 2.5 times 2, which is 5 cups. That seems straightforward. But wait, let me make sure I'm not missing anything. Is there any other ingredient or factor that might affect the number of loaves? The problem doesn't mention anything else, so I think it's just about the flour.Maybe I should write it out as a division problem to be clear. If she has 5 cups and each loaf uses 2.5 cups, then the number of loaves is 5 divided by 2.5. Let me do that calculation: 5 ÷ 2.5 equals 2. So, that confirms it. She can make 2 loaves with 5 cups of flour.I wonder if there's another way to think about it. Maybe by converting the cups into a different unit or something. But since both the total flour and the flour per loaf are in cups, I don't think that's necessary. It's a straightforward division problem.Wait, could there be a mistake in my calculation? Let me double-check. 2.5 cups times 2 loaves is indeed 5 cups. Yep, that matches the total flour she has. So, I think I'm confident that the answer is 2 loaves.Just to be thorough, maybe I'll visualize it. If one loaf is 2.5 cups, then two loaves would be 2.5 plus 2.5, which is 5. That's exactly the amount she has. So, no flour is left over, and she can make exactly 2 loaves.I guess another way to look at it is by simplifying the numbers. If 2.5 cups is half of 5 cups, then she can make half of 2 loaves? Wait, that doesn't make sense. No, actually, 2.5 is half of 5, so she can make 2 loaves because 2.5 times 2 is 5. Yeah, that makes sense.I think I've thought it through enough. The answer should be 2 loaves of bread.

💡Alright, so I have this problem here: "Given that the coefficient of the ( x^3 ) term in the expansion of ( (ax - 1)^5 ) is 80, find the coefficient of the ( x^2 ) term in the expansion." Hmm, okay, let's break this down step by step.First, I remember that when expanding a binomial like ( (ax - 1)^5 ), we can use the binomial theorem. The binomial theorem states that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ). In this case, our binomial is ( (ax - 1)^5 ), so ( a ) is ( ax ) and ( b ) is ( -1 ), and ( n ) is 5.So, applying the binomial theorem, each term in the expansion will look like ( binom{5}{k} (ax)^{5 - k} (-1)^k ). Simplifying that, each term becomes ( binom{5}{k} a^{5 - k} (-1)^k x^{5 - k} ).Now, the problem mentions the coefficient of the ( x^3 ) term is 80. So, I need to find the value of ( k ) that gives me ( x^3 ). Since the exponent on ( x ) is ( 5 - k ), I set ( 5 - k = 3 ). Solving for ( k ), I get ( k = 2 ).So, the term with ( x^3 ) is ( binom{5}{2} a^{5 - 2} (-1)^2 x^3 ). Simplifying that, it's ( binom{5}{2} a^3 (1) x^3 ). The coefficient here is ( binom{5}{2} a^3 ).Given that this coefficient is 80, I can write the equation:[binom{5}{2} a^3 = 80]Calculating ( binom{5}{2} ), which is ( frac{5!}{2!(5 - 2)!} = frac{120}{2 times 6} = 10 ). So, the equation becomes:[10 a^3 = 80]Solving for ( a ), I divide both sides by 10:[a^3 = 8]Taking the cube root of both sides:[a = 2]Okay, so ( a = 2 ). Now, I need to find the coefficient of the ( x^2 ) term in the expansion. Let's go back to the general term in the binomial expansion:[binom{5}{k} a^{5 - k} (-1)^k x^{5 - k}]I need the term where the exponent on ( x ) is 2, so ( 5 - k = 2 ). Solving for ( k ), I get ( k = 3 ).So, the term with ( x^2 ) is ( binom{5}{3} a^{5 - 3} (-1)^3 x^2 ). Simplifying that, it's ( binom{5}{3} a^2 (-1)^3 x^2 ). The coefficient here is ( binom{5}{3} a^2 (-1)^3 ).First, calculate ( binom{5}{3} ). That's ( frac{5!}{3!(5 - 3)!} = frac{120}{6 times 2} = 10 ).Next, ( a = 2 ), so ( a^2 = 4 ).And ( (-1)^3 = -1 ).Putting it all together:[10 times 4 times (-1) = 10 times (-4) = -40]So, the coefficient of the ( x^2 ) term is -40.Wait, let me double-check my steps to make sure I didn't make a mistake. First, I found ( a = 2 ) correctly by using the coefficient of ( x^3 ). Then, for the ( x^2 ) term, I correctly identified ( k = 3 ) and calculated the binomial coefficient, ( a^2 ), and the sign. Everything seems to add up. Yeah, I think that's right.