answer：First, I need to determine the amount of the first sales based on the given information about royalties and the change in the royalty rate.I'll start by defining the variables:- Let ( S ) represent the amount of the first sales.- Let ( R ) represent the royalty rate for the first sales.From the problem, I know that the company received 2 million in royalties on the first ( S ) amount in sales. This gives me the equation:[ 2 text{ million} = R times S ]Next, the company received 8 million in royalties on the next 100 million in sales. The royalty rate for this second period is 60% less than the initial rate ( R ), so the new rate ( R' ) is:[ R' = R - 0.60R = 0.40R ]Using this new rate, I can set up the equation for the second period:[ 8 text{ million} = 0.40R times 100 text{ million} ]Now, I'll solve for ( R ) using the second equation:[ 8 text{ million} = 0.40R times 100 text{ million} ][ 8 = 0.40R times 100 ][ 8 = 40R ][ R = frac{8}{40} ][ R = 0.20 ]With the royalty rate ( R ) determined, I can now use the first equation to find ( S ):[ 2 text{ million} = 0.20 times S ][ S = frac{2 text{ million}}{0.20} ][ S = 10 text{ million} ]Therefore, the amount of the first sales was 10 million.

answer：Alright, so I have this problem here about balls of different colors in a bag, and I need to figure out the minimum number of balls I must draw to ensure that I have at least 10 balls of one color. Let me try to break this down step by step.First, the bag has a total of 70 balls. Out of these, 20 are red, 20 are blue, 20 are yellow, and the remaining are black and white. So, let me calculate how many black and white balls there are. If there are 70 balls in total and 20 are red, blue, and yellow each, that's 20 * 3 = 60 balls. So, the remaining 10 balls must be black and white. I'm assuming that means 5 black and 5 white, but the problem doesn't specify, so maybe it's just a total of 10 without specifying the exact number of each. Hmm, maybe that's not too important for solving the problem.Okay, so the main point is that there are four colors: red, blue, yellow, and then black and white, which I'll treat as separate colors for now. So, in total, there are five colors: red, blue, yellow, black, and white, with red, blue, and yellow each having 20 balls, and black and white having 10 balls combined.Now, the question is asking for the minimum number of balls I need to draw to ensure that I have at least 10 balls of one color. This sounds like a problem that involves the Pigeonhole Principle. The Pigeonhole Principle is a counting argument that says if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. In this case, the "pigeonholes" are the colors, and the "pigeons" are the balls we draw.But in this case, we're not just looking for more than one; we're looking for at least 10 balls of one color. So, how does the Pigeonhole Principle apply here? Well, I think it's about ensuring that no matter how the balls are distributed when we draw them, we'll have at least 10 of one color.Let me think about the worst-case scenario. That is, the scenario where I'm trying to avoid getting 10 balls of any single color for as long as possible. So, I want to draw as many balls as possible without getting 10 of any color. Once I exceed that number, I can be sure that I've got at least 10 of one color.So, if I want to avoid getting 10 of any color, I would try to draw as many balls as possible with only 9 of each color. But wait, there are different numbers of balls for each color. For example, red, blue, and yellow each have 20 balls, while black and white have fewer. So, I need to consider the maximum number of balls I can draw without getting 10 of any color.Let me try to calculate that. If I draw 9 red, 9 blue, and 9 yellow balls, that's 9 * 3 = 27 balls. Then, for black and white, since there are only 10 balls combined, I can draw all 10 of them. So, in total, that's 27 + 10 = 37 balls. So, in this case, I've drawn 37 balls without getting 10 of any single color.But wait, is that correct? Let me double-check. If I draw 9 red, 9 blue, 9 yellow, and all 10 black and white, that's indeed 37 balls. So, in this case, I haven't yet reached 10 balls of any single color because red, blue, and yellow each have only 9, and black and white are all drawn but they're only 10 in total.So, if I draw one more ball after that, which would be the 38th ball, I have to consider what color that ball could be. Since I've already drawn all the black and white balls, the next ball must be either red, blue, or yellow. But I've already drawn 9 of each of those colors, so drawing one more would give me 10 of that color.Therefore, the minimum number of balls I need to draw to ensure that I have at least 10 balls of one color is 38.Wait a minute, let me think again. If black and white are separate colors, does that affect the calculation? Because if I draw all 10 black and white balls, that's 10 balls, but they're two different colors. So, does that mean I could have drawn 5 black and 5 white, or some other combination? But in the worst-case scenario, I think it's safe to assume that I could draw all 10 of them, regardless of their distribution between black and white.So, in the worst case, I could have drawn 9 red, 9 blue, 9 yellow, and all 10 black and white balls, totaling 37 balls, without having 10 of any single color. Then, the next ball, the 38th, would have to be red, blue, or yellow, each of which would then give me 10 of that color.Therefore, I think 38 is the correct answer.But let me consider another angle. Suppose that black and white are not separate colors but just two different colors within the remaining 10 balls. So, if I draw 9 red, 9 blue, 9 yellow, and then 10 black and white, that's still 37 balls. Then, the next ball would have to be red, blue, or yellow, giving me 10 of that color.Alternatively, if I consider black and white as separate colors, then I could have drawn 9 red, 9 blue, 9 yellow, 5 black, and 5 white, totaling 37 balls. Then, the next ball would have to be red, blue, or yellow, giving me 10 of that color.Either way, the calculation seems consistent. So, I think 38 is the right answer.But just to be thorough, let me think about the total number of balls. There are 70 balls in total. If I draw 38 balls, that leaves 32 balls remaining in the bag. But I don't think that affects the calculation because we're only concerned with the number of balls drawn, not the number remaining.Also, it's important to note that the problem specifies that the balls differ only by color, so we don't have to consider any other distinguishing features. It's purely about the color distribution.In summary, by considering the worst-case scenario where I draw as many balls as possible without getting 10 of any single color, I determine that drawing 37 balls could still leave me without 10 of any color. Therefore, drawing one more ball, making it 38, ensures that I have at least 10 balls of one color.

answer：Alright, so I have this function ( f(x) = 2cos^2 x + sin 2x - 1 ), and I need to check four propositions about it. Let me try to understand each part step by step.First, maybe I should simplify the function to make it easier to analyze. I remember that ( cos^2 x ) can be rewritten using a double-angle identity. Specifically, ( cos^2 x = frac{1 + cos 2x}{2} ). So, substituting that into the function:[f(x) = 2 left( frac{1 + cos 2x}{2} right) + sin 2x - 1]Simplifying this:[f(x) = (1 + cos 2x) + sin 2x - 1 = cos 2x + sin 2x]Okay, so the function simplifies to ( f(x) = sin 2x + cos 2x ). That looks a bit simpler. Maybe I can write this as a single sine or cosine function using the amplitude-phase form. I recall that ( asin theta + bcos theta = Rsin(theta + phi) ) where ( R = sqrt{a^2 + b^2} ) and ( phi = arctanleft(frac{b}{a}right) ).In this case, ( a = 1 ) and ( b = 1 ), so:[R = sqrt{1^2 + 1^2} = sqrt{2}][phi = arctanleft(frac{1}{1}right) = frac{pi}{4}]Therefore, the function can be rewritten as:[f(x) = sqrt{2} sinleft(2x + frac{pi}{4}right)]Alright, that seems useful. Now, let's tackle each proposition one by one.**1. The function is decreasing in the interval ( left[frac{pi}{8}, frac{5pi}{8}right] ).**To determine if the function is decreasing, I need to look at its derivative. Let's compute ( f'(x) ):[f'(x) = frac{d}{dx} left[ sqrt{2} sinleft(2x + frac{pi}{4}right) right] = 2sqrt{2} cosleft(2x + frac{pi}{4}right)]The function is decreasing when the derivative is negative. So, we need to find where ( cosleft(2x + frac{pi}{4}right) < 0 ).Let me find the values of ( 2x + frac{pi}{4} ) when ( x ) is in ( left[frac{pi}{8}, frac{5pi}{8}right] ):- When ( x = frac{pi}{8} ): [ 2x + frac{pi}{4} = 2 cdot frac{pi}{8} + frac{pi}{4} = frac{pi}{4} + frac{pi}{4} = frac{pi}{2} ] - When ( x = frac{5pi}{8} ): [ 2x + frac{pi}{4} = 2 cdot frac{5pi}{8} + frac{pi}{4} = frac{5pi}{4} + frac{pi}{4} = frac{3pi}{2} ]So, ( 2x + frac{pi}{4} ) ranges from ( frac{pi}{2} ) to ( frac{3pi}{2} ). In this interval, ( costheta ) is negative because cosine is negative in the second and third quadrants (i.e., between ( frac{pi}{2} ) and ( frac{3pi}{2} )).Therefore, ( f'(x) = 2sqrt{2} cosleft(2x + frac{pi}{4}right) ) is negative throughout the interval ( left[frac{pi}{8}, frac{5pi}{8}right] ), meaning the function is indeed decreasing there. So, proposition 1 is correct.**2. The line ( x = frac{pi}{8} ) is an axis of symmetry for the graph of the function.**An axis of symmetry at ( x = a ) means that for any point ( (a + h, f(a + h)) ), there is a corresponding point ( (a - h, f(a - h)) ) such that ( f(a + h) = f(a - h) ).Given that ( f(x) = sqrt{2} sinleft(2x + frac{pi}{4}right) ), let's check if ( fleft(frac{pi}{8} + hright) = fleft(frac{pi}{8} - hright) ).Compute ( fleft(frac{pi}{8} + hright) ):[fleft(frac{pi}{8} + hright) = sqrt{2} sinleft(2left(frac{pi}{8} + hright) + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{4} + 2h + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{2} + 2hright)]Similarly, compute ( fleft(frac{pi}{8} - hright) ):[fleft(frac{pi}{8} - hright) = sqrt{2} sinleft(2left(frac{pi}{8} - hright) + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{4} - 2h + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{2} - 2hright)]Now, using the identity ( sinleft(frac{pi}{2} + thetaright) = costheta ) and ( sinleft(frac{pi}{2} - thetaright) = costheta ), we have:[fleft(frac{pi}{8} + hright) = sqrt{2} cos(2h)][fleft(frac{pi}{8} - hright) = sqrt{2} cos(2h)]Thus, ( fleft(frac{pi}{8} + hright) = fleft(frac{pi}{8} - hright) ), which means the function is symmetric about ( x = frac{pi}{8} ). So, proposition 2 is correct.**3. The graph of the function ( f(x) ) can be obtained by translating the graph of ( y = sqrt{2}sin 2x ) to the left by ( frac{pi}{4} ).**Let's analyze this. The original function is ( y = sqrt{2}sin 2x ). Translating this graph to the left by ( frac{pi}{4} ) would result in the function:[y = sqrt{2}sinleft(2left(x + frac{pi}{4}right)right) = sqrt{2}sinleft(2x + frac{pi}{2}right)]But our function is ( f(x) = sqrt{2}sinleft(2x + frac{pi}{4}right) ). So, the phase shift here is ( frac{pi}{4} ) instead of ( frac{pi}{2} ). Therefore, translating ( y = sqrt{2}sin 2x ) to the left by ( frac{pi}{8} ) would give:[y = sqrt{2}sinleft(2left(x + frac{pi}{8}right)right) = sqrt{2}sinleft(2x + frac{pi}{4}right)]Which matches our function ( f(x) ). So, the correct translation should be ( frac{pi}{8} ) to the left, not ( frac{pi}{4} ). Therefore, proposition 3 is incorrect.**4. If ( x in left[0, frac{pi}{2}right] ), then the range of ( f(x) ) is ( [0, sqrt{2}] ).**To find the range, let's analyze ( f(x) = sqrt{2}sinleft(2x + frac{pi}{4}right) ) over ( x in left[0, frac{pi}{2}right] ).First, determine the interval for ( 2x + frac{pi}{4} ):- When ( x = 0 ): [ 2x + frac{pi}{4} = 0 + frac{pi}{4} = frac{pi}{4} ]- When ( x = frac{pi}{2} ): [ 2x + frac{pi}{4} = pi + frac{pi}{4} = frac{5pi}{4} ] So, ( 2x + frac{pi}{4} ) ranges from ( frac{pi}{4} ) to ( frac{5pi}{4} ).The sine function ( sintheta ) has a maximum of 1 and a minimum of -1. However, over the interval ( frac{pi}{4} ) to ( frac{5pi}{4} ), the maximum occurs at ( theta = frac{pi}{2} ) and the minimum at ( theta = frac{3pi}{2} ). But ( frac{3pi}{2} ) is outside our interval, so we need to find the minimum within ( frac{pi}{4} ) to ( frac{5pi}{4} ).The sine function decreases from ( frac{pi}{2} ) to ( frac{3pi}{2} ), but since our interval stops at ( frac{5pi}{4} ), the minimum in this interval occurs at ( frac{5pi}{4} ):[sinleft(frac{5pi}{4}right) = -frac{sqrt{2}}{2}]Therefore, the range of ( sinleft(2x + frac{pi}{4}right) ) over ( x in left[0, frac{pi}{2}right] ) is ( left[-frac{sqrt{2}}{2}, 1right] ). Multiplying by ( sqrt{2} ), the range of ( f(x) ) is:[left[-frac{sqrt{2}}{2} times sqrt{2}, 1 times sqrt{2}right] = left[-1, sqrt{2}right]]But the proposition claims the range is ( [0, sqrt{2}] ), which is incorrect because the function can take negative values in this interval. Therefore, proposition 4 is incorrect.**Summary:**- Proposition 1: Correct- Proposition 2: Correct- Proposition 3: Incorrect- Proposition 4: IncorrectSo, out of the four propositions, two are correct.**Final Answer**boxed{B}

answer：Okay, so I have this polynomial ( Q(x) = 3x^3 + dx^2 + ex + f ). I need to find the value of ( e ). Let me see what information I have.First, the problem says that the mean of its zeros is twice the product of its zeros. Hmm, okay. Also, the sum of the coefficients equals the y-intercept of the graph, which is 12. Alright, let's break this down step by step.Starting with the y-intercept. The y-intercept of a polynomial is the constant term when ( x = 0 ). So, plugging in ( x = 0 ) into ( Q(x) ), we get ( Q(0) = f ). The problem states that this y-intercept is 12, so ( f = 12 ). Got that.Next, the sum of the coefficients equals the y-intercept, which is 12. The coefficients are 3, ( d ), ( e ), and ( f ). So, adding them up: ( 3 + d + e + f = 12 ). Since we already know ( f = 12 ), let me substitute that in: ( 3 + d + e + 12 = 12 ). Simplifying that, ( 15 + d + e = 12 ), so ( d + e = -3 ). Hmm, okay, so I have an equation relating ( d ) and ( e ): ( d + e = -3 ). I'll keep that in mind.Now, moving on to the zeros of the polynomial. For a cubic polynomial ( ax^3 + bx^2 + cx + d ), the sum of the zeros is ( -b/a ), the product of the zeros is ( -d/a ), and the sum of the products of the zeros taken two at a time is ( c/a ). Wait, in our case, the polynomial is ( 3x^3 + dx^2 + ex + f ), so ( a = 3 ), ( b = d ), ( c = e ), and ( d = f ). So, applying Vieta's formulas:1. Sum of zeros: ( -d/3 )2. Product of zeros: ( -f/3 )3. Sum of products two at a time: ( e/3 )The problem states that the mean of the zeros is twice the product of the zeros. Let me parse that. The mean of the zeros is the sum of the zeros divided by the number of zeros. Since it's a cubic polynomial, there are three zeros. So, mean of zeros = ( frac{-d/3}{3} = -d/9 ).The product of the zeros is ( -f/3 ). So, according to the problem, ( -d/9 = 2 times (-f/3) ). Let me write that equation:( -d/9 = 2 times (-f/3) )Simplify the right side: ( 2 times (-f/3) = -2f/3 ). So, the equation becomes:( -d/9 = -2f/3 )Multiply both sides by 9 to eliminate denominators:( -d = -6f )Multiply both sides by -1:( d = 6f )Since we know ( f = 12 ), substitute that in:( d = 6 times 12 = 72 )So, ( d = 72 ). Now, going back to the earlier equation we had from the sum of coefficients: ( d + e = -3 ). Plugging in ( d = 72 ):( 72 + e = -3 )Subtract 72 from both sides:( e = -3 - 72 = -75 )So, ( e = -75 ). Let me just double-check my steps to make sure I didn't make any mistakes.1. Found ( f = 12 ) from the y-intercept.2. Sum of coefficients: ( 3 + d + e + 12 = 12 ) led to ( d + e = -3 ).3. Using Vieta's formulas, set up the equation for mean of zeros being twice the product: ( -d/9 = -2f/3 ), solved for ( d = 72 ).4. Plugged ( d = 72 ) into ( d + e = -3 ) to find ( e = -75 ).Everything seems to check out. I think I'm confident that ( e = -75 ).**Final Answer**The value of ( e ) is boxed{-75}.

answer：Alright, so I've got this problem about a rhombus, and I need to figure out the value of ( m ). Let me try to break it down step by step.First, the problem says that the side length of the rhombus ( ABCD ) is 5. I remember that in a rhombus, all sides are equal, and the diagonals bisect each other at right angles. So, if the diagonals intersect at point ( O ), then ( AO ) and ( BO ) are half the lengths of the diagonals.It also mentions that ( OA ) and ( OB ) are the roots of the quadratic equation ( x^2 + (2m - 1)x + m^2 + 3 = 0 ). Okay, so I need to use the properties of quadratic equations to relate the roots to the coefficients.Let me recall that for a quadratic equation ( x^2 + bx + c = 0 ), the sum of the roots is ( -b ) and the product is ( c ). So, in this case, the sum of ( OA ) and ( OB ) should be ( -(2m - 1) ), and their product should be ( m^2 + 3 ).But wait, since ( OA ) and ( OB ) are parts of the diagonals, they should satisfy the Pythagorean theorem because the diagonals intersect at right angles. So, ( OA^2 + OB^2 = (AB)^2 ). Since each side of the rhombus is 5, this becomes ( OA^2 + OB^2 = 5^2 = 25 ).Hmm, okay, so I have two things:1. ( OA + OB = -(2m - 1) )2. ( OA cdot OB = m^2 + 3 )3. ( OA^2 + OB^2 = 25 )I think I can express ( OA^2 + OB^2 ) in terms of ( OA + OB ) and ( OA cdot OB ). Let me recall that ( (OA + OB)^2 = OA^2 + 2(OA cdot OB) + OB^2 ). So, rearranging that, ( OA^2 + OB^2 = (OA + OB)^2 - 2(OA cdot OB) ).Substituting the known values:( 25 = [-(2m - 1)]^2 - 2(m^2 + 3) )Let me simplify this step by step.First, square ( -(2m - 1) ):( [-(2m - 1)]^2 = (2m - 1)^2 = 4m^2 - 4m + 1 )Then, subtract ( 2(m^2 + 3) ):( 4m^2 - 4m + 1 - 2m^2 - 6 = 2m^2 - 4m - 5 )So, the equation becomes:( 25 = 2m^2 - 4m - 5 )Let me move 25 to the right side:( 0 = 2m^2 - 4m - 5 - 25 )( 0 = 2m^2 - 4m - 30 )I can divide the entire equation by 2 to simplify:( 0 = m^2 - 2m - 15 )Now, I need to solve this quadratic equation for ( m ). Let's factor it:Looking for two numbers that multiply to -15 and add to -2. Hmm, 3 and -5 work because 3 * (-5) = -15 and 3 + (-5) = -2.So, the equation factors as:( (m + 3)(m - 5) = 0 )Setting each factor equal to zero gives:1. ( m + 3 = 0 ) → ( m = -3 )2. ( m - 5 = 0 ) → ( m = 5 )So, the possible values for ( m ) are -3 and 5. But wait, I should check if these values make sense in the context of the problem.The problem mentions that ( OA ) and ( OB ) are the lengths of the segments of the diagonals. Since lengths can't be negative, I need to ensure that the roots ( OA ) and ( OB ) are positive.Looking back at the quadratic equation ( x^2 + (2m - 1)x + m^2 + 3 = 0 ), the roots must be positive. So, let's check for ( m = -3 ) and ( m = 5 ).First, for ( m = -3 ):The equation becomes:( x^2 + (2*(-3) - 1)x + (-3)^2 + 3 = x^2 + (-6 - 1)x + 9 + 3 = x^2 - 7x + 12 = 0 )Factoring this:( (x - 3)(x - 4) = 0 ), so roots are 3 and 4, which are positive. Good.Now, for ( m = 5 ):The equation becomes:( x^2 + (2*5 - 1)x + 5^2 + 3 = x^2 + (10 - 1)x + 25 + 3 = x^2 + 9x + 28 = 0 )Let me check the discriminant to see if there are real roots:Discriminant ( D = 9^2 - 4*1*28 = 81 - 112 = -31 )Since the discriminant is negative, there are no real roots. That means ( m = 5 ) doesn't give us real lengths for ( OA ) and ( OB ), which isn't possible because lengths must be real numbers.Therefore, ( m = 5 ) is not a valid solution, and the only valid solution is ( m = -3 ).Wait, but the options given are:A: -3B: 5C: 5 or -3D: -5 or 3So, according to my reasoning, only -3 is valid, which is option A. But let me double-check my work to make sure I didn't make a mistake.When I solved the quadratic equation for ( m ), I got ( m = -3 ) and ( m = 5 ). Then, I checked both in the original equation to see if they yield positive roots. For ( m = -3 ), the roots are 3 and 4, which are positive. For ( m = 5 ), the equation has no real roots, so it's invalid.Therefore, the correct answer should be A: -3.But wait, let me think again. Maybe I made a mistake in calculating the discriminant for ( m = 5 ). Let me recalculate:For ( m = 5 ):Quadratic equation: ( x^2 + 9x + 28 = 0 )Discriminant: ( 9^2 - 4*1*28 = 81 - 112 = -31 ). Yep, still negative. So, no real roots. So, ( m = 5 ) is invalid.Therefore, only ( m = -3 ) is valid. So, the answer is A: -3.But wait, the options include C: 5 or -3, which suggests that both could be possible. Did I miss something?Let me think again. Maybe I should check if the quadratic equation with ( m = 5 ) could have positive roots even if the discriminant is negative? But no, if the discriminant is negative, the roots are complex, not real. So, they can't represent lengths, which must be real numbers.Therefore, only ( m = -3 ) is valid. So, the answer is A: -3.But I'm a bit confused because the options include C: 5 or -3. Maybe I made a mistake in calculating the discriminant? Let me check again.Quadratic equation for ( m = 5 ):( x^2 + 9x + 28 = 0 )Discriminant: ( 81 - 112 = -31 ). Yep, still negative.So, I think the correct answer is A: -3.Wait, but let me think about the problem again. The side length is 5, and the diagonals intersect at O. So, ( OA ) and ( OB ) are the halves of the diagonals, right? So, ( OA = frac{d_1}{2} ) and ( OB = frac{d_2}{2} ), where ( d_1 ) and ( d_2 ) are the lengths of the diagonals.Then, in a rhombus, the area can be calculated as ( frac{d_1 cdot d_2}{2} ), and also as ( side^2 cdot sin(theta) ), where ( theta ) is one of the internal angles. But I don't think that's necessary here.Wait, but I used the Pythagorean theorem correctly, right? Since the diagonals intersect at right angles, ( OA^2 + OB^2 = (side)^2 ). So, ( OA^2 + OB^2 = 25 ). That seems correct.And then I used the sum and product of roots to express ( OA^2 + OB^2 ) in terms of ( m ). That seems correct too.So, I think my reasoning is solid. Therefore, the answer should be A: -3.But just to be thorough, let me consider if there's another way to approach this problem.Alternatively, since ( OA ) and ( OB ) are the roots of the equation, and they must satisfy ( OA^2 + OB^2 = 25 ), I can use the relationships:( OA + OB = -(2m - 1) )( OA cdot OB = m^2 + 3 )And ( OA^2 + OB^2 = (OA + OB)^2 - 2(OA cdot OB) = 25 )So, substituting:( [-(2m - 1)]^2 - 2(m^2 + 3) = 25 )Which simplifies to:( (4m^2 - 4m + 1) - 2m^2 - 6 = 25 )So, ( 2m^2 - 4m - 5 = 25 )Wait, no, that's not correct. Wait, let me redo that.Wait, ( [-(2m - 1)]^2 = (2m - 1)^2 = 4m^2 - 4m + 1 )Then, subtract ( 2(m^2 + 3) ):( 4m^2 - 4m + 1 - 2m^2 - 6 = 2m^2 - 4m - 5 )Set equal to 25:( 2m^2 - 4m - 5 = 25 )So, ( 2m^2 - 4m - 30 = 0 )Divide by 2:( m^2 - 2m - 15 = 0 )Which factors to ( (m - 5)(m + 3) = 0 ), so ( m = 5 ) or ( m = -3 )Then, as before, checking ( m = 5 ) gives a quadratic with no real roots, so only ( m = -3 ) is valid.Therefore, the answer is A: -3.I think I've double-checked everything, and my conclusion remains the same. So, I'm confident that the answer is A: -3.

answer：Alright, so I'm trying to figure out how many moles of hydrogen are formed when 1 mole of ethane reacts with 2 moles of oxygen to form 2 moles of carbon dioxide. Okay, let's start by recalling what I know about chemical reactions and stoichiometry.First, ethane is C₂H₆, and oxygen is O₂. The products mentioned are carbon dioxide (CO₂) and, I assume, water (H₂O) since hydrogen is involved. The balanced chemical equation for the combustion of ethane is typically:2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂OFrom this equation, 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water. Each mole of water contains 2 moles of hydrogen atoms, so 6 moles of water would contain 12 moles of hydrogen atoms.But in the problem, we have only 1 mole of ethane reacting with 2 moles of oxygen. That seems like a smaller amount of both reactants compared to the balanced equation. I need to determine if there's enough oxygen to completely react with the ethane or if it's a limiting reagent situation.Let's see, in the balanced equation, 2 moles of ethane require 7 moles of oxygen. So, for 1 mole of ethane, the required oxygen would be 7/2 = 3.5 moles. But we only have 2 moles of oxygen available. That means oxygen is the limiting reagent here.Since oxygen is limiting, the reaction won't proceed as per the balanced equation. Instead, the amount of products formed will be based on the available oxygen. I need to adjust the stoichiometry accordingly.Let me set up the mole ratios. From the balanced equation:2 C₂H₆ : 7 O₂ : 4 CO₂ : 6 H₂OSo, the mole ratio of C₂H₆ to O₂ is 2:7, and the mole ratio of O₂ to CO₂ is 7:4, and O₂ to H₂O is 7:6.Given that we have 2 moles of O₂, let's find out how much CO₂ and H₂O can be formed.First, for CO₂:Moles of CO₂ = (4/7) * moles of O₂ = (4/7) * 2 ≈ 1.14 molesSimilarly, for H₂O:Moles of H₂O = (6/7) * moles of O₂ = (6/7) * 2 ≈ 1.71 molesNow, each mole of H₂O contains 2 moles of H atoms, so total moles of H atoms would be:Moles of H = 2 * moles of H₂O ≈ 2 * 1.71 ≈ 3.43 molesBut wait, the problem asks for moles of hydrogen formed, not hydrogen atoms. Since hydrogen typically forms H₂ molecules, I need to adjust for that.Moles of H₂ = moles of H atoms / 2 ≈ 3.43 / 2 ≈ 1.71 molesSo, approximately 1.71 moles of H₂ are formed when 1 mole of ethane reacts with 2 moles of oxygen to produce 2 moles of carbon dioxide.But let me double-check my calculations to ensure accuracy.Starting with 2 moles of O₂:Moles of CO₂ = (4/7)*2 ≈ 1.14 molesMoles of H₂O = (6/7)*2 ≈ 1.71 molesMoles of H atoms = 2*1.71 ≈ 3.43 molesMoles of H₂ = 3.43/2 ≈ 1.71 molesYes, that seems consistent.However, I'm a bit confused because the problem states that 2 moles of CO₂ are formed. In my calculation, I got approximately 1.14 moles of CO₂. There's a discrepancy here. Maybe I need to adjust the stoichiometry to match the given 2 moles of CO₂.Let's try that approach.If 2 moles of CO₂ are formed, then using the mole ratio from the balanced equation:Moles of O₂ required = (7/4)*moles of CO₂ = (7/4)*2 = 3.5 molesBut we only have 2 moles of O₂ available. This suggests that the amount of CO₂ formed cannot exceed what's possible with 2 moles of O₂, which is approximately 1.14 moles, as calculated earlier.Therefore, the problem might have an inconsistency. It specifies forming 2 moles of CO₂, but with only 2 moles of O₂, that's not possible based on the stoichiometry.Perhaps the problem assumes complete combustion despite the limited oxygen, or there's a different reaction pathway. Without further information, it's challenging to reconcile the given data with the stoichiometry.In conclusion, based on the provided amounts and standard combustion reactions, approximately 1.71 moles of H₂ are formed when 1 mole of ethane reacts with 2 moles of oxygen, producing about 1.14 moles of CO₂. However, the problem's statement of forming 2 moles of CO₂ introduces an inconsistency that needs clarification.