answer：Okay, so I have this problem here where there are two quadratic equations: (x^2 + ax + b = 0) and (x^2 + cx + d = 0). It's given that (a neq c), and both equations have the same root, which is denoted as (alpha). The question is asking me to find (alpha).Alright, let me start by recalling what it means for a number to be a root of a quadratic equation. If (alpha) is a root of both equations, then plugging (alpha) into both equations should satisfy them. That means:1. (alpha^2 + aalpha + b = 0)2. (alpha^2 + calpha + d = 0)So, I have these two equations:[begin{align*}alpha^2 + aalpha + b &= 0 quad text{(Equation 1)} alpha^2 + calpha + d &= 0 quad text{(Equation 2)}end{align*}]Since both equations equal zero, maybe I can set them equal to each other or subtract one from the other to eliminate (alpha^2). Let me try subtracting Equation 2 from Equation 1:[(alpha^2 + aalpha + b) - (alpha^2 + calpha + d) = 0 - 0]Simplifying this, the (alpha^2) terms cancel out:[aalpha + b - calpha - d = 0]Combine like terms:[(a - c)alpha + (b - d) = 0]So, I have:[(a - c)alpha = d - b]Wait, actually, let me check the signs again. When I subtract (calpha) and (-d), it should be:[(a - c)alpha + (b - d) = 0]So, moving the constants to the other side:[(a - c)alpha = d - b]Yes, that's correct. So, solving for (alpha), I get:[alpha = frac{d - b}{a - c}]But wait, the problem states that (a neq c), which is good because if (a = c), we would be dividing by zero, which is undefined. So, since (a neq c), this expression is valid.Let me just verify my steps to make sure I didn't make a mistake. I started by plugging (alpha) into both equations, subtracted them to eliminate (alpha^2), which left me with a linear equation in terms of (alpha). Then, I solved for (alpha) by isolating it on one side. The algebra seems straightforward, and I didn't skip any steps that could potentially introduce errors.Is there another way to approach this problem? Maybe by using the concept of common roots in quadratic equations. I remember that if two quadratic equations have a common root, then the determinant of their coefficients matrix is zero. But I'm not sure if that's necessary here since subtracting the equations worked out fine.Alternatively, I could use the idea that if two quadratics share a common root, then their resultant is zero. The resultant can be found using the determinant of the Sylvester matrix, but that might be overcomplicating things for this problem.Another thought: if (alpha) is a common root, then the other roots of each quadratic must be different because (a neq c). So, each quadratic has two roots, one common and one different. But I don't think that's directly helpful here since the question only asks for the common root.Wait, maybe I can use Vieta's formulas. For the first equation, the sum of the roots is (-a) and the product is (b). For the second equation, the sum is (-c) and the product is (d). Let me denote the other roots as (beta) and (gamma) for the first and second equations respectively.So, for the first equation:[alpha + beta = -a quad text{and} quad alphabeta = b]For the second equation:[alpha + gamma = -c quad text{and} quad alphagamma = d]Hmm, so I have these four equations:1. (alpha + beta = -a)2. (alphabeta = b)3. (alpha + gamma = -c)4. (alphagamma = d)I can try to express (beta) and (gamma) in terms of (alpha):From equation 1: (beta = -a - alpha)From equation 3: (gamma = -c - alpha)Now, substitute these into equations 2 and 4:From equation 2: (alpha(-a - alpha) = b)Which simplifies to:[-aalpha - alpha^2 = b]Similarly, from equation 4: (alpha(-c - alpha) = d)Which simplifies to:[-calpha - alpha^2 = d]Wait a minute, these are the same as the original equations I started with. So, this approach just brings me back to where I was before. It seems like subtracting the two equations was a more straightforward method.So, going back to my initial solution, I had:[alpha = frac{d - b}{a - c}]Is there a way to express this differently? Maybe factor out a negative sign:[alpha = frac{-(b - d)}{a - c} = frac{b - d}{c - a}]But both expressions are equivalent, just written differently. Depending on how the answer is expected, either form could be acceptable.Let me test this with some numbers to make sure it works. Suppose I have two quadratic equations that share a common root. Let's choose specific values for (a), (b), (c), and (d) and see if the formula gives the correct (alpha).Let me pick (a = 2), (b = 1), (c = 3), and (d = 3). So, the equations are:1. (x^2 + 2x + 1 = 0)2. (x^2 + 3x + 3 = 0)Wait, does the first equation have real roots? Let me check the discriminant: (2^2 - 4(1)(1) = 4 - 4 = 0). So, it has a repeated root at (x = -1). The second equation has discriminant (3^2 - 4(1)(3) = 9 - 12 = -3), which is negative, so it doesn't have real roots. Hmm, that's not helpful because they don't share a common real root.Let me choose different values. Let's take (a = 1), (b = -2), (c = 2), and (d = 0). So, the equations are:1. (x^2 + x - 2 = 0)2. (x^2 + 2x + 0 = 0)First equation: (x^2 + x - 2 = 0). Factoring: ((x + 2)(x - 1) = 0), so roots are (x = -2) and (x = 1).Second equation: (x^2 + 2x = 0). Factoring: (x(x + 2) = 0), so roots are (x = 0) and (x = -2).So, the common root is (-2). Let's plug into the formula:[alpha = frac{d - b}{a - c} = frac{0 - (-2)}{1 - 2} = frac{2}{-1} = -2]Perfect, it matches. So, the formula works in this case.Let me try another example. Let (a = 3), (b = 4), (c = 5), (d = 6). So, equations:1. (x^2 + 3x + 4 = 0)2. (x^2 + 5x + 6 = 0)First equation: Discriminant (9 - 16 = -7), so complex roots.Second equation: Discriminant (25 - 24 = 1), so roots are (-2) and (-3).Since the first equation has complex roots and the second has real roots, they don't share a common real root. But they might share a complex root. Let's see.First equation: (x = frac{-3 pm sqrt{-7}}{2}).Second equation: (x = -2) and (x = -3).So, no common roots. Therefore, this example isn't helpful because they don't share a common root. Maybe I should choose equations that actually share a common root.Let me pick (a = 4), (b = 3), (c = 5), (d = 6). So, equations:1. (x^2 + 4x + 3 = 0)2. (x^2 + 5x + 6 = 0)First equation factors as ((x + 1)(x + 3) = 0), roots at (x = -1) and (x = -3).Second equation factors as ((x + 2)(x + 3) = 0), roots at (x = -2) and (x = -3).So, common root is (-3). Let's apply the formula:[alpha = frac{d - b}{a - c} = frac{6 - 3}{4 - 5} = frac{3}{-1} = -3]Again, it works. So, the formula seems consistent.Wait, another thought: what if both quadratics have two common roots? That would mean they are identical equations, but since (a neq c), that's not possible. So, they can only have one common root, which is (alpha), and the other roots are different.So, in summary, by setting the two equations equal to each other and subtracting, I eliminated the quadratic term and was left with a linear equation in (alpha), which I solved to get (alpha = frac{d - b}{a - c}). Testing with specific examples confirmed that this formula works.I think I've covered all the bases here. I started with the given information, applied algebraic manipulation, verified the steps, tested with examples, and considered potential edge cases. I don't see any mistakes in my reasoning, so I'm confident that this is the correct solution.**Final Answer**The common root (alpha) is boxed{dfrac{d - b}{a - c}}.

answer：First, I need to determine how many baseball games Benny missed by finding the difference between the total number of games played and the number of games he attended.I'll subtract the number of games Benny attended (14) from the total number of games played (39).This calculation will give me the number of games Benny did not attend.

answer：First, I need to determine the cost per pound of apples based on the given rate of 4 for 5 pounds.Next, I'll calculate the cost for 15 pounds by multiplying the cost per pound by 15.Finally, I'll present the final answer clearly.

answer：First, I need to identify all two-digit numbers that end with the digit 3. These numbers are 13, 23, 33, 43, 53, 63, 73, 83, and 93.Next, I will check each of these numbers to determine if they are prime. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.Starting with 13:- 13 is a prime number because it has no divisors other than 1 and 13.Next, 23:- 23 is also a prime number as it has no divisors other than 1 and 23.Then, 33:- 33 is not a prime number because it can be divided by 1, 3, 11, and 33.Moving on to 43:- 43 is a prime number since it has no divisors other than 1 and 43.Next, 53:- 53 is a prime number as it has no divisors other than 1 and 53.Then, 63:- 63 is not a prime number because it can be divided by 1, 3, 7, 9, 21, and 63.Next, 73:- 73 is a prime number since it has no divisors other than 1 and 73.Then, 83:- 83 is a prime number as it has no divisors other than 1 and 83.Finally, 93:- 93 is not a prime number because it can be divided by 1, 3, 31, and 93.After evaluating all the numbers, the two-digit prime numbers that end with the digit 3 are 13, 23, 43, 53, 73, and 83. This gives us a total of 6 prime numbers.

answer：Okay, so I have this triangle ABC where AC is 4 units, BC is 4√3 units, and the angle at A, which is angle BAC, is 2π/3 radians. I need to find the measure of angle ABC for part (I) and then for part (II), there's a point D on BC such that AD is √7, and I need to find the length of DC.Starting with part (I). I remember that in a triangle, if I know two sides and the included angle, I can use the Law of Sines or the Law of Cosines to find other sides or angles. Since I know AC, BC, and angle BAC, maybe the Law of Sines will help here.The Law of Sines states that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, in triangle ABC, that would be:BC / sin(angle BAC) = AC / sin(angle ABC) = AB / sin(angle ACB)I know BC is 4√3, angle BAC is 2π/3, and AC is 4. So, plugging these into the Law of Sines:4√3 / sin(2π/3) = 4 / sin(angle ABC)I need to compute sin(2π/3). 2π/3 is 120 degrees, and sin(120°) is √3/2. So, substituting that in:4√3 / (√3/2) = 4 / sin(angle ABC)Simplify the left side: 4√3 divided by √3/2 is 4√3 * 2/√3. The √3 cancels out, so it's 4*2 = 8.So now we have:8 = 4 / sin(angle ABC)Solving for sin(angle ABC):sin(angle ABC) = 4 / 8 = 1/2So, sin(angle ABC) is 1/2. The angle whose sine is 1/2 is either π/6 (30 degrees) or 5π/6 (150 degrees). But in a triangle, the sum of angles is π, and we already have angle BAC as 2π/3 (120 degrees). So, angle ABC can't be 5π/6 because 120 + 150 is already more than 180. Therefore, angle ABC must be π/6.So, part (I) is solved, angle ABC is π/6.Moving on to part (II). Now, D is a point on BC such that AD is √7. I need to find DC.First, let me visualize triangle ABC. AC is 4, BC is 4√3, angle at A is 120 degrees, angle at B is 30 degrees, so angle at C must be π - 2π/3 - π/6 = π/6 as well. So, triangle ABC has angles 120°, 30°, 30°, making it an isosceles triangle with sides opposite equal angles equal. Wait, but AC is 4, BC is 4√3, so maybe it's not isosceles? Wait, hold on.Wait, in triangle ABC, sides opposite equal angles are equal. Since angles at B and C are both π/6, then sides opposite them should be equal. But side opposite angle B is AC, which is 4, and side opposite angle C is AB. So, AB should be equal to AC, which is 4. But BC is 4√3. Hmm, that seems contradictory.Wait, maybe I made a mistake earlier. Let me double-check. If angle BAC is 120°, and angles at B and C are both 30°, then sides opposite 30° angles should be equal. So, side AC is opposite angle B, which is 30°, and side AB is opposite angle C, which is also 30°, so AC and AB should be equal. So, AC is 4, so AB is also 4. Then, side BC is opposite angle A, which is 120°, and it's given as 4√3. So, that makes sense because in a triangle with two sides equal to 4 and included angle 120°, the third side can be calculated using the Law of Cosines.Wait, let me verify that. If AB = AC = 4, angle BAC = 120°, then BC² = AB² + AC² - 2*AB*AC*cos(angle BAC). So, BC² = 16 + 16 - 2*4*4*cos(120°). Cos(120°) is -1/2, so:BC² = 32 - 2*16*(-1/2) = 32 + 16 = 48. So, BC = √48 = 4√3. Perfect, that matches the given BC. So, triangle ABC is isosceles with AB = AC = 4, BC = 4√3, angles at B and C are both 30°, angle at A is 120°.Now, point D is on BC such that AD = √7. I need to find DC.So, let's denote DC as x. Then, BD would be BC - DC = 4√3 - x.Now, in triangle ADC, we have sides AC = 4, AD = √7, and DC = x. We also know angle at C is 30°, so maybe we can use the Law of Cosines here.Law of Cosines in triangle ADC:AD² = AC² + DC² - 2*AC*DC*cos(angle C)Plugging in the values:(√7)² = 4² + x² - 2*4*x*cos(30°)Simplify:7 = 16 + x² - 8x*(√3/2)Simplify further:7 = 16 + x² - 4√3 xBring all terms to one side:x² - 4√3 x + 16 - 7 = 0So:x² - 4√3 x + 9 = 0This is a quadratic in x. Let's solve it using the quadratic formula.x = [4√3 ± √( (4√3)² - 4*1*9 )]/2Compute discriminant:(4√3)² = 16*3 = 484*1*9 = 36So discriminant is 48 - 36 = 12Thus,x = [4√3 ± √12]/2Simplify √12 = 2√3So,x = [4√3 ± 2√3]/2Factor out 2√3:x = [2√3(2 ± 1)]/2 = √3(2 ± 1)Thus, two solutions:x = √3*(2 + 1) = 3√3orx = √3*(2 - 1) = √3So, DC can be either 3√3 or √3.But let's verify if both solutions make sense in the context.Since BC is 4√3, and DC is either √3 or 3√3, both are less than 4√3, so both are valid. So, both solutions are acceptable.Therefore, DC can be √3 or 3√3.**Final Answer**(I) The measure of angle ABC is boxed{dfrac{pi}{6}}.(II) The length of DC is boxed{3sqrt{3}} or boxed{sqrt{3}}.

answer：First, I need to determine the length of the mid-size model. Since it is 1/10th the size of the full-size Mustang, I can calculate it by dividing the full size length by 10.Next, I know the smallest model is 12 inches long. To find the ratio of the smallest model's length to the mid-size model's length, I will divide the length of the smallest model by the length of the mid-size model.Finally, by performing the division, I can determine the ratio and present it in the simplest form.