💡First, I need to determine how much money Todd spent on the candy bars. He started with 20 and has 12 left, so the amount spent is 20 minus 12, which equals 8.Next, to find the cost of each candy bar, I divide the total amount spent by the number of candy bars. Todd bought 4 candy bars, so I divide 8 by 4, which equals 2 per candy bar.

💡Okay, so I have this problem here where I need to find the number of distinct ordered pairs (x, y) where x and y are positive integers. They have to satisfy two equations: (xy)^4 - 2x^2y^2 + 1 = 0 and x + y = 4. The options are from A to E, with E being more than 3. Hmm, let me try to figure this out step by step.First, let me look at the first equation: (xy)^4 - 2x^2y^2 + 1 = 0. That looks a bit complicated, but maybe I can simplify it. Let me see... If I let a = (xy)^2, then (xy)^4 would be a^2. So substituting that in, the equation becomes a^2 - 2a + 1 = 0. Oh, that's a quadratic equation in terms of a. Let me solve that.So, a^2 - 2a + 1 = 0. This factors into (a - 1)^2 = 0, which means a = 1. Since a = (xy)^2, that means (xy)^2 = 1. Taking the square root of both sides, we get xy = 1 or xy = -1. But since x and y are positive integers, xy can't be negative. So, xy = 1.Alright, so now I know that the product of x and y is 1. Since x and y are positive integers, the only pairs that satisfy this are (1,1). Wait, but hold on, the second equation is x + y = 4. If x and y are both 1, then their sum is 2, not 4. That doesn't work. Hmm, so maybe I made a mistake somewhere.Let me double-check. The first equation simplifies to (xy)^4 - 2x^2y^2 + 1 = 0. I set a = (xy)^2, so the equation becomes a^2 - 2a + 1 = 0, which factors to (a - 1)^2 = 0, so a = 1. Therefore, (xy)^2 = 1, so xy = 1. That seems correct.But then, if xy = 1, and x + y = 4, the only positive integers that multiply to 1 are 1 and 1, but 1 + 1 = 2 ≠ 4. So, does that mean there are no solutions? But wait, maybe I missed something.Let me think again. Maybe I should consider the original equation without substitution. Let's write it out:(xy)^4 - 2x^2y^2 + 1 = 0.Hmm, that can be rewritten as (x^2y^2)^2 - 2x^2y^2 + 1 = 0. Let me let z = x^2y^2. Then the equation becomes z^2 - 2z + 1 = 0, which is the same as (z - 1)^2 = 0, so z = 1. Therefore, x^2y^2 = 1, which again gives us xy = 1 or xy = -1. Since x and y are positive, xy = 1.So, it seems that regardless of substitution, we end up with xy = 1. But then, as before, x + y = 4 and xy = 1 don't have positive integer solutions because the only pair is (1,1), which sums to 2, not 4. So, does that mean there are no solutions?Wait, hold on. Maybe I should consider that x and y don't have to be the same. Let me think about the equation x + y = 4. The possible positive integer pairs (x, y) that add up to 4 are (1,3), (2,2), and (3,1). Let me check each of these pairs to see if they satisfy the first equation.First, (1,3): Let's compute (xy)^4 - 2x^2y^2 + 1. So, x = 1, y = 3. Then, xy = 3. So, (3)^4 - 2*(1)^2*(3)^2 + 1 = 81 - 2*1*9 + 1 = 81 - 18 + 1 = 64. 64 ≠ 0, so this pair doesn't work.Next, (2,2): x = 2, y = 2. Then, xy = 4. So, (4)^4 - 2*(2)^2*(2)^2 + 1 = 256 - 2*4*4 + 1 = 256 - 32 + 1 = 225. 225 ≠ 0, so this pair doesn't work either.Lastly, (3,1): This is the same as (1,3), just reversed. So, we already saw that it gives 64, which isn't 0.Hmm, so none of the pairs that satisfy x + y = 4 also satisfy the first equation. Does that mean there are no solutions? But wait, the options include 0 as an option (A). So, is the answer 0?But wait, let me double-check my calculations because sometimes I might make a mistake. Let's recalculate for (1,3):(xy)^4 = (1*3)^4 = 3^4 = 81.2x^2y^2 = 2*(1^2)*(3^2) = 2*1*9 = 18.So, 81 - 18 + 1 = 64. Yep, that's correct.For (2,2):(xy)^4 = (2*2)^4 = 4^4 = 256.2x^2y^2 = 2*(2^2)*(2^2) = 2*4*4 = 32.So, 256 - 32 + 1 = 225. Correct again.And (3,1) is the same as (1,3), so same result.So, none of these pairs satisfy the first equation. Therefore, there are no solutions. So, the number of distinct ordered pairs is 0. So, the answer should be A) 0.Wait, but the initial substitution led me to xy = 1, which only gives (1,1), which doesn't satisfy x + y = 4. So, that also suggests there are no solutions. So, yeah, I think the answer is 0.But just to be thorough, let me consider if there are any other possible pairs. Since x and y are positive integers, and x + y = 4, the only possible pairs are (1,3), (2,2), and (3,1). We've checked all of them, and none satisfy the first equation. So, there are indeed no solutions.Therefore, the number of distinct ordered pairs is 0.

💡First, I need to determine the total number of matches Jimmy has. Since each matchbook contains 24 matches and Jimmy has 5 matchbooks, I multiply 5 by 24 to get the total number of matches.Next, I need to find out how many stamps Jimmy's matches are worth. Given that one stamp is worth 12 matches, I divide the total number of matches by 12 to determine the number of stamps Jimmy can receive.Tonya arrives with 13 stamps and trades 10 of them to Jimmy in exchange for his matchbooks. To find out how many stamps Tonya has left, I subtract the 10 stamps she traded from her initial 13 stamps.

💡First, I need to determine the unfilled volume for each hole by calculating the percentage that remains after the specified amount has been filled.For each hole, I'll subtract the filled percentage from 100% to find the unfilled percentage. Then, I'll multiply the total volume of the hole by this unfilled percentage to get the unfilled volume.After calculating the unfilled volume for all eight holes, I'll sum these values to find the total unfilled volume across all holes.

💡Alright, so I've got this geometry problem here, and I need to figure out the distance between point B and point D. Let me try to visualize what's going on here. There's an angle ∠BDA that's 60 degrees. A person starts at vertex D, walks 10 meters along side DA, then turns and walks another 14 meters to reach point B on side BD. I need to find the distance from B to D.Hmm, okay. So, let me try to sketch this out in my mind. There's a vertex D, and from there, two sides: DA and DB, forming a 60-degree angle. The person walks 10 meters from D to A, then turns and walks 14 meters to reach B on DB. So, points D, A, and B form a triangle, right? Triangle ABD.Wait, but actually, since the person starts at D, walks to A, then turns and walks to B, which is on BD. So, the path is D to A to B, with A somewhere along DA and B somewhere along DB. So, triangle ABD is formed with sides DA = 10 meters, AB = 14 meters, and angle at D is 60 degrees.I think I can use the Law of Cosines here. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:c² = a² + b² - 2ab cos(C)Where C is the angle opposite side c.In this case, I need to find BD, which is side opposite angle A, but wait, angle at D is 60 degrees. Let me make sure I get the sides and angles right.Let me label the triangle properly. Let's say point D is the vertex with the 60-degree angle. So, sides DA and DB are the two sides forming the 60-degree angle. The person walks from D to A, which is 10 meters, then turns and walks 14 meters to B. So, AB is 14 meters.So, in triangle ABD, we have:- DA = 10 meters- AB = 14 meters- Angle at D (∠BDA) = 60 degreesWe need to find BD.So, using the Law of Cosines, we can write:AB² = DA² + BD² - 2 * DA * BD * cos(∠BDA)Plugging in the known values:14² = 10² + BD² - 2 * 10 * BD * cos(60°)Let me compute each part step by step.First, 14 squared is 196.10 squared is 100.cos(60°) is 0.5.So, substituting these values in:196 = 100 + BD² - 2 * 10 * BD * 0.5Simplify the equation:196 = 100 + BD² - 10 * BDBecause 2 * 10 * 0.5 is 10.Now, let's rearrange the equation to form a quadratic equation:BD² - 10 * BD + 100 - 196 = 0Simplify 100 - 196:BD² - 10 * BD - 96 = 0So, the quadratic equation is:BD² - 10 * BD - 96 = 0Now, I need to solve for BD. Let's denote BD as x for simplicity.So, the equation becomes:x² - 10x - 96 = 0To solve this quadratic equation, I can use the quadratic formula:x = [10 ± sqrt( (-10)² - 4 * 1 * (-96) )] / (2 * 1)Compute the discriminant:D = (-10)² - 4 * 1 * (-96) = 100 + 384 = 484So, sqrt(484) is 22.Therefore, the solutions are:x = [10 + 22] / 2 = 32 / 2 = 16andx = [10 - 22] / 2 = (-12) / 2 = -6Since distance can't be negative, we discard x = -6.Therefore, BD = 16 meters.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting from the Law of Cosines:AB² = DA² + BD² - 2 * DA * BD * cos(60°)14² = 10² + BD² - 2 * 10 * BD * 0.5196 = 100 + BD² - 10 * BDRearranged to:BD² - 10 * BD - 96 = 0Quadratic formula:x = [10 ± sqrt(100 + 384)] / 2 = [10 ± 22] / 2So, x = 16 or x = -6. Discarding the negative, x = 16.Yes, that seems correct. So, the distance between point B and D is 16 meters.

💡Alright, so I have this problem here about finding the number of distinct common points between two curves. The curves are given by the equations (9x^2 + y^2 = 9) and (x^2 + 16y^2 = 16). The options are A) 0, B) 1, C) 2, D) 4, and E) 8. Hmm, okay, let me try to figure this out step by step.First, I recognize that both equations are ellipses. The first one, (9x^2 + y^2 = 9), can be rewritten as (frac{x^2}{1} + frac{y^2}{9} = 1), which is an ellipse centered at the origin with semi-major axis length 3 along the y-axis and semi-minor axis length 1 along the x-axis. The second equation, (x^2 + 16y^2 = 16), can be rewritten as (frac{x^2}{16} + frac{y^2}{1} = 1), which is another ellipse centered at the origin with semi-major axis length 4 along the x-axis and semi-minor axis length 1 along the y-axis.So, both ellipses are centered at the origin, but one is elongated along the y-axis and the other along the x-axis. I need to find how many points they have in common. That means I need to solve these two equations simultaneously.Let me write down the two equations again:1. (9x^2 + y^2 = 9)2. (x^2 + 16y^2 = 16)I think the best way to solve this system is by elimination. Maybe I can solve one equation for one variable and substitute into the other. Let me try solving the first equation for (y^2):From equation 1: (y^2 = 9 - 9x^2)Now, substitute this expression for (y^2) into equation 2:(x^2 + 16(9 - 9x^2) = 16)Let me simplify this:(x^2 + 144 - 144x^2 = 16)Combine like terms:(-143x^2 + 144 = 16)Subtract 144 from both sides:(-143x^2 = 16 - 144)(-143x^2 = -128)Divide both sides by -143:(x^2 = frac{128}{143})Okay, so (x^2 = frac{128}{143}). Let me compute that fraction:128 divided by 143 is approximately 0.895. So, (x^2) is approximately 0.895, which is positive, so real solutions exist for x. Therefore, (x = pm sqrt{frac{128}{143}}).Now, let's find (y^2) using the expression we had earlier:(y^2 = 9 - 9x^2 = 9 - 9 times frac{128}{143})Compute that:First, (9 times frac{128}{143} = frac{1152}{143})So, (y^2 = 9 - frac{1152}{143})Convert 9 to a fraction with denominator 143: (9 = frac{1287}{143})Therefore, (y^2 = frac{1287}{143} - frac{1152}{143} = frac{135}{143})So, (y^2 = frac{135}{143}), which is approximately 0.944. Again, positive, so real solutions exist for y. Therefore, (y = pm sqrt{frac{135}{143}}).So, from this, it seems like there are four points of intersection: ((sqrt{frac{128}{143}}, sqrt{frac{135}{143}})), ((sqrt{frac{128}{143}}, -sqrt{frac{135}{143}})), ((- sqrt{frac{128}{143}}, sqrt{frac{135}{143}})), and ((- sqrt{frac{128}{143}}, -sqrt{frac{135}{143}})). So, four points in total.But wait, let me double-check my calculations because sometimes when solving systems, especially with quadratics, you might introduce extraneous solutions or make a mistake in algebra.Starting again:Equation 1: (9x^2 + y^2 = 9)Equation 2: (x^2 + 16y^2 = 16)Solving equation 1 for (y^2): (y^2 = 9 - 9x^2)Substitute into equation 2:(x^2 + 16(9 - 9x^2) = 16)Simplify:(x^2 + 144 - 144x^2 = 16)Combine like terms:(-143x^2 + 144 = 16)Subtract 144:(-143x^2 = -128)Divide:(x^2 = frac{128}{143})Yes, that's correct. Then (y^2 = 9 - 9 times frac{128}{143} = 9 - frac{1152}{143} = frac{1287 - 1152}{143} = frac{135}{143}). That's correct too.So, both (x^2) and (y^2) are positive, so we have real solutions. Therefore, four points of intersection.But wait, let me think about the geometry of the ellipses. The first ellipse is taller, stretching along the y-axis, while the second is wider, stretching along the x-axis. It's possible that they intersect at four points, two in the first quadrant and two in the third, or maybe two in the second and fourth. But in any case, four points.But hold on, sometimes ellipses can intersect at more points, but in this case, since both are centered at the origin and symmetric, the maximum number of intersection points is four. So, if they intersect, it's four points.But wait, another thought: could they intersect at more points? I don't think so, because two ellipses can intersect at up to four points. So, if they intersect, it's four points.But just to be thorough, let me check if these solutions actually satisfy both equations.Take (x = sqrt{frac{128}{143}}) and (y = sqrt{frac{135}{143}}).Plug into equation 1:(9x^2 + y^2 = 9 times frac{128}{143} + frac{135}{143} = frac{1152}{143} + frac{135}{143} = frac{1287}{143} = 9). Correct.Plug into equation 2:(x^2 + 16y^2 = frac{128}{143} + 16 times frac{135}{143} = frac{128}{143} + frac{2160}{143} = frac{2288}{143} = 16). Correct.So, yes, these points satisfy both equations. Therefore, there are indeed four points of intersection.Wait, but the options are A) 0, B) 1, C) 2, D) 4, E) 8. So, D) 4 is the answer.But hold on, another thought: sometimes when solving quadratic systems, especially with ellipses, you might get multiple solutions, but sometimes they might coincide or be complex. But in this case, both (x^2) and (y^2) are positive, so we have real solutions, and since both ellipses are symmetric about both axes, the solutions come in pairs, so four points in total.Therefore, the number of distinct common points is 4.Wait, but in the initial problem, the user wrote "Determine the number of distinct common points between the curves...", and the options are A) 0, B) 1, C) 2, D) 4, E) 8. So, D) 4 is the correct answer.But just to make sure, let me visualize the ellipses.First ellipse: (9x^2 + y^2 = 9) or (frac{x^2}{1} + frac{y^2}{9} = 1). So, it's an ellipse with semi-major axis 3 along y-axis and semi-minor axis 1 along x-axis.Second ellipse: (x^2 + 16y^2 = 16) or (frac{x^2}{16} + frac{y^2}{1} = 1). So, semi-major axis 4 along x-axis and semi-minor axis 1 along y-axis.Plotting these, the first ellipse is tall and narrow, the second is wide and short. They should intersect at four points: two in the upper half-plane and two in the lower half-plane, symmetric about both axes.Therefore, the answer is D) 4.But wait, in the initial problem, the user wrote "Determine the number of distinct common points...", and the options are A) 0, B) 1, C) 2, D) 4, E) 8. So, D) 4 is the correct answer.But just to make sure, let me check if there's any possibility that these ellipses don't intersect. For example, if one ellipse is entirely inside the other, they might not intersect. But in this case, the first ellipse has a semi-minor axis of 1 along x, and the second has a semi-major axis of 4 along x. So, the first ellipse is entirely within the second ellipse along the x-axis? Wait, no, because the first ellipse has a semi-major axis of 3 along y, which is larger than the semi-minor axis of the second ellipse, which is 1 along y. So, they overlap in both x and y directions.Therefore, they must intersect at four points.Wait, but let me think again. The first ellipse has a semi-minor axis of 1 along x, so it extends from x = -1 to x = 1. The second ellipse has a semi-major axis of 4 along x, so it extends from x = -4 to x = 4. Similarly, the first ellipse extends from y = -3 to y = 3, while the second extends from y = -1 to y = 1. So, in the x-direction, the first ellipse is entirely within the second ellipse, but in the y-direction, the second ellipse is entirely within the first ellipse. Therefore, they must intersect at four points.Yes, that makes sense. So, the answer is D) 4.