💡Okay, so I have this problem where there's a 4x4x4 cube made up of 64 dice. Each die is a normal one where opposite sides add up to 7. I need to find the smallest possible sum of all the visible faces on the six faces of this large cube. The options are 128, 136, 144, 152, and 160. Hmm, let's break this down step by step.First, I should visualize the cube. It's a 4x4x4 cube, so each edge has 4 dice. When you look at one face of the cube, you see a 4x4 grid of dice faces. Since it's a cube, there are six faces, each with 16 dice faces visible. So, in total, there are 6 * 16 = 96 dice faces visible on the entire cube.But wait, each die is part of the cube, and depending on its position, it might have more than one face visible. For example, corner dice have three faces visible, edge dice (not on the corner) have two faces visible, and the dice in the middle of each face have only one face visible. So, I need to figure out how many dice have one, two, or three faces visible.Let me think. In a cube, there are 8 corners, each with three faces visible. Then, each edge has dice that are not corners. Since it's a 4x4x4 cube, each edge has 4 dice, but two of them are corners, so the remaining two on each edge are edge dice with two faces visible. There are 12 edges on a cube, so 12 edges * 2 edge dice = 24 edge dice.Then, the faces have dice that are not on the edges. Each face is a 4x4 grid, so there are 16 dice per face. Subtracting the edge dice, which are the outermost layer, each face has (4-2)*(4-2) = 4 dice that are only on the face, with one face visible. Since there are 6 faces, that's 6 * 4 = 24 face-center dice.So, in total, we have 8 corner dice, 24 edge dice, and 24 face-center dice. Let me check: 8 + 24 + 24 = 56. Wait, but the cube has 64 dice. So, where are the remaining 8 dice? Oh, those are the internal dice, completely inside the cube, not visible from any face. So, 64 - 56 = 8 internal dice. But since we're only concerned with the visible faces, we can ignore these internal dice.Now, each die is a standard die where opposite faces add up to 7. So, 1 opposite 6, 2 opposite 5, and 3 opposite 4. To minimize the sum of the visible faces, we need to minimize the numbers on the visible faces. However, since opposite faces sum to 7, if we minimize one face, the opposite face will be maximized. But since the internal dice are not visible, their opposite faces can be maximized without affecting the visible sum.So, for each die, depending on how many faces are visible, we need to assign the smallest possible numbers to the visible faces, and the largest possible numbers to the hidden faces. But we have to do this in a way that the opposite faces still sum to 7.Let's consider the corner dice first. Each corner die has three faces visible. To minimize the sum, we need the three smallest numbers on those faces. The smallest numbers on a die are 1, 2, and 3. However, we have to make sure that these three numbers are on adjacent faces, not opposite each other. On a standard die, 1, 2, and 3 are adjacent, so that's possible.So, for each corner die, the visible faces can be 1, 2, and 3, summing to 6. Since there are 8 corner dice, the total contribution from the corners is 8 * 6 = 48.Next, the edge dice. Each edge die has two faces visible. To minimize the sum, we should have the two smallest numbers on those faces. The smallest numbers are 1 and 2. Again, we have to ensure that these two numbers are on adjacent faces. On a die, 1 and 2 are adjacent, so that's fine.So, each edge die contributes 1 + 2 = 3 to the total sum. There are 24 edge dice, so the total contribution from the edges is 24 * 3 = 72.Now, the face-center dice. Each of these has only one face visible. To minimize the sum, we just need the smallest number on that face, which is 1. Since there are 24 face-center dice, the total contribution from the face centers is 24 * 1 = 24.Adding all these up: 48 (corners) + 72 (edges) + 24 (face centers) = 144.Wait, but let me double-check. Is there a way to have an even smaller sum? For example, can we arrange the dice so that some of the visible faces have even smaller numbers? But since 1 is the smallest possible number on a die, we can't go lower than that. So, 1 is the minimum for any single face.For the edges, having 1 and 2 is the minimum sum for two adjacent faces. Similarly, for the corners, 1, 2, and 3 are the smallest three adjacent numbers. So, I think this is the minimal configuration.Therefore, the smallest possible sum of all the visible faces on the six faces of the large cube is 144. Looking back at the options, that's option C.**Final Answer**boxed{C}

💡First, I need to find the other number when one of the numbers is 24, the reciprocal of the HCF is 1/12, and the reciprocal of the LCM is 1/312.I know that the reciprocal of the HCF is 1/12, so the HCF itself must be 12. Similarly, the reciprocal of the LCM is 1/312, meaning the LCM is 312.There's a relationship between the HCF, LCM, and the two numbers: the product of the HCF and LCM of two numbers equals the product of the numbers themselves. So, I can set up the equation:HCF × LCM = A × BPlugging in the known values:12 × 312 = 24 × BNow, I can solve for B:B = (12 × 312) / 24Simplifying the equation:B = (12/24) × 312B = (1/2) × 312B = 156Therefore, the other number is 156.

💡First, I need to determine the monthly payment for the trailer. Given that the trailer costs 120,000 and is paid over 20 years, which is 240 months, I can calculate the monthly payment by dividing the total cost by the number of months.Next, I know that the monthly payment for the house is 1,500 more than that of the trailer. By adding 1,500 to the trailer's monthly payment, I can find the monthly payment for the house.Finally, to find the total cost of the house, I multiply the monthly payment for the house by the total number of months (240). This will give me the total amount Jane would pay for the house over the 20-year period.

💡Okay, so I have this problem here: If x minus y equals 6 and x plus y equals 12, what is the value of x? Hmm, let's see. I've seen similar problems before, so maybe I can figure this out step by step.First, I know that there are two equations given:1. x - y = 62. x + y = 12I need to find the value of x. Since there are two equations and two variables, I can probably solve this using either substitution or elimination. I think elimination might be easier here because if I add the two equations together, the y terms might cancel out.Let me try adding the two equations:(x - y) + (x + y) = 6 + 12Simplifying the left side:x - y + x + y = 2xAnd the right side:6 + 12 = 18So now I have:2x = 18To solve for x, I just need to divide both sides by 2:x = 18 / 2x = 9Okay, that seems straightforward. But wait, let me double-check to make sure I didn't make any mistakes. Maybe I can plug the value of x back into one of the original equations to find y and then verify both equations.Let's use the first equation: x - y = 6If x is 9, then:9 - y = 6Subtracting 9 from both sides:-y = 6 - 9-y = -3Multiplying both sides by -1:y = 3Now, let's check the second equation: x + y = 12If x is 9 and y is 3:9 + 3 = 12Which is true. So both equations are satisfied with x = 9 and y = 3.I think that's correct. But just to be thorough, maybe I can try solving it using substitution instead to see if I get the same answer.Starting with the first equation: x - y = 6I can solve for x in terms of y:x = y + 6Now, substitute this expression for x into the second equation: x + y = 12(y + 6) + y = 12Simplify:2y + 6 = 12Subtract 6 from both sides:2y = 6Divide by 2:y = 3Now, substitute y = 3 back into x = y + 6:x = 3 + 6x = 9Again, I get x = 9. So both methods give me the same result, which makes me more confident that x is indeed 9.I guess another way to think about it is graphically. If I plot these two equations on a graph, they should intersect at the point (x, y) = (9, 3). The first equation, x - y = 6, is a straight line with a slope of 1 and y-intercept at -6. The second equation, x + y = 12, is also a straight line with a slope of -1 and y-intercept at 12. These two lines should intersect at (9, 3), which confirms the algebraic solution.Alternatively, I could use matrices or determinants to solve this system, but that might be overcomplicating things for such a simple system. The elimination and substitution methods seem sufficient here.Just to recap:1. I added the two equations to eliminate y and solve for x.2. I found x = 9.3. I substituted x back into one of the original equations to find y = 3.4. I verified both solutions in both original equations to ensure they work.5. I also considered solving the system using substitution and got the same result.6. I thought about the graphical interpretation to reinforce the solution.All these steps lead me to conclude that x = 9 is the correct answer. I don't see any mistakes in my reasoning, and all methods point to the same solution. So I'm pretty confident that x is indeed 9.

💡Okay, so I have this problem here about angle α being in the fourth quadrant. The terminal side of angle α passes through the point P(4, y), and it's given that sin α equals y over 5. I need to find the value of tan α, and the options are A: -4/3, B: -3/4, C: 3/4, D: 4/3.Alright, let me start by recalling some basic trigonometry. In the unit circle, any angle has a terminal side that passes through a point (x, y). The sine of the angle is y divided by the radius (which is the distance from the origin to the point), and the tangent is y divided by x.Given that α is in the fourth quadrant, I remember that in the fourth quadrant, sine is negative, cosine is positive, and tangent is negative because tangent is sine over cosine. So, since α is in the fourth quadrant, sin α is negative, which means y must be negative. That's useful information.The point P is given as (4, y), so x is 4, and y is unknown. It's also given that sin α equals y over 5. Wait, sin α is y over the radius, right? So the radius here must be 5 because sin α is y over 5. That means the distance from the origin to the point P(4, y) is 5.So, using the distance formula, the radius r is sqrt(x² + y²). Plugging in x = 4 and r = 5, we have sqrt(4² + y²) = 5. Let me write that down:sqrt(16 + y²) = 5If I square both sides to eliminate the square root, I get:16 + y² = 25Subtracting 16 from both sides gives:y² = 9Taking the square root of both sides, y is either 3 or -3. But since α is in the fourth quadrant, y must be negative. So y = -3.Now, I need to find tan α. Tangent is y over x, so that's (-3)/4. So tan α is -3/4.Looking back at the options, that's option B: -3/4.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I know that sin α = y / r, and it's given as y / 5, so r = 5.2. Using the distance formula, sqrt(4² + y²) = 5, which simplifies to sqrt(16 + y²) = 5.3. Squaring both sides: 16 + y² = 25, so y² = 9.4. Therefore, y = ±3, but since it's in the fourth quadrant, y is negative, so y = -3.5. Then, tan α = y / x = -3 / 4.Everything seems to check out. So, I think I did it correctly. The answer should be B: -3/4.**Final Answer**boxed{B}

💡Okay, so I've got this statistics problem here, and I need to figure out what happens to the mean and variance after correcting two errors in the data. Let me try to break it down step by step.First, the problem says that someone collected a sample of 50 data points. The mean of this sample was 70, and the variance was 75. Then, they found out that two data points were recorded incorrectly: one was recorded as 60 instead of 80, and another was recorded as 90 instead of 70. Now, after correcting these errors, we need to find out how the mean and variance change.Alright, let's start with the mean. The mean is the average of all the data points. So, if we have 50 data points, the total sum of all these points would be 50 times the mean, which is 50 * 70 = 3500.Now, the two incorrect data points were 60 and 90. Their sum is 60 + 90 = 150. But the correct values should have been 80 and 70, whose sum is 80 + 70 = 150. Wait a minute, the total sum of these two points is the same before and after correction. So, the overall sum of all 50 data points doesn't change. That means the mean should stay the same, right? So, the new mean, which is denoted as X̄, should still be 70.Okay, so the mean doesn't change. Now, what about the variance? Variance measures how spread out the data is. The formula for variance is the average of the squared differences from the mean. So, if we correct the two data points, we need to see how this affects the squared differences.Let's think about the incorrect data points first. The incorrect points were 60 and 90. The mean is 70, so the squared differences for these points are (60 - 70)^2 = (-10)^2 = 100 and (90 - 70)^2 = 20^2 = 400. So, the total squared difference for these two points is 100 + 400 = 500.Now, the correct data points are 80 and 70. Let's calculate their squared differences. For 80, it's (80 - 70)^2 = 10^2 = 100. For 70, it's (70 - 70)^2 = 0^2 = 0. So, the total squared difference for these two points is 100 + 0 = 100.Comparing the two, the incorrect data points contributed 500 to the total squared differences, while the correct ones contribute only 100. That means after correction, the total squared differences decrease by 500 - 100 = 400.Since variance is the average of these squared differences, and we're dealing with a sample of 50 points, the total squared differences are divided by 49 (since it's sample variance, we use n - 1). But regardless, the total sum of squared differences decreases, which means the variance will decrease as well.So, putting it all together, the mean remains the same at 70, and the variance decreases from 75 to something less than 75. Therefore, the correct answer should be that the mean is still 70, and the variance is less than 75.