answer：Alright, so I need to figure out how many miles Mary ran for her half-marathon. The problem says she ran a total of (2847_9) miles, and I need to convert that from base nine to base ten. Hmm, okay, let me think about how base conversion works.First, I remember that when you have a number in a different base, each digit represents a power of that base. So, for example, in base nine, the rightmost digit is (9^0), which is 1, then moving left, the next digit is (9^1), then (9^2), and so on. So, each digit is multiplied by 9 raised to the power of its position, starting from 0 on the right.Let me write down the number: (2847_9). I think it's helpful to break this down digit by digit. So, starting from the right, the digits are 7, 4, 8, and 2. Each of these corresponds to a power of nine. Let me assign each digit its place value:- The rightmost digit is 7, which is in the (9^0) place.- Next, moving left, we have 4, which is in the (9^1) place.- Then, 8 is in the (9^2) place.- Finally, the leftmost digit is 2, which is in the (9^3) place.So, to convert (2847_9) to base ten, I need to calculate each digit multiplied by its corresponding power of nine and then add them all together. Let me write that out step by step.First, let's calculate each power of nine:- (9^0 = 1)- (9^1 = 9)- (9^2 = 81)- (9^3 = 729)Now, multiply each digit by its corresponding power:- The 7 in the (9^0) place: (7 times 1 = 7)- The 4 in the (9^1) place: (4 times 9 = 36)- The 8 in the (9^2) place: (8 times 81 = 648)- The 2 in the (9^3) place: (2 times 729 = 1458)Okay, so now I have these four values: 7, 36, 648, and 1458. I need to add them all together to get the total in base ten.Let me add them step by step:First, add 7 and 36:(7 + 36 = 43)Next, add 648 to that result:(43 + 648 = 691)Then, add 1458 to 691:(691 + 1458 = 2149)So, adding all those together, I get 2149. Therefore, (2847_9) in base ten is 2149.Wait, let me double-check my calculations to make sure I didn't make a mistake. It's easy to slip up with multiplication or addition.Starting again, the digits are 2, 8, 4, 7 in base nine. Let me write each digit with its place value:- 2 is in the (9^3) place: (2 times 729 = 1458)- 8 is in the (9^2) place: (8 times 81 = 648)- 4 is in the (9^1) place: (4 times 9 = 36)- 7 is in the (9^0) place: (7 times 1 = 7)Adding them up:1458 + 648 = 21062106 + 36 = 21422142 + 7 = 2149Hmm, same result. So, that seems consistent.Just to be thorough, let me verify each multiplication:- (2 times 729): 2 times 700 is 1400, and 2 times 29 is 58, so 1400 + 58 = 1458. Correct.- (8 times 81): 8 times 80 is 640, and 8 times 1 is 8, so 640 + 8 = 648. Correct.- (4 times 9): 36. Correct.- (7 times 1): 7. Correct.Adding them again:1458 + 648: Let's break this down. 1458 + 600 = 2058, then 2058 + 48 = 2106.2106 + 36: 2106 + 30 = 2136, then 2136 + 6 = 2142.2142 + 7: 2149. Yep, that's consistent.So, I think I can be confident that (2847_9) converted to base ten is 2149 miles.Just to make sure, let me think if there's another way to approach this. Maybe using the expansion method where each digit is multiplied by 9 raised to the power of its position from the right, starting at 0.So, for (2847_9):- The rightmost digit is 7, which is (7 times 9^0 = 7 times 1 = 7)- Next digit is 4, which is (4 times 9^1 = 4 times 9 = 36)- Next is 8, which is (8 times 9^2 = 8 times 81 = 648)- Leftmost digit is 2, which is (2 times 9^3 = 2 times 729 = 1458)Adding them up: 7 + 36 + 648 + 1458 = 2149. Same result.Alternatively, I can use the positional notation method where each digit is processed from left to right, multiplying the current total by the base and adding the next digit.Let me try that method as a cross-check.Starting with the leftmost digit: 2.Multiply by 9: 2 * 9 = 18.Add the next digit: 18 + 8 = 26.Multiply by 9: 26 * 9 = 234.Add the next digit: 234 + 4 = 238.Multiply by 9: 238 * 9. Let me compute that: 200*9=1800, 38*9=342, so 1800 + 342 = 2142.Add the last digit: 2142 + 7 = 2149.Same answer again. So, that's another way to get to the same result, which makes me more confident that 2149 is correct.Just to recap, the two methods I used:1. Expanded form: Each digit multiplied by its place value, then summed.2. Positional processing: Starting from the left, multiplying by the base and adding the next digit each time.Both gave me 2149, so I think that's solid.I could also use a calculator to verify, but since I don't have one handy, I can compute each multiplication step carefully.Wait, let me compute 238 * 9 again to make sure.238 * 9:Break it down:200 * 9 = 180030 * 9 = 2708 * 9 = 72So, 1800 + 270 = 20702070 + 72 = 2142Yes, that's correct.Then, 2142 + 7 = 2149.Alright, so all steps check out.Another way to think about it: Maybe convert each digit separately and add.But I think that's essentially the same as the expanded form method.Alternatively, I can think about the number in base nine and how it relates to base ten.But I think I've covered the main methods, and both lead to the same result.Therefore, I feel confident that Mary ran 2149 miles in base ten.**Final Answer**Mary ran boxed{2149} miles.

answer：Okay, so I have this algebra problem here. It says that if (-3a^{2x-1}b) and (5ab^{y+4}) can be combined into a single term, then I need to find (left(x-2right)^{2016}+left(y+2right)^{2017}). Hmm, let me try to figure this out step by step.First, I remember that in algebra, when you have like terms, you can combine them by adding or subtracting their coefficients. For terms to be like terms, they must have the same variables raised to the same exponents. So, in this case, the two terms given are (-3a^{2x-1}b) and (5ab^{y+4}). Let me write them down again:1. (-3a^{2x-1}b)2. (5ab^{y+4})I need to check if these are like terms. For them to be like terms, the exponents of the corresponding variables must be equal. That means the exponent of (a) in the first term must be equal to the exponent of (a) in the second term, and similarly for (b).Looking at the exponents of (a):In the first term, the exponent is (2x - 1). In the second term, the exponent is 1 because (a) is just (a^1). So, to make them equal, I can set up the equation:(2x - 1 = 1)Similarly, looking at the exponents of (b):In the first term, the exponent is 1 because it's just (b). In the second term, the exponent is (y + 4). So, setting those equal:(1 = y + 4)Alright, so now I have two equations:1. (2x - 1 = 1)2. (1 = y + 4)Let me solve the first equation for (x):(2x - 1 = 1)Add 1 to both sides:(2x = 2)Divide both sides by 2:(x = 1)Okay, so (x = 1). Now, let's solve the second equation for (y):(1 = y + 4)Subtract 4 from both sides:(y = 1 - 4)(y = -3)Great, so (y = -3). Now, the problem asks for (left(x - 2right)^{2016} + left(y + 2right)^{2017}). Let me substitute the values of (x) and (y) into this expression.First, compute (x - 2):(x - 2 = 1 - 2 = -1)Then, raise that to the power of 2016:((-1)^{2016})Since 2016 is an even number, any negative number raised to an even power becomes positive. So:((-1)^{2016} = 1)Next, compute (y + 2):(y + 2 = -3 + 2 = -1)Then, raise that to the power of 2017:((-1)^{2017})Since 2017 is an odd number, a negative number raised to an odd power remains negative. So:((-1)^{2017} = -1)Now, add these two results together:(1 + (-1) = 0)Wait, so the sum is 0? Hmm, that seems straightforward. Let me just double-check my steps to make sure I didn't make any mistakes.First, I set the exponents equal because the terms need to be like terms to be combined. That gave me (2x - 1 = 1) and (1 = y + 4). Solving those gave me (x = 1) and (y = -3). Then, substituting into the expression, I got ((-1)^{2016} = 1) and ((-1)^{2017} = -1), which add up to 0. I think that's correct. The exponents being even and odd play a crucial role here because they determine the sign of the result. Since 2016 is even, it becomes positive, and 2017 is odd, so it stays negative. Adding them together cancels out to zero.Just to be thorough, let me verify the original terms with (x = 1) and (y = -3):First term: (-3a^{2(1)-1}b = -3a^{2 - 1}b = -3a^{1}b = -3ab)Second term: (5ab^{(-3)+4} = 5ab^{1} = 5ab)Yes, both terms are (-3ab) and (5ab), which are indeed like terms. So, they can be combined into (2ab). That checks out.Therefore, my calculations seem correct, and the final answer should be 0.**Final Answer**boxed{0}

answer：Alright, so I have this probability problem here. It says: Six distinct integers are picked at random from {1,2,3,…,10}. What is the probability that, among those selected, the second smallest is 3? The options are A through E, with E being none of these. Hmm, okay.First, I need to understand the problem. We're selecting six distinct numbers from 1 to 10. So, the total number of possible ways to do this is the combination of 10 choose 6. Let me calculate that. The formula for combinations is n choose k, which is n! / (k! (n - k)!). So, 10 choose 6 is 10! / (6! 4!) = (10×9×8×7) / (4×3×2×1) = 210. So, there are 210 possible sets of six numbers.Now, the question is about the probability that the second smallest number in the selected six is 3. So, I need to figure out how many such sets have 3 as their second smallest number.Let me think about what it means for 3 to be the second smallest. That would imply that the smallest number is less than 3, right? So, the smallest number has to be either 1 or 2. Then, the second number is 3, and the remaining four numbers have to be larger than 3.So, to break it down:1. The smallest number is either 1 or 2.2. The second smallest is 3.3. The remaining four numbers are chosen from the numbers greater than 3, which are 4,5,6,7,8,9,10. That's seven numbers.So, let's compute how many such sets there are.First, the smallest number can be either 1 or 2. So, there are 2 choices for the smallest number.Once the smallest number is chosen, the second number is fixed as 3. Then, we need to choose the remaining four numbers from the seven numbers greater than 3.So, the number of ways to choose these four numbers is 7 choose 4. Let me compute that: 7! / (4! 3!) = (7×6×5) / (3×2×1) = 35.Therefore, for each choice of the smallest number (either 1 or 2), there are 35 ways to choose the remaining four numbers. So, total number of favorable sets is 2 × 35 = 70.Wait, hold on. Let me make sure I didn't make a mistake here. So, the total number of favorable sets is 70. The total number of possible sets is 210. So, the probability is 70 / 210, which simplifies to 1/3. Hmm, 1/3 is one of the options, option C.But hold on, let me think again. Is this correct? Because sometimes in probability, especially with combinatorics, it's easy to miscount.So, the second smallest is 3. That means that exactly one number is less than 3, and the rest are greater than 3. So, the numbers selected must include exactly one number from {1,2}, the number 3, and four numbers from {4,5,6,7,8,9,10}.So, the number of favorable sets is indeed the number of ways to choose 1 number from {1,2}, times the number of ways to choose 4 numbers from {4,...,10}. So, that's 2 choose 1 times 7 choose 4, which is 2 × 35 = 70. So, that seems correct.Therefore, the probability is 70 / 210 = 1/3. So, the answer should be C, 1/3.Wait, but let me make sure I didn't overlook something. Is there another way to think about this?Alternatively, maybe I can think about the position of 3 in the ordered set. When we pick six numbers, we can order them as a1 < a2 < a3 < a4 < a5 < a6. We want a2 = 3.So, the number of such sets is equal to the number of ways where a2 is 3. So, a1 must be less than 3, which can be 1 or 2, and a2 is 3, and a3 to a6 are greater than 3. So, same as before.So, same calculation: 2 choices for a1, and 7 choose 4 for a3 to a6, so 2 × 35 = 70.Thus, 70 / 210 = 1/3. So, that's consistent.Alternatively, maybe I can think in terms of probabilities step by step.What's the probability that 3 is the second smallest? So, first, we need to have exactly one number less than 3 and the rest greater than 3.So, the probability is equal to the number of ways to choose 1 number from {1,2} and 4 numbers from {4,...,10}, divided by the total number of ways to choose 6 numbers from 10.Which is exactly what we did earlier: (2 choose 1 × 7 choose 4) / (10 choose 6) = 70 / 210 = 1/3.So, seems consistent.Wait, but another thought: sometimes, in probability, when dealing with equally likely outcomes, we can also think about the probability that 3 is the second smallest is equal to the number of favorable cases over total cases.But perhaps another way is to fix the position of 3 and compute the probability accordingly.Alternatively, maybe using hypergeometric distribution or something, but I think the combinatorial approach is straightforward here.So, in summary, the total number of favorable cases is 70, total cases is 210, so probability is 1/3, which is option C.But hold on, wait, the answer choices are:(A) 1/60(B) 1/6(C) 1/3(D) 1/2(E) none of theseSo, 1/3 is option C, so that should be the answer.But wait, just to make sure, let me think if I made any mistake in counting.So, the second smallest is 3. So, in the selected six numbers, 3 is the second smallest. So, that means that exactly one number is less than 3, and the remaining four numbers are greater than 3.So, numbers less than 3 are 1 and 2, so two numbers. So, we have to choose exactly one of them. So, 2 choose 1.Numbers greater than 3 are 4 through 10, which is seven numbers. So, we have to choose four of them: 7 choose 4.So, total favorable is 2 × 35 = 70.Total possible is 10 choose 6 = 210.So, 70 / 210 = 1/3.Yes, that seems correct.Alternatively, maybe another way: think about all possible positions of 3.In the six selected numbers, 3 can be in any position from 2 to 6, since it's the second smallest, so it's the second position when ordered.Wait, actually, 3 can be the smallest, second smallest, up to sixth smallest, but in our case, we want it specifically as the second smallest.So, the number of ways where 3 is the second smallest is equal to the number of ways where exactly one number is less than 3 and the rest are greater.Which is the same as before.Alternatively, maybe I can think of it as arranging the numbers.But perhaps that complicates things.Alternatively, maybe think about the probability that in the six numbers, 3 is included, and exactly one number less than 3 is included, and the rest are greater.So, the probability is:(Number of ways to choose 1 from {1,2} and 4 from {4,...,10}) divided by total number of ways.Which is again 2 × 35 / 210 = 70 / 210 = 1/3.So, same result.Therefore, I think the answer is 1/3, which is option C.Wait, but hold on, just to make sure, let me think about another approach.Suppose we fix 3 as the second smallest. So, the smallest has to be either 1 or 2, as we said.So, the probability that the smallest is 1 or 2, and the second is 3.But actually, perhaps another way is to consider the probability that 3 is the second smallest.So, the probability that 3 is in the set, and exactly one number less than 3 is in the set, and the rest are greater.So, the probability is:[Number of ways to choose 1 from {1,2} and 4 from {4,...,10}] / [Number of ways to choose 6 from 10].Which is again 2 × 35 / 210 = 1/3.So, same result.Alternatively, maybe I can think in terms of conditional probability.What's the probability that 3 is the second smallest given that 3 is selected.Wait, but that might complicate it more.Alternatively, maybe think about all possible positions of 3.But I think that's overcomplicating.Alternatively, maybe think about all possible pairs for the two smallest numbers.So, the two smallest numbers must be some pair where the first is less than 3, and the second is 3.So, the number of such pairs is 2 (since the first can be 1 or 2), and the second is 3.Then, the remaining four numbers are chosen from the numbers greater than 3.So, same as before, 2 × 35 = 70.So, 70 / 210 = 1/3.So, same result.Alternatively, maybe think about the probability that 3 is the second smallest is equal to the probability that exactly one of 1 or 2 is selected, and 3 is selected, and the rest are greater.So, the number of favorable sets is C(2,1) × C(7,4) = 2 × 35 = 70.Total sets is C(10,6) = 210.So, 70 / 210 = 1/3.So, same result.Therefore, I think 1/3 is correct, so the answer is C.Wait, but hold on, let me think again.Is there a way that I might have overcounted or undercounted?Wait, for example, if I fix 3 as the second smallest, does that necessarily mean that exactly one number less than 3 is chosen?Yes, because if two numbers less than 3 are chosen, then 3 would be the third smallest, not the second.Similarly, if no numbers less than 3 are chosen, then 3 would be the smallest.Therefore, to have 3 as the second smallest, exactly one number less than 3 must be chosen.Therefore, the number of favorable sets is indeed 2 × C(7,4) = 70.Therefore, 70 / 210 = 1/3.Therefore, the answer is C, 1/3.Wait, but hold on, another thought: is the selection of numbers without replacement, so once you pick 1 or 2, you can't pick them again.But in our calculation, we considered that, because we are choosing 1 from {1,2}, and 4 from {4,...,10}, so 3 is included as the second smallest, so that's correct.Alternatively, maybe another way is to think about the probability step by step.So, the probability that 3 is in the set is C(9,5) / C(10,6) = 126 / 210 = 3/5.But that's the probability that 3 is selected.But we need more than that: we need 3 to be the second smallest.So, given that 3 is selected, what's the probability that exactly one number less than 3 is selected.So, that would be:Number of favorable sets where 3 is selected and exactly one of 1 or 2 is selected, divided by the number of sets where 3 is selected.So, number of favorable sets is C(2,1) × C(7,4) = 70.Number of sets where 3 is selected is C(9,5) = 126.So, the conditional probability is 70 / 126 = 5/9.But that's not the same as the overall probability.Wait, so the overall probability is 70 / 210 = 1/3.Alternatively, the conditional probability given that 3 is selected is 5/9, but the overall probability is 1/3.So, that seems consistent.Alternatively, maybe I can think in terms of the expectation or something else, but that might complicate.Alternatively, maybe think about the position of 3.In the six selected numbers, 3 can be in any position from 1 to 6.But we want it specifically in position 2.So, the number of ways where 3 is in position 2 is equal to the number of ways where exactly one number is less than 3, and the rest are greater.Which is again 2 × 35 = 70.So, same as before.Therefore, I think 1/3 is correct.Wait, but hold on, another thought: is 3 necessarily in the set?Yes, because if 3 isn't in the set, then the second smallest can't be 3.So, in our favorable cases, 3 must be included.So, that's why we have to include 3 in the set.So, that's why we have 1 number from {1,2}, 1 number which is 3, and 4 numbers from {4,...,10}.So, that's correct.Alternatively, maybe I can think about the probability as follows:The probability that 3 is the second smallest is equal to the probability that exactly one of 1 or 2 is selected, and 3 is selected, and the rest are greater.So, the number of favorable sets is C(2,1) × C(7,4) = 70.Total sets is C(10,6) = 210.So, 70 / 210 = 1/3.Therefore, the answer is C.Alternatively, maybe I can think about the problem as arranging the numbers.But I think that complicates it.Alternatively, maybe think about the problem as first selecting the six numbers, and then looking at their order statistics.But in any case, the combinatorial approach seems solid.So, I think the answer is 1/3, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this problem where I need to find the distance from point M₀ to a plane that passes through three other points M₁, M₂, and M₃. The coordinates are given as:M₁(1, -1, 2)M₂(2, 1, 2)M₃(1, 1, 4)M₀(-3, 2, 7)Hmm, I remember that the distance from a point to a plane can be found using a specific formula, but first, I need to figure out the equation of the plane passing through M₁, M₂, and M₃. Once I have the plane's equation, I can plug in the coordinates of M₀ into the distance formula.Alright, let me recall how to find the equation of a plane given three points. I think it involves finding two vectors that lie on the plane and then computing their cross product to get the normal vector of the plane. Once I have the normal vector, I can write the plane equation using one of the points.So, first, let me find two vectors on the plane. I can do this by subtracting the coordinates of M₁ from M₂ and M₃.Vector M₁M₂ = M₂ - M₁ = (2 - 1, 1 - (-1), 2 - 2) = (1, 2, 0)Vector M₁M₃ = M₃ - M₁ = (1 - 1, 1 - (-1), 4 - 2) = (0, 2, 2)Okay, so now I have two vectors:Vector a = (1, 2, 0)Vector b = (0, 2, 2)Next, I need to find the cross product of these two vectors to get the normal vector n.The cross product formula is:n = a × b = |i   j   k|             1   2   0             0   2   2Calculating the determinant:i*(2*2 - 0*2) - j*(1*2 - 0*0) + k*(1*2 - 2*0)= i*(4 - 0) - j*(2 - 0) + k*(2 - 0)= 4i - 2j + 2kSo, the normal vector n is (4, -2, 2). Hmm, I can also write this as (4, -2, 2). Maybe I can simplify this by dividing by 2, but I think it's okay as it is.Now, the general equation of a plane is given by:n₁(x - x₀) + n₂(y - y₀) + n₃(z - z₀) = 0Where (x₀, y₀, z₀) is a point on the plane, which in this case can be M₁(1, -1, 2). So plugging in:4(x - 1) - 2(y - (-1)) + 2(z - 2) = 0Let me simplify this step by step.First, expand each term:4x - 4 - 2(y + 1) + 2(z - 2) = 0Now distribute the -2 and the 2:4x - 4 - 2y - 2 + 2z - 4 = 0Combine like terms:4x - 2y + 2z - 4 - 2 - 4 = 04x - 2y + 2z - 10 = 0Hmm, so the equation of the plane is 4x - 2y + 2z - 10 = 0.Wait, maybe I can simplify this equation by dividing all terms by 2 to make it simpler:2x - y + z - 5 = 0Yes, that looks better. So the plane equation is 2x - y + z = 5.Alright, now that I have the plane equation, I can use the distance formula from a point to a plane. The formula is:Distance = |Ax + By + Cz + D| / sqrt(A² + B² + C²)But wait, let me make sure I have the plane equation in the standard form Ax + By + Cz + D = 0. Currently, my plane equation is 2x - y + z - 5 = 0, so A = 2, B = -1, C = 1, and D = -5.But in the distance formula, it's Ax + By + Cz + D, so in this case, it's 2x - y + z - 5. So, the numerator will be |2*(-3) + (-1)*2 + 1*7 - 5|.Wait, let me double-check that. The formula is |A x₀ + B y₀ + C z₀ + D|, right? So plugging in the values:A = 2, B = -1, C = 1, D = -5x₀ = -3, y₀ = 2, z₀ = 7So, numerator = |2*(-3) + (-1)*2 + 1*7 + (-5)|Let me compute that step by step:2*(-3) = -6(-1)*2 = -21*7 = 7Adding D: -5So, adding all together: -6 - 2 + 7 - 5Compute step by step:-6 - 2 = -8-8 + 7 = -1-1 - 5 = -6So, the absolute value is | -6 | = 6Now, the denominator is sqrt(A² + B² + C²) = sqrt(2² + (-1)² + 1²) = sqrt(4 + 1 + 1) = sqrt(6)So, the distance is 6 / sqrt(6)Hmm, can I simplify that? Yes, 6 divided by sqrt(6) is equal to sqrt(6). Because 6 / sqrt(6) = (6 * sqrt(6)) / (sqrt(6)*sqrt(6)) = (6 sqrt(6)) / 6 = sqrt(6)So, the distance is sqrt(6)Wait, let me verify my calculations again because sometimes I make mistakes.First, the plane equation: 2x - y + z = 5Distance formula: |2*(-3) -1*(2) +1*(7) -5| / sqrt(2² + (-1)² +1²)Compute numerator:2*(-3) = -6-1*(2) = -21*(7) = 7-5 is just -5So, adding them: -6 -2 +7 -5 = (-6 -2) + (7 -5) = (-8) + (2) = -6Absolute value: 6Denominator: sqrt(4 +1 +1) = sqrt(6)So, 6 / sqrt(6) = sqrt(6). Yes, that's correct.Alternatively, I can rationalize the denominator:6 / sqrt(6) = (6 sqrt(6)) / 6 = sqrt(6). Yep, same result.So, the distance is sqrt(6). Hmm, that seems straightforward.Wait, but let me think again if I did everything correctly. Sometimes, when dealing with vectors and cross products, signs can be tricky.Let me recheck the cross product:Vector a = (1, 2, 0)Vector b = (0, 2, 2)Cross product:i*(2*2 - 0*2) - j*(1*2 - 0*0) + k*(1*2 - 2*0)= i*(4 - 0) - j*(2 - 0) + k*(2 - 0)= 4i - 2j + 2kYes, that's correct. So the normal vector is (4, -2, 2). Then, when I divided by 2, I got (2, -1, 1). So the plane equation is 2x - y + z = 5. That seems correct.Then, plugging in M₀(-3, 2, 7):2*(-3) -1*(2) +1*(7) -5 = -6 -2 +7 -5 = -6. Absolute value is 6.Divide by sqrt(2² + (-1)² +1²) = sqrt(6). So 6 / sqrt(6) = sqrt(6). So, yes, that seems correct.Wait, but let me think about another way to compute the distance. Maybe using vectors.Alternatively, the distance from M₀ to the plane can be found by projecting the vector from M₁ to M₀ onto the unit normal vector of the plane.So, vector M₁M₀ = M₀ - M₁ = (-3 -1, 2 - (-1), 7 - 2) = (-4, 3, 5)Then, the unit normal vector n is (2, -1, 1) divided by its magnitude, which is sqrt(2² + (-1)² +1²) = sqrt(6). So, unit normal vector is (2/sqrt(6), -1/sqrt(6), 1/sqrt(6))Then, the distance is the absolute value of the dot product of M₁M₀ and the unit normal vector.So, compute the dot product:(-4)*(2/sqrt(6)) + 3*(-1/sqrt(6)) + 5*(1/sqrt(6))Compute each term:-8/sqrt(6) -3/sqrt(6) +5/sqrt(6) = (-8 -3 +5)/sqrt(6) = (-6)/sqrt(6)Absolute value is 6/sqrt(6) = sqrt(6). Same result.So, that confirms the distance is sqrt(6). Okay, so I think that's correct.Therefore, the distance from M₀ to the plane is sqrt(6).**Final Answer**The distance from point ( M_{0} ) to the plane is boxed{sqrt{6}}.

answer：Okay, so I've got this problem here: there are 5 distinct red balls and 4 distinct white balls. I need to figure out the number of different ways to extract two balls of different colors. The options are A: 9, B: 16, C: 20, D: 32. Hmm, let's break this down step by step.First, let me make sure I understand the problem correctly. We have two groups of balls: red and white. The red balls are distinct, meaning each one is unique, and the same goes for the white balls. So, when they say "extracting two balls of different colors," that means one red and one white ball. Got it.Now, the question is asking for the number of different ways to do this. So, essentially, how many unique pairs can I form where one is red and the other is white. Since the balls are distinct, the order might matter, but I think in this case, since we're just extracting two balls, the order doesn't matter. So, it's a combination problem, not a permutation.Wait, hold on. Let me think again. If the balls are distinct, does that mean that each pair is unique regardless of the order? For example, if I pick red ball 1 and white ball 1, is that different from picking white ball 1 and red ball 1? Hmm, in combinations, the order doesn't matter, so those two would be considered the same. But in permutations, they would be different. But in this case, since we're just extracting two balls, I think it's a combination.But wait, actually, no. Wait, hold on. Let me clarify. When extracting two balls, if the problem doesn't specify order, it's usually combinations. So, the number of ways to choose one red and one white ball would be the product of the number of red balls and white balls.So, if there are 5 red balls and 4 white balls, the number of different pairs would be 5 multiplied by 4, which is 20. So, that would be option C: 20 kinds.But let me double-check. Maybe I'm oversimplifying. Let's think about it another way. If I have 5 red balls, each one can pair with 4 white balls. So, for each red ball, there are 4 possible white balls it can be paired with. So, 5 times 4 is 20. That seems right.Alternatively, if I think about it as combinations, the number of ways to choose one red ball is 5, and the number of ways to choose one white ball is 4. Since these are independent choices, the total number of ways is 5 multiplied by 4, which is 20. So, that still gives me 20.Wait, but let me consider if the problem is considering the order. If it is, then it would be permutations, and the number would be higher. But the problem says "extracting two balls of different colors." It doesn't specify order, so I think it's combinations. So, 20 is the right answer.But hold on, let me think if there's another way to approach this. Maybe using the combination formula. The total number of ways to choose two balls of different colors is equal to the number of ways to choose one red and one white. So, it's C(5,1) multiplied by C(4,1). C(5,1) is 5, and C(4,1) is 4, so 5*4=20. Yep, that's consistent.Alternatively, if I think about the total number of ways to choose any two balls, regardless of color, that would be C(9,2), which is 36. But since we want only two balls of different colors, we can subtract the number of ways to choose two red balls and the number of ways to choose two white balls.So, the number of ways to choose two red balls is C(5,2)=10, and the number of ways to choose two white balls is C(4,2)=6. So, total same color pairs are 10+6=16. Therefore, the number of different color pairs would be total pairs minus same color pairs: 36-16=20. So, that also gives me 20. So, that's another way to confirm.Therefore, I think the correct answer is 20, which is option C.But wait, let me just make sure I didn't make a mistake in the second method. So, total number of ways to choose two balls is C(9,2)=36. Number of ways to choose two red is C(5,2)=10, and two white is C(4,2)=6. So, 10+6=16. Then, 36-16=20. Yep, that seems correct.Alternatively, if I think about it as ordered pairs, meaning that the order matters, then the number of ways would be 5*4 + 4*5=40. But that would be if order matters, but since the problem doesn't specify, I think it's combinations, so 20 is correct.Wait, but let me check the problem statement again. It says "extracting two balls of different colors, the number of different ways to do this is ( )". So, it's about the number of different ways, which in combinatorics usually refers to combinations, not permutations. So, yeah, 20 is the right answer.So, I think I'm confident now that the answer is C: 20 kinds.**Final Answer**boxed{C}

answer：Alright, so I have this problem here where there's a semi-circle with diameter endpoints at P(-4, 0) and Q(16, 0). There's a point R(0, t) on the circle, and I need to find the value of t where t is positive. The options are A) 6, B) 10, C) 8, D) 9, E) 7.First, let me visualize this. The diameter is from (-4, 0) to (16, 0), so that's a horizontal line on the x-axis. The semi-circle must be either above or below this diameter. Since the point R is given as (0, t) with t > 0, it must be on the upper semi-circle.Okay, so to find t, I need to figure out the equation of the circle and then plug in x=0 to find the corresponding y-coordinate, which is t.To write the equation of a circle, I need the center and the radius. The center is the midpoint of the diameter PQ. So, let me find the midpoint first.The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). Plugging in the coordinates of P and Q:Midpoint M = ((-4 + 16)/2, (0 + 0)/2) = (12/2, 0) = (6, 0). So, the center is at (6, 0).Next, the radius is half the length of the diameter. Let me find the length of PQ first. Since both points are on the x-axis, the distance between them is just the difference in x-coordinates.Distance PQ = 16 - (-4) = 20 units. So, the radius is half of that, which is 10 units.Now, with the center at (6, 0) and radius 10, the equation of the circle is:(x - 6)^2 + (y - 0)^2 = 10^2Simplifying, that's:(x - 6)^2 + y^2 = 100Now, the point R(0, t) lies on this circle, so plugging x=0 and y=t into the equation:(0 - 6)^2 + (t)^2 = 100Calculating (0 - 6)^2: that's (-6)^2 = 36So, 36 + t^2 = 100Subtract 36 from both sides:t^2 = 64Taking the square root of both sides:t = sqrt(64) = 8Since t is positive, t = 8.Wait, let me double-check that. So, the center is at (6, 0), radius 10. Plugging in (0, t):(0 - 6)^2 + t^2 = 100 => 36 + t^2 = 100 => t^2 = 64 => t = 8. Yep, that seems right.Looking at the options, C is 8. So, the answer should be C) 8.But just to make sure I didn't make any mistakes, let me go through the steps again.1. Midpoint between P(-4, 0) and Q(16, 0): average of x-coordinates: (-4 + 16)/2 = 12/2 = 6. Y-coordinate is 0. So, center is (6, 0). Correct.2. Length of diameter: distance between P and Q. Since they are on the x-axis, it's 16 - (-4) = 20. So, radius is 10. Correct.3. Equation of the circle: (x - 6)^2 + y^2 = 100. Correct.4. Plugging in R(0, t): (0 - 6)^2 + t^2 = 100 => 36 + t^2 = 100 => t^2 = 64 => t = 8. Correct.So, yeah, I think 8 is the right answer. So, option C.**Final Answer**boxed{C}