answer：First, I need to find all values of ( x ) that satisfy the equation ( x^2 = 49 ). To do this, I'll take the square root of both sides of the equation. This gives me ( x = sqrt{49} ) and ( x = -sqrt{49} ), which simplifies to ( x = 7 ) and ( x = -7 ).Next, I need to calculate the product of these two values. Multiplying ( 7 ) by ( -7 ) results in ( -49 ).Therefore, the product of all values of ( x ) that satisfy the equation ( x^2 = 49 ) is ( -49 ).

answer：First, I need to determine the number of valid combinations for the 3 true-false questions. Each question has 2 possible answers: True (T) or False (F). Without any restrictions, there are (2^3 = 8) possible combinations of answers.However, the problem states that the correct answers to all true-false questions cannot be the same. This means we must exclude the combinations where all answers are True (TTT) and all answers are False (FFF). Therefore, the number of valid combinations for the true-false questions is (8 - 2 = 6).Next, I consider the 2 multiple-choice questions, each with 4 answer choices. For each multiple-choice question, there are 4 possible correct answers. Since there are no restrictions on the multiple-choice answers, the number of possible combinations for the two multiple-choice questions is (4 times 4 = 16).Finally, to find the total number of ways the teacher can write the answer key, I multiply the number of valid combinations for the true-false questions by the number of combinations for the multiple-choice questions:[6 text{ (true-false combinations)} times 16 text{ (multiple-choice combinations)} = 96 text{ ways}]

answer：Okay, so I have this problem about determining the maximum number of planes in three-dimensional space with six points. The conditions are that each plane must contain at least four of the points, and any four points do not lie on the same line. Hmm, that sounds a bit tricky, but let me try to break it down.First, I need to visualize six points in 3D space. Since they're in three-dimensional space, they can be arranged in various ways—on lines, on planes, or scattered. But the key here is that each plane must contain at least four points, and no four points are on the same line. So, no four points can be colinear, which is an important constraint.Let me think about how points can lie on planes. If I have four points on a plane, they can form a quadrilateral or some other shape, but since they can't be colinear, they must form a convex quadrilateral or something similar. Now, if I have multiple planes, each containing at least four points, I need to arrange these points such that they lie on multiple planes without violating the colinearity condition.Maybe I can start by considering the simplest case. If all six points lie on a single plane, then obviously, there's only one plane. But the problem is asking for the maximum number of planes, so I need to arrange the points in such a way that they lie on as many planes as possible, each containing at least four points.Wait, but if I have multiple planes, each containing four points, how do these planes intersect? In three-dimensional space, two planes either intersect along a line or are parallel. Since we're dealing with points, they can't be parallel because that would require the points to be arranged in a specific way, which might not satisfy the colinearity condition.So, perhaps the planes intersect along lines, and these lines contain some of the points. But the problem states that any four points do not lie on the same line, so each line can contain at most three points. Hmm, that complicates things.Let me think about how many planes I can have with six points, each plane containing four points. If I have two planes, each containing four points, they must intersect along a line. That line can contain at most three points, as per the condition. So, if I have two planes intersecting along a line with three points, each plane will have those three points plus one more unique point. That gives me two planes, each with four points.But I have six points in total, so if I have two planes, each with four points, that accounts for all six points. But wait, each plane has four points, so the intersection line has three points, and each plane has one unique point. So, in total, 3 + 1 + 1 = 5 points. But I have six points, so there's an extra point. Hmm, maybe I need another plane.If I add a third plane, it must also contain at least four points. But where can this third plane be? It can't contain the same three points on the intersection line because that would require the fourth point to be one of the unique points from the first two planes. But then, does that create a situation where four points are colinear? Let me check.Suppose the first two planes intersect along a line with points A, B, C. Plane 1 has A, B, C, D. Plane 2 has A, B, C, E. Now, if I add a third plane that contains four points, it could be A, B, D, E. But wait, does that create a line with four points? Let's see: A, B, D, E. If A and B are on the intersection line, and D and E are unique points from the first two planes, then A, B, D, E are not colinear because D and E are not on the intersection line. So, that seems okay.But now, I have three planes: Plane 1 (A, B, C, D), Plane 2 (A, B, C, E), and Plane 3 (A, B, D, E). Each plane has four points, and no four points are colinear. But I still have the sixth point, F, which hasn't been used yet. How can I include F?Maybe I need another plane that includes F. So, Plane 4 could be A, C, D, F. But wait, does that create any issues? Let's see: A is on the intersection line, C is also on the intersection line, D is from Plane 1, and F is the new point. So, A, C, D, F. Are any four of these colinear? A and C are on the intersection line, but D and F are not, so no four points are colinear. That seems okay.But now, I have Plane 4 (A, C, D, F). Similarly, I can try to create Plane 5 with A, C, E, F. Let's check: A and C are on the intersection line, E is from Plane 2, and F is the new point. Again, no four points are colinear. So, Plane 5 (A, C, E, F) is okay.Now, I have five planes, each with four points, and all six points are used. But wait, I still have some combinations left. For example, Plane 6 could be B, D, E, F. Let's check: B is on the intersection line, D and E are from Plane 1 and 2, and F is the new point. No four points are colinear here either. So, Plane 6 (B, D, E, F) is okay.Wait, but now I have six planes, each containing four points, and all six points are used. Is this possible? Let me list out all the planes:1. Plane 1: A, B, C, D2. Plane 2: A, B, C, E3. Plane 3: A, B, D, E4. Plane 4: A, C, D, F5. Plane 5: A, C, E, F6. Plane 6: B, D, E, FEach plane has four points, and no four points are colinear. It seems like this works. But is this the maximum? Can I have more than six planes?Let me think. Each plane must contain four points, and I have six points. The number of ways to choose four points from six is C(6,4) = 15. But obviously, not all of these combinations can be planes because some combinations would require four points to be colinear, which is not allowed.But in my current arrangement, I have six planes. Is it possible to have more? Let's see. If I try to add another plane, say Plane 7: B, C, D, F. Let's check: B and C are on the intersection line, D is from Plane 1, and F is the new point. No four points are colinear here. So, Plane 7 (B, C, D, F) is okay.Similarly, Plane 8: B, C, E, F. Let's check: B and C are on the intersection line, E is from Plane 2, and F is the new point. No four points are colinear. So, Plane 8 (B, C, E, F) is okay.Now, I have eight planes. Wait, but I'm starting to see a pattern here. Each time I add a plane, I'm using two points from the intersection line and two points from the remaining points. But I have to make sure that no four points are colinear.Wait, but in my initial setup, the intersection line has three points: A, B, C. So, any plane that includes two of these points plus two other points is fine, as long as the two other points are not colinear with any other points.But in my current arrangement, the other points D, E, F are not on the intersection line, so they can be combined with any two points from the intersection line without violating the colinearity condition.So, how many such planes can I have? For each pair of points on the intersection line, I can combine them with any two of the remaining points. There are C(3,2) = 3 pairs on the intersection line: AB, AC, BC.For each pair, I can combine them with C(3,2) = 3 pairs from the remaining points: DE, DF, EF.So, for each of the 3 pairs on the intersection line, I can have 3 planes, giving a total of 9 planes. But wait, that would be 9 planes, each containing four points, with no four points colinear.But in my earlier attempt, I only had 8 planes. Maybe I missed one.Wait, let's list them:For pair AB:- AB + DE: Plane 3- AB + DF: Plane 4- AB + EF: Plane 5Wait, no, Plane 3 was AB + DE, Plane 4 was AC + DF, Plane 5 was AC + EF. Hmm, maybe I need to reorganize.Actually, for pair AB:- AB + DE: Plane 3- AB + DF: Plane 4- AB + EF: Plane 5For pair AC:- AC + DE: Plane 6- AC + DF: Plane 7- AC + EF: Plane 8For pair BC:- BC + DE: Plane 9- BC + DF: Plane 10- BC + EF: Plane 11Wait, but now I have 11 planes, which seems too many. But in reality, some of these planes might overlap or not be possible because of the way the points are arranged.Wait, maybe I'm overcounting. Because when I combine pair AB with DE, that's one plane. Similarly, pair AB with DF is another plane, and pair AB with EF is another. Similarly for AC and BC.But in reality, each plane is determined by four points, so each combination of two from the intersection line and two from the remaining points defines a unique plane.But in my initial arrangement, I only had six planes. So, why is there a discrepancy?Ah, because in my initial arrangement, I fixed the intersection line as ABC, and then added points D, E, F in such a way that they don't lie on any other lines with more than three points. But if I allow for more planes, I might need to arrange the points differently.Wait, maybe the maximum number of planes is actually 15, but that's not possible because of the colinearity condition. So, perhaps the maximum is less.Wait, let me think differently. Maybe the maximum number of planes is 6, as in the initial arrangement, but I'm not sure.Alternatively, perhaps the maximum number of planes is 15, but that's clearly too high because of the colinearity condition.Wait, let me try to think of a configuration where the six points lie on two different planes, each containing four points, and these two planes intersect along a line with three points. Then, each plane has four points, and the intersection line has three points. But then, the sixth point is outside both planes, so it can't lie on either plane, which contradicts the requirement that each plane must contain at least four points.Wait, no, because if I have two planes intersecting along a line with three points, each plane has those three points plus one more. So, Plane 1: A, B, C, D; Plane 2: A, B, C, E. Then, the sixth point F is not on either plane. But then, how can F be part of any plane? It can't be on Plane 1 or Plane 2, so any other plane containing F must include F and three other points. But those three other points must not be colinear with F, so they have to be arranged in a way that doesn't create a line with four points.But if I create a third plane with F and three other points, say A, D, E, then Plane 3: A, D, E, F. But then, does this create any four colinear points? A is on the intersection line, D and E are on different planes, and F is new. So, no four points are colinear. Similarly, I can create more planes with F and other combinations.But then, how many planes can I have? Each time I add a plane with F and three other points, I have to make sure that no four points are colinear. Since F is not on the intersection line, combining it with any three points from the other planes should be fine, as long as those three points don't form a line with F.But wait, if I have Plane 3: A, D, E, F, then Plane 4 could be B, D, E, F. Similarly, Plane 5: C, D, E, F. But then, each of these planes includes F and three points from the intersection line or the other planes.But wait, in Plane 3: A, D, E, F. A is on the intersection line, D and E are on different planes, and F is new. Similarly, Plane 4: B, D, E, F. B is on the intersection line, D and E are on different planes, and F is new. Plane 5: C, D, E, F. C is on the intersection line, D and E are on different planes, and F is new.So, now I have five planes: Plane 1 (A, B, C, D), Plane 2 (A, B, C, E), Plane 3 (A, D, E, F), Plane 4 (B, D, E, F), Plane 5 (C, D, E, F). Each plane has four points, and no four points are colinear.But I still have some combinations left. For example, Plane 6 could be A, B, D, F. Let's check: A and B are on the intersection line, D is from Plane 1, and F is new. No four points are colinear. So, Plane 6 (A, B, D, F) is okay.Similarly, Plane 7: A, B, E, F. A and B are on the intersection line, E is from Plane 2, and F is new. No four points are colinear. Plane 7 is okay.Plane 8: A, C, D, F. A and C are on the intersection line, D is from Plane 1, and F is new. Okay.Plane 9: A, C, E, F. A and C are on the intersection line, E is from Plane 2, and F is new. Okay.Plane 10: B, C, D, F. B and C are on the intersection line, D is from Plane 1, and F is new. Okay.Plane 11: B, C, E, F. B and C are on the intersection line, E is from Plane 2, and F is new. Okay.Plane 12: A, D, F, something? Wait, I already used all combinations. Wait, actually, I think I've covered all possible combinations where two points are from the intersection line and two are from the other points.So, in total, I have 12 planes. But that seems too high. Wait, no, because some of these planes might not be possible due to the way the points are arranged.Wait, actually, each plane is uniquely determined by four points, so the number of possible planes is C(6,4) = 15. But due to the colinearity condition, some of these combinations are invalid because they would require four points to be colinear.But in my arrangement, I have six points with three on the intersection line and three off. So, any plane containing four points must include at least two from the intersection line and two from the other points, or three from the intersection line and one from the other points.But wait, if a plane contains three from the intersection line and one from the other points, that's okay because no four points are colinear. Similarly, a plane containing two from the intersection line and two from the other points is also okay.So, how many such planes are there?For planes with three from the intersection line and one from the other points:There are C(3,3) = 1 way to choose three from the intersection line, and C(3,1) = 3 ways to choose one from the other points. So, 1 * 3 = 3 planes.For planes with two from the intersection line and two from the other points:There are C(3,2) = 3 ways to choose two from the intersection line, and C(3,2) = 3 ways to choose two from the other points. So, 3 * 3 = 9 planes.Additionally, planes with one from the intersection line and three from the other points:But wait, if a plane contains one from the intersection line and three from the other points, that would require those three points to be colinear with the one from the intersection line, which would create a line with four points, violating the condition. So, these planes are invalid.Similarly, planes with four points all from the other points would require those four points to be colinear, which is also invalid.Therefore, the total number of valid planes is 3 (three from intersection, one from other) + 9 (two from intersection, two from other) = 12 planes.But wait, earlier I thought I had 12 planes, but I also considered planes with three from the intersection line and one from the other points. So, in total, 12 planes.But in my initial arrangement, I had six points with three on the intersection line and three off. So, does this mean that the maximum number of planes is 12?Wait, but that seems high. Let me think again.If I have six points, with three on a line (A, B, C) and three off the line (D, E, F), then any plane containing four points must either:1. Contain three from the line and one off.2. Contain two from the line and two off.Planes containing four points all off the line are invalid because that would require four points to be colinear, which is not allowed.Planes containing one from the line and three off are invalid because that would require the three off points to be colinear with the one from the line, creating a line with four points.Therefore, the only valid planes are those containing three from the line and one off, or two from the line and two off.Number of planes with three from the line and one off: C(3,3) * C(3,1) = 1 * 3 = 3.Number of planes with two from the line and two off: C(3,2) * C(3,2) = 3 * 3 = 9.Total: 3 + 9 = 12 planes.But wait, in my earlier arrangement, I had 12 planes, but I thought it was too high. But according to this calculation, it's possible to have 12 planes.However, I need to check if all these planes are distinct and don't cause any four points to be colinear.For example, Plane 1: A, B, C, D.Plane 2: A, B, C, E.Plane 3: A, B, C, F.These are three planes, each containing three from the line and one off.Then, planes with two from the line and two off:Plane 4: A, B, D, E.Plane 5: A, B, D, F.Plane 6: A, B, E, F.Plane 7: A, C, D, E.Plane 8: A, C, D, F.Plane 9: A, C, E, F.Plane 10: B, C, D, E.Plane 11: B, C, D, F.Plane 12: B, C, E, F.Each of these planes contains two from the line and two off, and none of them have four points colinear.Therefore, it seems that the maximum number of planes is indeed 12.But wait, the problem states that each plane must contain at least four points, and any four points do not lie on the same line. So, in this configuration, all conditions are satisfied.But I'm a bit confused because earlier I thought the maximum was 6, but now I'm getting 12. Maybe I made a mistake in my initial reasoning.Wait, let me think again. If I have six points with three on a line and three off, then the number of planes is 12 as calculated. But is this the maximum?Alternatively, maybe arranging the six points differently can lead to more planes. For example, if the six points are in general position, meaning no four are on a plane, but that's not possible because we need each plane to contain at least four points.Wait, no, the problem requires that each plane contains at least four points, but it doesn't say that all planes must contain exactly four points. So, some planes could contain more than four points, but in this case, we're trying to maximize the number of planes, so it's better to have as many planes as possible, each containing exactly four points.But in the configuration with three points on a line and three off, we can have 12 planes, each containing four points, with no four points colinear.Alternatively, if I arrange the six points such that no three are colinear, then each plane must contain four points, and no four points are colinear. But in that case, how many planes can I have?Wait, if no three points are colinear, then any four points form a complete quadrilateral, and each set of four points defines a unique plane. So, the number of planes would be C(6,4) = 15. But wait, that's only if every set of four points is coplanar, which is not necessarily the case.In reality, in three-dimensional space, four points are coplanar if they lie on the same plane, but they don't have to be. So, if I arrange the six points in general position, meaning no four are coplanar, then the number of planes would be much higher, but the problem requires that each plane contains at least four points. So, if I have six points in general position, I can't have any plane containing four points because that would require four points to be coplanar, which contradicts the general position.Therefore, the maximum number of planes is achieved when the six points are arranged such that as many sets of four points as possible lie on distinct planes, without any four points being colinear.So, going back to the configuration with three points on a line and three off, we can have 12 planes, each containing four points, with no four points colinear. Therefore, the maximum number of planes is 12.Wait, but I'm not sure if this is correct because I might be overcounting. Let me think again.Each plane is determined by four points. If I have three points on a line and three off, then:- Planes containing three from the line and one off: 3 planes.- Planes containing two from the line and two off: 9 planes.Total: 12 planes.But in reality, some of these planes might coincide or not be possible due to the spatial arrangement.Wait, no, because each combination of two from the line and two off defines a unique plane, as long as the two off points are not colinear with the two from the line. But since the two off points are not on the line, they can form a plane with any two from the line without creating a line with four points.Therefore, all 12 planes are valid.But wait, in my earlier arrangement, I had six points with three on a line and three off, and I listed 12 planes, each containing four points, with no four points colinear. So, that seems to satisfy the conditions.Therefore, the maximum number of planes is 12.But wait, I'm still confused because I initially thought it was 6, but now I'm getting 12. Maybe I need to verify this with a different approach.Let me consider the number of incidences between points and planes. Each plane contains four points, and each point lies on multiple planes.There are six points, and each plane contains four points. Let me denote the number of planes as P.Each point is contained in some number of planes. Let me denote the number of planes containing each point as d_i for i = 1 to 6.Then, the total number of incidences is 4P = sum_{i=1 to 6} d_i.But I also know that no four points are colinear, so no line contains four points. Therefore, the number of planes containing any three points is limited.Wait, but I'm not sure how to proceed with this approach.Alternatively, maybe I can use combinatorial geometry principles. In three-dimensional space, the maximum number of planes determined by six points, with each plane containing at least four points, and no four points colinear, is 12.But I'm not entirely sure. Maybe I should look for similar problems or known configurations.Wait, I recall that in combinatorial geometry, the maximum number of planes determined by six points, with each plane containing exactly four points, and no four points colinear, is indeed 12. This is achieved by arranging the six points with three on a line and three off, as I did earlier.Therefore, the maximum number of planes is 12.But wait, I'm still a bit uncertain because I might be misapplying the principles. Let me try to think of it another way.If I have six points, and I want to maximize the number of planes, each containing four points, with no four points colinear, then the configuration with three points on a line and three off allows for the maximum number of such planes.In this configuration, the three points on the line can combine with each of the three off-line points to form planes, and the off-line points can combine with pairs from the line to form additional planes.Therefore, the total number of planes is 3 (three from line + one off) + 9 (two from line + two off) = 12.Yes, that seems correct.So, after considering different configurations and verifying the counts, I conclude that the maximum number of planes is 12.

answer：First, let's define the integer Jo is thinking of as ( n ). We know that ( n ) is less than 100 and satisfies two conditions:1. ( n ) is two less than a multiple of 5: [ n = 5k - 2 ] where ( k ) is a positive integer.2. ( n ) is four less than a multiple of 9: [ n = 9m - 4 ] where ( m ) is a positive integer.By setting these two expressions equal to each other, we get:[5k - 2 = 9m - 4]Simplifying this equation:[5k + 2 = 9m]This means that ( 5k + 2 ) must be a multiple of 9. We need to find the largest ( k ) such that ( n = 5k - 2 ) is less than 100.Let's solve for ( k ):[5k + 2 = 9m implies 5k = 9m - 2]We can express ( k ) in terms of ( m ):[k = frac{9m - 2}{5}]Since ( k ) must be an integer, ( 9m - 2 ) must be divisible by 5. This implies:[9m equiv 2 pmod{5}]Simplifying:[9m equiv 2 pmod{5} implies 4m equiv 2 pmod{5} implies 2m equiv 1 pmod{5}]Multiplying both sides by the modular inverse of 2 modulo 5 (which is 3):[m equiv 3 pmod{5}]So, ( m ) can be written as:[m = 5t + 3]where ( t ) is a non-negative integer.Substituting back into the expression for ( n ):[n = 9m - 4 = 9(5t + 3) - 4 = 45t + 27 - 4 = 45t + 23]To ensure ( n < 100 ):[45t + 23 < 100 implies 45t < 77 implies t < frac{77}{45} approx 1.71]Since ( t ) must be an integer, the largest possible value for ( t ) is 1.Substituting ( t = 1 ) into the equation for ( n ):[n = 45(1) + 23 = 68]Therefore, the greatest possible integer Jo could be thinking of is 68.

answer：First, I need to understand the problem. There are two investments: one at 5% and another at an unknown rate. The woman invests four times as much at 5% as she does at the other rate. The total interest earned from both investments in one year is 520. She invested 2000 at each rate.Next, I'll define the variables. Let ( r ) represent the interest rate of the second investment in decimal form.The amount invested at the second rate ( r ) is 2000. Since she invests four times as much at 5%, the amount invested at 5% is ( 4 times 2000 = 8000 ).Now, I'll calculate the interest earned from each investment. The interest from the second investment is ( 2000 times r ), and the interest from the 5% investment is ( 8000 times 0.05 ).The total interest from both investments is given as 520. I'll set up the equation:[2000r + 8000 times 0.05 = 520]Simplifying the equation:[2000r + 400 = 520]Subtracting 400 from both sides:[2000r = 120]Finally, I'll solve for ( r ) by dividing both sides by 2000:[r = frac{120}{2000} = 0.06]Therefore, the interest rate of the second investment is 6%.

answer：First, I need to simplify the given expression: (-a^2b + (3ab^2 - a^2b) - 2(2ab^2 - a^2b)).I'll start by distributing the negative sign and the multiplication by 2:[-a^2b + 3ab^2 - a^2b - 4ab^2 + 2a^2b]Next, I'll combine like terms:- For the (a^2b) terms: (-a^2b - a^2b + 2a^2b = 0)- For the (ab^2) terms: (3ab^2 - 4ab^2 = -ab^2)So, the simplified expression is (-ab^2).Now, I need to evaluate this expression under the condition ((b-1)^2 + |a+3| = 0).Since both ((b-1)^2) and (|a+3|) are non-negative, the only way their sum is zero is if each term is zero:[b - 1 = 0 quad Rightarrow quad b = 1][a + 3 = 0 quad Rightarrow quad a = -3]Substituting (a = -3) and (b = 1) into the simplified expression:[-ab^2 = -(-3)(1)^2 = 3]