course mth 174 02-06-2008
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03:07:02 query problem 7.2.12 (3d edition #11) antiderivative of sin^2 x
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RESPONSE --> clicknig on to questions.
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03:08:32 what is the requested antiderivative?
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RESPONSE --> sin(x) - cos(x)+C
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03:12:54 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> I used the breakdown method: uv'= uv - u'v sin(x)sin(x)= sin(x) (-cos(x)) - cos(x)(-cos(x)) = sin(x) - cos(x)+C
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03:13:29 query problem 7.2.16 antiderivative of (t+2) `sqrt(2+3t)
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RESPONSE --> -> questions
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04:02:27 what is the requested antiderivative?
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RESPONSE --> (t+2) (1/1.5 (2t + 3t^2 / 2) ^ 1.5) - (t^2 /2 + 2t) (1/1.5 (2t + 3t^2 / 2) ^ 1.5) +c
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04:04:42 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> I broke the function into parts u= t+2 and v'= (2 +3t)^.5 however, I'm pretty shaky on my pick of variables I think I made it harder than necessary.
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04:05:05 query problem 7.2.27 antiderivative of x^5 cos(x^3)
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RESPONSE --> -> questions
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04:46:29 what is the requested antiderivative?
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RESPONSE --> (1/3) x^3 sin(x^3) + (1/3) cos(x^3) + C
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04:49:16 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> Breakdown the function into x^5 and Cos(x)^3 and use the formula for integrating by parts.
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04:51:22 query problem 7.2.52 (3d edition #50) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).
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RESPONSE --> ->questions
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05:04:07 What is the value of the requested integral?
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RESPONSE --> THe first integral which gives us f'= x^2/2 + 1.5 by integrating again we get f= 1/6 X^3 + 1.5x +6
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05:08:46 How did you use integration by parts to obtain this result? Be specific.
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RESPONSE --> for the first integration is a constant which integrates to x^2 /2. I then broke this into parts by saying it was x^2 * .5 and plugged it into the formula which integrated to 1/6x^3 then the 1.5x was added and then C was evaluated for at x=0.
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05:12:17 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> overall I mostly understood the concepts of what was happening but I still thought it was more difficult than prevous ways. I had a similar reaction when learning to derivate when we did implicit derivitves.
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05:12:30 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> okay
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17:28:07 `questionNumber 30000 Query 2.3.15 (Y ^ Z')U X, univ={a,..g}, X={a,c,e,g}, Y = {a,b,c}, Z = {b, ..., f} What is the set (Y ^ Z')U X?
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RESPONSE --> a is the only one in the set. confidence assessment: 1
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17:28:54 `questionNumber 30000 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'. So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. **
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RESPONSE --> self critique assessment: 3
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17:30:19 `questionNumber 30000 Give the intersection of the two sets Y and Z'
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RESPONSE --> b and c. confidence assessment: 3
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17:31:26 `questionNumber 30000 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.**
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RESPONSE --> self critique assessment: 3
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17:33:10 `questionNumber 30000 Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')
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RESPONSE --> Not in A or B. confidence assessment: 1
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17:33:49 `questionNumber 30000 ** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'. STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A. INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. **
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RESPONSE --> self critique assessment: 3
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17:37:43 `questionNumber 30000 ** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_. Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection. } MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true. The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B. The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5. The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6. The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. **
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RESPONSE --> . self critique assessment: 3
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17:40:14 `questionNumber 30000 Query 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'.
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RESPONSE --> 5 and 2. confidence assessment: 1
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17:41:19 `questionNumber 30000 ** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}. (X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5} The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. **
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RESPONSE --> self critique assessment: 3
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17:53:19 `questionNumber 30000 2.3.72 A = {3,6,9,12}, B = {6,8}.
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RESPONSE --> A X B = (( 3, 6), (3,8), (6,6), (6, 8) , (9,6), (9,8) , (12,6) , (12,8)) B X A = (( 6, 3), (6, 6), ( 6, 9) (6, 12), (8,3) (8,6), (8,9) , (8,12)) confidence assessment: 3
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17:53:31 `questionNumber 30000 ** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)} (B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)} How is n(A x B) related to n(A) and n(B)? n(S) stands for the number of elements in the set S, i.e., its cardinality. n(A x B) = n(A) * n(B) **
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RESPONSE --> self critique assessment: 3
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17:55:36 `questionNumber 30000 2.3.84 Shade A U B
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RESPONSE --> everything but u would be shaded. confidence assessment: 2
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17:56:19 `questionNumber 30000 ** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. **
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RESPONSE --> self critique assessment: 3
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17:58:20 `questionNumber 30000 Query 2.3.100 Shade (A' ^ B) ^ C
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RESPONSE --> All of B is shaded except the overlap with A. Everything is shaded in C except its overlap with A confidence assessment: 2
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17:58:40 `questionNumber 30000 ** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A **
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RESPONSE --> self critique assessment: 3
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17:59:36 `questionNumber 30000 Describe the shading of the set (A ^ B)' U C.
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RESPONSE --> Nothing would be shaded in the regions. confidence assessment: 2
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18:00:27 `questionNumber 30000 ** All of C would be shaded because we have a union with C, which will include all of C. Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded. Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. **
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RESPONSE --> self critique assessment: 3
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18:02:03 `questionNumber 30000 2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region.
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RESPONSE --> (A ^ B' ) U C' confidence assessment: 2
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18:02:40 `questionNumber 30000 ** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. **
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RESPONSE --> self critique assessment: 3
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XϓqFZ assignment #004 004. `Query 4 College Algebra 01-30-2008
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18:06:35 `questionNumber 40000 2.4.12 n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10
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RESPONSE --> A = 10 B = 10, 28 A' U B' = 25, 40 confidence assessment: 2
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18:07:02 `questionNumber 40000 ** In terms of the picture (2 circles, linked, representing the two sets) there are 28 in B and 10 in A ^ B so there are 18 in the region of B outside of A--this is the region B-A. There are 25 outside of A, and 18 of these are accounted for in this region of B. Everything else outside of A must therefore also be outside of B, so there are 25-18=7 elements in the region outside of both A and B. A ' U B ' consists of everything that is either outside of A or outside of B, or both. The only region that's not part of A ' U B ' is therefore the intersection A ^ B, since everything in this region is inside both sets. A' U B' is therefore everything but the region A ^ B which is common to both A and B. This includes the 18 elements in B that aren't in A and the 7 outside both A and B. This leaves 40 - 18 - 7 = 15 in the region of A that doesn't include any of B. This region is the region A - B you are looking for. **
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RESPONSE --> self critique assessment: 3
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18:08:56 `questionNumber 40000 query 2.4.18 wrote and produced 2, wrote 5, produced 7 &&&& How many did he write but not produce?
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RESPONSE --> 5 confidence assessment: 3
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18:09:18 `questionNumber 40000 ** You need to count the two he wrote and produced among those he wrote, and also among those he produced. He only wrote 5, two of which he also produced. So he wrote only 3 without producing them. In terms of the circles you might have a set A with 5 elements (representing what he wrote), B with 7 elements (representing what he produced) and A ^ B with 2 elements. This leaves 3 elements in the single region A - B and 5 elements in the single region B - A. The 3 elements in B - A would be the answer to the question. **
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RESPONSE --> self critique assessment: 3
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}x~ƿ߿| assignment #005 005. `Query 5 College Algebra 02-06-2008
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11:01:08 `questionNumber 50000 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> There are aleph-null of them. confidence assessment: 2
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11:01:40 `questionNumber 50000 ** There are 10 numbers in the set: 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 **
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RESPONSE --> self critique assessment: 3
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11:02:21 `questionNumber 50000 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> There are aleph-null of them. confidence assessment: 1
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11:02:38 `questionNumber 50000 ** {x | x is an even integer } indicates the set of ALL possible values of the variable x which are even integers. Anything that satisfies the description is in the set. This is therefore the set of even integers, which is infinite. Since this set can be put into 1-1 correspondence with the counting numbers its cardinality is aleph-null. **
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RESPONSE --> self critique assessment: 3
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11:03:33 `questionNumber 50000 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> 3 confidence assessment: 2
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11:05:27 `questionNumber 50000 ** Listing them in order, according to the order of listing in the set. We have: [ {S,D},{B,R},{Dic.,BL}] , [{S,bl},{B,D},{Dic.,R}], [{S,R},{B,Bl},{dic.,D}] [ {S,D},{B,DL},{Dic.,R}], [{S,bl},{B,R},{Dic.,D}], [{S,R},{B,D},{dic.,B1}] for a total of six. Reasoning it out, there are three choices for the character paired with Stallone, which leaves two for the character to pair with Bogart, leaving only one choice for the character to pair with diCaprio. **
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RESPONSE --> self critique assessment: 2
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11:10:45 `questionNumber 50000 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> (-17,1) (-22, 2) (-27, 3) (-5n) confidence assessment: 2
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11:11:00 `questionNumber 50000 **You have to describe the 1-1 correspondence, including the rule for the nth number. A complete description might be 1 <-> -17, 2 <-> -22, 3 <-> -27, ..., n <-> -12 + 5 * n. You have to give a rule for the description. n <-> -12 - 5 * n is the rule. Note that we jump by -5 each time, hence the -5n. To get -17 when n=1, we need to start with -12. THE REASONING PROCESS TO GET THE FORMULA: The numbers in the first set decrease by 5 each time so you need -5n. The n=1 number must be -17. -5 * 1 = -5. You need to subtract 12 from -5 to get -17. So the formula is -5 n - 12. **
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RESPONSE --> self critique assessment: 3
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11:12:30 `questionNumber 50000 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> confidence assessment: 0
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11:13:29 `questionNumber 50000 ** This is a pretty tough question. One way of describing the correspondence (you will probably need to do the construction to understand): Sketch a straight line from the top of the blue line at the right to the top of the blue line at the left, extending this line until it meets the dotted line. Call this meeting point P. Then for any point on the shorter blue line we can draw a straight line from P to that point and extend it to a point of the longer blue line, and in our 1-1 correspondence we match the point on the shorter line with the point on the longer. From any point on the longer blue line we can draw a straight line to P; the point on the longer line will be associated with the point we meet on the shorter. We match these two points. If the two points on the long line are different, the straight lines will be different so the points on the shorter line will be different. Thus each point on the longer line is matched with just one point of the shorter line. We can in fact do this for any point of either line. So any point of either line can be matched with any point of the other, and if the points are different on one line they are different on the other. We therefore have defined a one-to-one correspondence. **
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RESPONSE --> self critique assessment: 3
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רT֍qΥ[zz assignment #006 006. `Query 6 College Algebra 02-06-2008
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11:15:53 `questionNumber 60000 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> Inductive confidence assessment: 2
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11:16:01 `questionNumber 60000 ** The argument is inductive, because it attempts to argue from a pattern. **
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RESPONSE --> self critique assessment: 3
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11:17:33 `questionNumber 60000 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> Deductive confidence assessment: 2
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11:17:45 `questionNumber 60000 ** this argument is deductive--the conclusions follow inescapably from the premises. 'all men' is general; 'Socrates' is specific. This goes general to specific and is therefore deductive. COMMON ERROR: because it is based on a fact, or concrete evidence. Fact isn't the key; the key is logical inevitability. The argument could be 'all men are idiots, Socrates is an man, therefore Socrates is an idiot'. The argument is every bit as logical as before. The only test for correctness of an argument is that the conclusions follow from the premises. It's irrelevant to the logic whether the premises are in fact true. **
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RESPONSE --> self critique assessment: 3
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11:18:31 `questionNumber 60000 Query 1.1.20 1 / 3, 3 / 5, 5/7, ... Probable next element.
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RESPONSE --> 7/9 confidence assessment: 3
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11:18:48 `questionNumber 60000 **The numbers 1, 3, 5, 7 are odd numbers. We note that the numerators consist of the odd numbers, each in its turn. The denominator for any given fraction is the next odd number after the numerator. Since the last member listed is 5/7, with numerator 5, the next member will have numerator 7; its denominator will be the next odd number 9, and the fraction will be 7/9. There are other ways of seeing the pattern. We could see that we use every odd number in its turn, and that the numerator of one member is the denominator of the preceding member. Alternatively we might simply note that the numerator and denominator of the next member are always 2 greater than the numerator and denominator of the present member. **
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RESPONSE --> self critique assessment: 3
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11:20:01 `questionNumber 60000 Query 1.1.23 1, 8, 27, 64, ... Probable next element.
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RESPONSE --> 216 confidence assessment: 3
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11:20:29 `questionNumber 60000 ** This is the sequence of cubes. 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125. The next element is 6^3 = 216. Successive differences also work: 1 8 27 64 125 .. 216 7 19 37 61 .. 91 12 18 24 .. 30 6 6 .. 6 **
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RESPONSE --> self critique assessment: 3
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11:22:11 `questionNumber 60000 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> 11111 X 11111 = 1,234,554,321 confidence assessment: 3
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11:22:19 `questionNumber 60000 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> self critique assessment: 3
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11:23:46 `questionNumber 60000 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> I think it would always add another one. confidence assessment: 2
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11:24:05 `questionNumber 60000 ** You could think forward to the next few products: What happens after you get 12345678987654321? Is there any reason to expect that the sequence could continue in the same manner? The middle three digits in this example are 8, 9 and 8. The logical next step would have 9, 10, 9, but now you would have 9109 in the middle and the symmetry of the number would be destroyed. There is every reason to expect that the pattern would also be destroyed. **
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RESPONSE --> self critique assessment: 3
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11:26:44 `questionNumber 60000 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> confidence assessment: 0
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11:27:36 `questionNumber 60000 ** Pair up the first and last, second and second to last, etc.. You'll thus pair up 1 and 2000, 2 and 1999, 3 and 1998, etc.. Each pair of numbers totals 2001. Since there are 2000 numbers there are 1000 pairs. So the sum is 2001 * 1000 = 2,001,000 **
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RESPONSE --> self critique assessment: 3
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11:30:55 `questionNumber 60000 Query 1.1.57 142857 * 1, 2, 3, 4, 5, 6. What happens with 7? Give your solution to the problem as stated in the text.
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RESPONSE --> 1,42857 ; 285714 ; 428,571 ; 571,428 ; 714,285 ; 857,142 ; 999,999 They all contain the same numbers until you get to seven. confidence assessment: 2
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11:31:15 `questionNumber 60000 ** Multiplying we get 142857*1=142857 142857*2= 285714 142857*3= 428571 142857*4=571428 142857*5= 714285 142857*6=857142. Each of these results contains the same set of digits {1, 2, 4, 5, 7, 8} as the number 1428785. The digits just occur in different order in each product. We might expect that this pattern continues if we multiply by 7, but 142875*7=999999, which breaks the pattern. **
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RESPONSE --> self critique assessment: 3
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11:31:47 `questionNumber 60000 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> It doesn't always work confidence assessment: 3
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11:31:57 `questionNumber 60000 ** Inductive reasoning would have led us to expect that the pattern continues for multiplication by 7. Inductive reasoning is often correct it is not reliable. Apparent patterns can be broken. **
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RESPONSE --> self critique assessment: 3
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