course Mth 152 ÃûÌÔü‚þâ¡ÑpÌÂÏß±²xÃx¢assignment #004
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11:01:07 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9,6) = 84
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11:02:26 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> o.k
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11:13:07 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> C(9,4) = 126 possible ways to get combinations of 4 classrooms. There would have to be at least 8 possible combinations of 4 to get it wrong or C(8,4) If (9,4) could get it right and (8,4) could get it wrong subtract the wrong from the right and get 126 - 70 = 56 would get it right and 44 wrong.
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11:13:15 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> o.k.
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11:13:56 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> I don't understand what the question is asking. Sequence of what?
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11:14:43 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> O.K. As I said before I didn't understand the question.
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11:20:29 Query 11.4.42 (x+y)^8
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RESPONSE --> (x+y)^8= x^8 + y^8
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11:24:44 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE --> I don't understand this one at all. I tried looking back through the text but with no luck. Why is it worked this way.
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11:25:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The last two questions were hard for me. But I will go back and go over the processes to arrive at the given answers.
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11:28:01 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> Will have to review all of section 11-4
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ÃûÌÔü‚þâ¡ÑpÌÂÏß±²xÃx¢ assignment #004 004. `query 4 Liberal Arts Mathematics II 09-14-2008
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11:01:07 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9,6) = 84
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11:02:26 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> o.k
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11:13:07 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> C(9,4) = 126 possible ways to get combinations of 4 classrooms. There would have to be at least 8 possible combinations of 4 to get it wrong or C(8,4) If (9,4) could get it right and (8,4) could get it wrong subtract the wrong from the right and get 126 - 70 = 56 would get it right and 44 wrong.
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11:13:15 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> o.k.
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11:13:56 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> I don't understand what the question is asking. Sequence of what?
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11:14:43 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> O.K. As I said before I didn't understand the question.
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11:20:29 Query 11.4.42 (x+y)^8
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RESPONSE --> (x+y)^8= x^8 + y^8
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11:24:44 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE --> I don't understand this one at all. I tried looking back through the text but with no luck. Why is it worked this way.
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11:25:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The last two questions were hard for me. But I will go back and go over the processes to arrive at the given answers.
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11:28:01 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> Will have to review all of section 11-4
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ÃûÌÔü‚þâ¡ÑpÌÂÏß±²xÃx¢ assignment #004 004. `query 4 Liberal Arts Mathematics II 09-14-2008
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11:01:07 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C(9,6) = 84
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11:02:26 ** You need to go to the n=9 row, the r=6 position, which is the 10th row and the 7th number in the row. Looks like you went to the 6th position. Note that C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. Note also that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left. **
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RESPONSE --> o.k
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11:13:07 Query 11.4.18 clueless check of four of nine possible classroomsHow many of the possible selections will fail to locate the classroom?
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RESPONSE --> C(9,4) = 126 possible ways to get combinations of 4 classrooms. There would have to be at least 8 possible combinations of 4 to get it wrong or C(8,4) If (9,4) could get it right and (8,4) could get it wrong subtract the wrong from the right and get 126 - 70 = 56 would get it right and 44 wrong.
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11:13:15 ** There are C(9,4) possible combinations of four classrooms. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. Note that the chance of ending up in the right classroom is 56 / 126, a little less than 50-50, with 56 of the 126 possible ways being successful. **
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RESPONSE --> o.k.
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11:13:56 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> I don't understand what the question is asking. Sequence of what?
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11:14:43 ** The numbers 1, 1, 2, 3, 5, 8, 13, 21, ... form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The sums of the diagonals are all Fibonacci numbers. **
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RESPONSE --> O.K. As I said before I didn't understand the question.
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11:20:29 Query 11.4.42 (x+y)^8
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RESPONSE --> (x+y)^8= x^8 + y^8
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11:24:44 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. **
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RESPONSE --> I don't understand this one at all. I tried looking back through the text but with no luck. Why is it worked this way.
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11:25:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The last two questions were hard for me. But I will go back and go over the processes to arrive at the given answers.
.................................................
......!!!!!!!!...................................
11:28:01 ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. **
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RESPONSE --> Will have to review all of section 11-4
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