course Mth 152 Mr. Smith,I am enclosing my Math 152, Lesson 11.4 homework answers. I had some troubles with a few of these. As you will notice I had a question or two inside the homework as well.
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19:40:55 Query 11.4.6 Find C(9,6) on Pascal's triangle.
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RESPONSE --> C (9,4) is equal to 126 according to Pascal's triangle (9th row down, 4th number over (past the 1))
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19:42:05 ** 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 --> Okay, my mistake there... I was reading it as it was out of the book. My particular text book shows C (9,4) and I was going over my homework area. I also need to practice more on the multiplication.
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19:45:04 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 --> This is a selection 4 possible classes out of 9. The answer is 126 different combinations possible.
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19:45:31 ** 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 --> Okay
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19:49:35 Query 11.4.30 What sequence by totaling diagonals of Pascal's Triangle?
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RESPONSE --> The next 5 numbers (by the standards of the problem in my book), is 15, 21, 28, 36, and 45. I could not remember the exact name of the sequence but I remember its principles. The sum is added to the digit to get the next.
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19:51:03 ** 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 --> Okay, I've copied that to a word file for future studying. I was confused about this problem. Question for class: Where, beyond our text books, does the Fibonacci Sequence pop up? (it says that it seems to occur in all sorts of unexpected places, so I am curious.)
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19:57:06 Query 11.4.42 (x+y)^8
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RESPONSE --> The Binomial Theorem expansion for (x+y)^8 is: (x+y)^8= x8+8x7y+28x6y2+56x5y3+70x4y4+56x3y5+28x2y6+ 8xy7+y8
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19:58:13 **(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 --> Okay, I see what I need to add... the ^. I should also critque myself more regarding positioning on the Pascal's Triangle. Question: On tests, will we actually get to see the Pascal Triangle or do we need to memorize its structure?
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19:59:12 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I learned alot from this section but I feel that I need to go over it again a few times. Pascal's Triangle is interesting but difficult for me to master. I was doing well with the first to parts of this chapter 11.1 and 11.2, but I feel that I need to really study 11.3 and 11.4 before I am ready for my test.
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19:59:30 ** 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 --> Okay
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