course Mth 151 ???G???????w·????
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10:32:32 5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.
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RESPONSE --> (1 + squ root of 5) / 2 = (1 + 2.2361) / 2 = 3.2361 / 2 = 1.618 confidence assessment: 2
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10:33:05 ** The Golden Ratio is [ 1+`sqrt(5) ] /2 [ 1+`sqrt(5) ] /2=[ 1+2.2361 ] /2=3.2361/2=1.618 **
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RESPONSE --> I wasn't sure that that was what the question was asking. self critique assessment: 2
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10:39:17 5.5.12 2^3 + 1^3 - 1^3 = 8; 3^3 + 2^3 - 1^3 = 34; 5^3 + 3^3 - 2^3 = 144; 8^3 + 5^3 - 3^3 = 610. What are the next two equations in this sequence?
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RESPONSE --> 2^3+1^3-1^3=8 3^3+2^3-1^3=34 5^3+3^3-2^3=144 8^3+5^3-3^3=610 13^3+8^3-5^3=2584 21^3+13^3-8^3=10946 confidence assessment: 2
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10:43:14 ** The numbers 1, 1, 2, 3, 5, 8, ... are the Fibonacci numbers f(1), f(2), ... The left-hand sides are f(3)^3 + f(2)^3 - f(1)^3, f(4)^3 + f(3)^3 - f(2)^3, f(5)^3 + f(4)^3 - f(3)^3, f(6)^3 + f(5)^3 - f(4)^3 etc.. The right-hand sides are f(5) = 8, f(8) = 34, f(11) = 144, f(14) = 610. So the equations are f(3)^3 + f(2)^3 - f(1)^3 = f(5) f(4)^3 + f(3)^3 - f(2)^3 = f(8) f(5)^3 + f(4)^3 - f(3)^3 = f(11) f(6)^3 + f(5)^3 - f(4)^3 = f(14) etc.. The next equation would be f(7)^3 + f(6)^3 - f(5)^3 = f(17). Substituting f(7) = 13, f(6) = 8 and f(5) = 5 we get 13^3 + 8^3 - 5^3 = f(17). The left-hand side gives us result 2584, which is indeed f(17), so the pattern is verified in this instance. **
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RESPONSE --> I got the answer but I arrived at it in differently. I looked and saw a pattern and continued the pattern. self critique assessment: 2
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10:49:17 5.5.18 show whether F(p+1) or F(p-1) is divisible by p. Give your solution to this problem.
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RESPONSE --> F(3+1)/3 =4/3 = 1 remainding 1 F(3-1)/3= 2/3 F(7+1)/7 = 8/7 = 1remaining 1 F(7-1)/7= 6/7 F(11+1)=12/11 - 1 remaining 1 F(11-1)= 10/11 confidence assessment: 2
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11:14:06 ** For p=3 we get f(p-1) = f(2) = 1 and f(p+1) = f(4)= 3; f(p+1) = f(4) = 3 is divisible by p, which is 3 So the statement is true for p = 3. For p=7 we get f(p-1) = f(6) = 8 and f(p+1) = f(8) = 21; f(p+1) = 21 is divisible by p = 7. So the statement is true for p = 7. For p = 11 we get f(p-1) = f(10)= 55 and f(p+1) = f(12) = 144. f(p-1) = 55 is divisible by p = 11. So the statement is true for p = 11. So the conjecture is true for p=3, p=7 and p=11.**
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RESPONSE --> I didn't understand how to work it but after looking at how you worked it and in the book, I see that f(2) is in the 2nd place in the Fibonacci sequence which is 1, f(4) is the 4th place which is 3. Now I understand self critique assessment: 2
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11:24:20 5.5.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated in your text.
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RESPONSE --> 3+7=10 3+7+18=28 3+7+18+47=75 3+7+18+47+123=198 1st and 3rd are divisible by 5 and 2nd & 3rd are divsible by 2. confidence assessment: 2
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11:25:13 ** The Lucas sequence is 1 3 4 7 11 18 29 47 76 123 199 etc. So L2 + L4 = 3 + 7 = 10; L2 + L4 + L6 = 3 + 7 + 18 = 28; L2 + L4 + L6 + L8 = 3 + 7 + 18 + 47 = 75, and L2 + L4 + L6 + L8 + L10 = 3 + 7 + 18 + 47 + 123 = 198. Note that 10 is 1 less than 11, which is L5; 27 is 1 less than 28, which is L7; and 198 is 1 less than 199, which is L9. So L2 + L4 = L5 - 1, L2 + L4 + L6 = L7 - 1, etc.. So we can conjecture that the sum of a series of all evenly indexed members of the Lucas sequence, starting with index 2 and ending with index 2n, is 1 less than member 2n+1 of the sequence. **
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RESPONSE --> I didn't see that but I see it now that you have pointed it out. self critique assessment: 2
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