course Mth 151 ?????|m?????x?assignment #026026. `query 26
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20:55:44 5.4.12 What is [ (10+7) * (5+3) ] mod 10 and how did you get your result?
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RESPONSE --> x mod is the remainder when xis divided by 10 [ (10+7) * (5*3)] mod 10=(17*8) mod 10= 136/10 mod 10=6 136/10 r6 confidence assessment: 3
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20:57:11 query 5.4.20 2 / 3 on 5-hour clock What is 2 divided by 3 on a 5-hour clock, and how did you obtain this result?
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RESPONSE --> 3*4=2 2/3=4 confidence assessment: 2
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20:58:12 query 4.4.42 (3 - 27) mod 5 What is (3 - 27) mod 5, and how did you reason out your result?
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RESPONSE --> (3-27) mod 5=-24 mod 5 confidence assessment: 1
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21:02:24 query 5.4.20 Pos Integer solns (5x-3) = 7 (mod 4)
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RESPONSE --> x=1,2,3,4 5x-3 values: 2,7,12,17,22,27,32,37 when div. by 4 leaves a remainder of 2 3,0,1,2,3,0,1,2,3,0,1,2,3,0, x=2 2,6,10,14, x=2+4n n=0,1,2,3 confidence assessment: 1
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21:06:19 query 5.4.30 table for addition mod 7 and properties of operation
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RESPONSE --> 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 confidence assessment: 3
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21:09:42 ** The correct table is 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 For example the row across from 2 is obtained as follows: 2 * 0 = 0 and 2 * 1 = 2, as always. Then 2 * 2 mod 4 = 4 mod 4, which is 0 and 2 * 3 mod 4 = 6 mod 4, which is 2. the operation is closed because the results all come from the set {0, 1, 2, 3} being operated on 1 is the identity because the row and column for 1 both have 0,1,2,3 in that order, so 1 doesn't change a number when multiplied by that number. 0 and 2 lack inverses--they can't be combined with anything else to get 1--so the operation lacks the inverse property. symmetry about the main diagonal implies commutativity associativity follows from associativity of multiplication of real numbers**
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RESPONSE --> wrong button 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 self critique assessment: 3
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21:15:43 query 5.4.70 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc.
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RESPONSE --> 2002 + [[2002-1/4]] - [[2002-1/100]] + [[2002-1/400]] 2002 + [[500.20]] - [[20.01]] + [[5.0025]] =2002+500-20+5 =2487 2487 mod 7 is 2 monday=1 sunday = 0 tuesday = 2S??z????????assignment #027 027. `query 27 College Algebra 12-14-2007
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21:20:50 5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.
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RESPONSE --> golden ratio is [ 1+'sqrt(5)]/2 [1+'sqrt(5)]/2= [1+2.2361]/2=3.2361/2=1.618 confidence assessment: 1
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21:35:25 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 --> left 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 , right 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= f14 , next seq. is f 7^3 + f 6^3-f 5^3=f17 f 7=13 f 6=8 f 5=5 13^3+8^3-5^3=f 17 confidence assessment: 1
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21:44:59 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 --> part 1 p=3 f(p-1)=f(2)=1 & f(p+1)= f (4)=3 f(p+1)= f(4)=3 divided by p part 2 p=7 f(p-1)= f(6)=8 & f(p+1)= f(8)=21 f (p+1) =21 is divided by 7 part 3 p=11 f(p-1)=f(10)=55 & f(p+1)=f(12)=144 f(p+1)=55 is divided by 11 the conj. is true for all equationsp=3,p=7, p=11 confidence assessment: 2
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21:53:58 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 --> L2 + L4=3+7=10, L2 + L4 + L6= 3+7+18=28, L2 + L4 + L6 +L8= 3+7+18+47=75, L2 + L4 + L6 + L 8+ L10= 3+7+18+47+123=198 if you compare these answers to the luccas you can see they there is 1 number differnt between the numbers. ex: L9 is 198 which is one less. confidence assessment: 1
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confidence assessment: 2
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