They are in alphabetical order in their word forms:
eight
five
four
.
.
three
two
zero
21000 = (2 x 10000) + (1 x 1000) + (0 x 100) + (0 x 10) + (0 x 1) OR (2 x 10^4) + (1 x 10^3) + (0 x 10^2) + (0 x 10^1) + (0 x 10^0)
100,203 in expanded form is (1 x 100000) + (0 x 10000) + (0 x 1000) + (2 x 100) + (0 x 10) + (3 x 1)(1x1,000)+(2x3) or (1 x 10^5) + (0 x 10^4) + (0 x 10^3) + (2 x 10^2) + (0 x 10^1) + (3 x 10^0)100000 + 200 + 3
10.205 = (1 x 10) + (0 x 1) + (2/10) + (0/100) + (5/1000)
0 . . . . . 0 0 0 0 1 . . . . . 0 0 0 1 2 . . . . . 0 0 1 0 3 . . . . . 0 0 1 1 4 . . . . . 0 1 0 0 5 . . . . . 0 1 0 1 6 . . . . . 0 1 1 0 7 . . . . . 0 1 1 1 8 . . . . . 1 0 0 0 9 . . . . . 1 0 0 1 10 . . . . 1 0 1 0
7x10^3 + 4x10^2 + 3x10^1 + 9x10^0
0 = 0/10, 1/2 = 5/10 and 1 = 10/10 so 6/10 is nearest to 5/10 or 1/2.
Pattern: +7,-4,+8,-3,+9,-2... Pattern continuation: +10,-1,+11,*0,+12,+1 17+10=27
100 = (1 x 10^2) + (0 x 10^1) + (0 x 10^0) or 1 x 10^2
[(4 * 10^2) + (0 * 10^1) + (0 * 10^0)] + [(1 * 10^1) + (0 * 10^0)]
21000 = (2 x 10000) + (1 x 1000) + (0 x 100) + (0 x 10) + (0 x 1) OR (2 x 10^4) + (1 x 10^3) + (0 x 10^2) + (0 x 10^1) + (0 x 10^0)
(73)10 = (1 0 0 1 0 0 1)2
In the decimal system, the places are 10^0 (1), 10^1 (10), 10^2 (100), and 10^3 (1000) In the binary system, the places are 2^0 (1), 2^1 (2), 2^2 (4), and 2^3 (8) 1101 has one 1, one 4 and one 8. That makes 13, base ten.
100,203 in expanded form is (1 x 100000) + (0 x 10000) + (0 x 1000) + (2 x 100) + (0 x 10) + (3 x 1)(1x1,000)+(2x3) or (1 x 10^5) + (0 x 10^4) + (0 x 10^3) + (2 x 10^2) + (0 x 10^1) + (3 x 10^0)100000 + 200 + 3
The vector representation of a convolutional code is g1=[1 1] and g2=[1 0]. If the received sequence is 11 11 01 10, we can use Viterbi Algorithm to decode it. The full trellis diagram, including the updated trellis state metrics, is shown below: Time t1: State 0 (00): Metric = 0 State 1 (10): Metric = 0 Time t2: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 0 + (1-1)^2 + (1-0)^2 = 1 Time t3: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 1 + (1-1)^2 + (1-0)^2 = 2 State 0 (01): Metric = 0 + (1-1)^2 + (0-1)^2 = 1 State 1 (11): Metric = 0 + (1-1)^2 + (0-0)^2 = 0 Time t4: State 0 (00): Metric = 0 + (1-1)^2 + (1-1)^2 = 0 State 1 (10): Metric = 2 + (1-1)^2 + (1-0)^2 = 3 State 0 (01): Metric = 1 + (1-1)^2 + (0-1)^2 = 2 State 1 (11): Metric = 0 + (1-1)^2 + (0-0)^2 = 0 Therefore, the decoded sequence is 11 11 01 10.
Normally 1 - 1 = 0 the binary number for 1 is 1 the binary number for 2 is 10 the binary number for 3 is 11 3 - 2 = 1 The binary form of that equation is 11 - 10 = 1 The binary inverse operation would be 1 + 10 = 11 The rest is binary math 11 + 10 = 101 10 + 10 = 100 101 - 1 = 100 100 - 1 = 11 11 - 1 = 10 10 - 1 = 1 1 - 1 = 0 Therefore according to the pattern being displayed, the binary code for zero is 0.
10.205 = (1 x 10) + (0 x 1) + (2/10) + (0/100) + (5/1000)
There are 12 ways to make change for a 50 dollar bill using 5, 10's and 20's. $20's $10's $5's 2 1 0 2 0 2 1 3 0 1 2 2 1 1 4 1 0 6 0 5 0 0 4 2 0 3 4 0 2 6 0 1 8 0 0 10