It depends on the precise nature of the coder and there are many possibilities. Here are three:
A: f(x) = 2x
B: f(x) = 4*(x - 1)
C: f(x) = 2*(x2 - 3x + 4)
The simplest of these is f(x) = 2x.
In that case the decoder is y = ln(x)/ln(2) where ln is the natural logarithm. Logs to base 10, or any other base can also be used and, if the chosen base is 2, then the decoder becomes:
y = log2(x)
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4 and 2/3 - 2/3 = 4 4/8 + 3 and 2/8 is the same as 3 + 2/4 +1/4 which is 3 and 3/4 4 > 3 + 3/4 (aka, yes)
-2 3/4 - 1/4 = -(2 3/4 + 1/4) = -[2 (3+1)/4] = -(2 4/4) = -(2 + 1) = -3 or -2 3/4 - 1/4 = - (4*2 + 3)/4 - 1/4 = -11/4 - 1/4 = (-11 - 1)/4 = -12/4 = -3
There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .
((3/3)/4)*2 = (1/4)*2 = 2/4 = 1/2.
2/3 x 4= 2/3 x 4/1 = 8/3 = 2 2/3 Therefore 2/3 of 4 ft. = 2 2/3