It was chaos.
Firstly, all the currency had to be changed from BC to AD.
Then the calendars needed to be changed from counting down to counting up.
Not to mention the "Year 0k" bug.
(b b b)( b b b )(b d g a)(b....)(c c c c)(c b b b)(a a a b)(a...d)(b b b)(b b b)(b d g a)(b....)(c c c c)(c b b b)(d d c a)(g.....)
a b c c c c b a g g a b g a b c c c c b a g b a a b c c c c b a g g a b a b c d b c e c b a b a g g
B b b b b b b d g a b c c c c c b b b a a b a d b b b b b b b d g a b c c c c c b b b d d c a g
PLEASE NOTE ~ |= MEASURE SEPARATION ALL OF THE Ds ARE HIGH D AND OPEN D WILL BE WRITTEN IN ITALICS ( D )4/4 B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D|B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G| D B A G D | D B A G E | E C# B A F | D D C# A B| D B A G D | D B A G E | E C# B A D D D D | E-(high) D C# A G D| B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D| B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G|
A string; D (3) C# (2) B (1) rest (crochets) D (3) C# (2) B (1) rest (crochets) B (1) B (1) B (1) B (1) C# (2) C# (2) C# (2) C# (2) (quavers) D (3) C# (2) B (1) rest (crochets) D- note (2) - finger Music is set out in 4/4 time. 1 bar per line.
Y=C+I C=C°+bY I=I° Y=C°+bY+I° Y-bY=C°+I° Y(1-b)=C°+I° Y=(C°+I°)/(1-b) Y+ΔY = (C°+I°+ΔI;ΔC)/(1-b) Y+ΔY = (C°+I°)/(1-b) + ΔI;ΔC/(1-b) = Y + ΔI;ΔC/(1-b) ΔY=ΔI;ΔC/(1-b) ΔY/ΔI;ΔC=1/(1-b) ΔY/ΔI=1/(1-b) ΔY/ΔC=1/(1-b)
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
The properties of addition are: * communicative: a + b = b + a * associative: a + b + c = (a + b) + c = a + (b + c) * additive identity: a + 0 = a * additive inverse: a + -a = 0 The properties of multiplication: * communicative: a × b = b × a * associative: a × b × c = (a × b) × c = a × (b × c) * distributive: a × (b + c) = a × b + a × c * multiplicative identity: a × 1 = a * multiplicative inverse: a × a^-1 = 1
Multiplicative identitya*1 = aReciprocalitya * b = 1then a and b are reciprocals: a = 1/b and b = 1/aAssociativitya * (b * c) = (a * b) * cCommutativitya * b = b * aDistributivitya * (b + c) = a*b + a*c
use this strategy: integral of (b^x) dx = (b^x)/ln(b) + K [K is integration constant, b is not a variable]rewrite (1/c)^(1-x) = ((1/c)^1)*((1/c)^(-x)) = (1/c)*(c^x). (1/c) is a constant, so bring outside the integral, then let b = c in the formula above, and you have (1/c)*(c^x)/ln(c) + K
1. Commutative - a * b = b * a 2. Associative - (a * b) * c = a * (b * c) 3. Distributive - a * (b + c) = (a * b) + (a * c)
Well, isn't that just a happy little math problem! If A is less than B and B plus C equals 10, then it must be true that A plus C is less than 10. Just remember, in the world of numbers, everything adds up beautifully in the end.
a,b,c,d
b d d b b c b b a b a c c d d b c b a d c a b c d a b c c a a d b d d b a a d b c a c d d c b b a
Presumably you mean is it true that: ( A nor B ) nor C == A nor ( B nor C ) ? No. Let's make a table: A B C (A nor B) (B nor C) [ (A nor B ) nor C ] [ A nor ( B nor C ) ] 0 0 0 1 1 0 0 0 0 1 1 0 0 1 .... So you see right away for A=0, B=0, and C=1 it doesn't work.
Yes because A > B, B > C, so A has to be > C.ExampleA=5B=3C=1A (5) > B (3)B (3) > C (1)A (5) > C (1)
A B C + 1 2 3= 357