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|
C B A C B A C C C B B B A B c
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)
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
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
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
#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; }
1. Commutative - a * b = b * a 2. Associative - (a * b) * c = a * (b * c) 3. Distributive - a * (b + c) = (a * b) + (a * c)
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