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Gemiola [76]
2 years ago
13

Is using the associative property of addition a good strategy for simplifying (4/3 + 2/3) + 2? Why or why not?

Mathematics
2 answers:
Scilla [17]2 years ago
7 0

Answer:

Easy way

Step-by-step explanation:

(4/3 + 2/3) + 2

= 2 + 2

=4

Oviously (4/3 + 2/3) is directly 2.

Kitty [74]2 years ago
3 0

Answer:

I say yes because when u solve it and flip the ( )'s, you get the same answer.

Step-by-step explanation:

Work:

(4/3 + 2/3) + 2

6/3 + 2

2 + 2

= 4

4/3 + (2/3 + 2)

4/3 + 8/3

12/3

= 4

Hope this helps!!

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Rewrite <img src="https://tex.z-dn.net/?f=%5Cfrac%7B200x%20-%20300%7D%7Bx%7D" id="TexFormula1" title="\frac{200x - 300}{x}" alt=
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Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.
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Step-by-step explanation:

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R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

R is a partial ordering, because R is reflexive, antisymmetric and transitive.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).  

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R is not a partial ordering. because R is not transitive .

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

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R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

R is not transitive, because ( 1 , 2 )∈R and ( 2 , 0 )∈R, while ( 1 . 0 )∉R.

R is not a partial ordering, because R is not transitive,

e):  R = { ( 0 , 0 ) , ( 0, 1 ) , ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 0 ) , ( 2 , 2 ) , ( 3 , 3 ) }

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R is not antisymmetric, because ( 1 , 0 )∈R and ( 0 , 1 )∈R while 0 is not equal to 1.

R is not transitive, because ( 2 , 0 )∈Rand ( 0 , 3 )∈R, while ( 2 , 3 )∉R .

R is not a partial ordering, because R is not the antisymmetric and not the transitive.

3 0
3 years ago
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