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Serhud [2]
3 years ago
5

Two fractions are Equal.They also have the same donominator.What must be true of the numerators of the fractions?Explain

Mathematics
1 answer:
Kay [80]3 years ago
8 0
Well u cant have a fracion that is equal and have the same dinominator .

2/4=1/2 cause the donominator is times 2 and the numerator is time 2 so thats what makes them = cause they are both times 2
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The two coefficients are 2 and 32, and one factor common to both of these is 2.     Thus, <span>2x^2-32y^2 = 2(x^2-16y^2).

You could stop here.  But it'd make more sense to continue, since x^2-16y^2 is easy to factor:

The three factors of </span><span>2x^2-32y^2  are 2, x-4 and x+4</span>
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Find the least number by which 1470 must be divided to get a number which is the perfect square?
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Ty, who weighs 120 lbs, sits 5 ft to the left side on a seesaw. In order to balance the seesaw, Charley has to sit 6 ft to the r
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(a) Let R = {(a,b): a² + 3b &lt;= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
2 years ago
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