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Ipatiy [6.2K]
3 years ago
5

HelpPlease I don’t understand

Mathematics
1 answer:
Simora [160]3 years ago
6 0
That is certainly confusing, what type of math is that?
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Mrs. King has a stack of papers to grade that contains 12 boy’s papers and 8 girl’s papers. What is the probability that the fir
Artist 52 [7]

Answer:3/5

Step-by-step explanation:

20 papers in total

12 boy papers

12/20

simplified to 3/5

7 0
3 years ago
Read 2 more answers
(a) Let R = {(a,b): a² + 3b <= 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
Given the following system of equations: 2x + 6y = 12 4x + 3y = 15 Which action creates an equivalent system that will eliminate
ozzi

Answer:

Multiply the second equation by −2 to get −8x − 6y = −30.

Step-by-step explanation:

{2x + 6y = 12

{4x + 3y = 15

{2x + 6y = 12

{−8x − 6y = −30 >> New Equation

* Doing this will give you <em>additive</em><em> </em><em>inverses</em><em> </em>of −6y and 6y, which result in 0, so they are both ELIMINATED.

** [3, 1] is your solution.

I am joyous to assist you anytime.

6 0
3 years ago
Read 2 more answers
How do you draw the unit circle in radians when the denominator is 5?
pochemuha
The radian is with a denominator of 5 is 2.5
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3 years ago
Question 16<br> 5 pts<br> Convert the Improper Fraction to a Mixed Number.<br> 45/12
AlekseyPX

Answer: 3 3/4

Step-by-step explanation:

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2 years ago
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