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Artyom0805 [142]
3 years ago
11

State the domain and range of the relation. Determine whether the relation represents a function.

Mathematics
1 answer:
dangina [55]3 years ago
4 0

Answer:

Domain {-2,0,2}

Range {-2,0,2}

Relation is a Function

Step-by-step explanation:

We are given a relation:

{ (-2,-2) , (0,0) , (2,2) }

Domain can be defined as the all possible values of x for a relation. It is considered as a set of all first values of the ordered pairs of a given relation.

Domain of the given relation is {-2,0,2}

Range can be defined as all possible value of y which corresponds to the values of x in the domain. It is considered as a set of all second values of the ordered pairs of a given relation.

Range of the given relation is {-2,0,2}

A relation is a function if only there is one value of y for each value of x. If in the set of ordered pair of the relation, the value of x gets repeated, then the relation is not a function.

As no values of x are getting repeated, the relation is a function.

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The answer would be the sequence has a common ratio of 4.
The explanation for this would be:
To get the common ratio for get the simplest form of the two numbers, so 12/3 = 4 and 48/12 = 4So we know that their common ratio is 4.
To check:
3 x 4 = 12
12 x 4 = 48
48 x 4 = 192
192 x 4 = 768
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A student was conducting a study to determine how many pagos he would need for the book he is writing. So, he found that the fol
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224 pages

Step-by-step explanation:

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If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

5 0
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Select the correct answer.<br> What is the solution to this equation <br><br> (1/4)x+1=32
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Step-by-step explanation:

6 0
1 year ago
4(3x^2y^4)^3 / (2x^3y^5)^4
tatyana61 [14]

Solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Step-by-step explanation:

We need to Solve the expression: \frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

Solving:

\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}

Applying exponent rule: \frac{x^a}{x^b}=x^{a-b}

=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}

Another exponent rule says: x^{-a}=\frac{1}{x^a}=

=\frac{27}{4x^{6}y^{8}}

So, solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Keywords: Solving Exponents  

Learn more about Solving Exponents at:

  • brainly.com/question/13174260
  • brainly.com/question/13174254
  • brainly.com/question/13174259

#learnwithBrainly

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