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.

You might be interested in

Simplify 4(x-y)-8(-7x+6y)

aliya0001 [1]

The answer is 60x-52y

Hope that helped!

8 0

3 years ago

Read 2 more answers

Which expression is it equivalent to?

horrorfan [7]

Option A) Is the answer. $\boxed{\mathbf{\dfrac{3f^3}{g^2}}}$

For this question; You are needed to expose yourselves to popular usages of radical rules. In this we distribute the squares as one-and-a-half fractions as the squares eliminate the square roots. So, as per the use of fraction conversion from roots. It becomes relatively easy to solve and finish the whole process more quicker than everyone else. More easier to remember.

Starting this with the equation editor interpreter for mathematical expressions, LaTeX. Use of different radical rules will be mentioned in between the steps.

Radical equation provided in this query.

$\mathbf{\sqrt{\dfrac{900f^6}{100g^4}}}$

Divide the numbered values of 900 and 100 by cancelling the zeroes to get "9" as the final product in the next step.

$\mathbf{\sqrt{\dfrac{9f^6}{g^4}}}$

Imply and demonstrate the rule of radicals. In this context we will use the radical rule for fractions in which a fraction with a denominator of variable "a" representing a number or a variable, and the denominator of variable "b" representing a number or a variable are square rooted by a value of "n" where it can be a number, variable, etc. Here, the radical of "n" is distributed into the denominator as well as the numerator. Presuming the value of variable "a" and "b" to be greater than or equal to the value of zero. So, by mathematical expression it becomes:

$\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}, \: \: a \geq 0 \: \: \: b \geq 0}}$

$\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{\sqrt{g^4}}}$

Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

$\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}$

$\mathbf{Since, \quad \sqrt{g^4} = g^{\frac{4}{2}}}$

$\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{g^2}}$

Exhibit the radical rule for two given variables in this current step to separate the variable values into two new squares of variables "a" and "b" with a radical value of "n". Variables "a" and "b" being greater than or equal to zero.

$\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}, \: \: a \geq 0 \: \: \: b \geq 0}}$

So, the square roots are separated into root of 9 and a root of variable of "f" raised to the value of "6".

$\mathbf{\therefore \quad \dfrac{\sqrt{9} \sqrt{f^6}}{g^2}}$

Just factor out the value of "3" as 3 × 3 and join them to a raised exponent as they are having are similar Base of "3", hence, powered to a value of "2".

$\mathbf{\therefore \quad \dfrac{\sqrt{3^2} \sqrt{f^6}}{g^2}}$

The radical value of square root is similar to that of the exponent variable term inside the rooted enclosement. That is, similar exponential values. We apply the following radical rule for these cases for a radical value of variable "n" and an exponential value of "n" with a variable that is powered to it.

$\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^n} = a^{\frac{n}{n}} = a}}$

$\mathbf{\therefore \quad \dfrac{3 \sqrt{f^6}}{g^2}}$

Again, Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

$\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}$

$\mathbf{Since, \quad \sqrt{f^6} = f^{\frac{6}{2}} = f^3}$

$\boxed{\mathbf{\underline{\therefore \quad Required \: \: Answer: \dfrac{3f^3}{g^2}}}}$

Hope it helps.

8 0

3 years ago

solo el 20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal. Si

prisoha [69]

Usando la distribución binomial, hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

<h3>¿Qué es la distribución binomial?</h3>

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$