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Darina [25.2K]
3 years ago
5

Determine if the relation below is a function and state the domain and range.

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Answer:

Please check the explanation.

Step-by-step explanation:

Given the relation

{(1, -2), (-2, 0), (-1, 2), (1, 3)}

Please note that there is a repetition of x-values.

i.e. x=1 appears twice times.

This relation does not define a function because every x-value should have been unique and is associated with only one value of y.

Here, the x-values are repeated. Hence, this relation is not a function.

Finding the Domain of a relation:

We know that the domain is the complete set of all first elements of ordered pairs often called x-coordinates.

Domain of a relation: {-1, -2, 1}

<u>Note: Domain is in ascending order and repeated elements are written only once.</u>

Finding the Range of a relation:

We know that the domain is the complete set of all second elements of ordered pairs often called y-coordinates.

Range of a relation: {-2, 2, 3}

<u>Note: Range is in ascending order and repeated elements are written only once.</u>

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3 years ago
A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base
Olenka [21]

Answer:

The base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = x\times\frac{1}{2} =\frac{x}{2} \ m.

The height of triangle = x\times \frac{3}{4} =\frac{3x}{4}\ m.

The area of triangle = 16\ m^2.

Now, we put the formula of area to solve:

Area=\frac{1}{2} \times base\times height

16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}

16=\frac{3x^2}{16}

<em>Multiplying both sides by 16 we get:</em>

<em />256=3x^2<em />

<em>Dividing both sides by 3 we get:</em>

<em />\frac{256}{3} =x^2<em />

<em>Using square root on both sides we get:</em>

\frac{16}{\sqrt{3}}=x

x=\frac{16}{\sqrt{3}}

Now, by substituting the value of x to get the base and height:

Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.

<em>So, the base of triangle = </em>\frac{8}{\sqrt{3}} \ m.<em />

Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.

<em>Thus, the height of triangle =  </em>\frac{12}{\sqrt{3}} \ m.<em />

Therefore, the base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

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4 years ago
What are the values of X for which the denominator is equal to zero for y=x^2+3/x^2+2​
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Answer:

16=x

Step-by-step explanation:

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