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olga2289 [7]
3 years ago
13

Find the domain and range of the relation. {(−8,−4),(15,14),(−2,15),(14,−11),(19,−19)}

Mathematics
2 answers:
slega [8]3 years ago
7 0

Domain = {-8,15,-2,14,19}

Range = {-4,14,15,-11,-19}

Step-by-step explanation:

When a function is given in the form of a relation containing ordered pairs

The first element of each ordered pair is the member of the domain while the second element of each ordered pair is the member of the range.

Given relation is:

{(−8,−4),(15,14),(−2,15),(14,−11),(19,−19)}

We have to combine the first elements of each ordered pair to make the domain set and second element of each ordered pair to make the range set.

So,

Domain = {-8,15,-2,14,19}

Range = {-4,14,15,-11,-19}

Keywords: sets, domain, range

Learn more about sets at:

  • brainly.com/question/4168505
  • brainly.com/question/4228574

#LearnwithBrainly

Sladkaya [172]3 years ago
4 0

disregard my answer.

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frutty [35]

Answer: 3

x

−

2

y

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15

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Explanation:

We know that,

the slope of the line  

a

x

+

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+

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is  

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∴

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The slope of the line perpendicular to  

2

x

+

3

y

=

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m

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=

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1

−

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3

=

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2

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Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

x

1

)3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

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m

is

y

−

y

1

=

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(

x

−

Explanation:

the equation of a line in  

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is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

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3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

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=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular lineExplanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular line

7 0
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OlgaM077 [116]
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