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Gelneren [198K]
2 years ago
5

Determine whether each relation is a function. Give the domain and range for each relation.

Mathematics
1 answer:
brilliants [131]2 years ago
3 0

Answer:

Not a function

Domain: {3,4}

Range: {4,5}

Step-by-step explanation:

A function is a relation where each input has its own output. In other words if the x value has multiple corresponding y values then the relation is not a function

For the relation given {(3, 4), (3, 5), (4, 4), (4, 5)} the x value 3 and 4 have more than one corresponding y value therefore the relation shown is not a function

Now let's find the domain and range.

Domain is the set of x values in a relation.

The x values of the given relation are 3 and 4 so the domain is {3,4}

The range is the set of y values in a relation

The y value of the given relation include 4 and 5

So the range would be {4,5}

Notes:

The values of x and y should be written from least to greatest when writing them out as domain and range.

They should be written inside of brackets

Do not repeat numbers when writing the domain and range

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3 years ago
a total of 695 tickets were sold for the school play. They were either adult or student tickets. There was 55 fewer students tic
FinnZ [79.3K]

Answer:

375 + 320

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4 0
2 years ago
Marla bought 12 books at a garage sale. Some of them were hardback and the rest were paperback. She paid $0.50 for each paperbac
olga nikolaevna [1]

Answer: The number of books bought were 9 paperbacks and 3 hardbacks

Step-by-step explanation: We shall start by assigning letters to the unknown variables, hence let the hardback be called h, while the paperback shall be called p.

If Marla bought 12 books at the garage sale, that means she bought

h + p  = 12 ------(1)

Then she paid 0.5 dollars for paperback and 0.75 dollars for hardback and the total spent was 6.75 dollars for all of them, then we can express these as follows;

0.5p + 0.75h = 6.75 ------(2)

We now have a pair of simultaneous equations which are

h + p = 12 ------(1)

0.5p + 0.75h = 6.75 ------(2)

From equation (1), make h the subject of the equation,

h = 12- p

Substitute for h into equation (2)

0.5p + 0.75(12 - p) = 6.75

0.5p + 9 - 0.75p = 6.75

0.5p - 0.75p = 6.75 - 9

-0.25p = -2.25

Divide both sides of the equation by -0.25

p = 9

Now, substitute for p into equation (1)

h + p = 12

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Subtract 9 from both sides of the equation

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Therefore Marla bought 9 paperbacks and 3 hardbacks

6 0
3 years ago
Read 2 more answers
An equation parallel and perpendicular to 4x+5y=19
UNO [17]

Answer:

Parallel line:

y=-\frac{4}{5}x+\frac{9}{5}

Perpendicular line:

y=\frac{5}{4}x-\frac{1}{2}

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

4x+5y=19

we can change it into y=mx+b form

5y=-4x+19

y=-\frac{4}{5}x+\frac{19}{5}

so,

m=-\frac{4}{5}

Parallel line:

we know that slope of two parallel lines are always same

so,

m'=-\frac{4}{5}

Let's assume parallel line passes through (1,1)

now, we can find equation of line

y-y_1=m'(x-x_1)

we can plug values

y-1=-\frac{4}{5}(x-1)

now, we can solve for y

y=-\frac{4}{5}x+\frac{9}{5}

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

m'=\frac{5}{4}

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

y-y_1=m'(x-x_1)

we can plug values

y-2=\frac{5}{4}(x-2)

now, we can solve for y

y=\frac{5}{4}x-\frac{1}{2}


4 0
3 years ago
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