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3 years ago

Find any values of the variable for which each rational expression is undefined. Write answers with the symbol ≠.

Mathematics

1 answer:

astra-53 [7]3 years ago

4 0

Answer:

x $\neq$ 3

Step-by-step explanation:

if x = 3 then the rational expression 4 / (x - 3) is undefined

because we get 4/0 = undefined .

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For for every ten sheets of stickers you buy at a craft store the total cost increases $20.50

Len [333]

The equation which relates the change in price per sticker purchased is y = 2.05x

The number of stickers purchased = 10

Price change = $20.50

The change in price per sticker purchased can be expressed as :

Price change / number of stickers

Change in price per sticker = $20.50 / 10

Change in price per sticker purchased = 2.05

Therefore, the required equation which related y and x is :

y = 2.05x

Learn more :brainly.com/question/13218948

3 0

2 years ago

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .

3 0

3 years ago

In how many ways can the letters of the word "CHINA" be arranged?

juin [17]

It is 120 ways, surprisingly

4 0

3 years ago

HELLLP MEEEEEEEE PLEASE

jolli1 [7]

Answer:

Each side is 15 ft

Step-by-step explanation:

The area of a square is given by

A = s^2 where s is the side length

225 ft^2 = s^2

Take the square root of each side

sqrt ( 225 ft^2) = sqrt ( s^2)

15 ft = s

Each side is 15 ft

6 0

2 years ago

Read 2 more answers

PLEASE HELP! Triangle inequality!! WILL GIVE BRAINLIEST

Nitella [24]

A is possible because the two other sides combined are longer then six therefore can reach each other to create a triangle. b is not possible because 4 is the same length of 3 and 1 combined therefore will not create a triangle shape. hope this is what u where looking for(:

6 0

2 years ago