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3 years ago

Alexis reads 125 pages or about 45% of her book. How many total pages are in the book? Please explain

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

3 0

278. I divided 125 by .45 and it gave me roughly 278 :)

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The sales tax in your state is 5% if a car costs $24,600 what is the total cost including tax?

Svetach [21]

<h2>Greetings!</h2>

Answer:

$25830

Step-by-step explanation:

If 100% of the car price is 24,600 , then you can imagine that extra 5% being added to this

100 + 5 = 105

So you need to find what 105% of the cost was.

To do this you can use the percentage formula:

Amount x $\frac{percentage}{100}$

Now, you can simply plug the values into this

24,600 x $\frac{105}{100}$

24600 x 1.05 = 25830

<h3>So the total cost including tax would be $25830!</h3><h3 /><h2>Hope this helps!</h2>

7 0

3 years ago

Solve x 3(x+1)=-2(x-1)-4

Nataly [62]

Answer:

x= -1/3 or -2

Step-by-step explanation:

3x^2 +3x= -2x +2-4

3x^2 + 5x +2=0

by factorization method

3x(x+2) 1(x+2)=0

(3x+1)(x+2)

3 0

3 years ago

Read 2 more answers

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 $3 \sim 2$ .