What is 90% in the simplest form?

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

At CD Shack, all used CDs are the same price. Jesse spent $54 and Mimi spent $66 on used CDs. What is the most each used CD coul

Alinara [238K]

Hey there
__________
The correct answer is
Whatever each CD costs, what each person paid is that cost times the number of CDs purchased (no sales tax for this problem).
So, the price of one CD is a factor of $66 (a number of $ that divides $66 evenly).
In theory, it could be $1, $2, $3, $6, $11, $22, $66.
It could even be $0.50, $0.25, $0.20, $0.10, $0.05,...
Also, the price of one CD must be a factor of $54. such as $54,$27,$18,$9,$6,$3,$2,$1,... .
You are looking for the most that price could be.

The grew greatest price that is in both lists is $6.

How can you make those lists?
You can start with the total price, then the price divided by 2, by 3, by whatever whole number you can divide it.

Otherwise, you could find the greatest common factor of 66 and 54
from the prime factorization of both numbers.
___________________
Hope this helps you

5 0

3 years ago

3a.) Today, everything at a store is on sale. The store offers a 20% discount. 1 poin

Novay_Z [31]

Answer:

14.40

Step-by-step explanation:

100%-> 18

100-20=80

80%-> 14.40

6 0

3 years ago

Given the diagram below, what is the length of segment EF

bekas [8.4K]

Let x = length of segment EF.

Assume that
(a) line segments AD, EF, and BC are parallel, and
(b) the vertical distance between AD and EF is equal to the vertical distance between EF and BC.

Then from similarity between geometric shapes, we can write
$\frac{2.2}{x} = \frac{x}{4.8}$
$x^{2} =(2.2)(4.8)=10.56$
x=3.2 (nearest tenth)

Answer: B. 3.2

6 0

3 years ago

Read 2 more answers

Write (n+4)(n+3)(n+2) as a permutation

kenny6666 [7]

Answer:

Step-by-step explanation:

yo

7 0

3 years ago