Write an expression with parentheses (don't simplify!) to represent the following quantity:

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

Triangle J is shown below. James drew a scaled version of Triangle J using a scale factor of 4 and labeled it

Novay_Z [31]

Answer:

The area of triangle K is 16 times greater than the area of triangle J

Step-by-step explanation:

we know that

If Triangle K is a scaled version of Triangle J

then

Triangle K and Triangle J are similar

If two triangles are similar, then the ratio of its areas is equal to the scale factor squared

Let

z -----> the scale factor

Ak ------> the area of triangle K

Aj -----> the area of triangle J

so

$z^{2}=\frac{Ak}{Aj}$

we have

$z=4$

substitute

$4^{2}=\frac{Ak}{Aj}$

$16=\frac{Ak}{Aj}$

$Ak=16Aj$

therefore

The area of triangle K is 16 times greater than the area of triangle J

4 0

3 years ago

Read 2 more answers

A line goes through the origin and the point (6, 14). The point (2, y) is also on the line. Calculate y and justify that your va

guajiro [1.7K]

Answer:

Step-by-step explanation:

The key here is knowing that the equation of the line that passes through these points is same

Thus having (6,14) and (0,0), the slope is as follows;

m = y2-y1/x2-x1 = 0-14/0-6 = -14/-6 = 7/3

Now we can use this slope here to get the value of y in the question

All we need to do is tie write the equation of the line for between the points (6,14) and (2,y)

That would be;

7/3 = y-14/1-6

7/3 = y-14/-5

cross multiply;

-35 = 3(y-14)

-35 = -3y + 42

-3y = -35-42

-3y= -77

y = -77/3

y = 77/3

3 0

3 years ago

How to cross multiply 9/12 =x/100?

Ilia_Sergeevich [38]

The answer is 75. 1. You simplify which is 3/4
2. Change to decimal then number you will get .75 move decimal two to the right to change to number then you will get 75!

6 0

3 years ago

Ink cartridges are heavy 1/4 pound. The total weight of the cartridges in a box is 2 1/2 pounds. How many cartridges does the bo

-BARSIC- [3]

There are 10 cartridges in the box.

6 0

3 years ago

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