What must be added to -15 to -8<br>what do you thinck

notsponge [240]

F' $( \frac{y-1}{y+1})-y$ =
f' $( \frac{y-1}{y+1})$ -f'(y) [/tex]=
$\frac{f'(y-1)(y+1)-f'(y+1)(y-1)}{(y+1)^2} -1$ =
$\frac{(f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1)}{(y+1)^2} -1$ =
$\frac{(1-0)(y+1)-(1-0)(y-1)}{(y+1)^2} -1$ =
$\frac{(1)(y+1)-(1)(y-1)}{(y+1)^2} -1$ =
$\frac{(y+1)-(y-1)}{(y+1)^2} -1$ =
$\frac{y+1-y+1}{(y+1)^2} -1$ =
$\frac{2}{(y+1)^2} -1$

8 0

3 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

Why is 6√3 irrational or rational?

Oduvanchick [21]

Answer:

irrational.

Step-by-step explanation:

√3 is irrational.

product of rational and irrational number is always irrational.

6×√3=6√3

6 0

3 years ago

Read 2 more answers

What number should be placed in the box to help complete the division calculation?

Katarina [22]

Answer:

180

Step-by-step explanation:

its 180 cos its the closest number to 193 that 18 can go in, your overall answer to the equation is 310 remainder 13. But I dont know if u need that..

5 0

3 years ago

Five years ago ,a father was four times as old as his son.In five years time the father will be twice as old as the son.If x and

andrew-mc [135]

B 4y-y=15;x-2y=5 is the correct one

5 0

3 years ago

What must be added to -15 to -8what do you thinck​

What must be added to -15 to -8what do you thinck