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3 years ago

The sum of a number and its additive inverse is equal to ?

Mathematics

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

Zero

Step-by-step explanation:

To find the additive inverse, you merely change the sign of the number

For Example 5 {5 is the number} + (-5) {-5 is the additive inverse} = 0.

i.e. 6 + (-6) = 0

i.e (-7) + 7 = 0

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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

Collin did the work to see if 10 is a solution to the equation r/4=2.5

Natasha2012 [34]

Answer:

Yes, because if you substitute 10 for r in the equation and simplify, you find that the equation is true.

Step-by-step explanation:

$\frac{r}{4} = 2.5 \\ lhs = \frac{r}{4} \\ = \frac{10}{4} \\ = 2.5 \\ = rhs \\ \therefore \: \frac{r}{4} = 2.5 \\$

3 0

3 years ago

Lexi is making an orange drink for party. She puts 4 pints of orange drink and 2 pints of water into a punch bowl. Her friend Ol

Nina [5.8K]

I think she can add 5p more of orange which is 9p and 6p more water which is 8p. I'm pretty sure this works because 9 is the next closest multiple of 3 and 8 is one number less than that. Hope this helps!

8 0

3 years ago

A thermometer shows a temperature of Negative 20 and three-fourths degrees. A chemist recorded this temperature in her notebook

fredd [130]

Answer:

The answer is C.

Step-by-step explanation:

3/4 is equal to 0.75. If you combine 0.75 with -20, then it is 20.75.

4 0

3 years ago

Read 2 more answers

Triangle CDE is an isosceles triangle with angle d congruent to angle E if CD= 4x+9, DE= 7x-5, CE= 16x-27, find x and the measur

Montano1993 [528]

ANSWER

$x = 3$
$CD= 21 \: units$
$DE= 16 \: units$

$CE= 21 \: units$

EXPLANATION

It was given that ∆CDE is an isosceles triangle.

The length of the sides are given in terms of x as follows.

$CD=4x+9$

$DE=7x-5$

$CE=16x-27$

It was also given that, angle D is congruent to angle E.

This implies that, side CD and CE are equal.

Thus,

$|CD| = |CE|$
In terms of x, we have;

$4x + 9 = 16x - 27$

We group like terms to get,

$16x - 4x = 9 + 27$

$12x = 36$

Divide both sides by 12 to get,

$x = \frac{36}{12}$

$\therefore \: x = 3$

The length of the sides are;

$CD=4(3)+9 = 21 \: units$

$DE=7(3)-5 = 16 \: units$

$CE=16(3)-27 = 21 \: units$

3 0

3 years ago