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3 years ago

PLEASE SOMEONE HELP ILL MARK BRAINLIEST AND 5 STARS 2.1/x+ 1 4/5= 3/10x

Mathematics

2 answers:

enot [183]3 years ago

8 0

Answer:

the answer is .3 or 3/10

Step-by-step explanation:

steposvetlana [31]3 years ago

7 0

9514 1404 393

Answer:

x = -1

Step-by-step explanation:

Putting the entire equation in decimal form, we have ...

2.1x + 1.8 = 0.3x

Subtracting 0.3x gives ...

1.8x +1.8 = 0

Dividing by 1.8, we get ...

x + 1 = 0

So, the solution is ...

x = -1 . . . . . . subtract 1 from both sides.

_____

<em>Additional comment</em>

Sometimes, when all of the numbers in the equation are divisible by the coefficient of x, it works out well to do that division before separating variables and constants. The other way this could have been solved is ...

subtract 1.8 ⇒ 1.8x = -1.8

divide by 1.8 ⇒ x = -1

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natka813 [3]

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Step-by-step explanation:

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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 $3 \sim 2$ .