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3 years ago

-5(-3x+10)=30 what is x

Mathematics

1 answer:

Feliz [49]3 years ago

3 0

Answer:

5.33333333...

Step-by-step explanation:

it will be like -5 out the bracket x - 3 x which gives you 15 then we'll check again - 5 then you times it by there 10 in the bracket and then it gives you 50 and then it will be there in 15 x - 50 equals to 30 then you take the -50 to the other side where there is 30

then it will be 15 x equals to 30 + 50 and then then 15 x equals to 8 and then you divide both of the sides with 15 then get your answer

-5 (-3x+10)=30

15x-50=30

15x=30+50

15x/15=80/15

x=5.33333...

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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 $S\!$ .

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A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

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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 $R\!$ .

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$R$ 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 $3 \sim 2$ .

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