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3 years ago

Use the general solution to solve 5 – 6x = 8x + 17.

Mathematics

2 answers:

Jlenok [28]3 years ago

7 0

<span> 5 – 6x = 8x + 17
</span>⇔ 5 - 17 = 8x + 6x
⇔ -12 = 14x
⇔ x = -6/7

Lilit [14]3 years ago

5 0

Answer:

Step-by-step explanation:

-6/7

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On saturday, you bowl at mar vista bowl, where renting shoes costs $2 and each game bowled is $3.50. on sunday, you bowl at pinz

ExtremeBDS [4]

Given, at Mar Vista bowl the cost of renting shoes = $2.

The cost for each game bowled = $3.50

There are g games given. So the cost for g games bowled = $ $(3.50)(g)$ = $ $(3.50g)$

The total amount soent at Mar Vista Bowl = $ $(2+3.50g)$

At Pinz, the cost for shoe rental = $5.

The cost for each game bowled = $3.25.

So the cost for g game bowled = $ $(3.25)(g)$ = $ $(3.25g)$

Total amount spent at Pinz = $ $(5+3.25g)$

Given, the amount spent ateach of the places is same.

so we can write the equation as,

$2+3.50g = 5+3.25g$

We will move 3.25 to the left side by subtracting it from both sides. we will get,

$2+3.50g-3.25g = 5+3.25g-3.25g$

$2+0.25g = 5$

Now we will move 2 to the right side by subtracting 2 from both sides. We will get,

$2+0.25g-2 = 5-2$

$0.25g = 3$

Now to get g we will divide 0.25 to both sides. We will get,

$\frac{0.25g}{0.25}= \frac{3}{0.25}$

$g =\frac{3}{0.25}$

$g = 12$

So we have got the required answer.

The number of games bowled = 12.

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3 years ago

Functions. Type the correct domain in the box

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Function:

y = -2/3 x + 7

Substitute the given values to their respective variable and solve.

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3 years ago

Uhm ian askin fo a lot buh ,,, uh someone add me on Rblx , XxColdTyXx is da user

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Answer:

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Step-by-step explanation:

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2 years ago

How many triangles can be constructed with angles measuring 10º, 80º, and 90º?

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C! 80 + 10 is 90, 90+90 is 180. There are 180 degrees in one triangle

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3 years ago

Read 2 more answers

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 $3 \sim 2$ .

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3 years ago