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3 years ago

Solve 18n - 7p-15n = 5p for n. A n = 1 B. n=2p C. n = 1/2 D. n=4p

Mathematics

2 answers:

andrey2020 [161]3 years ago

5 0

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

18n - 7p-15n = 5p

To solve for n first make n the subject

Group like terms

That's

18n - 15n = 5p + 7p

Simplify

3n = 12p

Divide both sides by 3 to make n stand alone

That's

3n/3 = 12p/3

We have the final answer as

Hope this helps you

azamat3 years ago

3 0

Answer:

D) n = 4p

Step-by-step explanation:

18n - 7p - 15n = 5p

18n - 15n = 5p + 7p

Add like terms

3n = 12p

Divide both sides by 3

3n/3 = 12p/3

n = 4p

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Plz help I need to know how many bags of snacks they bought in all

spin [16.1K]

Answer:

7x - 2

Step-by-step explanation:

let x be the number of bags of chips, then

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3x - 2 is the number of bags of nachos ← 2 less than bags of pretzels

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6 0

3 years ago

In x² , what is the 2 called?

alexandr402 [8]

Answer:

<h2>an exponent</h2>

Step-by-step explanation:

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7 0

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 $3 \sim 2$ .