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Ira Lisetskai [31]

3 years ago

15

Matt is climbing a mountain when his elevation is higher than 1600 he has trouble breathing write an inequality that describes h

the elevation at which breathing is difficult for Matt

1 answer:

11111nata11111 [884]3 years ago

6 0

Answer is $h > 1600$

The convention is to write the variable first on the left side, then the inequality sign, followed by the other side of the inequality.

Writing $h > 1600$ means that h is larger than 1600. Think of an alligator mouth that is represented by the "greater than sign". The mouth opens up to the larger side. In this case, h could be something like 1700 which is larger than 1600. So we'd say $1700 > 1600$ for instance.

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A sailboat travels a distance of 2 1/2 miles in 1/6 of an hour.Which complex fraction represents the unit rate in miles per hour

Viktor [21]

2 1/2 miles = 1/6 hours
2 1/2 miles * 6 = 1/6 hours * 6
12 1/2 miles = 6/6 hours
12 1/2 miles = 1 hour
12 1/2 miles/hr

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3 years ago

What is the solution to this 20x+30y=1000

PilotLPTM [1.2K]

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2 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 $3 \sim 2$ .

3 0

2 years ago

3. The box-and-whisker plot below shows the heights,in inches, of the students in a 7th grade class.What percentage of the heigh

Fantom [35]

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2 years ago

Solve the equation. Check your answer:<br><br> 14+6a-8=18

Ivanshal [37]

A=2...
Subtract 14 from both sides,then subtract 8 from both sides. Your left with
6a=12
6x1=6
6x2=12
A=2

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2 years ago