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3 years ago

A skydiver falls 144 feet in 3 seconds. How far does the skydiver fall per second?

Mathematics

2 answers:

RUDIKE [14]3 years ago

6 0

I would answer this question but your name is too vsco

Mekhanik [1.2K]3 years ago

5 0

If a skydiver falls 144 feet in 3 seconds in you need to find the time per second that they’re falling, you would simply have to get the time per one second. You have to get the 3 to one, so you would divide by 3 on both sides, getting 48 feet per second. (assuming they fell at the same rate all 3 seconds)

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Aisha is a sales clerk at Macy's. She paid $8.00 per hour plus a commission of 4% on all sales. Assuming Aisha works 39 hours an

steposvetlana [31]

Answer:

Aisha gross pay equals $472

Step-by-step explanation:

If Aishat is paid $8.00 per hour, then for 39 hours, she will get

39 × 8 = $312

She gets 4% commission on all sales. If her sale is $4000, the her commission will be

4000 × 4/100

= $160

Then her total gross income will be $312 + $160 = $472

6 0

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 $3 \sim 2$ .