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2 years ago

The length of the box is x

Mathematics

1 answer:

Romashka [77]2 years ago

3 0

A polynomial that represents the volume of the Amazon box is; V(x) = x³ - 6x² + 5x

<h3>How to calculate the volume of a box</h3>

Length of box = x
Width of the box is 4 inches less than the length. Thus; width = x - 4
Height is said to be 1 inch less than the width. Thus; height = x - 4 - 1 = x - 5

Now, the formula for volume of a box is;

V = length × width × height

Thus;

V = x • (x - 1) • (x - 5)

V = x(x² - x - 5x + 5)

V = x(x² - 6x + 5)

V(x) = x³ - 6x² + 5x

In conclusion, the polynomial that represents the volume of the Amazon box is V(x) = x³ - 6x² + 5x

Read more on volume of a box at; brainly.com/question/15532376

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712yd 2115yd The area of the figure is

natima [27]

Answer:

1505880 sq yards

- emocliffordc

6 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 $3 \sim 2$ .