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3 years ago

85 points!! | All of the following expressions have the same value, when x= -2 and y= 4, except

Mathematics

1 answer:

slavikrds [6]3 years ago

4 0

Answer:

They have two sets of equal answers...

Step-by-step explanation:

-2 * 2 * 4 = -16

0 - 4 * -2 * -2 = -16

0 * -2 * -2 * 4 = 0

0 * 4 * 4 = 0

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6. A prism has bases that are equilateral triangles with sides lengths of 10 inches and a length of 30 inches.

eimsori [14]

Answer:

The volume of the prism is $1,299\ in^{3}$

Step-by-step explanation:

we know that

The volume of the prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$L=30\ in$

The area of a equilateral triangle is equal to

$B=\frac{1}{2}(10)^{2} sin(60\°)$

$B=25\sqrt{3}\ in^{2}$

substitute

$V=(25\sqrt{3})(30)=1,299\ in^{3}$

8 0

3 years ago

I need help on #'s 2,3,5, and 9 due tomorrow

Mashcka [7]

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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 $3 \sim 2$ .