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soldi70 [24.7K]

2 years ago

12

Booker and 85 video games he has three shelves to put the games on each shelf can hold 40 games how many more video games does h

e have room for

1 answer:

Luda [366]2 years ago

5 0

Answer:

Booker has room for 35 new games.

Step-by-step explanation:

multiply the amount of games that can be held on one bookshelf by 3.

40 times 3 = 120

he has room for 120 games, so subtract the amount of video games he already has with the amount he has room for.

120 - 85 = 35

Booker has room for 35 new games.

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please help I am really stuck on this one!! I would like an explanation with the answer please so I can learn how to do it from

dexar [7]

ANSWER:

x° = 24°

y° = 72°

Analysis on picture hope it helps

4 0

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

3 years ago

If f(x)equals 2xminus5 and k(x)equals f(xplus3), write the equation for k(x).

ololo11 [35]

F(x)=2x-5
k(x)=f(x+3)=2(x+3)-5=2x+6-5=2x+1

k(x)=2x+1

7 0

2 years ago

PLEASE ANSWER ASAP!!!

julsineya [31]

Answer:

The answer to your question is A) x = 1

Step-by-step explanation:

We know that f(x) = 2x + 3 and g(x) = -x + 6

But f(x) = g(x)

Then, 2x + 3 = - x + 6

2x + x = 6 - 3

3x = 3

x = 3/3

x = 1

The solution is circle in the graph below.

8 0

3 years ago

The slope of the line is ?

aleksandr82 [10.1K]

Answer:

slope= 4,

Step-by-step explanation:

y = 4x-6

7 0

2 years ago