15 points answer ,,,,,,,,,,,,,,,,,,,,,,,,

Help me create a creative math meme please (has to be related to math learned in Algebra 1)

Zielflug [23.3K]

Not really a meme but hope it helps.

7 0

3 years ago

The average distance from Mars to the sun is about one hundred forty-one million, six hundred twenty thousand miles. How is this

OleMash [197]

41,620,000 is the answer

5 0

4 years ago

Read 2 more answers

HELP FIRST TO ANSWER ALL IS BRAINLEIST!!!!!!!!!

andreev551 [17]

Answer:

1. D) 1 9/20

2. B) 7/12

4. D) $50.00

Step-by-step explanation:

The working is as follows :

1. Mrs. Lang bought 3/4 pound of red grapes and 7/10 pound of green grapes. To find out how many are in total we have to add both values total.

3/4 + 7/10

The LCM of 4 and 10 is 20.

3/4 * 5 = 15/20

7/10 * 2 = 14/20

15/20 + 14/20 = 29/20 = 1 9/20

2. 1/3 + 1/4

The LCM of 3 and 4 is 12

1/3 * 4 = 4/12

1/4 * 3 = 3/12

4/12 + 3/12 = 7/12

4. Sharon spent 2/3 of her clothing allowance; she was given $75. All we have to do is find 2/3 of $25 :

2/3 * $75 = $50.00

4 0

3 years ago

The answers

bazaltina [42]

Answer:

C

Step-by-step explanation:

First cuz the y-intercept is 3 so not B or D

And the slope is -2/3

so the answer is C.

i'm not really sure about the slope part.

3 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 $3 \sim 2$ .

3 0

3 years ago