All categories

3 years ago

In hours Mariah can rake 3 backyards. how much of a backyard can she rake in 1hour?

Mathematics

1 answer:

Leviafan [203]3 years ago

7 0

Answer: 3 or the whole backyard

Step-by-step explanation: In a hour she can rack 3 backyard. The question asked that how much of a backyard can she rake in 1 hour. She can rake the whole backyard in a hour.

You might be interested in

A bicycle wheel has a diameter of 63centermeter.the wheel.during a journey the wheel makes a 510revolation turn.how many meter d

olga55 [171]

63 × 510 = 32130

anything to ask please pm me

3 0

3 years ago

What is the missing reason in step 6? SSS congruency theorem CPCTC definition of a parallelogram opposite sides in a parallelogr

NARA [144]

For the two parallelogram to be congruent, their corresponding sides must be equal

<h3>Congruent figures</h3>

Two figures are said to be congruent if they are of the same shape and their corresponding sides and angles are congruent to each other. The SSS congruency theorem states that two figures are congruent of all their sides are congruent.

For the two parallelogram to be congruent, their corresponding sides must be equal

Find out more on similar figures at: brainly.com/question/26173060

7 0

2 years ago

F(x) = 9x-7. If f(x)= -7, find x.

svlad2 [7]

Answer: the answer is 56

Step-by-step explanation: since the f(x) is -7 it would be 9 times -7 then minus 7 and get the answer.

3 0

3 years ago

Help Please.. Write out mixed decimals using words.

Tatiana [17]

Twenty point twelve
ten point fourhundred and two

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 $3 \sim 2$ .

3 0

3 years ago

In hours Mariah can rake 3 backyards. how much of a backyard can she rake in 1hour?​

In hours Mariah can rake 3 backyards. how much of a backyard can she rake in 1hour?