All categories

3 years ago

There are 2.54 centimeters in an inch how many inches are in 51.78 centimeters round to the nearest tenths

1 answer:

3 0

2.54 cm in 1 inch......how many inches in 51.78 cm...

51.78 / 2.54 = 20.4 inches <==

or this way...
2.54 / 1 = 51.78 / x....2.54 cm to 1 in = 51.78 cm to x inches
cross multiply
(2.54)(x) = 51.78)(1)
2.54x = 51.78
x = 51.78 / 2.54
x = 20.4 inches <==

You might be interested in

2 gal 1 qt - 2 qt 1 pt

jekas [21]

Make all of the units equal.
1 gal = 4 quarts = 8 pints
1 quart = 2 pints

18 pt - 5 pt

13 pints or

1 gal 2 qt 1 pt

7 0

3 years ago

Read 2 more answers

5x + 7y = -29, y = x + 1

Afina-wow [57]

Answer:d

Step-by-step explanation:

6 0

3 years ago

100x56 Show work for brsinly

ratelena [41]

Answer:

5600

Step-by-step explanation:

100 times 56 = 5600 because you add the two zeros from the 100 to 56, etc.

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

Can you help me with this question i dont get it

Marizza181 [45]

The answer is c or A

3 0

3 years ago

Read 2 more answers