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2 years ago

What is 1/3 plus 1/6 in simplest terms

Mathematics

2 answers:

djyliett [7]2 years ago

6 0

1/2 in simplified term

Step-by-step explanation:

Tomtit [17]2 years ago

3 0

First find lowest common denominator (LCD) for all the fractions you are adding. That number is 6. Then multiply the numerator (top number) by whatever number you multiplied the denominator with to get the LCD. Then you can add the fractions once all the denominators are the same. At the end, reduce the fraction if necessary if the numerator and denominator are both divisible by a common number.

Answer to your question:
1/3 + 1/6
2/6 + 1/6 = 3/6
Reduce 3/6 to simplest terms
3/6 = 1/2 (Because both the numerator and denominator are divisible by the number 3).
Hope this helps a little. Thanks.

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20) Use the distributive property to factor the expression. 10xz − 20yz

Paraphin [41]

Answer:

C. 10z(x - 2y)

Step-by-step explanation:

10xz − 20y

Factor the terms

Factors of 10xz = 2 × 5 × x × z

Factors of -20yz = (-1) × 2 × 2 × 5 × y × z

Identify their greatest common factor

The common factors are 2, 5, and z. The greatest common factor is 10z.

Rewrite them with the GCF as a factor

10xz = 10z × x

-20yz = 10z × (-2y)

Replace the terms in the original expression with the factored terms

10xz − 20yz = 10z(x) + 10z(-2y)

Use the distributive property to isolate the greatest common factor

10z(x) + 10z(-2y) = 10z(x - 2y)

3 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 $3 \sim 2$ .