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shtirl [24]
3 years ago
13

How to write all real numbers in set notation?

Mathematics
1 answer:
laila [671]3 years ago
7 0
Hi, this is the notation for inclusion of all real numbers.

x \in \mathbb{R}

We can also state numbers which are positive or negative with a sign on top of the R notation like so:

x \in \mathbb{R}^{+} \text{ or } x \in \mathbb{R}^{-}
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<u>Part A)</u> 7(a+b)=7a+

Using the distributive property

7(a+b)=7a+7b

therefore

<u>the answer Part A is</u>

9-----> A

<u>Part R)</u>  4(5+x)=20+

Using the distributive property

4(5+x)=20+4x

therefore

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<u>Part Y)</u> 3(2x+9)=6x+

Using the distributive property

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therefore

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4-----> Y

Part S)  8(3x+1)=+8

Using the distributive property

8(3x+1)=24x+8

therefore

<u>the answer Part S is</u>

23-----> S

<u>Part O)</u>  a(4+b)=+ab

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a(4+b)=4a+ab

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17-----> O  

<u>Part E)</u> x(y+10)=+10x

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x(y+10)=xy+10x

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Using the distributive property

2(7x+4y)=14x+8y

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<u>the answer Part I is</u>

20-----> I

<u>Part D)</u> 6(9+5x)=54+

Using the distributive property

6(9+5x)=54+30x

therefore

<u>the answer Part D is</u>

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<u>Part W)</u> x(a+3b)=+3bx

Using the distributive property

x(a+3b)=ax+3bx

therefore

<u>the answer Part W is</u>

18-----> W

<u>Part E)</u> a(8x+2y)=8ax+

Using the distributive property

a(8x+2y)=8ax+2ay

therefore

<u>the answer Part E is</u>

7-----> E

<u>Part T)</u> 0.5(4a+10)=2a+

Using the distributive property

<u>Part T)</u> 0.5(4a+10)=2a+5

therefore

<u>the answer Part T is</u>

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<u>Part R)</u> (2/3)(12+9y)=8+

Using the distributive property

(2/3)(12+9y)=8+6y

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5x+5y=5(x+y)

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<u>Part T)</u> 9a+9b=9(\ +b)

9a+9b=9(a+b)

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Part W) 4m+4n=\ (m+n)

4m+4n=4(m+n)

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<u>Part H)</u> ab+3a=a(b+\ )

ab+3a=a(b+3)

therefore

<u>the answer Part H is</u>

2-----> H  

<u>Part E)</u> xy+15x=\ (y+15)

xy+15x=x(y+15)

therefore

<u>the answer Part E is</u>

11-----> E

Part A) bu+uv=\ (b+v)

bu+uv=u(b+v)

therefore

<u>the answer Part A is</u>

5-----> A

<u>Part F)</u> (2/5)m+(2/5)n=(2/5)( \ +n)

(2/5)m+(2/5)n=(2/5)(m+n)

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7ax+2ay=a(7x+2y)

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Part T) 4kx+11ky=\ (4x+11y)

4kx+11ky=k(4x+11y)

therefore

<u>the answer Part T is</u>

15-----> T

<u>Part R)</u> 3ay+8by=y(\ +8b)          

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therefore

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