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Ganezh [65]
3 years ago
10

X is less than or equal to 9 and greater than or equal to 5

Mathematics
2 answers:
OverLord2011 [107]3 years ago
6 0
I believe it will be 5< x<9 tell me if its wrong or not! :)
<span />
topjm [15]3 years ago
3 0
5≤x≤9 it can also be reversed as 9≥x≥5 which is the same thing
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(x+y)^2 (x2+2xy+y2) pleas show all work
lakkis [162]

Answer:

{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x

​4

​​ +4x

​3

​​ y+6x

​2

​​ y

​2

​​ +4xy

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​​ +y

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​​  

Step-by-step explanation:

1 Use Square of Sum: {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}(a+b)

​2

​​ =a

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​​ +2ab+b

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​​ .

({x}^{2}+2xy+{y}^{2})({x}^{2}+2xy+{y}^{2})(x

​2

​​ +2xy+y

​2

​​ )(x

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​​ +2xy+y

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​​ )

2 Expand by distributing sum groups.

{x}^{2}({x}^{2}+2xy+{y}^{2})+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x

​2

​​ (x

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​​ )+2xy(x

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​​ +2xy+y

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​​ )+y

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​​ (x

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3 Expand by distributing terms.

{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2xy({x}^{2}+2xy+{y}^{2})+{y}^{2}({x}^{2}+2xy+{y}^{2})x

​4

​​ +2x

​3

​​ y+x

​2

​​ y

​2

​​ +2xy(x

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​​ +2xy+y

​2

​​ )+y

​2

​​ (x

​2

​​ +2xy+y

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​​ )

4 Expand by distributing terms.

{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}({x}^{2}+2xy+{y}^{2})x

​4

​​ +2x

​3

​​ y+x

​2

​​ y

​2

​​ +2x

​3

​​ y+4x

​2

​​ y

​2

​​ +2xy

​3

​​ +y

​2

​​ (x

​2

​​ +2xy+y

​2

​​ )

5 Expand by distributing terms.

{x}^{4}+2{x}^{3}y+{x}^{2}{y}^{2}+2{x}^{3}y+4{x}^{2}{y}^{2}+2x{y}^{3}+{y}^{2}{x}^{2}+2{y}^{3}x+{y}^{4}x

​4

​​ +2x

​3

​​ y+x

​2

​​ y

​2

​​ +2x

​3

​​ y+4x

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​​ y

​2

​​ +2xy

​3

​​ +y

​2

​​ x

​2

​​ +2y

​3

​​ x+y

​4

​​  

6 Collect like terms.

{x}^{4}+(2{x}^{3}y+2{x}^{3}y)+({x}^{2}{y}^{2}+4{x}^{2}{y}^{2}+{x}^{2}{y}^{2})+(2x{y}^{3}+2x{y}^{3})+{y}^{4}x

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​​ y)+(x

​2

​​ y

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​​ +4x

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​​ y

​2

​​ +x

​2

​​ y

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​​ )+(2xy

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​​ +2xy

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​​ )+y

​4

​​  

7 Simplify.

{x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}x

​4

​​ +4x

​3

​​ y+6x

​2

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​​ +4xy

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