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artcher [175]
3 years ago
11

E2%7D%20" id="TexFormula1" title=" \frac{3}{2n} + \frac{1}{n^2} = \frac{n-2}{2n^2} " alt=" \frac{3}{2n} + \frac{1}{n^2} = \frac{n-2}{2n^2} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Sergio [31]3 years ago
8 0
Make denomenators the same
common denom is 2n²

multiply 3/2n by n/n
mulitply 1/n² by 2/2

we get
\frac{3n}{2n^2}+\frac{2}{2n^2}=\frac{n-2}{2n^2}
combine fractions
\frac{3n+2}{2n^2}=\frac{n-2}{2n^2}
multiply both sides by 2n²
2n+2=n-2
minus n both sides
n+2=-2
minus 2 both sides
n=-4
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(

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∘

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x

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is known as a composite function. Here's how composite functions work:

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)

=

1

3

(

1

)

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g

(

1

)

=

1

3

⇒

f

(

1

3

)

=

17

3

⇒

h

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17

3

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=

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(

h

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f

∘

g

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(

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Based on this, to find the function for  

(

h

∘

f

∘

g

)

(

x

)

(combined from right to left, by the way), simply replace  

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in  

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x

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(

x

)

, and replace  

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(

x

)

with the function of  

(

f

∘

g

)

(

x

)

, to get  

(

h

∘

f

∘

g

)

(

x

)

.

This, simplified, is equal to  

2

x

+

15

, and therefore  

(

h

∘

f

∘

g

)

(

1

)

=

2

(

1

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+

15

=

17

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