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Questionf(2)Find R(2) where f(2)g(2)x² – a- 2 - 3010x + 100and g(2)-22 – 5x + 6611x + 110(Simplify your answer.)Provide your answer below:

Sagot :

Answer:

Given that,

To find,

[tex]R(x)=\frac{f(x)}{g(x)}[/tex]

where,

[tex]f(x)=\frac{x^2-x-30}{10x+100}[/tex][tex]g(x)=\frac{-x^2-5x+66}{11x+110}[/tex]

Simplifing f(x) and g(x), we get

[tex]f(x)=\frac{x^2-x-30}{10x+10)}=\frac{x^2-6x+5x-30}{10(x+10)}[/tex][tex]=\frac{x(x-6)+5(x-6)}{10(x+10)}=\frac{(x-6)(x+5)}{10(x+10)}[/tex][tex]f(x)=\frac{(x-6)(x+5)}{10(x+10)}-----(1)[/tex]

This is the simplified form of f(x).

For g(x) we get,

[tex]g(x)=\frac{-x^2-5x+66}{11x+110}=\frac{x^2+5x-66}{-11(x+10)}[/tex][tex]=\frac{x^2+11x-6x-66}{-11(x+10)}=\frac{x(x+11)-6(x+11)}{-11(x+10)}[/tex][tex]g(x)=\frac{(x+11)(x-6)}{-11(x+10)}------(2)[/tex]

[tex]\frac{i}{g(x)}=\frac{-11(x+10)}{(x+11)(x-6)}[/tex]

Now To find R(x), we get

[tex]R(x)=\frac{f(x)}{g(x)}=f(x)\times\frac{1}{g(x)}[/tex][tex]=\frac{(x-6)(x+5)}{10(x+10)}\times\frac{-11(x+10)}{(x+11)(x-6)}[/tex][tex]=\frac{-11(x+5)}{10(x+11)}[/tex]

we get,

[tex]R(x)=\frac{-11(x+5)}{10(x+11)}[/tex]

Answer is:

[tex]R(x)=\frac{-11(x+5)}{10(x+11)}[/tex]