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3 years ago

If f(x)=×^2-81andg(x)

Mathematics

1 answer:

defon3 years ago

3 0

9514 1404 393

Answer:

the correct choice is marked

Step-by-step explanation:

$f(x)\cdot g(x)=(x^2-81)\dfrac{x+9}{x-9}=\dfrac{(x-9)(x+9)(x+9)}{(x-9)}\\\\=\boxed{(x+9)^2}$

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3 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

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**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

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3 years ago

If f(x)=×^2-81andg(x)​

If f(x)=×^2-81andg(x)