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3 years ago

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Mathematics

1 answer:

butalik [34]3 years ago

6 0

Answer:

Step-by-step explanation:

Put x = -8 to the formula of <em>h(x)</em>

$h(x)=\dfrac{x^2+3x}{4x+27}\\\\h(-8)=\dfrac{(-8)^2+3(-8)}{4(-8)+27}=\dfrac{64-24}{-32+27}=\dfrac{40}{-5}=-8$

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3 years ago

Leno4ka [110]

Answer:

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Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

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Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

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