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Fiesta28 [93]
3 years ago
6

Which table represents a nonlinear function? Help please!

Mathematics
2 answers:
motikmotik3 years ago
6 0
I think that the answer is the 2nd table
erma4kov [3.2K]3 years ago
5 0
The second one or the middle table mark Brainliest please
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I like a girl how do i tell her ?
ZanzabumX [31]

Answer:

Step-by-step explanation:

Just truely be open about your feelings, don't try and sugarcoat it, or leave things out, just be honest with her and tell her straight up, politely of course.

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2 years ago
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If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

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].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

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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3 years ago
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ratio of teachers : students = 12 : 240 ( divide both parts by 12 )

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dalvyx [7]
10.50 and the total cost will be 160.50.
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