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3 years ago

Help? 2+2-8+6-2+9+5-6+6-3+8-1+7-0+8-1+10-2+11+

Mathematics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:

Step-by-step explanation:

2+2 = 4

4 - 8 = -4

-4 + 6 = 2

2 - 2 = 0

0 + 9 = 9

9 + 5 = 14

14 - 6 = 8

8 + 6 = 14

14 - 3 = 11

11 + 8 = 19

19 - 1 = 18

18 + 7 = 25

25 - 0 = 25

25 + 8 = 33

33 - 1 = 32

32 + 10 = 42

42 - 2 = 40

40 + 11 = 51

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2 years ago

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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