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kirill115 [55]
3 years ago
8

Which of these is true a: 0.7 > 3/5 b: 0.7 < 3/5 c: 0.7 = 3/5

Mathematics
2 answers:
omeli [17]3 years ago
3 0
The answer is A. When 0.7 is put into fraction form it turns into 7/10. Then change the denominator of 3/5 to 10 which makes it 6/10. 7/10 is higher than 6/10 so the answer is A.
Ivanshal [37]3 years ago
3 0
The correct answer is A becoz 0.7 is larger than 3/5
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\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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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%20-%201%7D%20" id="TexFormula1" title=" \sqrt[3]{x - 1} " alt=" \sqrt[3]
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So the -1 is under the check mark too?

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