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pshichka [43]
3 years ago
10

What the answer and how do you solve it pls

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
4 0
A desk costs $44, but they boost the price 150%
44( \frac{150}{100} ) = 66



They are selling it at $66 (selling price)
Since Michelle sold it at $66 and she earns 20% on the selling price as commission, then it is simply

66( \frac{20}{100} ) = ?

That should be the commision amount michelle earns
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If
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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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Daniel's basic cell phone rate each month is $29.95. Add to that $5.95 for voice mail and $2.95 for text messaging. This past mo
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So Daniel spent $23.50 on long distance calling 
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