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Monica [59]
3 years ago
10

Pam and her brother both open savings accounts. Each begin with a balance of zero dollars.For every two dollars that pam saves i

n her account, her brother saves five dollars in his accounts. Determine a ratio to describe the money in pam's account to the money in her brother's account.
Mathematics
1 answer:
zheka24 [161]3 years ago
4 0
Pam and her brother both open savings accounts.  Each begin with a balance of zero dollars.  For every two dollars that Pam saves in her account, her brother saves five dollars in his account.
1. Determine a ratio to describe the money in Pam’s account to the money in her brother’s account.  :
 
2. If Pam has dollars in her account, how much money does her brother have in his account?  Use a tape diagram to support your answer.
 
Pam’s brother has dollars in his account. 
 
3. Record the equivalent ratio. :
 
4. Create another possible ratio that describes the relationship between the amount of money in Pam’s account and the amount of money in her brother’s account.  Answers will vary.  :, :, etc.
I hope I helped! :)
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Since 70% is allocated to the 1st floor, therefore the floor is receiving:

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What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
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Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

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Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

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\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

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Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

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