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Vlad [161]
1 year ago
5

Please help me with the below question.

Mathematics
1 answer:
Snezhnost [94]1 year ago
6 0

6a. By the convolution theorem,

L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}

6b. Similarly,

L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}

7. Take the Laplace transform of both sides, noting that the integral is the convolution of e^t and f(t).

\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau

\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)

where g(t) = e^t. Then G(s) = \frac1{s-1}, and

F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}

We have the partial fraction decomposition,

\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)

Then we can easily compute the inverse transform to solve for f(t) :

F(s) = -\dfrac1s + \dfrac4{s+3}

\implies \boxed{f(t) = -1 + 4e^{-3t}}

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Pls help :) determine which function has a greater rate and explain how you know.
Nesterboy [21]

Using exponential function concepts, it is found that the second function has a greater rate, as 0.8 > 0.2.

<h3>What is an exponential function?</h3>

It is modeled by:

y = ab^x

In which:

  • a is the initial value, that is, y when x = 0.
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Function 1 is given by:

y = 2(0.2)^x

Hence the rate is b = 0.2.

Considering the values on the table, function 2 is given by:

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4 0
2 years ago
1/3k+80=1/2k+120 what does k equal?
LenaWriter [7]

Answer:

k = -1/240

Step-by-step explanation:

to evaluate the value of k in the expression 1/3k+80=1/2k+120

we have

1/3k+80=1/2k+120

collect the like terms for easy evaluation

1/3k - 1/2k = 120 -80

1/3k - 1/2k = 40

find the lcm

2 - 3/6k = 40

-1/ 6k = 40

cross multiply

6k  x  40  = -1

240k = -1

divide both sides by 240

240k/240 = -1/ 240

k = -1/240

therefore the value of k in the expression 1/3k+80=1/2k+120 is equals to -1/240

8 0
3 years ago
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