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steposvetlana [31]
3 years ago
5

Help plz im sorry if im asking to many questions

Mathematics
2 answers:
denis-greek [22]3 years ago
8 0
The answer would be C, so the 3rd option
Zinaida [17]3 years ago
7 0

Answer:

The 3rd one 50 - 12p = 26

Step-by-step explanation:

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Answer:

X=20

Step-by-step explanation:

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2 years ago
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Please help me with the below question.
Snezhnost [94]

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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How many significant figures are in the measurement 33050 kilograms
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there are four significant figures

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zeros to the right of a non zero digit are ambiguous

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At summer camp, 150 kids had the opportunity to choose their activities. 1/3 of those kids selected archery for their morning ac
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