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inysia [295]
3 years ago
6

What's the area of the green highlighted face shown below??

Mathematics
1 answer:
GrogVix [38]3 years ago
5 0

Answer:

45.5 ft sqaured

Step-by-step explanation:

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6, 3/5 of 60minutes is 36. 1/10 of 60mins. is 6. 36/6 is 6.
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First one is the correct answer
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Re-order the premises in each of the arguments to show that the conclusion follows as a valid consequence from the premises. It
klemol [59]

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Step-by-step explanation:

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3 0
3 years ago
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}}

6 0
2 years ago
How to sovle <br> (x+1/x)^2
Slav-nsk [51]

(\frac{x + 1}{x} ) ^{2}  \\  =  \frac{ {(x + 1)}^{2} }{ {x}^{2} }  \\  = \frac{ (x + 1)(x + 1) }{ {x}^{2} } \\ = \frac{ {x}^{2}  + 2x + 1}{ {x}^{2} }

Hope you could get an idea from here.

Doubt clarification- use comment section.

4 0
3 years ago
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