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Alexeev081 [22]
3 years ago
12

Please help! Two math problems!!!

Mathematics
1 answer:
Fynjy0 [20]3 years ago
3 0

Answer:

Tell me if i am wrong. :)

Step-by-step explanation:

To solve for y:

Q.1   2x + y = -1

       y = -1 - 2x

Q. 2.  4x - 5y = 7

        y = 7 - 4x over -5

 

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Step-by-step explanation:

Just do:

3.984

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17 plus 3 equals 20 and .984 + 000 is .984

So your answer is 20.984

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Step-by-step explanation:

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3 years ago
The current in a circuit element is i(t) = 3(1 - e-2t) A when t ≥ 0 and i(t) = 0 when t &lt; 0. The total charge that has entere
lesantik [10]

Answer:

A=-3/2

B=3

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Step-by-step explanation:

Knowing that for t>0 i(t)=2\left(1-e^{-2t}\right) A, i(t)=0 when t<0 and using the definition of charge

q(t)=\int_{-\infty}^0i(t')dt'+\int_0^t i(t')dt'=0+\int_0^t i(t')dt'

The first term corresponds to q(0), the charge accumulated before t=0, in this case it luckily gives zero so we don't have to worry about it.

Let's proceed and integrate i(t) then when t>0

i(t)=3\int_0^t \left[ \int_0^tdt'-\inte_0^t e^{-2t'}dt' \right]dt'=3\left[ t+\frac{1}{2}\left( e^{-2t}-1 \right) \right]=3t+\frac{3}{2}e^{-2t}-\frac{3}{2}\,\, C

It is clear that:

A=-3/2

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4 0
3 years ago
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