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GenaCL600 [577]
3 years ago
12

The current flowing through an electric circuit is the derivative of the charge as a function of time. If the charge q is given

by the equation q(t)=3t^2+2t-5 what is the current at t = 0?
Mathematics
2 answers:
blsea [12.9K]3 years ago
6 0

Answer:

The current at t = 0 is

I (0) = 2 units of current

Step-by-step explanation:

If the current flowing through an electric circuit is defined as the derivative of the charge as a function of time and we have the equation of the charge as a function of time, then, to find the equation of the current, we derive the equation of the charge.

q (t) = 3t ^ 2 + 2t-5

\frac{dq(t)}{dt} = 3 (2) t ^ {2-1} +2t ^ {1-1} -0

Simplifying the expression we have:

\frac{dq(t)}{dt} = 6t + 2t ^0\\\\\frac{dq(t)}{dt} = 6t + 2 = I (t)

Finally, the equation that defines the current of this circuit as a function of time is:

I (t) = 6t +2

Now to find the current at t = 0 we make I (t = 0)

I (0) = 6 * 0 +2

I (0) = 2 units of current

LUCKY_DIMON [66]3 years ago
5 0

Answer: I just took the test and got 100%

1. C, 2

2. D, v(t) = 8t - 2

3. B, 902 mi/min

4. A, the car is slowing down at a rate 10 mi/h^2

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8 0
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vampirchik [111]

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5 0
3 years ago
Read 2 more answers
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ikadub [295]

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7 0
3 years ago
8/x-5 - 9/x-4 = 5/x^2-9x+20
adelina 88 [10]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

7 0
3 years ago
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