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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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Answer:

A ) y=-1/2x+2

Step-by-step explanation:

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3 years ago
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So, you just need to draw the points

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As for the trigonometric function, we have to start from the parent function y=\sin(x) and derive the graph of its child function via transformations:

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Write an equivalent expression by distributing the
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Answer:

-(-4.1b - 6c - 8) = 4.1b + 6c + 8

Step-by-step explanation:

Given

-(-4.1b - 6c - 8)

Required

Simplify

To solve this using distributive property, we simply open the bracket

-(-4.1b - 6c - 8)

This becomes

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Hence;

-(-4.1b - 6c - 8) = 4.1b + 6c + 8

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Soloha48 [4]
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