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vivado [14]
3 years ago
6

Which expression is equivalent to i 233? a)1 b)–1 c)i d)–i

Mathematics
2 answers:
Aneli [31]3 years ago
5 0

We have to find the value of the expression i^{233}

We know that the below values.

i^2=-1\\i^4=1

Hence, in order to find the value of the given expression, we can first rewrite it in terms of i^4

i^{233}=(i^4)^{58}\cdot i

Now, we know that i^4=1

Hence, we have

i^{233}=(1)^{58}\cdot i

i^{233}=1\cdot i

i^{233}=i

C is the correct option.

Alinara [238K]3 years ago
5 0
Choice c) I is the answer
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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
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Answer:

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Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

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