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alexdok [17]
4 years ago
13

√108

20} " alt="9 \times \frac{13}{20} " align="absmiddle" class="latex-formula">
2\pi + 3.4

Mathematics
1 answer:
ki77a [65]4 years ago
4 0
I'm assuming your allowed to use a calculator for this? That's what I would do.
\sqrt{108}  = 10.39
To go about the fraction, take the whole number and times it by the denominator, then add in the numerator to create a new fraction(a whole one not mixed). Then simply divide the numerator by the denominator.
9 \times 20 = 180
180 + 13 = 193
193 is the new numerator and the new whole fraction is this and dividing it out gives you,
\frac{193}{20}  = 9.65
For the third one, just input it into a calculator,
2\pi + 3.4 = 9.68
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3 years ago
Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

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

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

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

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

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

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

7 0
3 years ago
Greetings. As a beginner, I'm struggling a bit to learn calculus. May I know what is the derivative of x to the power 4 step by
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If you're just starting calculus, perhaps you're asking about using the definition of the derivative to differentiate x^4.

We have

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Expand the numerator using the binomial theorem, then simplify and compute the limit.

\dfrac{d}{dx} x^4 = \displaystyle \lim_{h\to0} \frac{(x^4+4hx^3 + 6h^2x^2 + 4h^3x + h^4) - x^4}h \\\\ ~~~~~~~~ = \lim_{h\to0} \frac{4hx^3 + 6h^2x^2 + 4h^3x + h^4}h \\\\ ~~~~~~~~ = \lim_{h\to0} (4x^3 + 6hx^2 + 4h^2x + h^3) = \boxed{4x^3}

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2 years ago
Arthur is conducting a study on the preferred study options of students from East County College. He randomly selected 32 studen
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Answer:

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What makes me happy irl is nothing really, I only exist to make others happy
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