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2 years ago

Make a scatter plot of the data. Then find an equation of the quadratic function that models the data.

Mathematics

1 answer:

nika2105 [10]2 years ago

5 0

The equation of the quadratic function that models the data is y = x^2 + 3x + 4

<h3>How to determine the scatter plot?</h3>

A quadratic function is represented as:

y = ax^2 +bx + c

Next, we enter the data values on the table to a graphing calculator.

From the graphing calculator, we have:

a = 1, b = 3 and c = 4

Substitute these values in y = ax^2 +bx + c

So, we have

y = x^2 + 3x + 4

Hence, the equation of the quadratic function that models the data is y = x^2 + 3x + 4 and the scatter plot is (a)

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Answer:

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Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

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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.

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

$500 = 8000e^{-13k}$

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

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

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

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

$k = 0.2133$

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

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.

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3 years ago

Simplify the (5a^2 b^3)^O

Rudiy27

If you would like to simplify <span>(5a^2 * b^3)^0, you can do this using the following steps:
</span>
<span>(5a^2 * b^3)^0</span> = 5^0 * (a^2)^0 * (b^3)^0 = 1 * 1 * 1 = 1

The correct result would be 1.

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3 years ago

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