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3 years ago

Plot -2 1/2 and 1 3/4 on a number line

Mathematics

1 answer:

Agata [3.3K]3 years ago

7 0

There you go. Sorry its blurry

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How many quarts are in 12 liters

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A bag has 4 orange scraps, 2 blue scraps, and 3 speckled scraps. You randomly pull a scrap from the bag, put it back, and then p

pashok25 [27]

Answer:

4/9

Step-by-step explanation:

4 plus 3 plus 2 = 9 total amount of scraps, goes on the denominator.

there are 4 orange scraps which is your numerator.

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2 years ago

Solve each system by graphing.

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(-1,4)

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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

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.

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

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

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

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

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

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

$k = 0.2133$

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

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

Two lines intersect to form the angles shown.

yawa3891 [41]

The answer i think it is ,is b

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