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3 years ago

Please help! I will try to mark brainliest

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

8 0

Answer:

1.23

0.067

25.34

Step-by-step explanation:

10^2 = 100

The point moves two places towards the right

Kitty [74]3 years ago

6 0

Answer:

1.23, 0.067, and 25.34

Step-by-step explanation:

Hope this helps!

brainliest?

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A. 12 ft², 16 ft², and 20 ft²

Step-by-step explanation:

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3 years ago

11. Max is designing a hot tub for a local spa.

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Answer:

Area required for circular hot tube = 6,218 inch² (Approx.)

Step-by-step explanation:

Given:

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Radius of hot tube = Diameter / 2

Radius of hot tube = 89 / 2 inches

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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.

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

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

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

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

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

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

$k = 0.2133$

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

Determine the half-life of the radioactive substance.

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

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

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

$e^{-0.2133t} = 0.5$

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

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

$t = 3.25$

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

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3 years ago

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I believe the answer is c.

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