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4 years ago

Can you please help me with pre-algebra?? :)

Mathematics

1 answer:

Lesechka [4]4 years ago

5 0

I will always be available to answer any of you're questions you can trust me I am the smartest in my class

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Diane is on her way home in her car. She has driven 12 miles so far, which is three-fourths of the way home. What is the total l

andrew-mc [135]

16miles left

Step-by-step explanation:

7 0

3 years ago

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Helen wrote the following facts to try to show that division by 0 results in 0. Explain her error.

earnstyle [38]

Helen should of put 0 first, because if you're gonna divide, you need to put the product first, then one of the factoes.

So instead it should be 0÷(-7)=0

3 0

3 years ago

Find the fourth root of 81 that<br> graphs on the positive y-axis.

klio [65]

Answer:

1.732

Step-by-step explanation:

cube root of 81 is

=1.732

3 0

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

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

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

$e^{-0.2133t} = 0.5$

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

$-0.2133t = \ln{0.5}$

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

$t = 3.25$

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

7 0

3 years ago

Evaluate.<br><br> 4 to the power of 3 + (-2) to the power of 4

finlep [7]

Answer:

Step-by-step explanation:

3+(-2) is the same as 3-2 which is 1. and since with multiple exponents you work from the top you have 4 to the power of 1^4. 1^4 is 1, so it's 4^1, which is 4.

3 0

2 years ago

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