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2 years ago

Pre-algebra book 12.3.4 page 468

Mathematics

1 answer:

Olin [163]2 years ago

8 0

What’s the question? i can help but uhh what’s the question

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13. What values of x satisfy the inequality 4x + 2 > 10? *<br> What is the answer

mojhsa [17]

Answer:

Hi! The answer to your question is x > 2

Step-by-step explanation:

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Hope this helps!!

- Brooklynn Deka

4 0

3 years ago

Read 2 more answers

23.160 divided by 10

Serjik [45]

Answer:

the answer is 2.316

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3 years ago

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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

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

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

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

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

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

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

Describe the domain of this function d=4300x/x^2 +40

Mashcka [7]

Step-by-step explanation:

The domain would be all real numbers

7 0

3 years ago

Answer the question in the attached picture please

tiny-mole [99]

Answer:

Option 1 is correct

5.345345... = 5340/999

Hope this helps!

8 0

3 years ago