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nikklg [1K]
3 years ago
15

The question is in the discription

Mathematics
2 answers:
Nostrana [21]3 years ago
8 0
If I’m not mistaken there are 2
user100 [1]3 years ago
5 0
It’s possible that it’s 3
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An isotope of cesium-137 has a half-life of 30 years. If 5.5 grams of cesium-137 disintigrates over a period of 105 years, how m
Olin [163]

Answer:

0.484 g

Step-by-step explanation:

N/No = (1/2)^t/t1/2

No = mass of cesium-137 originally present

N= mass of cesium-137 at time t

t1/2= half life of cesium-137

t = time taken

N/5.5=(1/2)^105/30

N/5.5=(1/2)^3.5

N/5.5= 0.088

N = 5.5 *  0.088

N= 0.484 g

5 0
3 years ago
PLEASE HELP ASAP! I WILL MARK YOU BRAINLIEST.
ololo11 [35]

Answer:

2x-1=5x

-1=3x

1.A. we take away 2x from both sides

2.A they are the same sine they both have the same solution of -1/3

Hope This Helps!!!

7 0
3 years ago
Read 2 more answers
A radioactive substance decays at a continuous rate of 17% per year, and 250 mg of the substance is present in the year 2009.
yaroslaw [1]

Answer:

a) A(t) = 250 (1-0.17)^t = 250(0.83)^t

b) A(t=14) = 250 (0.83)^{14}= 18.408 mg

c) For this case we want to find when the quantity is below 25 mg, so we can do this:

25 = 250 (0.83)^t

We can divide both sided by 250 and we got:

0.1 = 0.83^t

Now we can apply natural log on both sides and we got:

ln(0.1) = t ln (0.83)

And if we solve for t we got:

t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years

So the answer for this case would be 12.358 years after.

Step-by-step explanation:

Part a

For this case we know that the decay rate per year is about 17% or 0.17 in fraction and the initial amount is 250 mg for 2009, we can define the model like this:

A(t) = A_o (1-r)^t

Where A represent the amount of the substance in mg, r the decay rate = 0.17 and t the number of years after 2009.

Our model for this case would be:

A(t) = 250 (1-0.17)^t = 250(0.83)^t

Part b

For this case we have that t= 2023-2009=14 years, so then we can find the amount of substance like this:

A(t=14) = 250 (0.83)^{14}= 18.408 mg

Part c

For this case we want to find when the quantity is below 25 mg, so we can do this:

25 = 250 (0.83)^t

We can divide both sided by 250 and we got:

0.1 = 0.83^t

Now we can apply natural log on both sides and we got:

ln(0.1) = t ln (0.83)

And if we solve for t we got:

t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years

So the answer for this case would be 12.358 years after.

6 0
3 years ago
Simplify (2^3)^–2.<br> A: -1/64<br> B: 1/64<br> C: 2<br> D: 64
Airida [17]
Look on the app socratic.
3 0
3 years ago
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What is the sum of 7x/(x^2-4) and 2/(x+2)​
ioda

Answer:

The answer is 9x-4/(x+2)(x-2)

7 0
3 years ago
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