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If 800g of a radioactive substance are present initially and 8 years later only 450g remain, how much of the substance will be present after 16 years? (Round answer to a whole number)

Respuesta :

H1storyBuff

A=Pe^(rt)

P = 800g

t = 8 years

A = 450g

r = This is what we will try and find to start with

450=800e^(r*8)

After running the math through a calculator, we end with r = -0.07192

Now we just re-input this information into our equation: A=800e^(-0.07192*16)

A=800e^(1.15072)

Now we will re-write the equation using the negative exponent rule:

A = 800 1/e^1.15072

Combine right side:

A = 800/e^1.15072

Then do the math:

A = 253.12709836......

That will give us A = 253 (rounded to the whole number)

I hope this helps! :)

The substance that should be presented after 16 years is 253.

Given that,

  • If 800g of a radioactive substance are present initially and 8 years later only 450g remain.

Based on the above information, the calculation is as follows:

We know that

[tex]A=Pe^{rt}[/tex]

Here

P = 800g

t = 8 years

A = 450g

[tex]450=800e^{r\times 8}\\\\A=800e^{-0.07192\times 16}\\\\A=800e^{1.15072}\\\\A = 800 \ 1 \div e^{1.15072}\\\\A = 800\div e^{1.15072}[/tex]

A = 253

Therefore we can conclude that the substance that should be presented after 16 years is 253.

Learn more: brainly.com/question/16115373

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