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one hundred grams of radium are stored in a container. The amount R of radium present after t years can modeled by R=10e^-0.00043t. After how many years will only 5 grams of radium be present? round your answer to the nearest whole year.

Respuesta :

Using this equation, we get that it would take 1612 years for only 5 grams to be present.

Substituting 5 for R, we have
[tex]5=10e^{-0.00043t}[/tex]

Dividing both sides by 10,
[tex]\frac{5}{10}=\frac{10e^{-0.00043t}}{10} \\ \\0.5=e^{-0.00043t}[/tex]

Taking the natural log of both sides, we have
[tex]\ln{0.5}=\ln{e^{-0.00043t}} \\ \\ \ln{0.5}=-0.00043t[/tex]

Dividing both sides by -0.00043, we have
[tex]\frac{\ln{0.5}}{-0.00043}=t \\ \\1611.9=t \\ \\1612\approx t[/tex]
MrRoyal

It will take 700 years to remain only 5 grams of radium

The function is given as:

[tex]R=10e^{-0.00043t[/tex]

When the remaining gram is 5 grams, the equation of the function becomes

[tex]5=10e^{-0.00043t[/tex]

Divide both sides by 10

[tex]0.5=e^{-0.00043t[/tex]

Take the logarithm of both sides

[tex]\log(0.5)=-0.00043t[/tex]

Using a calculator, evaluate log(0.5)

[tex]-0.30102999566 =-0.00043t[/tex]

Divide both sides by -0.00043t

[tex]t = 700.069757349[/tex]

Approximate

[tex]t = 700[/tex]

Hence, it will take 700 years to remain only 5 grams of radium

Read more about exponential functions at:

https://brainly.com/question/11464095

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