Mathematics, 24.02.2020 00:47 eliza35

The half-life of a radioactive substance is 50 years. If you start with some amount of this substance, what fraction will remain in 200 years? What fraction will remain in 300 years?

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Answer from: nina7163

The fraction will remain in 200 years is 1/16 and the fraction will remain in 300 years is 1/64 and this can be determined by using the formula of half-life.

Given :

The half-life of a radioactive substance is 50 years.

The Half-life of any substance is given by the formula:

$\rm A = A_02^{\frac{-t}{h}}$ --- (1)

where $\rm A_0$ is the initial quantity of the substance, A is the quantity of the substance remaining, t is the time elapsed and h is the half-life of the substance.

Now put the value of $\rm A_0$ , t and h in the equation (1).

$\rm A = 1\times 2^{\frac{-200}{50}}$

$\rm A = 1\times 2^{-4}$

$\rm A = \dfrac{1}{16}$

The fraction will remain in 200 years = 1/16

Now, for t = 300 the quantity of the substance remaining is:

$\rm A = 1\times 2^{\frac{-300}{50}}$

$\rm A = 1\times 2^{-6}$

$\rm A = \dfrac{1}{64}$

The fraction will remain in 300 years = 1/64

For more information, refer to the link given below:

link

Answer from: Kdub4evaa

The fraction will remain in 200 years is $\frac{1}{16}$ and in 300 years is $\frac{1}{64}$ .

Step-by-step explanation:

Given:

The half-life of a radioactive substance is 50 years.

Now, to find the fraction will remain in 200 years and fraction will remain in 300 years.

So, to get the fraction in 200 years we put formula:

Let the amount initial be 1.

Ao = 1 (amount initial).

t = 200 years (time).

h = 50 (half-life of a substance)

A = remaining amount after time.

$A=Ao2^\frac{-t}{h}$

$A=Ao\times 2^\frac{-t}{h}\\$

$A=1\times 2^\frac{-200}{50}$

$A=1\times 2^{-4}$

$A=1\times \frac{1}{2^4}$

$A=1\times \frac{1}{16}$

$A=\frac{1}{16}$

The fraction will remain in 200 years = $\frac{1}{16} .$

Now, to get the fraction will remain in 300 years we put same formula but the time will change:

t = 300 years (time).

$A=Ao\times 2^\frac{-t}{h} \\\\A=1\times 2^\frac{-300}{50} \\\\A=1\times 2^{-6} \\\\A=1\times \frac{1}{2^6} \\\\A=\frac{1}{64}$

The fraction will remain in 300 years = $\frac{1}{64} .$

Therefore, the fraction will remain in 200 years is $\frac{1}{16}$ and in 300 years is $\frac{1}{64}$ .

Answer from: Quest

nothjing

step-by-step explanation:

Answer from: Quest

check above link.

graphically? well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

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