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tresset_1 [31]
3 years ago
12

Iodine-131, a radioactive substance that is effective in locating brain​ tumors, has a​ half-life of only eight days. a hospital

purchased 22 grams of the substance but had to wait seven days before it could be used. how much of the substance was left after seven ​days?
Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0
5 days/8 days = 0.625 half lives 

<span>fraction left after 0.625 half lives = (1/2)^0.625 </span>

<span>= 0.6484197773 </span>

<span>Amount left = 20*0.6484197773 gms </span>

<span>= 12.97 gms to dp <<<</span>
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Answer:

a

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b

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c

      \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

Step-by-step explanation:

From the question we are told that

    \frac{dy}{y} =  -\beta dt

Now integrating both sides

     ln y  =  \beta t + c

Now taking the exponent of both sides

       y(t) =  e^{\beta t + c}

=>     y(t) =  e^{\beta t} e^c

Let  e^c =  C

So

      y(t) = C e^{\beta t}

Now  from the question we are told that

      y(0) =  y_o

Hence

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=>     y_o = C

So

        y(t) = y_o e^{\beta t}

From the question we are told that

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substituting for y

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Now taking the exponent of both sides

        x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}

=>     x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c

Let  e^c  =  A

=>  x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

Now  from the question we are told that

      x(0) =  x_o

So  

      x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }

=>    x_o = K e^{\frac {\alpha y_o  }{\beta } }

divide both side  by    (K * x_o)

=>    K = x_o e^{\frac {\alpha y_o  }{\beta } }

So

    x(t) =x_o e^{\frac {-\alpha y_o  }{\beta } } *  e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

=>   x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }

=>    x(t) =  x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }

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so

    \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

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