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daser333 [38]
3 years ago
11

Forty grams of Iodine 131, a radioactive material, decays according to the equation A(t)=40e^(-0.087t) (t is in days and A is th

e amount present at time t)
a. How much iodine 131 is remaining after 5 days?


b. When will 10 grams of iodine 131 be left?
Mathematics
1 answer:
Hatshy [7]3 years ago
6 0
A.) After 5 days t=5 so:

a(5) = 40 {e}^{ - 0.087(5)}   = 25.89g
b.) Since we want to know how much time 40 grams will decay to 10 grams we will let A(t)=10 so:

10 = 40 {e}^{ - 0.087(t)}  \\  \\  \frac{10}{40}  =  {e}^{ - 0.087(t)}  \\  \\  ln( \frac{1}{4} )  =  - 0.087t \\  \\  \frac{ ln( \frac{1}{4} ) }{ - 0.087}  = t \\  \\ t = 15.93
10 grams will be left after 15.93 days.
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<u>Step-by-step explanation:</u>

Here we have following equation : x^{2}+(y-2.25)^{2} = \dfrac{196}{169}

We need to find the center & radius of this circle . Let's find out:

We know that , Equation of a circle is given by :

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Here , (h,k) are the co-ordinates of center & r is the radius of circle.Collectively called as a circle with radius r and center at (h,k) . Let's frame given equation in question :

⇒  x^{2}+(y-2.25)^{2} = \frac{196}{169}

⇒  (x-0)^{2}+(y-2.25)^{2} = (\frac{14}{13})^2

On comparing this equation with equation (1) we get :

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