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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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3 years ago
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Hi! Let me help you!

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Read 2 more answers
Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two
In-s [12.5K]
Let the original 2-digit number be xy.
Because 5 times the sum of the digits is 13 less than the number, therefore
5(x + y) = 10x + y - 13 
5x + 5y = 10x + y - 13
-5x + 4y = - 13             (1)

The number with reversed digits is yx.
Because 4 times the sum of the digits is 21 less than the reversed 2-digit number, therefore
4(x + y) = 10y + x - 21
4x + 4y = 10y + x - 21
3x - 6y = -21
x - 2y = -7
x = 2y - 7                 (2)

Substitute (2) into (1).
-5(2y - 7) + 4y = -13
-10y + 35 + 4y = -13
-6y = -48
y = 8

From (3), obtain
x = 2*8 - 7 = 9

Answers:
The original 2-digit number is 98
The reversed 2-digit number is 89
The difference between the original and the reversed 2-digit numbers is
98 - 89 = 9
4 0
3 years ago
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