Question

The half-life of a certain radioactive material is 37 days. An initial amount of the material has a mass of 477 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 6 days. Round your answer to the nearest thousandth.

a. y=477(1/2)^37x; 0kg
b. y=477(1/2)^1/37x; 426.288kg
c. y=2(1/477)^1/37x; 0.736kg
d. y=1/2(1/477)^1/37x; 0.184kg

Answer 1

...................................... well how do you do this? or how do you think you do this?

Answer 2

I am completely lost. I have no clue where to start.

Answer 3

... ok i have a bag of bread - it has x breads in it. how many breads do i have 'y days after, if every day my niece takes 1/2 of the remaining breads in the bag and eats them. :(. assume you can have decimals.

Answer 4

General formula is \[y=477a^x\]This means:

(x is the number of days that have passed)

at x=0 you have y = 477*a^0 = 477 kg.
at x=1 you have y = 477*a^1 kg
at x=2 you have y = 477*a^2 kg

Every day this new amount is multiplied with a.
After 37 days, you've multiplied the original amount (477) with a*a*a*...*a = a^37.
Now you have:\[477 \cdot a^{37}=477 \cdot \frac{ 1 }{ 2 }=238.5 kg\]
So: a^37=1/2.
Now look at the formulas in your answer options.

One has\[\left( \frac{ 1 }{ 2 } \right)^{\frac{ 1 }{ 37 }x}\]as a multiplication factor.
That's just what you need, because if you put in x=37, that factor becomes 1/2!

So the formula we are looking for is:\[y=477\left( \frac{ 1 }{ 2 } \right)^{\frac{ 1 }{ 37 }x}\]

Set x=6: y(6)=477*(1/2)^(6/37)=426.288 kg