Question

The radioactive substance strontium -90 has a half-life of 28 years. In other words, it takes 28 years for half of a given quantity of strontium -90 to decay to a non-radioactive substance. The amount of radioactive strontium -90 still present present after t years is modeled by the expression 80*2^-t/28 grams. Evaluate the expression for t=0, t=28 and t=56. What does each value of the expression represent?
NOTE: What values do you actually get?

Answer 1

\[80*2^{\frac{ -t }{ 28 }}\]

Answer 2

Is this the eqn?

Answer 3

yes it is

Answer 4

for t=0, it would be 80 as the multiplier becomes 1.

Answer 5

Assuming that is eqn, the answers will be 80, 40 and 20 respectively

Answer 6

When t=28 which is the half life, it would be 1/2 of 80 or 40 as pointed out by sama491.

Answer 7

thank you guys

Answer 8

You're welcome!

Answer 9

When t=56 it has done 2 halflifes 1/2 * 1/2 or 1/4the and again as sama491 has answered the solution is 20.

Answer 10

\[80*2^{-0/28}=80*1=80\]\[80*2^{-28/28}=80*2^{-1}=80*1/2=40\]\[80*2^{-56/28}=80*2^{-2}=80*1/2^{2}=80*1/4=20\]

Answer 11

Thanks @radar you are such a good helper

Answer 12

U R welcome. I will sign off now. cheers.