SOLUTION: THE AMOUNT OF RADIOACTIVE SUBSTANCE PRESENT, IN POUNDS, AT TIME T, IN DAYS IS GIVEN BY THE FORMULA Y=60(2)^0.2T FIND THE NUMBER OF POUNDS PRESENT IN 30 DAYS

Algebra ->  Equations -> SOLUTION: THE AMOUNT OF RADIOACTIVE SUBSTANCE PRESENT, IN POUNDS, AT TIME T, IN DAYS IS GIVEN BY THE FORMULA Y=60(2)^0.2T FIND THE NUMBER OF POUNDS PRESENT IN 30 DAYS      Log On


   

Question 1102662: THE AMOUNT OF RADIOACTIVE SUBSTANCE PRESENT, IN POUNDS, AT TIME T, IN DAYS IS GIVEN BY THE FORMULA Y=60(2)^0.2T FIND THE NUMBER OF POUNDS PRESENT IN 30 DAYS
Found 3 solutions by Boreal, greenestamps, josgarithmetic:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
want 60(2^(0.2t)) which for t=30, 0.2t=6, so it is 60*2^6=60*64=3840 pounds
graph%28300%2C300%2C-5%2C40%2C-1000%2C5000%2C60%2A2%5E%280.2x%29%29

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The problem is nonsense.

Radioactive decay is represented by an exponential decay function -- not an exponential growth function.

Is the formula supposed to have an exponent of -0.2T, instead of the 0.2T that you show?

And if you want a specific answer as to how much remains after 30 days, you need to tell us how much we started with....

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Your model is using 60 units or pounds for a starting quantity, and ASSUMING you forgot to include the NEGATIVE SIGN in your exponent expression, the model should be y=60%2A2%5E%28-0.2t%29. No usually any reason for using all upper case letters including the variables, unless you have some specific purpose for distinguishing between upper and lower cases.

Let t=30, and solve for y.

y=60%2A2%5E%28-0.2%2A30%29
y=60%2A2%5E%28-6%29
60%281%2F64%29
highlight%28y=0.9375%29 or y=15%2F16, or fifteen ounces