teller
  • teller
Someone mind checking my work? The radioactive substance cesium-137 has a half-life of 30 years. The amount A(t) (in grams) of a sample of cesium-137 remaining after t years is given by the following exponential function. A(t)=381(1/2)^t/30 Find the initial amount in the sample and the amount remaining after 50 years. Round your answers to the nearest gram as necessary. initial amount = 381 after 50 years = 1270
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
What do you mean? I see no work. A radioactive substance would have less grams after 50 years. In other words, radioactive substances decreases in grams every year.
teller
  • teller
I sent my work down below where it says initial amount = 381 after 50 years = 1270
teller
  • teller
I plugged the information into my online calculator and that's what I got after 50 years.

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anonymous
  • anonymous
You could at least say you plugged it in. There was no indicator to what you did except put in the answer.
anonymous
  • anonymous
Because I dont think you plugged it right.
teller
  • teller
A(50)=381(1/2)^50/30 got me 1270
anonymous
  • anonymous
\[381 \times (0.5)^{\frac{ 50 }{ 30 }}\]
teller
  • teller
But my thing says to use the formula A(t)=381(1/2)^t/30
anonymous
  • anonymous
Same thing.... (1/2)=0.5
teller
  • teller
120.00748
teller
  • teller
That's what I got.
teller
  • teller
so rounding it would make it 120?
anonymous
  • anonymous
Yea me too.
teller
  • teller
Alright thank you
anonymous
  • anonymous
No problem!
mathmate
  • mathmate
@teller Remember PEMDAS: A(t)=381(1/2)^t/30 equals A(t)=381[(1/2)^t]/30 which is not correct. You need to write A(t)=381(1/2)^(t/30) or else calculators will give wrong numbers. Calculators go strictly with PEMDAS, doesn't know what you want! xD
anonymous
  • anonymous
Yes exactly. I tried figuring out what @teller did wrong, but couldnt figure on how to get 1270.

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