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The half-life of a radioactive kind of praseodymium is 1717 minutes. If you start with 5656 grams of it, how much will be left after 3434 minutes?\newline____ grams

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Q. The half-life of a radioactive kind of praseodymium is 1717 minutes. If you start with 5656 grams of it, how much will be left after 3434 minutes?\newline____ grams
  1. Given values: We have:\newlineInitial amount: 5656 grams\newlineTotal time: 3434 minutes\newlineHalf-life period: 1717 minutes\newlineIdentify the values of aa, tt and hh.\newlineInitial amount (a)=56(a) = 56 grams\newlineTotal time (t)=34(t) = 34 minutes\newlineHalf-life period (h)=17(h) = 17 minutes
  2. Substitute values in formula: We found:\newlinea=56a = 56\newlinet=34t = 34\newlineh=17h = 17\newlineIdentify the formula of half-life after substituting the values.\newlineSubstitute a=56a = 56, t=34t = 34, and h=17h = 17 in y=a(12)thy = a \cdot \left(\frac{1}{2}\right)^{\frac{t}{h}}.\newliney=56(12)3417y = 56 \cdot \left(\frac{1}{2}\right)^{\frac{34}{17}}
  3. Simplify exponent: Let's simplify the exponent in y=56×(12)3417y = 56 \times (\frac{1}{2})^{\frac{34}{17}}. Simplify 3417\frac{34}{17}. 3417=2\frac{34}{17} = 2
  4. Calculate remaining quantity: We have:\newliney=56×(12)2y = 56 \times (\frac{1}{2})^2\newlineCalculate the remaining quantity of the radioactive substance.\newliney=56×(12)2y = 56 \times (\frac{1}{2})^2\newline=56×14= 56 \times \frac{1}{4}\newline=14= 14\newlineRemaining quantity of praseodymium after 3434 minutes: 1414 grams

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