document.write( "Question 452493: an element decays in such a way that every 50 years the amount of the element of the element has decreased by 15% in the year 1900,120mg of element was present.
\n" ); document.write( "a)what is amount in 2000
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Algebra.Com's Answer #311046 by nerdybill(7384)\"\" \"About 
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an element decays in such a way that every 50 years the amount of the element of the element has decreased by 15% in the year 1900,120mg of element was present.
\n" ); document.write( "a)what is amount in 2000
\n" ); document.write( ".
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( "where
\n" ); document.write( "A is amount after time t
\n" ); document.write( "P is the initial amount
\n" ); document.write( "r is the rate of growth/decline
\n" ); document.write( "t is time
\n" ); document.write( ".
\n" ); document.write( "first, we determine r...
\n" ); document.write( "Let x = initial amount
\n" ); document.write( "then
\n" ); document.write( "from: \"an element decays in such a way that every 50 years the amount of the element of the element has decreased by 15%\"
\n" ); document.write( "x - .15x = xe^(50r)
\n" ); document.write( "x(1 - .15) = xe^(50r)
\n" ); document.write( "(1 - .15) = e^(50r)
\n" ); document.write( ".85 = e^(50r)
\n" ); document.write( "ln(.85) = 50r
\n" ); document.write( "ln(.85)/50 = r
\n" ); document.write( "-0.0033 = r
\n" ); document.write( ".
\n" ); document.write( "Our formula then is:
\n" ); document.write( "A = Pe^(-0.0033t)
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\n" ); document.write( "Now, we can answer the question:\r
\n" ); document.write( "\n" ); document.write( "in the year 1900,120mg of element was present.
\n" ); document.write( "a)what is amount in 2000
\n" ); document.write( "t = 2000-1900 = 100
\n" ); document.write( "A = 120e^(-0.0033*100)
\n" ); document.write( "A = 120e^(-0.33)
\n" ); document.write( "A = 86.7 mg
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