Homework Help: Math - Algebra 2

Posted by AS on Saturday, December 2, 2006 at 12:48pm.

A chemical compound evaporates in a predictable way. In the first hour 1/2 of the amoun in the jar at the beginning of the hour evaporates. In the second hour 1/3 of the amount in the jar at the beginning of the second hour evaporates. In the third hour 1/4 of the amount in the jar at the beginning og the third hour evaporates, and so on. If a container started with 60L of the compound, after hour many hours would the amount of the comppound be less than 10L? SHOW ALL WORK.

60L x 1/2 (hour 1) = 30 L leaving 30L.
30L x 1/3 (hour 2) = 10 L leaving 20L.
20L x 1/4 (hour 3) = 5 L leaving 15L.
15L x 1/5 (hour 4) = 3 L leaving 12L.
12L x 1/6 (hour 5) = 2 L leaving 10L.

Note that the denominator of the fraction is 1 more than the time; therefore, t + 1.
60L x 1/(t+1) = 10 L
solve for t = 5 hours. Actually, you would need 1 second longer to have less than 10L. Check my thinking. Check my arithmetic.

3=60(30)^(x/1590)

3=60(30)^(x/1590)

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