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)
To solve the equation 3 = 60(30)^(x/1590), we can start by dividing both sides of the equation by 60:
3/60 = (30)^(x/1590)
Simplifying the left side:
1/20 = (30)^(x/1590)
To solve for x, we need to isolate the exponential term on the right side of the equation. We can do this by taking the logarithm (base 30) of both sides:
log30(1/20) = log30((30)^(x/1590))
Using the logarithm property loga(b^c) = c * loga(b), we can simplify the equation further:
log30(1/20) = (x/1590) * log30(30)
Since loga(a) = 1 for any base a, we can simplify further:
log30(1/20) = (x/1590) * 1
log30(1/20) = x/1590
Now, to solve for x, we can multiply both sides of the equation by 1590:
1590 * log30(1/20) = x
Using a calculator, evaluate log30(1/20) and multiply by 1590 to get the approximate value for x.