exponential functions

chassity

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Nov 14, 2010
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The area of a wetland drops by a sixth every four years. What percent of its total area disappears after twenty years?

Round your answer to two decimal places.


I cant figure this one out... i keep getting 99.987 which would be 99.99% and when i enter that it tells me its wrong. any help would be greatly appreciated.
 

Denis

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Feb 17, 2004
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HINT:
(1 - 1/6)^(20/4) = (5/6)^5 = .4018775....
 

soroban

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Jan 28, 2005
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Hello, chassity!

The area of a wetland drops by a sixth every four years.
What percent of its total area disappears after twenty years?
Round your answer to two decimal places.

\(\displaystyle \text{Every 4 years, the wetland has only }\tfrac{5}{6}\text{ of its previous area.}\)

\(\displaystyle \text{The function is: }\:A \:=\:A_o\left(\tfrac{5}{6}\right)^\frac{x}{4}}\;\;\text{ where: }\:\begin{Bmatrix}A_o &=& \text{initial area} \\ x &=& \text{no. of years} \end{Bmatrix}\)

\(\displaystyle \text{For }x = 20\text{, we have: }\:\frac{A}{A_o} \:=\:\left(\tfrac{5}{6}\right)^5\:=\:0.401877572 \;\approx\;40.19\%\)


\(\displaystyle \text{After 20 years, only 40.19\% of the wetland remains.}\)

\(\displaystyle \text{Therefore, about 59.81\% of the total area has disappeared.}\)

 

lookagain

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Aug 22, 2010
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2,373
chassity said:
The area of a wetland drops by a sixth every four years.
What percent of its total area disappears after twenty years?

Round your answer to two decimal places.
chassity,

for 20 years with a drop every 4 years, there are a total of 5 drops. After the first 4 years,

there is a drop by a \(\displaystyle \frac{1}{6}\) of the amount, leaving \(\displaystyle \frac{5}{6}\) of the amount remaining.

\(\displaystyle \frac{5}{6}A - \frac{1}{6}(\frac{5}{6}A)\) leaves a drop by \(\displaystyle \frac{1}{6}(\frac{5}{6})\) of the amount, and this is added to

the first drop by \(\displaystyle \frac{1}{6}\) of the amount. Then there is \(\displaystyle \frac{25}{36}A\) remaining after 8 years.

Continuing in this way, the sum of the drops for the total of 5 drops is

\(\displaystyle \frac{1}{6} + \frac{1}{6}( \frac{5}{6}) + . . . + \frac{1}{6}(\frac{5}{6})^4 \ = \\)

\(\displaystyle \frac{1}{6}[1 + \frac{5}{6} + . . . +(\frac{5}{6})^4] \ = \ ?\)


Please finish the calculation (including converting the result into a percent).
 
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