Nana has a water purifier that filters

Question

\[\dfrac{1}3\] of the contaminants each hour. She used it to purify water that had

\[\dfrac{1}2\] kilogram of contaminants.
Write a function that gives the remaining amount of contaminants in kilograms,
\[C(t)\],
\[t\] hours after Nana started purifying the water.
\[C(t)=\]

Answer 1

Im going to touch you lil bro!

Answer 2

To determine the amount of contaminants remaining after \( t \) hours, we can start with the initial amount of contaminants and see how the purifier's filtering process reduces that amount.

Given:

• The initial amount of contaminants is \( C(0) = \frac{1}{2} \) kg.
• Every hour, the amount of contaminants that remains after filtering is \(\frac{2}{3}\) of the previous hour's amount (since \(\frac{1}{3}\) of the contaminants is filtered out).

Thus, we can write a recursive formula for the remaining contaminants:

1. \( C(0) = \frac{1}{2} \)
2. After 1 hour: \[ C(1) = C(0) \cdot \frac{2}{3} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \]
3. After 2 hours: \[ C(2) = C(1) \cdot \frac{2}{3} = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9} \]
4. After 3 hours: \[ C(3) = C(2) \cdot \frac{2}{3} = \frac{2}{9} \cdot \frac{2}{3} = \frac{4}{27} \]

From this pattern, we see that each hour the remaining contaminants are multiplied by \(\frac{2}{3}\). Therefore, after \( t \) hours, we can express the remaining amount of contaminants as:

\[ C(t) = C(0) \cdot \left( \frac{2}{3} \right)^t = \frac{1}{2} \cdot \left( \frac{2}{3} \right)^t \]

Thus, the function that gives the remaining amount of contaminants in kilograms after \( t \) hours is:

\[ C(t) = \frac{1}{2} \cdot \left( \frac{2}{3} \right)^t \]