Question

Nana has a water purifier that filters 1/3 of the contaminants each hour. She used it to purify water that had 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

Answer:

$C(t)=\frac{1}{2}-\frac{1}{3}t$

Step-by-step explanation:

We know that the water purifier filters 1/3 kg of contaminant per hour, that can be expressed as

$\frac{1}{3}t$

Where $t$ is time in hours.

Then, the problem states that Nana needs to purify water that has 1/2 kg of contaminants. In other words, the initial condition of the function is 1/2.

Uniting all these, we can express the situation as

$C(t)=\frac{1}{2}-\frac{1}{3}t$

Where $C(t)$ is the contaminant remaining in kilograms.

The function has to have a subtraction, because the purifier does that, it subtract contaminant from the water.

Therefore, the function that modelates this situation is

$C(t)=\frac{1}{2}-\frac{1}{3}t$

Answer 2

Answer:

C(t)= 0.5kg* (2/3) ^ t

Step-by-step explanation:

The water purifier will filter 1/3 of the current contaminants each hour. In other words, it will reduce the contaminant into 2/3 of it current mass. If the contaminant 1 kg, after 1 hour will be 2/3 kg, then become 4/9 kg after 2 hours. You have to multiply the contaminant with 2/3 every hour.

The equation will be

C(t)= X * (2/3) ^ t  

C(t)= 0.5* (2/3) ^ t