00:03
All right, so here's the function that we're given in the problem, and it represents the cost in dollars for removing x percent of pollutants from a bayou where there was a spill.
00:14
And so for part a, we're asked to find the cost for removing 20 percent of the pollutants.
00:21
So i'm calling that f of 20.
00:23
So we're going to substitute 20 in for x in the function, and then we're going to simplify it.
00:31
And that works out to be 6 million, oops, how about 2 million divided by 80, and that is $25 ,000.
00:50
Okay.
00:51
And in part b, we're asked to find the cost of removing 60 % and the cost of removing 80 % of the pollutants.
00:59
So we do the same thing.
01:00
For f of 60, we put 60 in for x, and we simplify.
01:11
And that gives us 6 million divided by 4%.
01:16
And that works out to be $150 ,000.
01:21
So $150 ,000.
01:23
For f of 80, same idea.
01:26
Substitute the 80 in there and simplify.
01:36
Here we have $8 million divided by 20.
01:40
And that works out to be $400 ,000.
01:46
Okay, and then we have part c.
01:48
And the idea of part c is to get us to think about what's happening as the percent of pollutants gets higher and higher and closer to 100 percent.
01:56
So we're finding f of 90 and f of 99.
02:07
All right, so we'll plug 90 into our function for x, 100 ,000 times 90, divided by 100 minus 90...