Posted by Ann on Friday, August 6, 2010 at 10:18pm.
I am of need of HELP. I am working on the word problems for AXIA math 117 and I do not understand the first problem. I have listed the information below. I have tried many times and I have not figured this out. My assignment is due tomorrow and I am desperate for help with work shown so I can understand. There is multiple parts and I only copied part of it.
The cost, in millions of dollars, to remove x % of pollution in a lake modeled by the equation C=6,000/200-2x^2
What is the cost to remove 20% of the pollutant?
Please help! Thanks!
- Axia Math 117 - Reiny, Friday, August 6, 2010 at 11:26pm
First of all, I don't think you meant to type the equation as you did, or else why not reduce it to
C = 3000 - 2x^2
So I will assume you meant
C = 6000/(200-2x^2)
Secondly, the way you defined x % , we would have to replace x with 20, and then your result would be
C = 6000/(200 - 800) which will result in a negative cost, thus making no sense.
If we replace 20% with 0.2, we would get
C = 6000/(200 - .08) = 30.012
but the cost of removing 90% would then be
6000/(200 - 1.62) = 30.24
which is only slightly higher than removing 20%.
This does not make sense to me, so please check the typing of the equation, and the definition of x
Answer this Question
- Algebra 1 - Well we are doing factoring and I don't quite understand this ...
- Axia HIS-125 - Need axia SYLLABUS HIS-125 US HISTORY FROM 1865 TO 1945
- algebra - Find the value of y for a given value of x, if x varies directly with ...
- Fallacy - My assignment is to find fallacies in a newspaper editorial and a ...
- math/117 - Explain three rules for exponents listed in the chart on p. 239 (...
- math - A bike costs $117. This is $12 more than 3 times the amount Tom saved ...
- Axia College - Need help with solving the double equation problem. 3x + 2y = 3 ...
- algelbra 2 - I have a 3.50 but have 7 disks problem and can't think straight, my...
- english - Review the four paragraphs below. There is one paragraph matching each...
- com155 - can someone please review my answers to the questions below and let me ...