- Tue Feb 03, 2009 2:44 pm
- Forum: Beginning Algebra
- Topic: [SPLIT] A company installs underground cable for $500 for 1st 50
- Replies:
**8** - Views:
**5064**

at some point L has to be greater than 50

- Tue Feb 03, 2009 2:23 pm
- Forum: Beginning Algebra
- Topic: [SPLIT] A company installs underground cable for $500 for 1st 50
- Replies:
**8** - Views:
**5064**

my reasoning for the formula was simple. 500for the first 50 ft and 5 times for the extra length.

I see a pattern of which as the length increase by five. Is that what you referring to???

I see a pattern of which as the length increase by five. Is that what you referring to???

- Tue Feb 03, 2009 1:31 pm
- Forum: Beginning Algebra
- Topic: [SPLIT] A company installs underground cable for $500 for 1st 50
- Replies:
**8** - Views:
**5064**

hello eliz!!!

the calculation I came up with is

c=5l +500

I dont think that's quite right

the calculation I came up with is

c=5l +500

I dont think that's quite right

- Tue Feb 03, 2009 1:12 pm
- Forum: Beginning Algebra
- Topic: [SPLIT] A company installs underground cable for $500 for 1st 50
- Replies:
**8** - Views:
**5064**

hello eliz!!!!

I see your point!!! my problem is I am not sure how to set up the fuction.

I see your point!!! my problem is I am not sure how to set up the fuction.

- Tue Feb 03, 2009 12:12 pm
- Forum: Beginning Algebra
- Topic: [SPLIT] A company installs underground cable for $500 for 1st 50
- Replies:
**8** - Views:
**5064**

hello!!! I have a word problem of which I do not understand. A company installs underground cable at a cost of $500 for the first 50ft and $5 for each additional foot thereafter. Express the cost C as a function of the length L of underground cable if L >50ft do anyone have a master technique that I...