Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 Re: Help Me ! » percenntage question » 2007-04-03 08:37:49
pandapatrol26 wrote:umm if you just rewrite 20% in decimal form, which is .02 its much easier this way.
anyway, you need a variable since you're solving for a number, so .02 (OF) what number gives you 8?
so here: .02x=8technically you don't know what x is, that's what you're finding out. you know if you multiply .02 of a number you get 20% of it right? so this is the same thing, except you set it equal to eight because that's what its equal to. divide 8 by .02 and you should get 40.
Umm, 8 ÷ 0.02 ≠ 40.
I think you meant 20% = 0.2 and no 0.02, otherwise that'd be 2%.
So, 8 ÷ 0.2 = 40.
yeah my bad 
#2 Re: Help Me ! » percenntage question » 2007-04-03 08:00:15
umm if you just rewrite 20% in decimal form, which is .02 its much easier this way.
anyway, you need a variable since you're solving for a number, so .02 (OF) what number gives you 8?
so here: .02x=8
technically you don't know what x is, that's what you're finding out. you know if you multiply .02 of a number you get 20% of it right? so this is the same thing, except you set it equal to eight because that's what its equal to. divide 8 by .02 and you should get 40. 
#3 Help Me ! » calculus problem: need help » 2007-04-03 07:55:45
- pandapatrol26
- Replies: 3
Hi I have a calculus problem, and it would be great if you could give me some hints on how to solve it 
Supertankers off-load oil at a docking facility 4 miles offshore. The nearest refinery is 9 miles east of the shore point nearest the docking facility. A pipeline must be constructed connecting the docking facility with the refinery. The pipeline costs $300,000 per mile if constructed underwater and $200,000 per mile if overland.
a) Locate Point B to minimize the cost of the construction.
b) The cost of underwater construction is expected to increase, whereas the cost of overland construction is expected to stay constant. At what cost does it become optimal to construct the pipeline directly to Point A.
Docking Facility
|\
| \
^ l \
4 miles | l \
| l \ shore
V l--------\----------------------------------------------------
A B i put this as (9-x) Refinery
|--------------------------9 miles -------------|
(i put the distance from A to B as X)
* I solved question a by taking the hypotenuse = squareroot(16+X^2) and adding it to (9-x). Then I multiplied the hypotenuse by 300000 (since it's under water) and the (9-X) by 200000 since its on shore. I took the derivative of the entire equation:
1. 300000(sqroot(16+X^2) + (9-x)200000
2. f'(x) = (300000x)/(sqroot(X^2+16)) - 200000
Then I proceeded to solve for x:
X= sqroot(12.8)
X= 3.5777
Now I just need help with the second problem. Thanks
Pages: 1