VoyForums
[ Show ]
Support VoyForums
[ Shrink ]
VoyForums Announcement: Programming and providing support for this service has been a labor of love since 1997. We are one of the few services online who values our users' privacy, and have never sold your information. We have even fought hard to defend your privacy in legal cases; however, we've done it with almost no financial support -- paying out of pocket to continue providing the service. Due to the issues imposed on us by advertisers, we also stopped hosting most ads on the forums many years ago. We hope you appreciate our efforts.

Show your support by donating any amount. (Note: We are still technically a for-profit company, so your contribution is not tax-deductible.) PayPal Acct: Feedback:

Donate to VoyForums (PayPal):

Login ] [ Contact Forum Admin ] [ Main index ] [ Post a new message ] [ Search | Check update time | Archives: 1 ]
Subject: Re: please help me!


Author:
QUITTNER
[ Next Thread | Previous Thread | Next Message | Previous Message ]
Date Posted: 12:31:45 02/13/04 Fri
In reply to: Rawan Kilo 's message, "please help me!" on 18:47:48 02/09/04 Mon

>The volume of the rectangular solid is given by the
>expression 34.944x(cubed)+44.684x(squared)-8.29x-16.8.
>The surface area of one side of the rectangular solid
>is given by the expression 6.72x(squared)+12.47+5.6.
>The sum of the dimensions of the rectangular solid to
>the nearest hundredth.
..... What is the difficulty? You are given a volume of a rectangular solid in terms of an equation, call it y3 (3-dimensional).
..... You are also given the surface area of one side in terms of an equation, call it y2 (2-dimensional).
..... The volume of a rectangular solid is one side's length times the surface area of one side, so that y3 divided by y2 should give you the equation of the side's length, call it y1 (1-dimensional).
..... The graphs of these equations are partly negative, so you must first mention, then reject the negative parts. y1 appears to be sufficiently linear (depends on how many decimal places are used in the calculations), so that any two points can be used to get the equation of that line, such as the (approximate) root of y1 and its (-ve) intercept on the y-axis. Can you take it from here?

[ Next Thread | Previous Thread | Next Message | Previous Message ]

Replies:
[> [> [> Subject: Re: please help me!


Author:
geeknick (sad)
[ Edit | View ]

Date Posted: 23:59:29 11/04/11 Fri

1/1+a^n + 1/1+a^-n
My attempt:=1/1+a^n + 1+a^n/1=1+{1+a^n(1+a^n)}/1+a^n=1+1+2a^n+a^n^2/1+a^n=2+2a^n+a^n^2/1+a^nlet a^n=m=2+2m+2m^2/1+m=2(1+m+m^2)/1+m
I don't know what to do next, the answer is supposed to be 1.

will you please simplify the expression its very urgent... i tried on Many forum sites but didn't get the correct answer according to my mathematics textbook.


[ Post a Reply to This Message ]


Post a message:
This forum requires an account to post.
[ Create Account ]
[ Login ]
[ Contact Forum Admin ]


Forum timezone: GMT-8
VF Version: 3.00b, ConfDB:
Before posting please read our privacy policy.
VoyForums(tm) is a Free Service from Voyager Info-Systems.
Copyright © 1998-2019 Voyager Info-Systems. All Rights Reserved.