Google Answers Logo
View Question
 
Q: math equation ( Answered 5 out of 5 stars,   2 Comments )
Question  
Subject: math equation
Category: Reference, Education and News > Homework Help
Asked by: fizzleclizze-ga
List Price: $5.00
Posted: 25 Feb 2005 14:50 PST
Expires: 27 Mar 2005 14:50 PST
Question ID: 480923
how do i solve this equation?

(1.012)(1 + F)^2 = 1.0457
Answer  
Subject: Re: math equation
Answered By: skermit-ga on 25 Feb 2005 14:55 PST
Rated:5 out of 5 stars
 
First divide both sides by (1.012):

(1 + F)^2 = 1.0457/1.012

Then take the square root of both sides:

(1 + F) = sqrt(1.0457/1.012)

Then subtract 1 to get F alone:

F = sqrt(1.0457/1.012) - 1

Just for your convenience, I evaluated it already, and the answer is:
0.016513844105

Thank you for your question.

skermit-ga
fizzleclizze-ga rated this answer:5 out of 5 stars

Comments  
Subject: Re: math equation
From: jasonkoh-ga on 25 Feb 2005 16:38 PST
 
0.016513844105 is only half the answer.

When you take the square root of both sides, the +/- (plus or minus)
sign needs to be added :
 
 (1 + F) = +/- sqrt(1.0457/1.012)

therefore,
F = 0.016513844105 or -2.016513844105


good luck with your studies!
Subject: Re: math equation
From: skermit-ga on 25 Feb 2005 17:03 PST
 
Yup, forgot about that. Thanks jasonkoh-ga!

skermit-ga

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  


Google Home - Answers FAQ - Terms of Service - Privacy Policy