https://www.khanacademy.org/ Well, this site isn't like

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http://www.

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http://www.algebra.com/ http://www.

**mathway**.com/ http://www.analyzemath.com/Calculators.html http://www.algebra.com/services/rendering/simplifier.mpl Hope it helps!! :)9 Answers · Science & Mathematics · 07/03/2009

Well, I did an acronym search and the only thing that came up is :Quaker Youth Theater. Somehow that doesn't seem to match with a site named "

**mathway**.com", but you never know!2 Answers · Education & Reference · 06/12/2011

Not a scam. Legitimate website. http://www.mywot.com/en/scorecard/ http://safeweb.norton.com/report/show?url=

**Mathway**.com http://www.siteadvisor.com/sites/**mathway**.com4 Answers · Computers & Internet · 06/05/2010

**Mathway**is free my friend.4 Answers · Science & Mathematics · 26/10/2011

www.algebra.com www.mathsupport.com www.quickmath.com www.webmath.com www.algebra-answer.com www.solvemymath.com i hope these are enough!

1 Answers · Education & Reference · 08/07/2009

www.purplemath.com

1 Answers · Education & Reference · 25/02/2009

lim(x→∞) [√(0.25x^2 + 1) - 1]/(0.5x) = lim(x→∞) [√(0.25x^2 + 1) - 1] * [(√(0.25x^2 + 1) + 1] / {0.5x [(√(0.25x^2 + 1) + 1]} via conjugates = lim(x→∞) [(0.25x^2 + 1) - 1] / {0.5x [(√(0.25x^2 + 1) + 1]} = lim(x→∞) 0.5x / [(√(0.25x^2 + 1) + 1]. Now, divide each term by x = √x^2: lim(x→∞) 0.5 / [(√(0.25 + 1/x^2...

1 Answers · Science & Mathematics · 20/08/2012