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Old 05-15-2007   #1
alwaysalillost
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Post Algebra...Solve

One eighth of the square root of 7 less than a number is 2. What is the number?

What i get is the equation 1/8 sqrt7 - x = 2 but I get stuck on the solving part. Can anyone please help me? Thanks in advance!
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Old 05-15-2007   #2
Temperal
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Or,

Nothing but basic Algebra.
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Old 05-15-2007   #3
alwaysalillost
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n - (1/8)[sqrt(7)] = 2
n - (1/8)[sqrt(7)] + (1/8)[sqrt(7)] = 2 + (1/8)[sqrt(7)]
n = 2 + (1/8)[sqrt(7)]
The number is 2 + (1/8)[sqrt(7)]

That's what I got.
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Old 05-15-2007   #4
Temperal
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That's not simplest form. I got the same thing, but simplified.
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Old 05-17-2007   #5
Sillysidley

 
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Hmmm... Some people could read this and think it as:
=2
It's not exactly clear which one it is.
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Old 06-15-2007   #6
YvesBenoit
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I would see this as

(1/8)sqrt(7) - n = 2
...
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Old 06-16-2007   #7
soccerstar6997
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how do you guys get the forula to come up here?

Last edited by Scion; 06-17-2007 at 08:36 PM..
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Old 06-17-2007   #8
Scion

 
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Quote:
Originally Posted by soccerstar6997 View Post
how do you guys get the forula to come up here?

latex
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Old 06-22-2007   #9
hannah7

 
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Quote:
Originally Posted by somedude View Post
Hmmm... Some people could read this and think it as:
=2
It's not exactly clear which one it is.


how do you even make those things?oohhhhhhhhhhhhhh riiight...latex....is this algebra?
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Old 06-22-2007   #10
Temperal
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Look at the title.

http://www.artofproblemsolving.com/L...PS_L_About.php - Good LaTeX tutorial.
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