## Search found 11 matches

Fri Aug 19, 2011 11:06 am
Forum: Pre-Algebra
Topic: Solve, by steps: (1/sqrt[5]+1/sqrt[2]):(sqrt[5]+sqrt[2])
Replies: 4
Views: 8469

### Re: Solve, by steps: (1/sqrt[5]+1/sqrt[2]):(sqrt[5]+sqrt[2])

That is all of the problem, I meant simplify in that case. yes it is division
Fri Aug 19, 2011 10:28 am
Forum: Pre-Algebra
Topic: Solve, by steps: (1/sqrt[5]+1/sqrt[2]):(sqrt[5]+sqrt[2])
Replies: 4
Views: 8469

### Re: Solve, by steps

(\frac{1}{sqrt5}\, +\, \frac{1}{sqrt2}\,) :\, (\sqrt{5}\, +\, \sqrt{2}) (\frac{sqrt{2}\, +\, sqrt{5}}{sqrt10}\,) :\, (\sqrt{5}\, +\, \sqrt{2}) (\frac{sqrt{2}\, +\, sqrt{5}}{sqrt10}\,) :\, (\frac{\sqrt{5}\, +\, \sqrt{2}}{1}) \frac{sqrt{2}\, +\, sqrt{5}}{sqrt10}\, *\, \frac{1}{\sqrt{5}\, +\, \sqrt{2}...
Fri Aug 19, 2011 9:42 am
Forum: Pre-Algebra
Topic: Solve, by steps: (1/sqrt[5]+1/sqrt[2]):(sqrt[5]+sqrt[2])
Replies: 4
Views: 8469

### Solve, by steps: (1/sqrt[5]+1/sqrt[2]):(sqrt[5]+sqrt[2])

$(\frac{1}{sqrt5}\, +\, \frac{1}{sqrt2}\,) :\, (\sqrt{5}\, +\, \sqrt{2})$
Fri Jul 29, 2011 9:23 am
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re:

i still dont know what to do with this

$20(v\, \frac{v\, -\, 1}{4}\) \,$
Multiply it out. Cancel the four with the twenty.
$5(v^2\, -\, 1\,) \,$
Thu Jul 28, 2011 4:17 pm
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re: How to solve this problem

i still dont know what to do with this

$20(v\, \frac{v\, -\, 1}{4}\) \,$
Tue Jul 26, 2011 2:05 pm
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re:

What is the second term? a mixed number made up by variables? or is it v * (v-1)/4? or is it either? There are still no mixed numbers in this equation. To learn what "terms" are, please try this lesson . To learn how to simplify v(v - 1), please try this lesson or this lesson . :wink: is the soluti...
Mon Jul 25, 2011 6:35 pm
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re: How to solve this problem

What is the second term? a mixed number made up by variables? or is it v * (v-1)/4? or is it either?
Mon Jul 25, 2011 2:29 pm
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re: How to solve this problem

@stapel_eliz this is all the stages Im trying to solve the equation. Can you help? v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0 20v\, -\, 20v\, 20(\frac{v\, -\, 1}{4}\), +\, 20(\frac{v^2\, -\, 5}{5}\), +\, 20(\frac{1}{20}v^2\), =\, 20*0 20v\, -\, 20v\frac{20...
Mon Jul 25, 2011 6:51 am
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re: Re:

Im trying to learn algebra on the side...and Im have no idea how to solve this next equation.... IS the --v-1 a mixed number (the second division from the left)? No; mixed numbers are whole numbers together with fractions, like "three and a half". They do not contain variables. Ok, so its multiplie...
Sun Jul 24, 2011 11:18 am
Forum: Beginning Algebra
Topic: How to solve v - v(v - 1)/4 + (v^2 - 5)/5 + (1/20)v^2 = 0
Replies: 15
Views: 12162

### Re:

Is the equation as follows?

$v\, -\, v\frac{v\, -\, 1}{4}\, +\, \frac{v^2\, -\, 5}{5}\, +\, \frac{1}{20}v^2\, =\, 0$

Thank you.
Yes it is