# One MOre Help

• Mar 7th 2006, 08:52 PM
autopimp
One MOre Help

1. $\frac{ 6- \frac{a}{6}}{\frac{a}{3}}$

2. $a-\frac{1}{3}=-\frac{2}{5}$

3. Solve: $x-10=49$

4. $-(-\{-[-(-64)]\})$

5. Solve (use elimination):

$4x-5y=10$
$3x-2y=4$

6. $-|-3+4|+5-7$

7. Solve (use substitution)

$2x+3y=7$
$x-2y=7$

8. $\frac{18}{3} \times \frac{5}{4} \div \frac{3}{4}$

9. What fraction of $4\frac{2}{3}$ is $\frac{4}{9}$

10. $21\frac{7}{8}+3\frac{3}{4}$

11. $-4 \sqrt{27}+2\sqrt{12}$

12. $\frac{\frac{y}{b}+\frac{x}{b}}{\frac{y}{b}-\frac{x}{b}}$
• Mar 8th 2006, 02:02 AM
CaptainBlack
1. $\frac{ 6- \frac{a}{6}}{\frac{a}{3}}$

$\frac{ 6- \frac{a}{6}}{\frac{a}{3}}= \frac{ (6- \frac{a}{6})(\frac{3}{a})}{(\frac{a}{3})( \frac{3}{a})}$ (multiplying top and bottom by $\frac{3}{a}$)

So:

$
\frac{ 6- \frac{a}{6}}{\frac{a}{3}}= \frac{18}{a}-\frac{1}{2}=\frac{36-a}{2a}
$
• Mar 8th 2006, 05:11 AM
topsquark
Quote:

Originally Posted by autopimp

1. $\frac{ 6- \frac{a}{6}}{\frac{a}{3}}$

2. $a-\frac{1}{3}=-\frac{2}{5}$

3. Solve: $x-10=49$

4. $-(-\{-[-(-64)]\})$

5. Solve (use elimination):
$4x-5y=10$
$3x-2y=4$
6. $-|-3+4|+5-7$

7. Solve (use substitution)
$2x+3y=7$
$x-2y=7$
8. $\frac{18}{3} \times \frac{5}{4} \div \frac{3}{4}$

9. What fraction of $4\frac{2}{3}$ is $\frac{4}{9}$

10. $21\frac{7}{8}+3\frac{3}{4}$

11. $-4 \sqrt{27}+2\sqrt{12}$

12. $\frac{\frac{y}{b}+\frac{x}{b}}{\frac{y}{b}-\frac{x}{b}}$

Ummm...Auto? You are aware I already gave you solutions for 3, 4, and 7 right? If you couldn't understand them for some reason, just feel free to reply to my responses.

-Dan
• Mar 8th 2006, 01:26 PM
autopimp
oh i didnt se that my bad
• Mar 9th 2006, 01:54 PM
autopimp
pleace help me with the probs plz
• Mar 10th 2006, 04:53 AM
topsquark
Quote:

Originally Posted by autopimp
pleace help me with the probs plz

9. What fraction of $4\frac{2}{3}$ is $\frac{4}{9}$?

We are looking for a number x, such that $x \left (4\frac{2}{3} \right )=\frac{4}{9}$.

The first thing to do, as always, is to get rid of the mixed number $4\frac{2}{3}$.
$4\frac{2}{3}=4+\frac{2}{3}=\frac{12}{3}+\frac{2}{3 }=\frac{14}{3}$

So then we need a solution to $x \left ( \frac{14}{3} \right )=\frac{4}{9}$.

$x \frac{14}{3} =\frac{4}{9}$

$x \frac{14}{3} * \frac{3}{14}=\frac{4}{9} * \frac{3}{14}$

$x=\frac{4*3}{9*14}=\frac{2}{3*7}=\frac{2}{21}$.

So the answer is that $\frac{4}{9}$ is $\frac{2}{21}$ of $4\frac{2}{3}$.

-Dan

PS You have now had, I believe, examples of how to solve the rest of the problems on this post. If I may make a suggestion: It would probably help you more if you told us exactly what you are having a problem with, so we can focus our attention on that, rather than simply providing you with solutions.
• Mar 10th 2006, 02:17 PM
autopimp
i need help with this problems that i showed you i dont get them
• Mar 12th 2006, 02:52 PM
autopimp
plz some one anyone?
• Mar 12th 2006, 03:39 PM
topsquark
Quote:

Originally Posted by autopimp
plz some one anyone?

As I pointed out in the forum e-mail I sent to you when you asked me for help, it would be a good idea for you to tell us what it is specifically you need help on. You have posted several problems and gotten several solutions. Obviously you need more help than just simply seeing the solutions because you seem to be unable to apply those examples. If you can tell us what help you need, we can help you better.

-Dan
• Mar 12th 2006, 06:00 PM
autopimp
i just asking show me how to do those probs on the top and ill try to do other from my book
• Mar 13th 2006, 10:07 AM
topsquark
Quote:

Originally Posted by autopimp
i just asking show me how to do those probs on the top and ill try to do other from my book

Please let me fully explain my position on all of this. First I want to point out that these opinions and habits are mine alone. I am not a moderator at this forum, nor do I work for the site's owners in any way.

I would be quite pleased to do these problems for you. Heck, I'd do all of your homework for the rest of the school year, under the condition that it was apparent you were getting something out of it (besides merely a good grade, that is).

What is perplexing me here and making me wary is two related things. The level of complexity of the problems you have posted and asked solutions for is widely varied from #2 on your list, which solution is usually taught in Junior High or Freshman year to #1 and #12 which probably wouldn't be introduced until Junior or Senior year. In addition, anyone who has any chance at doing either #1 or #12 should reasonably have little difficulty with #2, especially since I already gave you the method in another post: http://www.mathhelpforum.com/math-he...=6823#post6823 see #10.

I apologize if it sounds snobbish, but I am left with the disturbing feeling that you are interested only in getting solutions and not learning the process. Until such time as you can provide me with more information about what specific difficulties you are having, that distrust will remain. Thus I will feel uncomfortable about providing solutions.

Obviously someone else on the forum may decide to provide those solutions for you, and I want to say that I will not judge them on that decision because it is entirely up to them what they post here. At this point in time, however, I cannot in good conscience do so. If you choose not to respond to this, I wish you the best.

Best Regards,
-Dan
• Mar 15th 2006, 04:26 PM
autopimp
Thanks man for helping me sorry about posting probs two times :( but i need help with this probs ASAP plz do i'm in 9Th Grade thats Alebgra 1 probs