# need help verifying answer ...

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• Apr 10th 2010, 06:42 AM
bringash
http://i43.tinypic.com/25auvfd.jpg

so constant = h/w = 25/10

constant = 2.5 , w = 10

2.5 x 10 = 25

would that be the answer ? or just 2.5?
• Apr 10th 2010, 06:57 AM
Sudharaka
Quote:

Originally Posted by bringash
http://i43.tinypic.com/25auvfd.jpg

so constant = h/w = 25/10

constant = 2.5 , w = 10

2.5 x 10 = 25

would that be the answer ? or just 2.5?

Dear bringash,

I think this is one of your online exam questions. Therefore you are to do this yourself and I hope that the moderators of this forum will look into this matter.
• Apr 10th 2010, 07:12 AM
bringash
Quote:

Originally Posted by Sudharaka
Dear bringash,

I think this is one of your online exam questions. Therefore you are to do this yourself and I hope that the moderators of this forum will look into this matter.

Will if you would really like to know ... my nephew who is in 9th grade asked me to help him with his algebra 2 homework that the teacher gives him online ... there are some that I am not sure about; thus, wanting to help my nephew understand this material to the fullest of his abilities i asked for some help ... you know what happens when you assume right sudharaka?
• Apr 10th 2010, 07:30 AM
Sudharaka
Dear bringash,

In that case (thinking my assumption is incorrect!!) I shall give you my ideas about the question. The answer 2.5 is correct.

$\displaystyle Constant=\frac{H(w)}{w}=\frac{25.00}{10.0}=2.50$ ; (using the rules of significant figures.)

Since it is given that the answer could be rounded to first decimal place 2.5 is also correct.

P.S: If you felt angry about my previous post I am really sorry about it. But MHF does not encourage any of it's members to give exam help.
• Apr 10th 2010, 07:42 AM
bringash
Quote:

Originally Posted by Sudharaka
Dear bringash,

In that case (thinking my assumption is incorrect!!) I shall give you my ideas about the question. The answer 2.5 is correct.

$\displaystyle Constant=\frac{H(w)}{w}=\frac{25.00}{10.0}=2.50$ ; (using the rules of significant figures.)

Since it is given that the answer could be rounded to first decimal place 2.5 is also correct.

P.S: If you felt angry about my previous post I am really sorry about it. But MHF does not encourage any of it's members to give exam help.

haha no i wasn't angry bro don't worry about it, it was me being a di*k not you ... thanks alot bro
• Apr 10th 2010, 08:38 AM
bringash
http://i42.tinypic.com/29qgydx.jpg

hey would the answer to this be x/2 (d) or (c) ? I am not completely sure although i think its d
• Apr 10th 2010, 08:44 AM
It can't be (d), try it with x = 2.

Generally $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$...
• Apr 10th 2010, 09:26 AM
bringash
Quote:

It can't be (d), try it with x = 2.

Generally $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$...

http://i41.tinypic.com/29woisp.jpg

i know it can either be (b) or (d), but im pretty sure it is B, am i wrong? if so how come ?
• Apr 10th 2010, 09:28 AM
You are correct. (I loled at "acceptable values of x".)
• Apr 10th 2010, 09:37 AM
bringash
Quote:

You are correct. (I loled at "acceptable values of x".)

yeah bro the wording for some of these problems kills me haha
• Apr 10th 2010, 09:40 AM
bringash
so if what i did for the other problem was correct ... the answer to this problem

http://i44.tinypic.com/2444ha9.jpg

would be (b) ?
• Apr 10th 2010, 09:51 AM
Ja.
• Apr 10th 2010, 09:59 AM
bringash
Quote:

Ja.

How would I solve this problem?
• Apr 10th 2010, 10:11 AM
Use $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$ to slit it up into $\displaystyle \sqrt{2}\sqrt{x^2}\sqrt{4}\sqrt{x}$, simplify, and then put the remaining radicals back together.
Use $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$ to slit it up into $\displaystyle \sqrt{2}\sqrt{x^2}\sqrt{4}\sqrt{x}$, simplify, and then put the remaining radicals back together.