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#52 Re: Help Me ! » I need your help » 2005-07-04 03:18:20

I just used a program, like what the administrator was talking about, that I made myself

#53 Re: Help Me ! » I need your help » 2005-07-03 19:20:04

ganesh I just used a calculater to figure it out so it is accurate

#55 Re: Help Me ! » can someone help me solve this riddle? » 2005-06-20 09:27:08

Umm just to let you konw this problem was already posted a few days ago under the name of  "I have a conundrum", the answer is there.

#56 Re: Help Me ! » i need help » 2005-06-19 14:37:35

What I meant is 4 and 6\7 as a mixed fraction simplyfied from the improper fraction of 34\7 where the 34 is divided by 7, 4 equal times with 6\7 left over, so it 4 and 6\7

#57 Re: Help Me ! » i need help » 2005-06-17 17:49:31

for the first part to find the mean just add all the numbers together like this:
8 + 7 + 7 + 5 + 3 + 2 + 2 = 34   then you divide that answer by the amount of terms or the amount of numbers, which is 7.   so then it would look like this:   34/7  which simplifys down to 4 6/7.

For the second problem think of f(x) to be just another way to put down a variable, "y"
so f(x) would just be y. so then for f (4) = 0 means that some function gives out the value of 0 when y = 4 and the second part f (6) = 6 means that the same function gives out a 6 when y = 6. so then you know that the function has to give out a 0 when y = 4, so then you can just put in 4 for all the x's untill one of them gives out the value of 0, in this case it is 3x - 12 where
3*4 - 12,  12 - 12, 0 , so that equation works for the value of 4. then you check it with the other part where y = 6 so 3*6 - 12,  18 - 12, which turns out a 6. so the equations works for both the values so it is the right answer.

#58 Re: Help Me ! » fourth dimension » 2005-06-16 09:43:31

These are all good ideas and yes I do agree with you mathsyperson.  The reason im interested in this topic is just for my own knowledge and curiosity. Right now I am reading something called flatland to try to help me understand more about other dimensions, When im done I will tell about anything interesting I learned. Oh and sorry if I started this in the wrong forum my mistake.

#59 Help Me ! » fourth dimension » 2005-06-15 12:49:39

im really bored
Replies: 8

Im looking for any ideas on how to explain the fourth dimesion to someone in a non mathmatical way

#60 Re: Help Me ! » Last one hopefully » 2005-05-21 07:28:20

Yes it is a 3 divided by a 2.   Now after following some of their advice I combined both terms to make one term which is:     (3x - 3)    +    (sqrt(2x^2 - 7x - 4))   All divided by 2.  But I can't get any further at the moment

#61 Re: Help Me ! » Last one hopefully » 2005-05-20 10:51:53

well anyways back to my original question,     I got some help from a friend and they said to start by making the (3/2)(x-1) into one term, and then multiply the sqrt by something that will allow you to combine it with the expanded fraction, then go from there. And they said that the final solution will end up with a square root inside of a square root, and that there wasnt a way that they knew to get around it.

#62 Re: Help Me ! » Some quick help please » 2005-05-20 10:46:39

Thanks, I was having some mental issues at the time...

#63 Help Me ! » Some quick help please » 2005-05-19 13:41:25

im really bored
Replies: 2

factor completely:   128x^6  -  2y^6

I can expand it to:   2(8x^3 + y^3)(8x^3 - y^3)

I know it can be expanded more but how would that go?

#64 Re: Help Me ! » Last one hopefully » 2005-05-18 09:30:30

So does anyone have any idea on how to do this problem?

#65 Re: Help Me ! » Last one hopefully » 2005-05-17 11:39:42

Im not having any better luck with this problem

#66 Help Me ! » Last one hopefully » 2005-05-17 09:43:53

im really bored
Replies: 18

Find the square root of (3/2)(x-1) + (sqrt(2x^2 - 7x - 4))

its not equal to anything so its all about manipulation, thats about all I can help lol.

#67 Re: Help Me ! » Another problem :( » 2005-05-15 15:30:51

thanks a bunch I hate logarithm stuff

The special property is i think just the property for logarithms with the same base:

where for positive mumbers b,x, and y where b != 1, log(b/x) = log(b/y) if and only if x = y.
I think...

#68 Re: Help Me ! » Another problem :( » 2005-05-15 12:35:04

and so x = -5, 10  I got that to

now I just need help on the second problem

#69 Re: Help Me ! » Another problem :( » 2005-05-15 12:02:49

I have another problem too, a logarithm problem

solve  2^(x +1) = 7^(x + 2)

#71 Help Me ! » Another problem :( » 2005-05-15 11:57:04

im really bored
Replies: 6

solve,    sqrt(11 - x)   -   sqrt(x + 6)  = 3

#72 Help Me ! » math help » 2005-05-15 05:37:19

im really bored
Replies: 1

Preform the indicated operations.  Simplify the answer in simplest form.


1 + _____1______               
                 
      1 + ____1____
                 1 - x


so far ive gotten here          1 + ______1_______
                                                 1 - x   +    1
                                                _____________
                                                      1 - x

#73 Help Me ! » Another Algebra Problem » 2005-05-12 13:47:03

im really bored
Replies: 1

if x = (c - ab) / (a - b)  , find the value of the expression a(x + b) in simplified form

So far I have gotten this far

substitute : a ( (c - ab) / (a - b) + b )

simplify:  ( a(c - ab) / a - b )  + ( ab/ 1 )

           (  a(c - ab) + ab(a - b) ) / (a - b)

        (  ca - a^2(b) + a^2(b) - ab^2  ) / (a - b)

       (  ca - ab^2 )   /  (a - b)


Is this all I could simplify it, or am I missing something?

#74 Re: Help Me ! » Algebra help » 2005-05-12 12:44:13

I know, but i can handle that.  It was just my approach that was getting me. I was trying to go at it in a more complicated way

#75 Re: Help Me ! » Algebra help » 2005-05-12 12:33:45

Wow it makes alot more sense putting it that way, I was tying to aproach it from some other way, which turned out horrible.Thanks for the help I understand it now

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