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Math Help Thread
Topic Started: May 5 2009, 06:18 PM (110 Views)
Kotetsu May 5 2009, 06:18 PM Post #1
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This is a math thread. Post your math problems (calculus and below) here and someone will try to solve them for you.

*try to use the symbol map as much as possible*


I'll start, I don't feel like working this one right now.

∫ 4/e^2x dx

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Trickster May 5 2009, 10:06 PM Post #2
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Wow, I use to get b's in math, but it has been so long that i don't even know what half that stuff is.
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supermanjack94 May 10 2009, 07:41 AM Post #3
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ok here is a question:

try and figure it out plz, i cant do it for the life of me

btw i have the answer, i just dont know how the teacher found it...
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nah nah
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Kotetsu May 10 2009, 11:51 AM Post #4
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There's multiple ways of doing this, but I can't quite figure out how you can find a value for Y, if you don't have a value for X.
X does not equal Y
And 5 does not equal x.

One way is 5+x=y. As it's the radius of the circle.
Another is x^2+x^2=Y^2 Using Pythagorean Therom.

But algebraically you need another value as the radius to figure out Y.
And you can't use a double value cancellation method because your answer would wind up being 0x+0Y=0

Unless the picture is drawn bad and 5 is = to x, and if that's the case then the answer is super simple.

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supermanjack94 May 11 2009, 05:44 AM Post #5
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actually i figured it out and your were right with the one with the powers, you can get y=sqrt 2 times x

the u use 5+x=y

then u can figure it out with similatious equations.
nah nah
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Rebel X May 11 2009, 12:16 PM Post #6
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Congratulations, Problem Solved. We should do problems to help the community more often lol.
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* NeciFiX May 15 2009, 05:05 AM Post #7
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Lol Kotetsu if you could help me with my homework I'd be extremely happy.
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Rebel X May 15 2009, 11:47 PM Post #8
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Kotetsu
May 5 2009, 06:18 PM
...I don't feel like working this one right now.

∫ 4/e^2x dx

y=2e^-2x
That seemed to easy so I'm probably wrong, and definitely outta practice with integration
I don't have a math problem to give but if you know or think I'm right please lemme know. tnx
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Kotetsu May 16 2009, 03:02 AM Post #9
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NeciFiX
May 15 2009, 05:05 AM
Lol Kotetsu if you could help me with my homework I'd be extremely happy.
Bring it on man.
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Rebel X May 16 2009, 11:32 AM Post #10
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May 16 2009, 03:02 AM
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May 15 2009, 05:05 AM
Lol Kotetsu if you could help me with my homework I'd be extremely happy.
Bring it on man.
To bad the school years nearly over for most lol. Had you done this thread months ago, who knows how much it could have affected grades
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+ green_480 May 16 2009, 08:26 PM Post #11
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so since this topic is already here i'll just ask my question here.
The question says "Solve the system of equations" 3x-2y=-6 and y=2x-3
I didn't know the final was going to be cumulative so i didn't keep the notes for this stuff.


Oh and the answer my teacher gave is (12,21)
Edited by green_480, May 16 2009, 08:36 PM.
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Kotetsu May 16 2009, 10:47 PM Post #12
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Sure no problem. Keep in mind there are two methods to doing these problems. Subtraction and addition.

First line your terms up. Remember that when you move the 2X over it becomes a -2X to place in front or back.

3X-2Y=-6
Y=2X-3

3X-2Y=-6
-2X+1Y=-3

Next, multiply either the top or bottom by (a one), this means that you will make either the top or bottom variable equal to itself, to cancel out either by addition or subtraction. At this point you should have an idea which will work better.

3X-2Y=-6
-2X+1Y=-3

3X-2Y=-6
2(-2X+1Y=-3)

3X-2Y=-6
-4X+2Y=-6
--------------
-1X+0Y=-12

Add or subtract down. In this case we are getting rid of Y, so we will add down.

-1X+0Y=-12
-1X=-12
X=-12/-1
X=12

Plug X back into the original equation

3(12)-2Y=-6
36-2Y=-6
-2Y=-6-36
-2Y=-42
Y=-42/-2

Y=21 X=12

Your answer was correct. The hardest part about these equations is figuring out whether to add them down or subtract them. If you were to subtract them, just make sure they Y's are equal so when you subtract you get 0.
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Rebel X May 16 2009, 10:54 PM Post #13
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damit i wanted to try one T_T lol
kidding nice work
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* NeciFiX May 17 2009, 05:25 PM Post #14
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Haha, we're working on Quadratic Models and Equations as the last unit in our freshman Algebra class, nothing nearly as hard as above but I'll let you know if I need help... most of us in class were struggling with the equations, including me. I got some help and now I understand it, but, still, who knows what's next.
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AngelicHottieDS May 17 2009, 05:35 PM Post #15
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Ahhhh math. I'm really bad at math so I'll be in this thread quite a lot when I start having math classes.
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