MetSox17

01-15-2010, 02:36 PM

2+2=5?

Discuss.

Discuss.

View Full Version : The Math Thread

MetSox17

01-15-2010, 02:36 PM

2+2=5?

Discuss.

Discuss.

CJSchneider

01-15-2010, 03:09 PM

http://blogs.saschina.org/bknowles/files/2009/09/divide-by-zero.jpg

bantx

01-15-2010, 03:59 PM

I def will be here with a lot of questions once I start class on Tuesday.

Cigaro

01-15-2010, 06:11 PM

No. ......................

Hines

01-15-2010, 06:42 PM

I hate math. It irritates the hell out of me.

TitleTown088

01-15-2010, 06:43 PM

Mmmm I love oriental.

CashmoneyDrew

01-15-2010, 06:45 PM

Only class I ever failed in high school was Algebra 2. Not only did I go on to ace it in summer school, but the next school year I took pre-cal/trig and got B grades.

Conspiracy much?

Conspiracy much?

BamaFalcon59

01-19-2010, 02:01 PM

Help needed.

How does

(DZ)e^-Z = (DT)T

go to

-e^-z = t^2/2 + c

I get the right part, but not the left, with the exponent a variable.

How does

(DZ)e^-Z = (DT)T

go to

-e^-z = t^2/2 + c

I get the right part, but not the left, with the exponent a variable.

GoRavens

01-19-2010, 02:38 PM

holy **** that looks hard

UKfan

01-19-2010, 02:44 PM

Help needed.

How does

(DZ)e^-Z = (DT)T

go to

-e^-z = t^2/2 + c

I get the right part, but not the left, with the exponent a variable.

Ok, take your first expression and divide through by DT, so you have

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

How does

(DZ)e^-Z = (DT)T

go to

-e^-z = t^2/2 + c

I get the right part, but not the left, with the exponent a variable.

Ok, take your first expression and divide through by DT, so you have

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

eaglesalltheway

01-19-2010, 03:13 PM

I love math... Not that ^^^^ **** though...

MetSox17

01-19-2010, 04:05 PM

Ok, take your first expression and divide through by DT, so you have

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

http://i106.photobucket.com/albums/m255/HarryBogard/mind-blown.jpg

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

http://i106.photobucket.com/albums/m255/HarryBogard/mind-blown.jpg

UKfan

01-19-2010, 04:11 PM

Haha, it is hard to explain Maths online. I have a degree in it so I couldn't help myself, but even just looking at that made me realise I am rusty as **** at it.

MetSox17

01-19-2010, 04:14 PM

What's funnier though is that i made this thread completely in jest, and now it's actually helping people with math.

BamaFalcon59

01-19-2010, 05:24 PM

Ok, take your first expression and divide through by DT, so you have

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

gracias!

.....

(DZ/DT)e^-z = T

Then integrate w.r.t. T.

There's probably more steps to get the DT over to the other side (it's been a couple years since I did math) but that is the basic concept...

gracias!

.....

vikes_28

01-19-2010, 05:34 PM

I enjoy Algebra and Geometry. Not algebra 2, just 1

brat316

01-19-2010, 06:31 PM

anyone for calc II or III

Wally03

01-19-2010, 07:35 PM

anyone for calc II or III

My bachelors was Applied Mathematics, so if you're asking for help with calc II or III I might be able to help you out. The biggest challenge is, as UKfan mentioned a few posts above, trying to explain/show the steps of solving a given problem without being able to really use the proper symbols on a forum.

My bachelors was Applied Mathematics, so if you're asking for help with calc II or III I might be able to help you out. The biggest challenge is, as UKfan mentioned a few posts above, trying to explain/show the steps of solving a given problem without being able to really use the proper symbols on a forum.

descendency

01-21-2010, 04:49 AM

My degrees says Mathematics but it means Espionage. (2 years of algebra and a year of analysis... and a bit of other stuff)

I also studied computer science. So I might be useful if you have a question or two.

I also studied computer science. So I might be useful if you have a question or two.

Addict

01-21-2010, 05:38 AM

2+2=5?

Discuss.

y-yes... of course it equals five.

http://chronicle.uchicago.edu/991104/1984.jpg

Discuss.

y-yes... of course it equals five.

http://chronicle.uchicago.edu/991104/1984.jpg

RaiderNation

01-21-2010, 02:01 PM

Im in Algebra 2 and its some of the most pointless stuff Ive ever learned

d34ng3l021

01-21-2010, 02:56 PM

Not really a huge fan of math. Chemistry and Biology are more of my things.

prock

01-21-2010, 03:00 PM

took calc last semester, was terrified of it in high school, so i waited til college. definitely not as hard as its made out to be. pretty easy actually. that being said, i ******* hate math.

vatech=accdomination

01-22-2010, 02:02 PM

anyone for calc II or III

I am in Calc 2, what that has to do with Accounting I will never know.

I am in Calc 2, what that has to do with Accounting I will never know.

NYGibril28

01-31-2010, 08:24 PM

(16/9)^-3/2

I know the answer, but idk how to show the work...help?

I know the answer, but idk how to show the work...help?

derza222

01-31-2010, 10:09 PM

(16/9)^-3/2

I know the answer, but idk how to show the work...help?

Not sure on the formal way to do it so these could be out of order but if you take the 16/9 and flip it to 9/16 then you have (9/16)^3/2 to get rid of the negative. Take the square root for the 1/2 and it's (3/4)^3, which is 27/64 and should be the answer you got unless I'm misinterpreting your expression.

I know the answer, but idk how to show the work...help?

Not sure on the formal way to do it so these could be out of order but if you take the 16/9 and flip it to 9/16 then you have (9/16)^3/2 to get rid of the negative. Take the square root for the 1/2 and it's (3/4)^3, which is 27/64 and should be the answer you got unless I'm misinterpreting your expression.

descendency

01-31-2010, 10:15 PM

y-yes... of course it equals five.

http://chronicle.uchicago.edu/991104/1984.jpg

For larger values of 2.

http://chronicle.uchicago.edu/991104/1984.jpg

For larger values of 2.

bantx

02-10-2010, 02:38 PM

studying for a test someone must be here all day to help me!

1st question

Evaluate each function at the given values of the independent variable and simplify.

1.) f(x) = 4x + 5

a.) f(6)

b.) f(x + 1)

c.) f(-x)

EDIT: I think I remember it now, correct if I'm wrong but I would just plug in 6 for X in the equation for A. and same goes for all the other ones right?

1st question

Evaluate each function at the given values of the independent variable and simplify.

1.) f(x) = 4x + 5

a.) f(6)

b.) f(x + 1)

c.) f(-x)

EDIT: I think I remember it now, correct if I'm wrong but I would just plug in 6 for X in the equation for A. and same goes for all the other ones right?

VoteLynnSwan

02-10-2010, 03:20 PM

studying for a test someone must be here all day to help me!

1st question

Evaluate each function at the given values of the independent variable and simplify.

1.) f(x) = 4x + 5

a.) f(6)

b.) f(x + 1)

c.) f(-x)

EDIT: I think I remember it now, correct if I'm wrong but I would just plug in 6 for X in the equation for A. and same goes for all the other ones right?

yes... that's exactly right.

a) f(6)=4*6+5=29

b) f(x+1)=4*(x+1)+5=4x+9

c) f(-x)=4*-x+5=-4x+5

I remember when that was a difficult concept for me... it is so incredibly easy though.

1st question

Evaluate each function at the given values of the independent variable and simplify.

1.) f(x) = 4x + 5

a.) f(6)

b.) f(x + 1)

c.) f(-x)

EDIT: I think I remember it now, correct if I'm wrong but I would just plug in 6 for X in the equation for A. and same goes for all the other ones right?

yes... that's exactly right.

a) f(6)=4*6+5=29

b) f(x+1)=4*(x+1)+5=4x+9

c) f(-x)=4*-x+5=-4x+5

I remember when that was a difficult concept for me... it is so incredibly easy though.

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