tutoring thread

[Brand New]

Thanks for your help MNFL!

 

[#1_War_Poet_ForLife]

Anytime.

 

[bbwchingy0007]

x^3+5x^2=1

(x^2)(x+5)=1

That might be factorised better than before. (could take 1 from both sides obviously)

 

[Brand New]

****

 

[JCB]

Can someone explain 4d?

I've looked at the solution online but can't figure out what was done to arrive at the answer. Necesito help, por favor.

 

[bbwchingy0007]

I think you will need to integrate f' to get f. Then plug in the two values.

By my reckoning f'(x):
y=-x-2 where -3\<x\<0
(x-2)^2+(y+2)^2=4 where 0\<x\<4 and y>/-2.

So try integrating those, I think

 

[Brand New]

Out of
Laos
Thailand
North Vietnam
South Vietnam
China

which ones were neutral, communist and the other one i forgot the name.

 

[lukewarmplay]

<div class='quotetop'>QUOTE (Bynumite @ Apr 1 2008, 10:42 PM) <{POST_SNAPBACK}></div><div class='quotemain'>Out of
Laos
Thailand
North Vietnam
South Vietnam
China

which ones were neutral, communist and the other one i forgot the name.</div>

www.wikipedia.org

 

[lukewarmplay]

<div class='quotetop'>QUOTE (chingy0007 @ Mar 14 2008, 11:43 AM) <{POST_SNAPBACK}></div><div class='quotemain'>I think you will need to integrate f' to get f. Then plug in the two values.

By my reckoning f'(x):
y=-x-2 where -3\<x\<0
(x-2)^2+(y+2)^2=4 where 0\<x\<4 and y>/-2.

So try integrating those, I think

</div>

nice. i guess they gave you f(0) because that way you can get both constants after the integrations.