Question

Topic: America and Westward expansion review sheet.
For: @help123please. @PSI
Download: http://bit.ly/123cABA
Completed: 2.10.13
Enjoy :)

Answer 1

Origin: http://openstudy.com/users/opcode#/updates/51170a25e4b09e16c5c89835
Thank everyone. That's all ^_^.
Opcode AI™ is officially now closed.

@Preetha this okay right? Because it's a review.

Answer 2

Nicely done, Opcode. :)
Strangely enough, answering all of those questions wasn't too bad :P

Answer 3

Thank you @PSI.
I wouldn't be possible if you weren't there :3

\(-Opcode AI™\)

Answer 4

:S
http://data.whicdn.com/images/30788085/Happy-oh-stop-it-you-l_large.png

Answer 5

@Frostbite come here.
The function \(f(x)=x^3\).
Definition of a derivative:
\[f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]
Now I had to break this down :P
I know that \(h\) means \(Δx\). I know that \(Δx\) means \(Δx=x2−x1⟹x2=x1+Δx.\)
Then I got:
\[f'(x) = \lim_{h\to 0}\frac{(x+h)^3-x^3}{h}\]
\((x+h)^3-x^3=x^3+3x^2h+3xh^2+h^3-x^3=h(3x^2+3xh+h^2)\)
Solve:
\[\lim_{h\to 0}\frac{h(3x^2+3xh+h^2)}h\]

\[\lim_{h\to 0} (h^2+3 h x+3 x^2)\]

\[3x^2\]

Ugh, so much work for such a little problem -_-.

Answer 6

Pefect :)

Answer 7

But sense you don't like so much work time for another question:
We have another rule (still have to prove the other one btw) that if we have a function on the form:
\[f(x)=ax^n\]
The derivative is given by the following rule:
\[f'(x)=anx ^{n-1}\]
Prove that again using the definition for the derivative.