Melody

Evaluate the following limit: lim h→0 ((3+h)^3−27)/h

\(\displaystyle\lim_{h \rightarrow 0}\;\frac{(3+h)^3-27}{h}\\~\\ =\displaystyle\lim_{h \rightarrow 0}\;\frac{(3+h)^3-3^3}{h}\qquad note:\;a^3-b^3=(a-b)(a^2+ab+b^2)\\~\\ =\displaystyle\lim_{h \rightarrow 0}\;\frac{((3+h)-3)((3+h)^2+(3+h)*3+3^2)}{h}\\~\\ =\displaystyle\lim_{h \rightarrow 0}\;\frac{h[(3+h)^2+3(3+h)+9]}{h}\\~\\ =\displaystyle\lim_{h \rightarrow 0}\;\:(3+h)^2+3(3+h)+9\\~\\ =9+9+9\\~\\ =27 \)

The position of a particle is given by x(t) = (t2 - 4)3 where x(t) is measured in centimeters and t is measured in seconds.

\(x(t)=(t^2-4)^3\)

Find where the particle is moving to the left.

velocity is negtive

\(x(t)=(t^2-4)^3\\ x'(t)=3(t^2-4)^2*2t\\ x'(t)=6t(t^2-4)^2\\ \)

Since t cannot be negative, velocity can never be negative.  

which ones :)

Ok - I better think about it better...   Thanks Alan :D

What happened to your equal sign Mathematician?

The question does not make sense .....

So does this mean that the answer cannot be determined without more information?

I stand corrected.

Thanks Alan :))

Sorry I just realized that you wnated it in Hexidecimal.  (It is answered on the middle row of the table)

\(11001110101110111_2= 19D77_{16}\)

If i have Q=(8/15) Ctan(Φ/2) if i move tan to the other side of the equal sign what does it become?

I'm confused.

why do you have a Q and a Phi   and what is Ctan?  I mean why is there a C in front of the tan?

Use the [2nd] in the top left hand corner   :)