Melody

$$\\(\sqrt3(2^7))^3\\\\
=(3^{0.5}2^7)^3\\\\
=3^{0.5*3}(2^{7*3})\\\\
=3^{1.5}*2^{21}\\\\$$

 Gino, my answers are a bit  different from yours - would you like to check?

f(x) = cos 4(x+5) + 3?

Amplitude is 1

vertical shift is 3 units up   +3

wavelength = 2pi/4 = pi/2

Phase shift is 5 to the left    (-5)   

To find the shift I say x+5=0   so x=-5

The green line is meant to be a arrow going left.

+ means in the positive direction, up and right.

There is a graphing calculator - Look at the bottom right of the web2 calculator.  

You can't solve it for a but p has to be 2.

If p is 2 then a can be anything.

$$$This is a factorisation that you need to memorise$\\\\
\boxed{a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)}\\\\
so\\
1+27x^3\\
=1^3+(3x)^3\\
=(1+3x)(1-3x+9x^2)$$

Thanks Chris and dragon.

I was just playing.  It is good to do that sometimes.  

Fri 3/10/14

1) The Ghost of Web2.0 calc has graced us with a visit, laying a lay of lavish wonderment at the feet of our dear Princess Kitty.

Yes I think that we are both right.   

$$\\-i=-\sqrt{-1}\\
or\\
-i=-1*\sqrt{-1}\\
-i=i^2*i\\
-i=i^3\\
or\\
-i=(-1)^3*i\\
-i=(i^2)^3*i\\
-i=i^6*i\\
-i=i^7\\
$continuing with this pattern I think$\\
-i=i^{4n+3}\qquad $Where n \in Z $ and $ n\ge 0$$

Now it has got me thinking more.

what if n was a negative integer?  Would that work?

$$\\i^{-4n+3}\\
=i^{-4n}*i^3\\
=\frac{1}{i^{4n}}*i^3\\
=\frac{1}{(i^{4})^n}*i^3\\
=\frac{1}{(1)^n}*i^3\\
=\frac{1}{1}*i^3\\
=i^3\\
=i^2*i\\
=-i\\\\
so\\\\
-i=i^{4n+3} \qquad $ where $ n \in Z\\
$n can be any integer, it does not have to be positive$$$