Melody

How about you give your thoughts parpar1? 

What do YOU think parpar1?

Here is how:

\((x+1)^3+x^2\cdot q(x)\\ =x^3+3x^2+3x+1+x^2\cdot q(x)\\ =x^2(x+3)+3x+1+x^2\cdot q(x)\\ =x^2(x+3+q(x))+3x+1\\ if\;\;q(x)=-3-x\\ \text{Then the degree of the resuling polynomial will be 1} \)

Sorry,

I started mine hours ago and got called away.

I did not know that Heureka and Hectictar had already answered it.

Lets see

\(z^2 + z + 1 = 0\\~\\ z=\frac{-1\pm \sqrt{1-4}}{2}\\ z=\frac{-1\pm \sqrt{-3}}{2}\\ z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\ z^2=\frac{-2\mp2\sqrt{-3}}{4}\\ z^2=\frac{-1\mp\sqrt{-3}}{2}\\ z^2=\bar z \)

\(z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\ so\\ z^{3n}=1\qquad \text{where n is a positive integer} so\\ z^{48}=z^{51}=1\)

\(z^{49}=z\\ z^{50}=\bar z\\ z^{51}=1\\ z^{52}=z\\ z^{53}=\bar z\\~\\ \text{Add them together and get}\\ 1+2(z+\bar z)\\ =1+2(\frac{-1}{2}+\frac{-1}{2})\\ =1+2(-1)\\ =1-2\\ =-1 \)

LaTex:

z^2 + z + 1 = 0\\~\\
z=\frac{-1\pm \sqrt{1-4}}{2}\\
z=\frac{-1\pm \sqrt{-3}}{2}\\
z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\
z^2=\frac{-2\mp2\sqrt{-3}}{4}\\
z^2=\frac{-1\mp\sqrt{-3}}{2}\\
z^2=\bar z

z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\
so\\
z^{3n}=1\qquad \text{where n is a positive integer}
so\\
z^{48}=z^{51}=1

z^{49}=z\\
z^{50}=\bar z\\
z^{51}=1\\
z^{52}=z\\
z^{53}=\bar z\\~\\
\text{Add them together and get}\\
1+2(z+\bar z)\\
=1+2(\frac{-1}{2}+\frac{-1}{2})\\
=1+2(-1)\\
=1-2\\
=-1

Don't know.... maybe it depends on what universe reality you are in   

YES  IT IS  !    

The approximate answer is   35.354%    

That is a link to something on your home computer. It is not going to work here.

That is not the graph.

Never mind.

Here is your graph, and the answer graph as well

1. What is the frequency of an orange light that has a wavelength of 620 x 10-9 m?

In your case the V is the speed of light.