Alternative picture comparison:
About 80:
\(2x^2-3x+1\rightarrow(2x-1)(x-1)\)
This equals zero when x = 1/2 and x = 1 so 2x2 - 3x + 1 is greater than zero when x is less than 1/2 and greater than 1.
What expression?
There is insufficient information here to provide an answer. You need something like an appropriate angle or another onbject of known size casting a shadow of known length, or some such.
Lots of places! See http://mathworld.wolfram.com/e.html for mention of several of them.
Like so?
\(\frac{3}{\sqrt{12}}\rightarrow\frac{3}{\sqrt{4\times3}}\rightarrow\frac{3}{2\sqrt{3}}\rightarrow\frac{\sqrt{3}\times\sqrt{3}}{2\sqrt{3}}\rightarrow\frac{\sqrt{3}}{2}\)
You can use \(v^2=u^2+2as\)
v = final velocity
u = initial velocity
a = acceleration
s = distance
\(v=\sqrt{9.6^2+2*4.2*450}=62.2 m/s\)
If you mean the magnitude of the vector in terms of the x, y and z components, then it is \(r=\sqrt{x^2+y^2+z^2}\)
1kg = 1000g so 3 kg + 125g = 3125 g or 3.125 kg
+33667 
