Alan

Good selection NinjaDevo; and, yes, forces should be in Physics.

I agree with all that kitty said except for the phase shift, which I think should be pi.

Careful with the order kitty.  Your first expression should be log(x,4).  i.e. log(number, base).

$${\sqrt{{\mathtt{3}}}} = {\mathtt{1.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$

$${\left|{\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right|} = {\mathtt{0.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$

$${\left|{\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right|} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

Add up the last two (which are only approximate, as √3 is irrational) to get 1.

Or:  |1 - √3| = √3 - 1 because √3 is bigger than 1

       |2 - √3| = 2 - √3 because 2 is bigger than √3

Add these up and the √3's cancel out, leaving 2 - 1 = 1.

You can use the calculator on the Home page here to do this.

$${{\mathtt{4}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{112}}$$

If you mean rotate a vector by 45° then multiply it by (1/2)i, where i = √(-1) to get a 45° anti-clockwise rotation.

Oops!  No, this will give a rotation by 90° and halve the magnitude!  You need to multiply by exp(i*pi/4).

"collinear" means along the same line; or if it refers to points, then it means the points lie along a straight line.

If you sum vectors A and B you get the same result as if you sum vectors B and A.

The picture with the chains means that the forces, F1 and F2, in two of the chains are added (vectorially) to give a resultant vector (F1+F2).  This is then added (vectorially) to the force F3 to give an overall resultant force FR.

Adding vectorially means you add the x-component values together, and add the y-component values together.

In the s***w eye you need to resolve F1 into it's components (F1*cos(theta1), F1sin(theta1)) and F2 into its components (F2cos(theta2), F2sin(theta2)), (where the thetas are angles between the vectors and the x-axis) and get the components of the resultant from

(FR_x, FR_y) = (F1*cos(theta1)+F2cos(theta2), F1sin(theta1)+F2sin(theta2))

The magnitude of the resultant is given by √( FR_x² + FR_y²) and its angle to the x-axis by tan^-1(FR_y/FR_x)

$$$$\sqrt{27x^2y^2z}=\sqrt{3*3^2*x^2*y^2*z}=3*x*y*\sqrt{3z}$$\\
or $$3xy\sqrt{3z}$$