Alan

What are the differences when using sum of vectors and multiplication of vectors? 

When you sum two vectors you just add corresponding components.  There are two sorts of vector multiplication - your example refers to cross-multiplication.

Given that multiplication of vectors is not cummulative; is the order always the fist of x,y,z multiplied with the later?Why is it called cross multiplication? ie, cross multiplication C=AxB and C= AxBsin(Theta) and so on. It seems straight multiplication to me.

You are correct that cross-mutiplication is not commutative.   If A and B are vectors then AxB is also a vector. It has magnitude ABsin(θ), where θ is the angle between them and A and B represent the magnitudes of A and B respectively, but it points out of the plane that contains A and B (in fact it is normal to that plane). The result of BxA is a vector of the same magnitude but it points in the opposite direction.

Your first image shows rxF is the vector M₀.  In this case r and F are at right-angles to each other so sin(θ) = 1 and M₀ = rF.  To find the direction of M₀ use the right-hand rule.  Hold your (right hand's) thumb, index and middle fingers at right-angles to each other, point  your index finger in the direction of r, your middle finger in the direction of F and your thumb will be pointing in the direction of M₀.

If it's a right-angled isosceles triangle the angle is 45°.  If it's a non-right-angled triangle, for which you know all the sides, you can use the cosine rule:  cosA = (b² + c² - a²)/(2*b*c)  where a is the side opposite angle A, and b and c are the other two sides.

There are 7 (= 2 + 5) segments in total, so 1 segment = 252/7 = 36.

This means 2 segments = 2*36 = 72 and 5 segments = 5*36 = 180.

So the two parts are 72 and 180.

Check:  72/180 = 2*36/(5*36) = 2/5,  and 72+180 = 252.

No harm in having more than one solution though!

If this is log to the base 10 then log n = c means n = 10^c

Rosala, this is just one of those silly 1+1 questions.  (x⁸/x⁸ = 1  and x⁰ = 1, so this is just 1 + 1).

I think the idea is to prove that the LHS (left-hand side) equals the RHS (right-hand side).

$$$$LHS = \sec^6 x - \tan^6 x$$
$$RHS = 1+3\tan^2 x +3\tan^4 x$$
\\
Divide $\sin^2 x + \cos^2 x = 1$ through by $\cos^2 x$ to get $\tan^2 x + 1 = \sec^2 x$ (remembering that $\sec x= 1/\cos x$)\\
So $\sec^6 x = (\sec^2 x)^3 = (\tan^2 x+1)^3 = \tan^6 x +3\tan^4 x + 3\tan^2 x + 1$\\\\
Therefore $$LHS = \tan^6 x +3\tan^4 x + 3\tan^2 x + 1 - \tan^6 x = 3\tan^4 x + 3\tan^2 x + 1 = RHS$$\\\\
Job done!$$

I see Melody beat me to it!

It's possible that log(5)x is meant to be log to the base 5 of x.  If so, and we are talking real numbers only then the following graph shows the x-intercept.

Yes, easy to overlook this, but it's always worth keeping an eye out for the difference of two squares, even when there isn't an explicit ² sign on the number 1. 

$$\frac{\tan^2a - 1}{1+\tan a }=\frac{(\tan a + 1)(\tan a -1)}{1+\tan a}=\tan a -1$$