Alan

R = 4*10^-3V/10^-10A → 4*10⁷ohms

.

Think of the 80 as 2*40. Then the first set of results from Wolfram Alpha would look like:

40*2c1

40*(2c1 - 1/2)

40*(2c1 + 1/2) 

etc.

where c1 is an integer.

.

The two red lines and the purple(?) line between them form an equilateral triangle. Find the area of this and multiply by six to get the area of the hexagon.

The area of a triangle, of course, is given by half base times height.

sqrt(10² - 5²) → sqrt(100 - 25) → sqrt(75) → 5*sqrt(3)

This is the height of the black(?) line going from the centre to the side in your picture.

Incidentally, in n dimensions, a vector space is the set of all possible values of (a₁, a₂, ... a_n) where the a's are the individual components (I.e. it's the set of all possible n dimensional vectors).

A regular hexagon is comprised of six equilateral triangles.  The "base" of each triangle is just 10 cm (in this case) and the "height" is sqrt(10² - 5²).

Can you take it from here?

You could certainly have functions of vectors. You could have functions that generate the components of vectors.  Without knowing exactly what your lecturer meant by "expressed by" though, I'm not going to be a great help!

Set y to be 50 and then take logs of both sides.

50 = 21.665x^-0.264

ln(50) = ln(21.665) - 0.264lnx   using the properties of logs.

Rearrange:   lnx = (ln(21.665) - ln(50))/0.264

x = e^{(ln(21.665) - ln(50))/0.264}

A vector in linear algebra is generally just a list of components.  In two dimensions these would represent the x and y components if you think of them graphically.  However, in general, there could be n terms, where n is much greater than 2, so these would be difficult to visualise! The components are usually listed within brackets. When these are aligned in a row you have a row vector; when they are aligned vertically, you have a column vector. There are specific rules for adding, multiplying etc.

Is this what you are looking for, or is there a specific feature you encountered that puzzled you?