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In order to work out the equation for a line, you need the gradient of the line and the coordinates of a point on the line. To find the gradient on the line, you look at two points on the line and calculate ...

If it's not sticking in your head; don't. I find it's easier to just remember what the point of a Taylor expansion is: to express a function as an infinite polynomial, often because polynomials are easy to d...

If you have a claim which says something about every element in a list of elements with each element depending on previous elements, induction might be a useful starting point. In your exams, that "list...

We can start by recognising one of our properties of log. That is log a (x) - log a (y) = log a (x/y). Performing this on our question we get: log 3 ((3b+1)/(a-2)) = -1. Now we can remove our log and rewrite...

To start, let us recall the form of a quadratic equation. y = ax 2 + bx + c. When making x the subject of the formula and setting y = 0, we get the quadratic formula, as you may have seen before. x = (-b +- ...