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Though this question initially appears complex, it can be broken down into logical steps that make the answer straightforward to find. The two statements should be approached individually, to give an equa...

y = x² - x - 6It is first important to realise that we are able to find the values of x when y = 0, that is where the parabola crosses the x axis. We can visually represent this and our solutio...

For two lines to be proven as parallel they must have the same gradient. The second equation needs to therefore be rearranged into y = mx + c format to see if this is the case.3y-9x+5 = 0 (subtract the 5)...

(x+14)(x+2)=0
x=-14, x=-2

Integrated: [x^3+2ln(x)] Sub in 3 and K: [3^3+2ln(3)-k^3-2ln(k)]Simplify: 27-K+2ln(3/K)—> K cannot = 0 otherwise ln(3/k) would have an undefined value, as would ln(k)