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To start with, we want to put this into quadratic form, where we have ax^2+bx+c=0. We notice that there are terms on both sides of the equation. So we can expand the right hand side and then bring all th...
Firstly, we need the equation to be in terms of one cosh argument if possible, which can be done by the identity cosh 2x = 2cosh2x - 1. Upon substitution and simplification we get 6cosh2JCAnswered by Jacob C. • Maths tutor4373 Views
First, we differentiate our equation using the power rule:dy/dx = 8x - 7This is the gradient of our tangent, to the original equation, at any point x. So, to calculate the gradient at x = 2, we substitute...
this question is a great example as we have many different aspects of a level maths workng together into one.initialy you will use use chain rule to find dy/dx = dy/du * du/dx.then you will either keep th...
f(x) = x3 + 3x V = π ∫ (f(x)2) dx V = π ∫02 (x3 + 3x)(x3 + 3x) dx V = π ∫02 (x6 + 6x4ZCAnswered by Zac C. • Maths tutor3087 Views
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