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The first step is to find out the value of x for which your bracket/value you think is a factor equals 0. So, x – 2 = 0, therefore x = 2. Now plug in this value into your original equation and if the resu...

To begin to solve this equation we must take natural logarithms of both sides of the equation. This gives: ln(3^x*e^4x) = lne^7 Then we can use the log rules on the left hand side to expand it slightly to...

Suppose we have a matrix M = [{a b} {c d}]

The inverse of a matrix M is any matrix, that when multiplied by M, gives the unit matrix (I) - in this case: [{1 0} {0 1}]

First, we have to d...

To make this equation easier to differentiate it would be easier to write it using index rules as y = 4x^3 - 5x^-2 From here we can begin to differentiate: dy/dx = 3*4x^(3-1) - (-2)*5x^(-2-1) Then finally...

Firstly, let's split the equation "xcosx" into two parts to integrate them separately. 1) Let u=x and dv/dx=cosx 2)As the integral of x is 1, du/dx=1 3)To find v, we integrate cosx to get v=sinx...