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To solve this inequality, we need to take steps to collect like terms on each side of the inequality. To start with, we can try getting all the numbers on one side of the inequality - on the left hand sid...

You need to remember some ‘rules’:if it has the ‘co-‘ prefix, the derivative will be negative (e.g. cos x -> -sin x)if it has the ‘-sec’ suffix, the derivative will only include itself (e.g. cosec x -&...

\frac{1}{x^2+x}=\frac{A}{x} + \frac{B}{x+1}, 1=A(x+1)+Bx, let x=-1: 1=-B, B=-1let x=0: 1=A, A=1Hence, \int_{1}^{2}{\frac{1}{x^2+x}dx} = \int_{1}^{2}{\frac{1}{x}dx} - \int_{1}^{2}{\frac{1}{x+1}dx}=[ln(x)-l...

Using the product rule :u = x, so du/dx = 1
v = sin(x), so dv/dx = cos(x)
Therefore dy/dx = v(du/dx) + u(dv/dx) So dy/dx = sin(x) + x.cos(x)

We start by trying to find out the values of a and c using the information about the minimum point. We know we can rearrange the right hand side by completing the square: y = a(x^2 + 3x)+cy = a(x+3/2a)^2+...